Conformal group, Conformal algebra and Conformal invariance in field theory

[samalkhaiat]

I have noticed that questions about this subject get either ignored or receive some confusing answers. So I decided to write a "brief" but self-contained introduction to the subject. I'm sure you will find it useful.
It is going to take about 13 or 14 post to complete the work. Be patient with me as my time allow me to post only 2 or 3 posts a day.
Equations and exercises will be numbered by the post number; for example Eq(1.6) means equation 6 in post#1 and Ex(5.7) stands for exercise 7 in post#5.

SO PLEASE DO NOT POST YOUR COMMENTS, QUESTIONS etc.,IN BETWEEN MY POSTS, AS THIS WOULD MESS UP THE NUMBERING.

CONFORMAL TRANSFORMATIONS

Consider a flat n-dimensional Minkowski spacetime $$(M^{n}, \eta)$$ . The conformal group C(1,n-1) can be formally realized as a group of (nonlinear) coordinate transformations:

$$f: x \rightarrow \bar{x} = \bar{x}(x)$$

which leaves the metric $$(f* \bar{g})_{ab}$$ , where

$$\bar{g} = \eta_{ab} d \bar{x}^{a} \otimes d\bar{x}^{b},$$

invariant up to a scale:

$$\bar{g}_{ab}(x) \left ( = \partial_{a} \bar{x}^{c} \partial_{b} \bar{x}^{d} \eta_{cd} \right ) = S(x) \eta_{ab} \ \ (1.1)$$

I.e.,

$$d \bar{s}^{2} \left ( = \eta_{ab} d\bar{x}^{a}d\bar{x}^{b} \right ) = S(x) ds^2 \ \ (1.1')$$

and we say that the conformal group preserves the light-cone structure. This excludes the conformal group as a symmetry of massive particle theories. If massive particles are included, the condition S=1 must be imposed which restricts the symmetry to the Poicare subgroup.
For spacetime with n>2, the conformal group is finite-dimensional. To see this, let us solve (1.1) for general infinitesimal coordinate transformation;

$$\bar{x}^{a} = x^{a} + f^{a}(x)$$

which leads to

$$\partial_{a} f_{b} + \partial_{b} f_{a} = \eta_{ab} (S-1),$$

or, taking the trace to obtain $$(S-1) = 2/n \partial .f \equiv F$$ ,

$$\partial_{a} f_{b} + \partial_{b} f_{a} = \eta_{ab} F(x) \ \ (1.2)$$

By applying an extra derivative $$\partial_{c}$$ on this (conformal Killing) equation, permuting the indices and taking a linear combination, we get

$$\partial_{c} ( \partial_{a} f_{b} - \partial_{b} f_{a}) = \eta_{cb} \partial_{a} F - \eta_{ac} \partial_{b} F \ \ (1.3)$$

or, after integration,

$$\partial_{a} f_{b} - \partial_{b} f_{a} = \int ( \partial_{a} F dx_{b} - \partial_{b} F dx_{a} ) + 2 \omega_{ab} \ \ (1.4)$$

for some constant antisymmetric tensor $$\omega_{ab}$$ .
Adding (1.2) to (1.4) and integrating again, we find

$$f^{a} = a^{a} + \omega^{ba}x_{b} + \frac{1}{2} \int dx^{a} F + \frac{1}{2} \int dx^{b} \int \left ( \partial_{b} F dx^{a} - \partial^{a} F dx_{b} \right ) \ \ (1.5)$$

where $$a^{a}$$ is a constant n-vector. Notice that the first two terms represent Poincare transformation. This is expected because F = 0 corresponds to a coordinate transformations which do not change the form of the metric, i.e., a general solution to the homogeneous differential equation $$\partial_{a}f_{b} + \partial_{b}f_{a} = 0$$ .

The integral equation (1.5) determines the conformal Killing vector f once the function F(x) is found. So let us find it; By contracting the indices (c,a) in eq.(1.3), we get

$$\partial^{2} f_{b} = (1 - n/2) \partial_{b} F \ \ \ (1.6)$$

Operate by $$\partial_{a}$$ and form the symmetric combination;

$$2(1 - n/2) \partial_{a} \partial_{b} F = \partial^{2} ( \partial_{a}f_{b} + \partial_{b}f_{a} )$$

now, use the conformal Killing equation to find

$$(2 - n) \partial_{a} \partial_{b} F = \eta_{ab} \partial^{2} F \ \ (1.7)$$

Finally, contracting with $$\eta^{ab}$$ , we end up with

$$(n -1) \partial^{2} F = 0 \ \ (1.8)$$

Therefore $$\partial^{2} F = 0$$ except for the trivial case n = 1.
Thus for n > 1, Eq(1.7) becomes

$$(2 - n) \partial_{a} \partial_{b} F = 0 \ \ (1.9)$$

When n > 2, this implies

$$\partial_{a} \partial_{b} F(x) = 0 \ \ (1.10)$$

I.e., F is at most linear in the coordinates:

$$F(x) = -2 \alpha + 4 c_{a}x^{a} \ \ (1.11)$$

with $$(\alpha , c_{a})$$ are constants.

Inserting (1.11) in (1.5), we find our conformal Killing vector

$$f^{a} = a^{a} + \omega^{ba}x_{b} - \alpha x^{a} + c_{b} (2x^{a}x^{b} - \eta^{ab} x^{2}) \ \ (1.12)$$

This depends on (n + 1)(n + 2)/2 parameters: n translations, n(n-1)/2 Lorentz transformations, one dilation and n special conformal transformations.

In n = 2, Eq(1.9) does not imply Eq(1.10), which was crucial for the finiteness of the group C(1,n-1) in the n>2 case, in this case (n=2) every harmonic function F determines a solution, i.e., the group C(1,1) becomes infinite-dimensional. This C(1.1), interesting for string theories, will not be discussed in this introduction.


more to come, please be patient...

 

[samalkhaiat]

Lie Groups and Algebras

The local structure of any Lie group contains the most essential feature of the group, namely its continuity, and it is studied by considering the group for values of the parameters in the neighbourhood of the identity (infinitesimal transformations). The remarkable property of Lie groups (discovered by S Lie himself) is that in spite of the continuity of the parameters, almost all information about the local structure can be obtained from finite discrete system, namely Lie algebras.
Lie algebras may be regarded as (finite-dimensional) vector spaces equipped with a multiplication law;
$$[ . , . ] : \cal{L} \times \cal{L} \rightarrow \cal{L}$$
which satisfies the conditions

i) [X,Y] = - [Y,X]

ii) [X,[Y,Z]] + [Z,[X,Y]] + [Y,[Z,X]] = 0

for all (X,Y,Z) in $$\cal{L}$$ .

Please note that the product [X,Y] (Lie bracket) need not necessarily be of the form (XY - YX), Poisson's bracket is another example of Lie bracket.

The 1-to-1 correspodence between Lie algebras and local Lie groups can be expressed explicitly by means of the exponential mapping, i.e.,(almost all) elements of a connected Lie group can be represented as exponentials of Lie algebra elements;

$$g = e^{X} = e^{a_{i}X^{i}} \ \ (2.1)$$

Where $$\{X^{i}\}$$ is a basis for $$\cal{L}$$ .The components $$a_{i}$$ of Lie algebra elements (the parameters) play the role of local coordinates (in the neighbourhood of the identity) of the group manifold. The fact that the union of elements of the form (2.1) presents a Lie group is a consequence of the Baker-Campbell-Hausdroff formula

$$e^{X} e^{Y} = e^{X + Y + \frac{1}{12}[X,[X,Y]] + ...} \ \ (2.2)$$

i.e.,
$$g(a).g(b) = g(f(a,b))$$


more to come, please don't reply to my posts just yet...

 

[samalkhaiat]

The conformal algebra

From infinitesimal transformations to Lie algebra

By considering the combined action of various infinitesimal transformations in different orders, the Lie algebra of the conformal group can be abstructed.
Let us write the conformal Killing vector (1.12) in the form;

$$f^{a} = \lambda_{A} \delta_{( \lambda )}^{A} x^{a} \ \ (3.1)$$

where

$$\lambda_{A} \equiv ( a^{a}, \frac{1}{2}\omega^{ab}, \alpha , c^{a})$$

and $$\delta^{A}$$ are the generators of the following infinitesimal coordinate transformations;

Translation;

$$\delta^{a}_{(T)} x^{c} = \eta^{ab} \ \ (3.2a)$$

Lorentz;

$$\delta^{ab}_{(L)} x^{c} = \eta^{bc}x^{a} - \eta^{ac}x^{b} \ \ (3.2b)$$

Dilation(scale);

$$\delta_{(D)}x^{c} = -x^{c} \ \ (3.2c)$$

Conformal;

$$\delta^{a}_{(C)}x^{c} = 2 x^{a}x^{c} - \eta^{ac}x^{2} \ \ (3.2d)$$

Ex(3.1) Show that;

$$\left[\delta^{a}_{(T)},\delta^{b}_{(T)}\right] = 0 \ \ (3.3a)$$

$$\left[\delta^{ab}_{(L)},\delta^{c}_{(T)}\right] = \eta^{bc}\delta^{a}_{(T)} - \eta^{ac}\delta^{b}_{(T)} \ \ (3.3b)$$

$$\left[\delta^{ab}_{(L)},\delta^{cd}_{(L)}\right] = \eta^{ac}\delta^{bd}_{(L)} - \eta^{bc}\delta^{ad}_{(L)} + \eta^{ad}\delta^{cb}_{(L)} - \eta^{bd}\delta^{ca}_{(L)} \ \ (3.3c)$$

$$\left[\delta_{(D)},\delta_{(D)}\right] = \left[\delta^{ab}_{(L)},\delta_{(D)}\right] = 0 \ \ (3.3d)$$

$$\left[\delta_{(D)},\delta^{a}_{(T)}\right] = \delta^{a}_{(T)} \ \ (3.3e)$$

$$\left[\delta_{(D)},\delta^{a}_{(C)}\right] = -\delta^{a}_{(C)} \ \ (3.3f)$$

$$\left[\delta^{a}_{(C)},\delta^{b}_{(C)}\right] = 0 \ \ (3.3g)$$

$$\left[\delta^{ab}_{(L)},\delta^{c}_{(C)}\right] = \eta^{bc}\delta^{a}_{(C)} - \eta^{ac}\delta^{b}_{(C)} \ \ (3.3h)$$

$$\left[\delta^{a}_{(C)},\delta^{b}_{(T)}\right] = 2 \eta^{ab}\delta_{(D)} - 2 \delta^{ab}_{(L)} \ \ (3.3i)$$


Thus the operators $$\delta^{A}_{(\lambda )}$$ span a vector space with multiplication law as in (3.3). It is the Lie algebra of the conformal group C(1,n-1).


more to come, please dot reply just yet...

 

[samalkhaiat]

From vector fields to Lie algebra

Let $$f_{(1)} = f^{a}_{(1)}\partial_{a}$$ and $$f_{(2)} = f^{a}_{(2)}\partial_{a}$$ be two conformal Killing fields in Minkowsiki spacetime. Their Lie bracket

$$[f_{(1)},f_{(2)}] = f_{(3)} = f^{a}_{(3)}\partial_{a} \ \ (4.1)$$

with

$$f^{a}_{(3)} = f^{b}_{(1)}\partial_{b}f^{a}_{(2)} - f^{b}_{(2)}\partial_{b}f^{a}_{(1)} \ \ (4.2)$$

gives the vector field $$f_{(3)}$$ which also satisfies the conformal Killing Eq(1.2). thus, the set of all conformal Killing vectors form a Lie algebra.

Now, we introduce a basis

$$G^{A}_{(\lambda )} \equiv \left( p^{a}, j^{ab}, d, k^{a}\right) \ \ (4.3)$$

for the algebra by the rule

$$f = \lambda_{A}\left( \delta^{A}_{(\lambda )} x^{a} \right) \partial_{a} = i \lambda_{A}G^{A}_{(\lambda )} \ \ (4.4)$$

From (3.2) we find

$$ip^{a} = \partial^{a} \ \ (4.5a)$$

$$ij^{ab} = x^{a}\partial^{b} - x^{b}\partial^{a} \ \ (4.5b)$$

$$id = -x^{a}\partial_{a} \ \ (4.5c)$$

$$ik^{a} = 2x^{a} x^{c}\partial_{c} - x^{2} \partial^{a} \ \ (4.5d)$$

And, by explicit calculations we find the algebra

$$[ip^{a} , p^{b}] = [id , d] = [ij^{ab}, d] = [ik^{a},k^{b}] = 0 \ \ (4.6a)$$

$$[id , p] = p \ \ (4.6b)$$

$$[id, k] = -k \ \ (4.6c)$$

$$\left[ ik^{a},p^{b}\right] = 2 \eta^{ab} d - 2 j^{ab} \ \ (4.6d)$$

$$\left[ ij^{ab}, k^{c}\right] = \eta^{ac} k^{b} - \eta^{bc} k^{a} \ \ (4.6e)$$

$$\left[ ij^{ab}, p^{c}\right] = \eta^{ac} p^{b} - \eta^{bc} p^{a} \ \ (4.6f)$$

$$\left[ ij^{ab}, j^{cd}\right] = \eta^{bc}j^{ad} - \eta^{ac}j^{bd} + \eta^{ad}j^{bc} - \eta^{bd}j^{ac} \ \ (4.6g)$$


Since every Lie algebra has a faithful matrix representation, we can forget about the explicit (differential) realization (4.5) and postulate the conformal algebra as an abstract real algebra subject to 2 requirements;
1) it has a basis

$$G^{A} = \left( P^{a}, J^{ab}=-J^{ba}, D, K^{a}\right) \ \ (4.7)$$
with multiplication law as in (4.6);
2) its general element is of the form

$$X = i \lambda_{A}G^{A} \ \ (4.8)$$

The conformal group C(1,n-1) is formally obtained by exponentiation of the conformal algebra

$$g(\lambda ) = \exp (i \lambda_{A} G^{A} ) \ \ (4.9)$$


more to come, please don't reply just yet,...

 

[samalkhaiat]

From unitary representation to Lie algebra

This time, we cosider various subgroups of the conformal group, find their multiplication laws and derive the corresponding Lie subalgebras using a unitary representations of the laws.

Scale plus Translation

Consider the transformation;

$$\bar{x} = T(\alpha , a) x = e^{\alpha} x + a \ \ (5.1)$$

It is easy to see that two such transformations is also a transformation of the same kind;

$$T(\alpha, a) T(\beta , b) = T\left( \alpha + \beta , a + be^{\alpha} \right) \ \ (5.2)$$

Ex(5.1): Prove the above "easy to see statement".

From (5.2), we find;

1)

$$\{T(\alpha , a) T(\beta , b) \} T(\gamma , c) = T\left(\alpha + \beta +\gamma , a + be^{\alpha} + ce^{\alpha + \beta}\right) = T(\alpha , a) \{ T(\beta , b) T(\gamma , c) \} \ \ (5.3a)$$

i.e., the multiplication law (5.2) is associative.

Ex(5.2): Prove Eq(5.3a).

2)

$$T(\alpha , a) T(0,0) = T(0,0) T(\alpha , a) = T(\alpha , a) \ \ (5.3b)$$

i.e., an identity element exists; $$E = T(0,0)$$

3)

$$T\left(\alpha , a \right) T\left(-\alpha , -ae^{-\alpha} \right) = T \left(-\alpha , -ae^{-\alpha} \right) T(\alpha , a) = T(0,0) \ \ (5.3c)$$

i.e., an inverse element exsists

$$T^{-1}(\alpha , a) = T\left( -\alpha , -ae^{-\alpha} \right) \ \ (5.4)$$


Thus, the set $$\{T(\alpha , a)\}$$ with the multiplication law (5.2) forms a group.

We know from QM that coordinate transformations, such as our T, induce a unitary linear transformation on vectors in the physical Hilbert space;

$$| \Psi_{T}\rangle = U(T) | \Psi \rangle \ \ (5.5)$$

i.e., the unitary operator U forms a representation of the group (multiplication law) in question;

$$U( \alpha , a) U(\beta , b) = U( \alpha + \beta , a + be^{\alpha} ) \ \ (5.6)$$

Since U(T(0,0)) = U(E) carries any vector into itself, it must be proportional to the unit operator. So, in at least a finite neighbourhood of the identity, U can be represented by a power series. For infinitesimal coordinate transformations, U(T) must then differ from 1 by terms linear in the parameters $$(\alpha , a)$$ , so we may write

$$U(\alpha , a) \approx 1 + i \alpha D + i a^{a}P_{a} \ \ (5.7)$$

where $$(D,P)$$ are Hermitian operators independent of the parameters $$(\alpha , a)$$ .
In order to find the transformations laws of D and P, you need to work out the following;

Ex(5.3): Show that

$$U(\beta , b)U(\alpha , a)U^{-1}(\beta , b) = U\left(\alpha , ae^{\beta} - \alpha b \right) \ \ (5.8)$$

where $$(\beta ,b)$$ are the parameters of new transformation unrelated to $$(\alpha ,a)$$

Next, write (5.8) to the 1st order in $$(\alpha ,a)$$ ,equate coefficients of $$\alpha$$ and $$a$$ on both sides and show that

$$\bar{P}_{a} = U^{-1}(\beta , b) P_{a} U(\beta , b) = e^{-\beta} P_{a} \ \ (5.9a)$$
$$\bar{D} = U^{-1} D U = D + b_{a}P^{a} e^{-\beta} \ \ (5.9b)$$

Now, let $$(\beta , b)$$ themselves be infinitesimals and find the commutation relations;

$$[iD , D] = 0 \ \ (5.10a)$$
$$[iP^{a} , P^{b}] = 0 \ \ (5.10b)$$
$$[iD , P^{a}] = P^{a} \ \ (5.10c)$$


This is the Lie algebra of the (semi-direct product) group of scale & translation.

****

Poincare' group

The same can be applied to the Poincare transformations

$$\bar{x} = T(\Lambda , a)x = \Lambda x + a \ \ (5.11)$$

which have the multiplication law

$$T_{1}T_{2} = T(\Lambda_{1} \Lambda_{2} , \Lambda_{1} a_{2} + a_{1}) \ \ (5.12)$$

For infinitesimal transformation, $$T(1 + \omega , \epsilon )$$ , we may write the corresponding unitary operator as

$$U(1+\omega ,\epsilon ) = 1 + i \epsilon . P - (i/2) \omega . J \ \ ((5.13)$$

where $$\omega . J \equiv \omega_{ab}J^{ab}$$


Ex(5.4): Use Eq(5.12) to show that

$$U(\Lambda ,a)U(1+\omega ,\epsilon )U^{-1}(\Lambda , a) = U(\Lambda \omega \Lambda^{-1} , \Lambda \epsilon - \Lambda \omega \Lambda^{-1} a) \ \ (5.14)$$

write this to 1st order in $$(\omega , \epsilon)$$ , then show that J is a Lorentz tensor;

$$\bar{J}^{ab} = U^{-1}J^{ab}U = \Lambda_{c}{}^{a}\Lambda_{d}{}^{b}( J^{cd} + a^{c}P^{d} - a^{d}P^{c}) \ \ (5.15a)$$

and P is a vector

$$\bar{P}^{a} = U^{-1}P^{a}U = \Lambda_{b}{}^{a}P^{b} \ \ (5.15b)$$

From the infinitesimal versions of Eq(5.15) deduce the Lie algebra of Poincare' group;

$$[iP_{a},P_{b}] = 0 \ \ \ \ (5.16a)$$
$$[iJ^{ab},P^{c}] = \eta^{ac}P^{b} - \eta^{bc}P^{a} \ \ (5.16b)$$
$$[iJ^{ab},J^{cd}] = \eta^{bc}J^{ad} - \eta^{ac}J^{bd} + \eta^{ad}J^{bc} - \eta^{bd}J^{ac} \ \ (5.16c)$$

****

Scale & Lorentz group

$$\bar{x} = T(\alpha ,\Lambda ) x = e^{\alpha} \Lambda x \ \ \(5.17)$$

For this group, we have

$$u(\alpha_{1} , \Lambda_{1})u(\alpha_{2} , \Lambda_{2}) = u(\alpha_{1} + \alpha_{2} , \Lambda_{1} \Lambda_{2} ) \ \ (5.18)$$

From this, it is seen;

$$u(\alpha , 1)u(0 , \Lambda )u(-\alpha , 1) = u(0 , \Lambda) \ \ (5.19)$$

By writting

$$u(\alpha , 1) = 1 + i \alpha D \ \ (5.20a)$$
$$u(0 , \Lambda) = 1 + (i/2) \omega . J \ \ (5.20b)$$

we find

$$[iJ^{ab}, D] = 0 \ \ (5.21)$$

Next, we postulate that the algebra of C(1,n-1) closes on the (n+1)(n+2)/2 generators (P,J,D,K) and that K is a Lorentz vecto

$$[iJ^{ab}, K^{c}] = \eta^{ac}K^{b} - \eta^{bc}K^{a} \ \ (5.22)$$

Surprisingly, this postulate together with what we already know about the subalgebra (P,J,D), is sufficient to determine the remaining commutators
[P,K], [D,K] and [K,K] from the Jacobi identities alone.

Ex(5.5): On grounds of Lorentz invariance, i.e., the Jacobi identities found by taking the above commutators with J, conclude that

$$[iD, K^{a}] = aK^{a} + bP^{a} \ \ (5.23a)$$
$$[iK^{a},K^{b}] = cJ^{ab} \ \ (5.23b)$$
$$[iK^{a},P^{b}] = d\eta^{ab} D - dJ^{ab} \ \ (5.23c)$$

where a,b,c and d are constants.

Now, use the J. identities

$$\left[iD ,[iK^{a},P^{b}] \right] + \left[iP^{b}, [iD,K^{a}] \right] + \left[iK^{a}, [iP^{b},D] \right] = 0 \ \ (5.24a)$$
$$\left[iD, [iK^{a},K^{b}] \right] + ... + ... = 0 \ \ (5.24b)$$
$$\left[iP^{a}, [iK^{b},K^{c}] \right] + ... + ... = 0 \ \ (5.24c)$$


to find that a = -1 and c = bd = 0 .

Notice that a vanishing [P,K] would also be compatible with the Jacobi identities, in that case there would be an arbitrary real factor in [D,K]. However the choice b = 0 and d = 2 is compatible with our previous work

$$[iD,K^{a}] = - K^{a} \ \ (5.25a)$$
$$[iK^{a},K^{b}] = 0 \ \ (5.25b)$$
$$[iK^{a},P^{b}] = 2\eta^{ab} D - 2J^{ab} \ \ (5.25c)$$



more to come...

 

[samalkhaiat]

From SO(2,n)-algebra to the conformal algebra

Here we will show that the algebra of C(1,n-1) is isomorphic to that of the "Lorentz" group SO(2,n). The latter may be cosidered as a set of pseudoorthognal transformations in a (n+2)-dimensional flat spacetime $$M^{n+2}$$ with the metric;

$$\eta^{\mu \nu} = (-1,1,\eta^{ab}) \ \ (6.1)$$

where

$$(\mu , \nu) = -2, -1, 0, 1, ...,n-1$$
$$(a , b) = 0, 1, ..., n-1$$

the algebra of SO(2,n) is given by

$$[iM^{\mu \nu},M^{\rho \sigma}] = \eta^{\nu \rho}M^{\mu \sigma} - \eta^{\mu \rho}M^{\nu \sigma} + \eta^{\mu \sigma}M^{\nu \rho} - \eta^{\nu \sigma}M^{\mu \rho} \ \ (6.2)$$

Ex(6.1) Define the following generators

$$D = M^{-2,-1} \ \ (6.3a)$$
$$J^{ab} = M^{ab} \ \ (6.3b)$$
$$\frac{1}{2}(P^{a} - K^{a}) = M^{-2,a} \ \ (6.3c)$$
$$\frac{1}{2}(P^{a} + K^{a}) = M^{-1,a} \ \ (6.3d)$$

then show that Eq(6.2) gives the conformal algebra.


The origin of the nonlinearity of conformal transformations can now be understood by considering the relation between C(1,n-1) and SO(2,n). In $$M^{n+2}$$ ,the SO(2,n) coordinate transformations are linear. However, by projecting these transformations on $$M^{n}$$ ,we get a nonlinear realization of the group SO(2,n), that coincides with the action of C(1,n-1) on $$M^{n}$$ .


more to come...

 

[samalkhaiat]

From conformal algebra to conformal transformations

Can we arrive at the conformal coordinate transformations from the conformal algebra? Yes, we can. This is because the Minkowski space $$M^{n}$$ can be identified with the coset manifold C(1,n-1)/H, with H the subgroup generated by $$(J^{ab},D,K^{a})$$ ,i.e., it can be parametrized by;

$$T(x) = \exp(iP.x) \ \ (7.1)$$

In general, a set of group elements T(x), labelled by as many parameters as necessary, parametrizes the manifold if each coset contains one of the T's. Once a parametrization T(x) has been chosen, each group element g can be uniquely decomposed into a product

$$g = T(x) . L \ \ (7.2)$$

where T is the representative member of the coset to which g belongs and L connect T to g within the coset. A product of g with an arbitrary element, and in particular with some T(x) will therefore give another T and an L according to

$$g . T(x) = T(\bar{x}) . L \ \ (7.3)$$

where
$$\bar{x} = \bar{x}(x,g) \ \ (7.4a)$$
$$L = L(x,g) \ \ (7.4b)$$

We will see that the conformal algebra is sufficient to determine $$\bar{x}$$ and L explicitly from

$$g . e^{iP.x} = e^{iP. \bar{x}} . L \ \ (7.3')$$


Translations:

$$g = T(a) = \exp(ia.P)$$

gives
$$\bar{x} = x + a$$
$$L = 1$$

Lorentz trans

$$g = \exp(\frac{i}{2} \omega . J)$$

For infinitesimal $$\omega$$ , we can write

$$(1 + \frac{i}{2} \omega . J) e^{ix.P} = e^{ix.P}\left[ e^{-ix.P}(1 + i/2 \omega . J) e^{ix.P}\right] \ \ (7.5)$$

The expression in [ ] can be expanded by use of the Hausdorff formula

$$e^{-A} B e^{A} = B + [B,A] + 1/2 [[B,A],A] + ... \ \ (7.6)$$

this gives

$$(1 + \frac{i}{2}\omega .J) e^{ix.P} = e^{ix.P}\left(1 + \frac{i}{2}\omega .J + \frac{i}{2}\omega_{ab}x_{c}[iJ^{ab},P^{c}] \right) \ \ (7.7)$$

upon inserting the algebra (5.16b), we find

$$(1 + i/2 \omega .J) e^{ix.P} = e^{i(x^{a} - \omega^{ab}x_{b})P_{a}} (1 + i/2 \omega .J) + o(\omega^{2}) \ \ (7.8)$$

For finite $$\omega$$ we find the expected Lorentz transformations

$$\bar{x}^{a} = \Lambda^{a}{}_{b}x^{b}$$

and
$$L = g =\exp(i/2 \omega .J) \ \ (7.9)$$


Ex(7.1): For infinitesimal D-transformations (dilations) and K-transformations (special conformal trans), show that

$$(1 - i \alpha D) e^{ix.P} = e^{i(1 - \alpha )x.P} (1 - i \alpha D) + o(\alpha^{2}) \ \ (7.10)$$

$$(1 + i c.K) e^{ix.P} = e^{iP_{a}(x^{a}+2c.x x^{a}-c^{a}x^{2})}\left(1 + ic.K + 2ic.xD - 2ic_{b}x_{a}J^{ab}\right) + o(c^{2}) \ \ (7.11)$$


Next, we will talk physics :!)

 

[samalkhaiat]

Conformal Invariance; The troubles!

[You don't have to read this post if you are not familiar with QFT]

Whereas conformal transformations are well understood in mathematics, this is not the case in physics! The physical meaning of the conformal transformations in n>2 spacetime dimensions is far from clear. Many believe that conformal invariance has no physical meaning and consequently should not play "any" role in physics!
So, what is the problem? Well, apart from the Poincare' algebra, we can not say that the conformal algebra (invariance) is realized in nature. For if

$$[iD,P^{a}] = P^{a} \ \ (8.1)$$

holds in nature, then it is also true that;

$$e^{-i \alpha D}P^{2}e^{i \alpha D} = e^{-2 \alpha}P^{2} \ \ (8.2)$$

So, if |P> is some 1-particle state of given positive mass m ;

$$P^{2}|P \rangle = m^{2}|P \rangle \ \ (8.3)$$

then the state

$$|\bar{P} \rangle = e^{-i \alpha D} |P \rangle \ \ (8.4)$$

carries the mass
$$\bar{m}^{2} = e^{2\alpha} m^{2} \ \ (8.5)$$

i.e.,

$$P^{2}|\bar{P}\rangle = e^{2\alpha}m^{2}|\bar{P}\rangle \ \ (8.6)$$

If we assume that scale invariance is not spontaneously broken, i.e.,

$$e^{-i\alpha D}|0 \rangle = |0 \rangle \ \ (8.7)$$

then we conclude that

$$|\bar{P}\rangle = e^{-i\alpha D}a^{\dagger}(p) |0 \rangle \propto a^{\dagger}(e^{\alpha}p) |0 \rangle \ \ (8.8)$$

This means that the state $$|\bar{P}\rangle$$ is a quantum of the same field as the state $$|P\rangle$$ but with a rescaled momentum, i.e., by virtue of conformal invariance, both states must belong to the same Hilbert space. Therefore, by Eq(8.5), conformal invariance implies that the mass spectrum is either continuous or all masses vanish! In order to avoid this physically absurd conclusion, conformal symmetry must be explicitly or spontaneously broken (mathematically this problem is equivalent to the fact that $$P^{2}$$ is not a Casmir's operator).

Notice though, if the vacuum was not unique then the state $$|\bar{P}>$$ would belong to a different Hilbert space than the state |P> and our conclusion would be avoided.

Another problem is that conformal invariance at the quantum level does not follow from conformal invariance at the classical level. Operationally this means that the renormalized energy-momentum tensor possesses a non-vanishing trace and there are anomalies in the dilation and conformal currents.
We will see that scale invariance requires that there be no dimensionfull parameters in the Lagrangian. But QFT does not make sense without a regularization prescription that introduces a scale in the theory. This scale, at which the theory is renormalized, breaks conformal invariance.

Ex(8.1): Derive (8.2) and (8.6).

more to come...

 

[samalkhaiat]

The action of generators on local fields; (I) General

Before considering conformal field theory and the conditions under which an action integral is conformally invariant, we must decide how the fields transform under the conformal transformations.
If the configuration variables are transformed according to

$$x \rightarrow \bar{x} = g x \ \ (9.1)$$

with g belongs to some Lie group, state vectors and ,therefore, local fields will be in general subject to a unitary transformations;

$$|\Phi \rangle \rightarrow |\Phi_{g} \rangle = u(g) |\Phi \rangle \ \ (9.2)$$

$$\bar{\phi}_{i}(x) = u^{-1}(g) \phi_{i} u(g) = \mathcal{D}_{i}{}^{j}\phi_{j}(g^{-1}x) \ \ (9.3)$$

where $$\mathcal{D}(g)$$ is a finite-dimensional (matrix) representation of the group, i.e., set of matrices satisfying the Lie algebra of the group. They need not necessarily be unitary. Their effect is a reshuffling of field components.

Eq(9.3) can be understood as follow;

Since the expectation values;

$$C_{i}(x) = \langle \Phi | \phi_{i}(x) | \Phi \rangle \ \ (9.4)$$

are C-numbers, we expect them to transform like classical fields (which belong to some representation D):

$$\bar{C}_{i}(\bar{x}) = \mathcal{D}_{i}{}^{j}C_{j}(x) \ \ (9.5)$$
and, by Eq(9.2), the expectation value of the transformed field operator $$\bar{\phi}(\bar{x})$$ in a state described by $$|\Phi \rangle$$ is the same as the expectation value calculated using the untransformed operator $$\phi(\bar{x})$$ and the transformed state $$|\Phi_{g}\rangle$$ :

$$\langle \Phi |\bar{\phi}(\bar{x}) |\Phi \rangle \equiv \langle \Phi |u^{-1}\phi(\bar{x}) u |\Phi \rangle = \mathcal{D} \langle \Phi |\phi(x) |\Phi \rangle \ \ (9.6)$$

Since this is true for any state vector, we conclude that;

$$\bar{\phi}(\bar{x}) = u^{-1}\phi (\bar{x}) u = \mathcal{D} \phi(x) \ \ (9.7)$$

or, by (9.1),

$$\bar{\phi}(\bar{x}) = u^{-1}\phi(\bar{x}) u = \mathcal{D} \phi (g^{-1}\bar{x}) \ \ (9.8)$$

This leads to Eq(9.3) after renaming the coordinates $$\bar{x}=x$$ .
Now I want you to work out the action of infinitesimal generators on local fi eld.
Ex(9.1): Let

$$g^{-1}x^{a} = x^{a} + \lambda_{A}f^{Aa}(x) \ \ (9.9)$$
$$u(g) = \exp(- i \lambda_{A}G^{A}) \ \ (9.10)$$

and

$$\mathcal{D}^{i}_{j} = \exp(\lambda_{A}D^{Ai}{}_{j}) \ \ (9.11)$$

where $$\lambda_{A}$$ [A = 1,2,...,dim(g)] are the infinitesimal

parameters and $$D^{A}$$ forms a matrix representation of the

generators $$G^{A}$$ .
Write Eq(9.3) to 1st order in lambda and show that;

$$\delta^{A}_{(\lambda)}\phi_{i}(x) = [iG^{A} , \phi_{i}(x)] = f^{Aa}\partial_{a}\phi_{i} + D_{i}^{Aj}\phi_{j}(x) \ \ (9.12)$$

where

$$\delta \phi_{i} = \lambda_{A}\delta^{A}_{(\lambda)}\phi_{i} = \bar{\phi}_{i}(x) - \phi_{i}(x) \ \ (9.13)$$

is the infinitesimal variation (change) in the form of the field function.


more to come soon...

 

[samalkhaiat]

The action of conformal generators II

Here, we will use Eq(9.3) or its infinitesimal version (9.12) to determine the action of the generators (P,J,D,K) on local field:

For translation;

$$g^{-1}x = x - a \ \ (10.1)$$
$$u(a) = \exp(i a.P) \ \ (10.2)$$
$$\mathcal{D}^{i}_{j} = \delta^{i}_{j} \ \ (10.3)$$

we find

$$\bar{\phi}_{i}(x) = \phi_{i}(x) - a_{a}[i P^{a},\phi_{i}] = \phi_{i}(x) - a_{a}\partial^{a}\phi_{i}$$
or
$$\delta^{a}_{(T)}\phi = [i P^{a},\phi (x)] = \partial^{a}\phi \ \ (10.4)$$

For Lorentz transformations;

$$g^{-1}x^{a} = x^{a} - \omega^{ab}x_{b} \ \ (10.5)$$
$$u(\omega) = \exp(-\frac{i}{2}\omega . J) \ \ (10.6)$$
$$\mathcal{D}^{i}_{j} = \exp(\frac{1}{2}\omega . \Sigma^{i}_{j}) \ \ (10.7)$$

where $$\Sigma$$ is the appropriate spin matrix for the field,

we get;

$$\delta^{ab}_{(L)}\phi_{i} = [i J^{ab},\phi_{i}] = (x^{a}\partial^{b} - x^{b}\partial^{a})\phi_{i} + \Sigma^{abj}_{i}\phi_{j} \ \ (10.8)$$

Ex(10.1): Check Eq(10.8).

For Scale transformations (Dilations);

$$g^{-1}x = x + \alpha x \ \ (10.9)$$
$$u(\alpha) = \exp(-i \alpha D) \ \ (10.10)$$
$$\mathcal{D} = \exp(d \alpha ) \ \ (10.11)$$

where d is (real number) called the canonical dimension of the field. We will see how to find its values for fermions and bosons.

Inserting (10.9), (10.10) and (10.11) in Eq(9.3), we find

$$\bar{\phi}_{i} = \phi_{i} + \alpha [i D, \phi_{i}] = \phi_{i} + \alpha (x.\partial + d ) \phi_{i}$$

or

$$\delta_{(D)}\phi = [i D, \phi ] = ( d + x . \partial ) \phi \ \ (10.12)$$


For free fields, we have the canonical equal-time (anti)commutation relations:

$$\{ \psi (x,t) , \psi^{\dagger}(y,t) \} = \delta^{3}(x-y) \ \ (10.13a)$$
$$[\phi(x,t) , \dot{\phi}(y,t) ] = i \delta^{3}(x-y) \ \ (10.13b)$$

The scale dimension d is defined so that (10.13) remain invariant under scale transformation. Transforming the fields in (10.13a) according to Eq(9.7) and (10.11), i.e.,

$$\psi(x,t) = e^{- d \alpha} \bar{\psi}(\bar{x},\bar{t}) \ \ (10.14)$$

one finds

$$\{\psi(x,t) , \psi^{\dagger}(y,t) \} = e^{-2d \alpha} \delta^{3}(\bar{x} - \bar{y}) = e^{\alpha (3 - 2d)}\delta^{3}(x-y) \ \ (10.15)$$

Thus the invariance of (10.13a) implies that the canonical scale dimension for fermion field is

$$d_{\psi} = \frac{3}{2} \ \ (10.16)$$

Ex(10.2): show that for boson field, the invariqance of (10.13b) under scale transformations, implies

$$d_{\phi} = 1 \ \ (10.17)$$

These values for d correspond to the natural dimention of the fields in units of mass.


Special conformal transformations;

In Eq(9.12), if we put A = a (a spacetime index),

$$f^{ab} = 2 x^{a}x^{b} - \eta^{ab} x^{2} \ \ (10.18)$$
$$D^{aj}_{i} = -2 d x^{a}\delta^{j}_{i} + 2 x_{b}\Sigma^{baj}_{i} \ \ (10.19)$$

and
$$G^{a} = K^{a} \ \ (10.20)$$

we find

$$\delta^{a}_{(C)}\phi_{i} = [i K^{a}, \phi_{i}] = (2x^{a}x^{b} - \eta^{ab}x^{2}) \partial_{b}\phi_{i} + 2d x^{a}\phi_{i} - 2 x_{b}\Sigma^{baj}_{i}\phi_{j} \ \ (10.21)$$

In offering Eq(10.19) we assumed that the field carries irreducible representation of Lorentz group. This will become clear in the next post when we rederive the equations (10.8),(10.12) and (10.21), using the theory of induced representations. At this moment in time, Eq(5.25c) and (7.11) are the only hints we have for (10.19).


more to come soon..

 

[samalkhaiat]

The actin of conformal generators III

Let $$\mathcal{G}$$ be a Lie group such that the Minkowski space can be identified with the coset space

$$M^{n} = \mathcal{G} / H \equiv \frac{(G^{A},P^{a})}{(G^{A})} \ \ (11.1)$$

where the subgroup H (generated by $$G^{A}$$ ) leaves the origin x = 0 invariant, and P generates the usual abelian group of translation.

Now, given any representation of the "little" group H on $$\phi (0)$$ , we can induce it to the whole group, i.e., to a representation of $$\mathcal{G}$$ on $$\phi (x)$$ . This is done using the definition

$$\phi (x) = e^{i P.x}\phi (0) e^{-i P.x} \ \ (11.2a)$$

i.e.,

$$\partial \phi (x) = [iP_{a}, \phi (x)] \ \ (11.2b)$$

and the algebra of $$\mathcal{G}$$ , i.e., the commutators [P,G], [P,P] and [G,G].

We take the action of H to be

$$[iG^{A},\phi_{i}(0)] = D^{Aj}_{i}\phi_{j}(0) \ \ (11.3)$$

where D's are matrix representation of the generators G's of the little group H.

Ex(11.1): Show that the D's in (11.3) form a matrix representation of the algebra

$$[iG^{A},G^{B}] = f^{AB}{}_{C}G^{C} \ \ (11.4)$$


Now, Eq(11.2a) and (11.3) can be used to write

$$[i (e^{iP.x} G^{A} e^{-iP.x}) , \phi_{i}(x)] = D^{Aj}_{i}\phi_{j}(x) \ \ (11.5)$$


Applying Hausdroff formula Eq(7.6) to the lefthand-side, gives

$$[iG^{A},\phi (x)] = D^{Aj}_{i}\phi_{j}(x) + \left[i \left( [iG^{A},x.P] -\frac{1}{2}\left[i[G^{A},x.P],x.P\right] + ...\right) , \phi_{i}(x) \right] \ \ (11.6)$$

Notice that the RHS can be evaluated using only the algebra of the group $$\mathcal{G}$$ .

Let us apply this method to the conformal group C(1,n-1).
Here the subgroup (H) that leaves the point x = 0 invariant is generated by;

$$G^{A} = (J^{ab}, D, K^{a}) \ \ (11.7)$$

If we remove the translation generator (P) from the conformal algebra, we get something identical to Poicare' algebra augmented by dilations (D), because of the similar role played by P and K.
Next we introduce a set of matrices;

$$D^{A} = (\Sigma^{ab}, \Delta , \mathcal{K}^{a}) \ \ (11.8)$$

to define the action of this little group on $$\phi (0)$$ ;

$$[iJ^{ab},\phi_{i}(0)] = \Sigma^{abj}_{i}\phi_{j}(0) \ \ (11.9a)$$

$$[iD,\phi_{i}(0)] = \Delta^{j}_{i}\phi_{j}(0) \ \ (11.9b)$$

$$[iK^{a},\phi_{i}(0) = \mathcal{K}^{aj}_{i}\phi_{j}(0) \ \ (11.9c)$$


these matrices must form arepresentation of the reduced algebra

$$[\Sigma,\Delta ] = [\Delta ,\Delta ] = [\mathcal{K},\mathcal{K}] = 0 \ \ (10.10a)$$

$$[\Delta , \mathcal{K}^{a}] = \mathcal{K}^{a} \ \ (11.10b)$$

$$[\Sigma^{ab},\mathcal{K}^{c}] = \eta^{bc}\mathcal{K}^{a} - \eta^{ac}\mathcal{K}^{b} \ \ (11.10c)$$

$$[\Sigma^{ab},\Sigma^{cd}] = \eta^{ac}\Sigma^{bd} - ... \ \ (11.10d)$$


Ex(11.2) use the algebra of (J, D, K) and Eq(11.9) to derive Eq(11.10).


If the field belongs to an irreducible representation of Lorentz group, then by Schur's lemma, any matrix that commutes with all the (irreducible) grnerators $$\Sigma$$ must be a multiple of the unit matrix. Consequently, the matrix $$\Delta$$ is

$$\Delta^{j}_{i} = d \delta^{j}_{i} \ \ (11.11)$$

where d is a real number, and the algebra (11.10b) forces all matrices $$\mathcal{K}$$ to vanish.

If we now put

$$D^{Aj}_{i} = \Sigma^{abj} , d\delta^{j}_{i} , 0$$

and

$$G^{A} = J^{ab} , D, K^{a}$$

in Eq(11.6), we arrive (after using the conformal algebra) at

$$[iJ^{ab},\phi (x)] = (x^{a}\partial^{b} - x^{b}\partial^{a} + \Sigma ) \phi (x) \ \ (11.12a)$$

$$[iD,\phi (x)] = (d + x. \partial ) \phi (x) \ \ (11.12b)$$

and

$$[iK^{a},\phi] = (2x^{a}x^{b} - \eta^{ab}x^{2})\partial_{b}\phi + (2dx^{a} + 2x_{b}\Sigma^{ab})\phi \ \ (11.12c)$$


These equations together with

$$[iP^{a},\phi ] = \partial^{a}\phi \ \ (11.12d)$$


determine the action of the conformal group on local fields. In the next post we will see that, when the conditions for conformal invariance are met, Noether theorem gives us a time independent and Lorentz covariant objects satisfying the conformal algebra and effect the proper transformations on the fields, i.e., satisfing (11.12). Therefore one can identify Noether charges with the conformal generators. However, we will see that the noether charges (P,J,D,K) satisfy (11.12) even when the (K,D)-symmetry is broken!


more to come tomorow...

 

[samalkhaiat]

Local field theory and invariance condition

An arbitrary field theory is described by a Lagrangian density which we take to be a real function of the field variables $$\phi(x)$$ and of their first derivative $$\partial_{a}\phi$$ . If $$\mathcal{L}$$ depends only on the state of the fields in an infinitely small neighbourhood of the point x ,i.e., on the values oh $$\phi$$ and of $$\partial \phi$$ evaluated at the same point x , then it is called a local Lagrangian, and the corresponding theory is said to be a local theory.
So, we may write;

$$\mathcal{L}(x) = \mathcal{L}\left(\phi (x), \partial_{a}\phi (x)\right) \ \ (12.1)$$

It is assumed that $$\phi (x) \rightarrow 0$$ sufficiently fast as $$|\vec{x}| \rightarrow \infty$$ .
The integral of the Lagrangian over bounded, arbitrary contractible, region in spacetime;
$$S = \int_{D}d^{4}x \mathcal{L}(x) \ \ (12.2)$$
where
$$D \equiv ( V \cup \partial V ) \subset \mathbb{R}^{4} \ \ (12.3)$$

is called the action.

From the variation principle of stationary action;

$$\delta \int_{D} d^{4}x \mathcal{L}(x) = 0 \ \ (12.4)$$

together with the assumption that the variation of the fields $$\delta \phi$$ vanish on $$\partial V$$ (the surface of the 4-volume over which the integral is taken) but arbitrary in $$V$$ , we obtain the field equations in V:

$$\frac{\delta \mathcal{L}}{\delta \phi} = 0 \ \ (12.5a)$$

where

$$\frac{\delta \mathcal{L}}{\delta \phi} \equiv \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{a}\left(\frac{\partial \mathcal{L}}{\partial \partial_{a}\phi}\right) \ \ (12.5b)$$

is the Euler derivative.
In the canonical formalism, the field and its conjugate momentum

$$\pi = \frac{\partial \mathcal{L}}{\partial \partial_{0}\phi} \ \ (12.6)$$

satisfy the equal-time commutation relations

$$[\phi(t,x) , \pi (t,y)] = i\delta^{3}(x-y) \ \ (12.7a)$$
$$[\phi , \phi ] = [\pi ,\pi ] = 0 \ \ (12.7b)$$

The connection between continuous symmetries and conserved quantities is highlighted by Noether's theorem which states: to every Lie group of transformations which leaves the action unchanged, there corresponds a definite combination of the Lagrangian derivatives that determines the field invariants when the fields satisfy the E-L equation, i.e., when the field equations are satisfied, a systematic procedure for obtaining conservation laws can be developed from a direct study of the invariance properties of the action integral.
Therefore, it is interesting to ask what conditions on $$\mathcal{L}$$ insure the invariance of the action? In other words, if G is a given Lie group, then the question will be; how to formulate a G-invariant theory?

To answer this question, let us consider an infinitesimal coordinate transformations
$$x^{a} \rightarrow \bar{x}^{a} = x^{a} + \lambda_{A}f^{Aa}(x) \ \ (12.8)$$
such that the fields transform according to some known representation of G :

$$\phi (x) \rightarrow \bar{\phi}(\bar{x}) = \phi (x) + \lambda_{A}\bar{\delta}^{A}\phi (x) \ \ (12.9)$$

where $$\lambda_{A}$$ are the infinitesimal parameters, A=1,2,..,dim(G),
and $$\bar{\delta}^{A}\phi$$ is the local variation in the field due to the change in the form of the field function and its argument:

$$\bar{\delta}^{A}\phi = \delta^{A}\phi + f^{Aa}\partial_{a}\phi \ \ (12.10)$$

where the variation in the form of the field is defined by

$$\delta \phi \equiv \lambda_{A}\delta^{A}\phi = \bar{\phi}(x) - \phi (x) \ \ (12.11)$$

Please note that

$$\partial_{a}\delta^{A}\phi = \delta^{A}\partial_{a}\phi \ \ (12.12a)$$

but

$$\partial_{a}\bar{\delta}^{A}\phi = \bar{\delta}^{A}\partial_{a}\phi + \partial_{a}f^{Ab}\partial_{b}\phi \ \ (12.12b)$$

Ex(12.1): Derive the above two equations.

Under these transformations (the action of the group G), the form of a local Lagrangian would change according to

$$\delta^{A}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}\delta^{A}\phi + \pi^{a}\partial_{a}\delta^{A}\phi \ \ (12.13)$$

where

$$\pi^{a} = \frac{\partial \mathcal{L}}{\partial \partial_{a}\phi} \ \ (12.14a)$$

$$\pi^{0}\equiv \pi \ \ (12.14b)$$

Since the explicit form for $$\bar{\delta}^{A}\phi$$ is assumed known (matrix representation of G), we can use (12.10) to put (12.13) in a form like;

$$\delta^{A}\mathcal{L} + \partial_{a}(f^{Aa}\mathcal{L}) = C^{A}\left(\mathcal{L},\frac{\partial \mathcal{L}}{\partial \phi},\pi^{a};D \right) \ \ (12.15)$$

with $$C^{A}$$ has no exiplicit x-dependence and $$D$$ is the matrix representation of G.

If we can construct a Lagrangian (i.e.,formulate a theory) in such a way that

$$C^{A} = 0 \ \ (12.16)$$

then our theory will be invariant under the given Lie group G. Mathematically speaking, the solution $$(\mathcal{L})$$ of the system of 1st order PDE (12.16), describes the most general G-invariant theory. Also, whether or not a given theory is G-invariant can be decided by examining the Lagrangian against (12.16).

To show that (12.16) is invariance condition, look at the variation of the action;

$$\bar{\delta}\int_{D}\mathcal{L} d^{4}x = \int_{\bar{D}} \bar{\mathcal{L}}(\bar{x}) d^{4}\bar{x} - \int_{D}\mathcal{L}(x) d^{4}x \ \ (12.17)$$

it is made of the sum of the local variation of $$\mathcal{L}$$ and of the variation in the region of integration:

$$\bar{\delta}^{A}\int_{D}\mathcal{L} d^{4}x = \int_{D}\left(\bar{\delta}^{A}\mathcal{L}\right) d^{4}x + \int_{D} \mathcal{L}(x) \bar{\delta}^{A}(d^{4}x) \ \ (12.18)$$

Symbolically, this can be understood in terms of the algebric property of the

local variation symbol $$\bar{\delta}$$ .
Indeed, to the 1st order, it is a derivation;

$$\bar{\delta}(fg) = (\bar{\delta}f)g + f(\bar{\delta}g)$$

Now, the local variation in $$\mathcal{L}$$ can be expressed as
a variation in the form of the lagrangian plus "drag";

$$\bar{\delta}^{A}\mathcal{L} = \delta^{A}\mathcal{L} + f^{Aa}\partial_{a}\mathcal{L} \ \ (12.19)$$

[see Eq(12.10)], and

$$\bar{\delta}(d^{4}x) \equiv d^{4}\bar{x} - d^{4}x = J(\frac{\bar{x}}{x}) d^{4}x - d^{4}x \ \ (12.20)$$

J-is the Jacobian of the transformations (12.8). To the 1st order we can write;

$$\bar{\delta}^{A}(d^{4}x) = \partial_{a}f^{Aa} d^{4}x \ \ (12.21)$$

Putting (12.19) and (12.21) in Eq(12.18) leads to;

$$\bar{\delta}^{A}\int_{D}\mathcal{L} d^{4}x = \int_{D} C^{A} d^{4}x \ \ (12.22)$$

Since $$D \subset \mathbb{R}^{4}$$ is an arbitrary contractible domain, we conclude that the action is invariant if and only if $$C^{A}= 0$$ .

So what is the link between the invariance condition (12.16) and Noether theorem?

In terms of the Euler derivative (12.5b), we can rewrite (12.13) as;

$$\delta^{A}\mathcal{L} = \partial_{a}(\pi^{a}\delta^{A}\phi ) + \frac{\delta \mathcal{L}}{\delta \phi}\delta^{A}\phi \ \ (12.23)$$

Putting this in (12.15) we find;

$$C^{A} = \frac{\delta \mathcal{L}}{\delta \phi}\delta^{A}\phi + \partial_{a}(\pi^{a}\delta^{A}\phi + f^{Aa}\mathcal{L}) \ \ (12.24)$$

Thus, our invariance condition $$C^{A} = 0$$ is equivalent to the Noether identity;

$$\frac{\delta \mathcal{L}}{\delta \phi}\delta^{A} = \partial_{a}\mathcal{J}^{aA} \ \ (12.25)$$

where

$$\mathcal{J}^{aA} = -\pi^{a}\delta^{A}\phi - f^{Aa}\mathcal{L} \ \ (12.26)$$

is the canonical Noether current. the conservation of this current;

$$\partial_{a}\mathcal{J}^{aA} = 0 \ \ (12.27)$$

follows from (12.25), when the field satisfies E-L equations.

Thus, in terms of the invariance condition (12.16), the statement of Noether theorem becomes

$$\left( C^{A}= 0, \frac{\delta \mathcal{L}}{\delta \phi}= 0 \right) \Rightarrow \left(\partial_{a}\mathcal{J}^{aA}= 0 \right) \ \ (12.28)$$


When G is a symmetry group (C = 0), we can use (12.27) to show that the Noether charges

$$Q^{A}= \int d^{3}x \mathcal{J}^{A0} \ \ (12.29)$$


1) are time-independent.
2) are G-covariant.
3) generate the correct transformation on the field;

$$\delta^{A}\phi = [i Q^{A}, \phi ] \ \ (12.30)$$

4) satisfy the Lie algebra of G:

$$[iQ^{A},Q^{B}] = f^{AB}{}_{C}Q^{C} \ \ (12.31)$$

Because of (3) and (4), Q is called the generator of the symmetry transformations.

Ex(12.2) Derive Eq(12.30).

It is clear that when G is not a symmetry operation, i.e., $$C^{A}\ne 0$$ , it is still possible to define the current (12.26) which is not conserved, and the charge (12.29). In this case, two situations are distinguished. If G is an internal Lie group, i.e., $$f^{Aa}= 0$$ , the Noether charges would still have the properties (3) & (4) but not (1) or (2). For Spacetime groups (1),(2) & (4) are fulse but (3) is still true.

Ex(12.3): Let G be a broken internal symmetry with

$$f^{Aa}=0 \ \ (12.32a)$$

$$\delta^{A}\phi = i T^{A}\phi \ \ (12.32b)$$

and

$$[iT^{A},T^{B}] = f^{AB}{}_{C}T^{C} \ \ (12.32c)$$

Define the non-conserved current

$$\mathcal{J}^{Aa} = \pi^{a}\delta^{A}\phi \ \ (12.33)$$

and its (time-dependent) charge

$$Q^{A}(t) = i \int d^{3}x \pi T^{A}\phi \ \ (12.34)$$

then show that Q satisfies the Lie algebra of G and generates the transformation (12.32b).



In the next post (the last I hope) we will describe the conditions of conformal invariance.

 

[samalkhaiat]

Conformal Field Theory (I)

In this and next post, we will apply the general formalism of post #12 to the conformal group C(1,3).
we will derive the explicit form of the invariance condition Eq(12.16) for translations, Lorentz, scale and special conformal transformations, and find the corresponding Noether currents and charges.

Translations

$$\bar{x}^{a}= x^{a} + \eta^{ab}a_{b} \ \ (13.1a)$$

$$\delta^{b}_{(T)}\phi = - \eta^{ba}\partial_{a}\phi \ \ (13.1b)$$

I.e., the notation of post #12 becomes

$$A = b \ \ (13.2a)$$

$$f^{Aa}\rightarrow \eta^{ba} \ \ (13.2b)$$

$$\delta^{A}\rightarrow \delta^{b}_{(T)} \ \ (13.2c)$$

[(T) is not an index, it stands for translation]

In this case Eq(12.13) becomes;

$$\delta^{b}_{(T)}\mathcal{L} = - \partial_{a}(\eta^{ab}\mathcal{L}) + \frac{\delta \mathcal{L}}{\delta x_{b}} \ \ (13.3)$$

From Eq(12.15) & (12.16), we see;

$$C^{b}_{(T)} \equiv \frac{\delta \mathcal{L}}{\delta x_{b}} = 0 \ \ (13.4)$$

Thus, as expected, the condition for translation invariance is equivalent to the statement that $$\mathcal{L}$$ has no explicit x-dependence.
From Eq(12.26) we find the conserved translation current;

$$T^{ab} = \pi^{a}\partial^{b}\phi - \eta^{ab}\mathcal{L} \ \ (13.5)$$

This is the canonical energy momentum tensor. The corresponding charge

$$P^{a} =\int d^{3}x T^{0a} \ \ (13.6)$$

represents the energy momentum 4-vector.

Ex(13.1): Given;

$$\partial_{a}T^{ab}= 0 \ \ (13.7)$$

show that $$P^{a}$$

1) is time-independent.
2) is 4-vector.
3) grnerates the transformation;

$$\delta^{a}_{(T)}\phi = [i P^{a},\phi] \ \ (13.8)$$

4) satisfies the Lie algebra of the translation group;

$$[i P^{a},P^{b}] = 0 \ \ (13.9)$$


Lorentz Transformation

Here, the index A of post #12 is a double spacetime index;

$$A = ab$$

$$f^{(bc)a}(x) = \eta^{ab}x^{c} - \eta^{ac}x^{b} \ \ (13.10a)$$

$$\delta^{bc}_{(L)}\phi = ( x^{b}\partial^{c} - x^{c}\partial^{b} + \Sigma^{bc}) \phi \ \ (13.10b)$$

Inserting these in (12.13) and using the translation invariance Eq(13.4), leads to;

$$\delta^{bc}_{(L)}\mathcal{L} = - \partial_{a}(f^{bca}\mathcal{L}) + C^{bc}_{(L)} \ \ (13.11a)$$

where

$$C^{bc}_{(L)} \equiv ( \pi^{b}\partial^{c} - \pi^{c}\partial^{b}) \phi + \pi^{a}\Sigma^{bc}\partial_{a}\phi + \frac{\partial \mathcal{L}}{\partial \phi}\Sigma^{bc}\phi \ \ (13.11b)$$

Hence the condition for Lorentz invariance of the theory is;

$$C^{bc}_{(L)} = 0 \ \ (13.12)$$

and the canonical Noether current is

$$M^{abc} = x^{b}T^{ac} - x^{c}T^{ac} + \pi^{a}\Sigma^{bc}\phi \ \ (13.13)$$

The integrated charge (the ungular momentum of the field) is

$$J^{bc} = \int d^{3}x M^{0bc} = L^{bc} + S^{bc} \ \ (13.14a)$$

where

$$L^{bc} = \int d^{3}x (x^{b}T^{0c} - x^{c}T^{0b}) \ \ (13.14b)$$

is the intrinsic "orbital" angular momentum of the field, and

$$S^{bc} = \int d^{3}x \pi \Sigma^{bc}\phi \ \ (13.14c)$$

describes the polarization properties of the field. It corresponds to the spin of particles described by the quantized field.

Warning: some textbooks call S; the spin "tensor", and L; the orbital angular momentum "tensor". This is not correct, in general, L and S are not covariant quantities, the sum of them is.

Ex(13.2): Check, by explicit calculation, that

$$\partial_{a}M^{abc} = 0 \ \ (13.15)$$

follows from the invariance condition (13.12) and E-L equation.

Ex(13.3): Show that

$$[i J^{ab},\phi] = \delta^{ab}_{(L)}\phi$$

$$[i J^{ab}, P^{c}] = \eta^{ac}P^{b} - \eta^{bc}P^{a}$$
and
$$[i J^{ab},J^{cd}] = \eta^{bc}J^{ad} - \eta^{ac}J^{bd} + \eta^{ad}J^{bc} - \eta^{bd}J^{ac}$$

Ex(13.4): Show that the Belinfante tensor, defined by

$$T^{ab}_{(B)} = T^{ab} + \frac{1}{2}\partial_{c}X^{cab} \ \ (13.16a)$$

where

$$X^{cab} = \pi^{c}\Sigma^{ab}\phi - \pi^{a}\Sigma^{cb}\phi - \pi^{b}\Sigma^{ca}\phi \ \ (13.16b)$$

is conserved and leads to the same momentum 4-vector $$P^{b}$$ . Hence it may be used instead of the canonical tensor $$T^{ab}$$ .
Use the Lorentz-invariance condition (13.12) as well as E-L equation to show that

$$T^{ab}_{(B)} = T^{ba}_{(B)} \ \ (13.17)$$


Ex(13.5): Define the tensor

$$M^{abc}_{(B)} = x^{b}T^{ac}_{(B)} - x^{c}T^{ab}_{(B)} \ \ (13.18)$$

then show that it is conserved when $$M^{abc}$$ is. Also show that it leads to the same angular momentum $$J^{ab}$$ as $$M^{abc}$$ does.

more to come...

 

[samalkhaiat]

Conformal field theory II: Scale & conformal Invariance

Cosider an infinitesimal scale transformation;

$$\bar{x}^{a} = x^{a} - \alpha x^{a} \ \ 14.1$$

In here, the parameter carries no index, so the index A of post #12 disapears:

$$f^{a}(x) = - x^{a} \ \ (14.2)$$

$$\delta_{(D)}\phi = ( d + x^{c}\partial_{c})\phi \ \ (14.3)$$

Following the general formalism of post #12, it is easy to obtain

$$\delta_{(D)}\mathcal{L} = \partial_{a}(x^{a}\mathcal{L}) + C_{(D)} \ \ (14.4)$$

with
$$C_{(D)} \equiv d \frac{\partial \mathcal{L}}{\partial \phi}\phi + (d+1) \pi_{c}\partial^{c}\phi - 4 \mathcal{L} = 0 \ \ (14.5)$$

is the condition for dilation invariance.
Since d(boson) = 1 and d(fermion) = 3/2, the kinetic term of the Lagrangian satisfies (14.5). Examples of scale invariant interactions are;

$$\phi^{4}\ \mbox{and}\ \bar{\psi}\ \psi \phi .$$

It is easy to see that (14.5) requires that the scale dimension of $$\mathcal{L}$$ be 4, i.e., that there be no dimensionful parameters in the lagrangian. Terms like $$m^{2}\phi^{2}$$ would of course break the scale symmetry explicitly as their scale dimension is 2. Notice the difference between scale dimension and ordinary dimention which is 4 for the term $$m^{2}\phi^{2}$$ . Scale transformation effects the dynamical variables but not the dimensionful parameters.

Ex(14.1) Show that the scale-invariance condition (14.5) implies

$$d(\mathcal{L}) = 4 \ \ (14.6)$$

In terms of the canonical energy-momentum tensor, the dilation Noether current is

$$D^{a} = T^{ab}x_{b} + d \pi^{a}\phi \ \ (14.7)$$

and the corresponding charge is

$$D = \int d^{3}x (T^{0b}x_{b} + d \pi \phi ) \ \ (14.8)$$

Ex(14.2) Use E-L equation to put the invariance condition (14.5) in the form

$$T^{a}_{a} + \partial_{c}( \pi^{c} d \phi ) = \partial_{c}D^{c} = 0 \ \ (14.9)$$

then use this to prove

$$[i D,P^{a}] = P^{a} \ \ (14.10)$$
$$[i J^{ab},D] = 0 \ \ (14.11)$$

Ex(14.3) Show that the dilation charge D generates the correct transformation on the field

$$\delta_{(D)}\phi = [i D ,\phi ] \ \ (14.12)$$

Ofcourse this is true even in the absence of scale symmetry, i.e., even if (14.5) is not true.
The trace of $$T^{ab}$$ lives in the invariance condition (14.9) for a reason. It did not appears there by accident! Let me explain this:
Under an arbitrary transformation of the coordinates

$$x \rightarrow x + f(x)$$

the action changes as

$$\delta S = \int d^{4}x \frac{\partial \mathcal{L}}{\partial\eta_{ab}} \delta \eta_{ab} = (1/2) \int d^{4}x \theta^{ab}(\partial_{a}f_{b} + \partial_{b}f_{a}) \ \ (14.13)$$

Where $$\theta_{ab}$$ is the symmetric energy-momentum tensor(we will call it E-M tensor from now on). This is true even if the field does not satisfy the E-L equation.
From the conformal Killing equation (1.2), it follows that

$$\delta S = (1/2) \int d^{4}x \theta_{a}^{a} F(x) \ \ (14.14)$$

and therefore, traceless E-M tensor implies that the action is invariant under the conformal group C(1,3). Of course the converse is not true because F(x) is not an arbitrary function.

more to come

I'm having problem with my PC, so I will break this post to 2 may be 3 smaller posts

 

[samalkhaiat]

Conformal field theory III: Special conformal invariance

The Poicare' group together with dilations forms a subgroup of the full conformal group. This means that a theory invariant under this subgroup is not necessarily invariant under special conformal transformations. The conditions under which it should be invariant are determined by the tracelessness of the E-M tensor.
Under certain condition, the E-M tensor of a theory, with scale invariance, can be made traceless much in the same way as it can be made symmetric in a Lorentz-invariant theory. If this can be done, then it follows from Eq(14.14) that full conformal invariance is a consequence of scale invariance and poincare' invariance.

Let us do this trick. We start with scale invariant theory, i.e., with;

$$\partial_{a}D^{a} = T_{a}^{a} + \partial_{a}( \pi^{a} d \phi ) = 0 \ \ (15.1)$$

Next we define the field virial by

$$V^{a} = \pi^{a} d \phi - \pi_{c}\Sigma^{ca}\phi \ \ (15.2)$$

and consider the following tensor

$$\theta^{ab} = T^{ab} + (1/2) \partial_{c}X^{cab} + (1/2) \partial_{c}\partial_{d}Y^{cdab} \ \ (15.3)$$

where the first 2 terms constitute the Belinfante tensor (13.16a).

1) from the conservation law

$$\partial_{a}\theta^{ab} = 0 \ \ (15.4a)$$

it follows that

$$\partial_{a}\partial_{b}\partial_{c}Y^{cdab} = 0 \ \ (15.4b)$$

Hence, Y must not be completely symmetric in the first 3 indices.

2) from the symmetry property

$$\theta_{ab} = \theta_{ba} \ \ \ (15.5a)$$

it follows that

$$\partial_{c}\partial_{d}Y^{cdab} = \partial_{c}\partial_{d}Y^{cdba} \ \ \ (15.5b)$$

Thus, the part of Y antisymmetric in (a,b) must also be antisymmetric in (c,d),i.e.,

$$Y^{cdab} - Y^{cdba} \propto ( \eta^{ca}\eta^{db} - \eta^{cb}\eta^{ad}) \ \ \ (15.5c)$$

3) finally, we take the trace of Eq(15.3):

$$\theta_{a}^{a} = T^{a}_{a} + (1/2) \partial_{c}X^{ca}{}_{a} + (1/2) \partial_{a}\partial_{b}Y^{abc}{}_{c} \ \ \ (15.6)$$

From Eq(13.16b) we see that

$$\partial_{c}X^{ca}{}_{a} = - 2 \partial_{c}( \pi_{a}\Sigma^{ca}\phi ) \ \ \ (15.7)$$

Using (15.2), this becomes

$$\partial_{c}X^{ca}{}_{a} = \partial_{a}( \pi^{a} d \phi - V^{a}) \ \ \ (15.8)$$

Putting this in (15.6) and using (15.1), leads to

$$\theta_{a}^{a} = \partial_{c}D^{c} - \partial_{a}V^{a} + (1/2)\partial_{a}\partial_{b}Y^{abc}{}_{c} \ \ \ (15.9)$$

If the virial is a total divergence

$$V^{a} = \partial_{c}\sigma^{ca} \ \ \ (15.10)$$

then Y can be constructed such that

$$\partial_{c}V^{c} = (1/2) \partial_{a}\partial_{b}X^{abc}{}_{c} \ \ \ (15.11)$$

This leads to our final result

$$\partial_{c}D^{c} = \theta_{a}^{a} \ \ \ (15.12)$$

This means that scale invariance implies that the "improved" E-M tensor (15.3) is traceless, provided that the field virial satisfies (15.10).
This shows that for conformal invariance two conditions must be met:
1) the theory should be scale invariant,i.e., (14.5) must be true; and
2) Eq(15.10) must hold.

more to come

 

[samalkhaiat]

Coformal field theory IV

In this (LAST) post, we will rederive the conditions for special conformal invariance using, this time, the general formalism of post #12.
From the infinitesimal conformal transformations:

$$f^{ba}(x) = 2 x^{b}x^{a} - \eta^{ba}x^{2} \ \ 16.1$$

$$\delta^{b}_{(C)} = f^{ba}\partial_{a}\phi + 2 d x^{b} \phi - 2 x_{a}\Sigma^{ab}\phi \ \ 16.2$$


it is easy to see that

$$\delta^{b}_{(C)}\mathcal{L} = \partial(f^{ba}\mathcal{L}) + 2 C_{(D)}x^{b} + V^{b} \ \ 16.3a$$

where

$$C_{(D)} = d \frac{\partial \mathcal{L}}{\partial \phi} \phi + (d+1) \pi_{a}\partial^{a}\phi - 4 \mathcal{L} \ \ 16.3b$$

Inderiving (16.3), Lorntz and translation invariance conditions are used.

From (16.3) we see that conformal invariance requires that

$$C_{(D)} = 0 \ \ 16.4a$$

$$V^{c} = \partial_{b}\sigma^{ba} \ \ 16.4b$$

You might ask why not $$C = 0 \mbox{and} V^{a} = 0$$ , after all this how it should be according to our general invariance condition Eq(12.16)? Well, explicit computation show that V vanishes identically for spin-(1/2) and spin-1 fields, but for spin-0 field, we have

$$V^{a} = \pi^{a}\phi = (1/2) \partial^{a}\phi^{2} \neq 0 \ \ 16.5$$

I have not been able to understand this unique role of the scalar field!
Remarkably,Eq(16.4b) turns out to be true for all renormalizable theories, even though scale invariance is of course broken. Eq(16.4b) is also true in all field theories (involving spins; 0, 1/2 & 1) without derivative coupling. Consequently, for these theories conformal invariance is equivalent to scale invariance.
In terms of the canonical E-M tensor, the conformal Noether current is

$$K^{ab} = f^{a}{}_{c}T^{bc} + 2 \pi^{b}(d x^{a} - x_{c}\Sigma^{ca}) \phi - 2 \sigma^{ab} \ \ 16.6$$

By inserting the formula for $$\theta^{ab}$$ in terms of $$T^{ab}$$ in Eq(16.6) and in Eq(14.7), one finds after many tedious steps:

$$D^{a} = \theta^{ab}x_{b} + \partial_{b}Z^{ba} \ \ 16.7a$$

$$K^{ab} = f^{a}{}_{c}\theta^{cb} + \partial_{c}Z^{cab} \ \ 16.7b$$

The total divergence terms in (16.7) are superpotentials; they may be droped and we are left with the final compact forms for the currents

$$D^{a} = \theta^{ab}x_{b} \ \ 16.8a$$

$$K^{ab} = (2 x^{a}x_{c} - \eta^{a}_{c}x^{2}) \theta^{cb} \ \ 16.8b$$

From these it follows that

$$\partial_{a}K^{ab} = 2 x^{b}\theta_{c}^{c} = 2 x^{b} \partial_{c}D^{c} \ \ 16.9$$

Thus in real life, we see that both scale and conformal symmetries are broken by the trace of the E-M tensor $$\theta$$ .

Finally, we remark that it is also possible to deduce the compact forms of the Noether currents;

$$(\theta_{ab}, M_{(B)}^{ab}, D^{a}, K^{ab})$$

directly from the action integral (14.13)

$$\delta S = \int d^{4}x \theta^{ab}\partial_{a}f_{b} \ \ 16.10$$
or, by factoring out the parameters,

$$\delta = \lambda_{A}\delta^{A}$$

$$f_{b} = f^{A}{}_{b}\lambda_{A}$$

we get

$$\delta^{A} S = \int d^{4}x \theta^{ab}\partial_{a}f^{A}_{b} \ \ 16.11$$

When the field equation is satisfied, the translation current is conserved

$$\partial_{a}\theta^{ab} = 0$$

so, we can write

$$\delta^{A} S = \int d^{4}x \partial_{a}(\theta^{ab}f^{A}_{b}) \ \ 16.12$$

and identify Noether current of the conformal group by

$$\mathcal{J}^{aA} = \theta^{ab}f^{A}{}_{b} \ \ 16.13$$


best regards


sam

I look forward for your inputs; comments questions etc.

 

[strangerep]

FANTASTIC series of posts! Great contribution.
I have plenty of older papers on the conformal group, and some
textbooks, but having it all together like you've done is great.
However, it will take me while to get familiar with your notation,
and some of the details. But I'll ask a couple of things immediately...

1st question:

Back in post #8 you wrote:

then we conclude that

$$|\bar{P}\rangle = e^{-i\alpha D}a^{\dagger}(p) |0 \rangle \propto a^{\dagger}(e^{\alpha}p) |0 \rangle \ \ (8.8)$$

This means that the state $$|\bar{P}\rangle$$ is a quantum of the same field as the state $$|P\rangle$$ but with a rescaled momentum, i.e., by virtue of conformal invariance, both states must belong to the same Hilbert space. Therefore, by Eq(8.5), conformal invariance implies that the mass spectrum is either continuous or all masses vanish! In order to avoid this physically absurd conclusion, conformal symmetry must be explicitly or spontaneously broken (mathematically this problem is equivalent to the fact that $$P^{2}$$ is not a Casmir's operator).

Notice though, if the vacuum was not unique then the state $$|\bar{P}>$$ would belong to a different Hilbert space than the state |P> and our conclusion would be avoided.

I'm familiar with unitarily inequivalent representations in QFT (i.e:
disjoint Fock spaces, and all that). But I didn't follow how you arrived
at the possibility that $$|\bar{P}>$$ could belong to a different
representation. Don't you have to construct the dilation operator D
explicitly in terms of creation/annihilation operators to show this?

2nd question: Do you know of anywhere that I could find all 3 conformal
Casimirs written out explicitly in terms of the generators P, J,D,K? Most
textbooks write them out in terms of operators in the SO(4,2) acting
in the linear space. But converting these expressions to explicit ones
in P,J,D,K is seriously tedious and error-prone, so I'm wondering
whether someone has already done it somewhere?

Thanks again,

- strangerep.

 

[samalkhaiat]

FANTASTIC series of posts! Great contribution.
I have plenty of older papers on the conformal group, and some
textbooks, but having it all together like you've done is great.
However, it will take me while to get familiar with your notation,
and some of the details. But I'll ask a couple of things immediately...

Thanks, I hope you will find it interesting!

1st question:


Back in post #8 you wrote:

I'm familiar with unitarily inequivalent representations in QFT (i.e:
disjoint Fock spaces, and all that). But I didn't follow how you arrived
at the possibility that $$|\bar{P}>$$ could belong to a different
representation. Don't you have to construct the dilation operator D
explicitly in terms of creation/annihilation operators to show this?

Yes, you are correct. If the vacuum is not scale-invariant, then

$$e^{-i \alpha D}|0> \equiv |\bar{0}> = e^{- \alpha f(a, a^{\dagger})}|0> \ \ (I)$$

where f depends on the (spontaneously-broken) model you are using.
In this case we have;

$$|\bar{P}> = e^{-i \alpha D} a^{\dagger}(p) |0> = e^{-i \alpha D} a^{\dagger}(p) e^{i \alpha D} e^{-i \alpha D} |0> = e^{d(a)}a^{\dagger}(e^{\alpha}p) |\bar{0}>$$

where d(a) is the scale dimension of the operator a. Clearly $$|\bar{P}>$$ belongs to a Hilbert space generated from the "new" vacuum $$||\bar{0}>$$ , i.e., it is a quantum of different field. This means that $$|P> \mbox{and}|\bar{P}>$$ do not live in the same multiplet (representation).

As simple example, consider a free massless scalar field with a Lagrangian

$$\mathcal{L} = (1/2)(\partial S)^{2}$$

invariant under the field translations

$$\bar{S}(x) = S(x) + b$$

The correspondig conserved current is

$$J^{\mu} = \partial^{\mu}S(x)$$

and the associated Noether charge is

$$Q = \int d^{3}x \partial_{0}S(x)$$

The vacuum is , of course, not invariant under these transformations;

$$<0|\delta S(x)|0> = b <0|0> \neq 0$$

Alternatively, by expanding the field, we could show that

$$|\bar{0}> = e^{i b Q}|0> = e^{(b/2)[a(0) - a^{\dagger}(0)]}|0>$$

which is a coherent superposition of zero-energy momentum state.
This is the explicit form (for this model) of Eq(I) which led to our conclusion about the inequivalent Hilbert spaces.

2nd question: Do you know of anywhere that I could find all 3 conformal
Casimirs written out explicitly in terms of the generators P, J,D,K? Most
textbooks write them out in terms of operators in the SO(4,2) acting
in the linear space. But converting these expressions to explicit ones
in P,J,D,K is seriously tedious and error-prone, so I'm wondering
whether someone has already done it somewhere?

I have no idea at all, Never seen it done! I have never tried to do this task as well. However, the isomorphism; SO(6) ~ SU(4) can certainly be usefull! The question now is what are the casimir's operators of the group SU(4)
{SU(2,2)to be exact}? Can the Mathmaticians on these forums help?


regards


sam

 

[strangerep]

[...] If the vacuum is not scale-invariant, then

$$e^{-i \alpha D}|0> \equiv |\bar{0}> = e^{- \alpha f(a, a^{\dagger})}|0> \ \ (I)$$

where f depends on the (spontaneously-broken) model you are using.

Yes, but... I'm wondering what is the explicit expression for the f(a,a*)
exponent in the case of dilations and special conformal transformations?
Was that what you intended to show with your field-translation example,
or was that just a separate example of inequivalent representations?

[...] the isomorphism; SO(6) ~ SU(4) can certainly be usefull!
The question now is what are the casimir's operators of the group SU(4)
{SU(2,2)to be exact}?

Plenty of textbooks give the casimirs of SO(4,2), SU(2,2), but they
do it as elegant formulae involving the L_ab generators in the 6-dimensional
hyperspace, where the L_ab act linearly. For example, Wybourne quotes
them as:

C2 = L_ab L^ab

C3 = epsilon_abcdef L^ab L^cd L^ef

C4 = L_ab L^bc L_cd L^da

though I find his 6D metric conventions a bit confusing. But anyway, using
these it's not too hard to re-express C2 in terms of P,J,D,K.However, C3
and C4 are far more tedious. I.e: it was really the boring task of doing
this re-expressing that I was hoping someone else had already done.
(I always seem to make heaps of dumb mistakes when doing such long
tedious straightforward stuff.)

BTW, the reason for my interest is that I'm wondering how one might
construct a QFT using irreps of the conformal group as the starting
point. I.e: not starting from trying to impose a conformal invariance
on a Minkowski-space classical theory, but rather constructing a
more general Fock-like space directly from conformal irreps.
Since mass^2 is not a conformal casimir, I presume the C2 above
is the closest thing, but the hard question then is how to restrict or
contract back to a Poincare-like group in a physically-sensible way,
to get back in touch with orthodox QFT.

 

[samalkhaiat]

Yes, but... I'm wondering what is the explicit expression for the f(a,a*)
exponent in the case of dilations and special conformal transformations?
Was that what you intended to show with your field-translation example,
or was that just a separate example of inequivalent representations?

The example (though not about conformal symmetry) does both jobs for you. I meant to hit to birds by one stone.
In the Lagrangian formalism of free fields, we can write any Noether charge (symmetry generator) in terms of $$(a,a^{\dagger})$$ . All we need is;

1) Lagrangian for a specific representation, say;

$$\mathcal{L} = \frac{1}{2} \partial_{a}S \partial^{a}S$$

2) Expansion in terms of a complete set of solutions of the E-L equation

$$S(x) = \int \frac{d^{3}\vec{p}}{(2 \pi )^{3}2p^{0}} [a(p) e^{-ip.x} + a^{\dagger}(p) e^{ip.x}] \ \ (I)$$

Now for example the $$(D,K^{a})$$ symmetry generators of the above model can be obtained from [see Eq(16.8)];

$$D = \int d^{3}x \theta_{0a}x^{a} \ \ (IIa)$$

$$K^{a} = \int d^{3}x \theta^{ca}(2x^{0}x_{c} - \eta^{0}_{c}x^{2}) \ \ (IIb)$$

with

$$\theta_{ab} = T_{ab} + \frac{1}{6}(\eta_{ab}\partial^{2} - \partial_{a}\partial_{b})S^{2}$$

[see Eq(15.3)]

and

$$T_{ab} = \partial_{a}S \partial_{b}S - \eta_{ab}\mathcal{L}$$

Insert the expansion (I) in (II), do the integrations and you will get D and K in terms of (a , a*)[choose a rainy afternoon to do it ]

You may need to use the mathematical trick of Normal ordering of operators, so that D|0> = 0 .
Notice that such explicit expressions (for Noether chrges) do not exist in the interactive QFT. To deal with interactions, we work with Ward's identities instead.

In the so-called axiomatic QFT, incorporating Noether's idea faces basic difficulty, namely there is no lagrangian available in this formalism. However, one is still able to define the symmetry of a system. How does one get from global conserved quantities (originating from the generators of the symmetry group) to the locally conserved currents? There is still no answer to this question. This is why the axiomatic formalism is just about hopeless for describing interactions. Some (only some) progress was achieved by;
Buchholz, Doplicher and Longo, Ann.Phys,170(1988)1.
Doplicher and Longo, Comm.Math.Phys, 88(1983)399.

C2 = L_ab L^ab

C3 = epsilon_abcdef L^ab L^cd L^ef

C4 = L_ab L^bc L_cd L^da

though I find his 6D metric conventions a bit confusing. But anyway, using
these it's not too hard to re-express C2 in terms of P,J,D,K.However, C3
and C4 are far more tedious. I.e: it was really the boring task of doing
this re-expressing that I was hoping someone else had already done.
(I always seem to make heaps of dumb mistakes when doing such long
tedious straightforward stuff.)

I am afraid, doing physics requires hard work and patience. One needs to do such boring calculation at least once in a life time.

BTW, the reason for my interest is that I'm wondering how one might
construct a QFT using irreps of the conformal group as the starting
point. I.e: not starting from trying to impose a conformal invariance
on a Minkowski-space classical theory, but rather constructing a
more general Fock-like space directly from conformal irreps.

This sounds like axiomatic (conformal) field theory! Nearly all forms of the axiomatic QFT are based upon the spinorial group, the universal covering group of the Poincare' group, i.e.,a semidirect product of the group SL(2,C) and the group of translations. As well as the above mentioned difficulty regarding Noether currents, the axiomatic conformal field theory faces another problem, namely the Buchholz-Fredenhagen theorem;
"If there exist massless particles in a model displaying dilation covariance then the S-matrix for these particles is trivial",i.e., the presence of massless particles (which is crucial for conformal symmetry) renders the theory trivial.
Some people claim that the story changes in 2D spacetime! and that an axiomatic approach to the C(1,1) field theory is possible! I can not comment about this claim as the C(1,1)-theory lies beyond my professional abilities.

Since mass^2 is not a conformal casimir, I presume the C2 above
is the closest thing, but the hard question then is how to restrict or
contract back to a Poincare-like group in a physically-sensible way,
to get back in touch with orthodox QFT.

If you are not (already) familiar with the axiomatic formalism of QFT, I recommend the following book for you

J. Lopuszanski
"An Introduction to Symmetry and Supersymmetry in Quantum Field Theory"
World Scientific, 1991.

The only bad thing about this book is its titel!


regards

sam (I am sorry for the delay June/July are the busiest months of the year)

 

[strangerep]

I am afraid, doing physics requires hard work and patience. One needs
to do such boring calculation at least once in a life time.

I've done a few of those. In this case, I'm more interested in the
destination than the journey, so I found an alternate approach for
getting an explicit form of the dilation generator in Fock space
(adapted from Stefanovich physics/0504062)...

For a finite dilation with parameter 'w' the
ordinary 3-momentum changes according to:

$$p \rightarrow p' = e^{w} p$$

In Fock space, applying such a dilation to a state of momentum
$$\vec{p}$$ is equivalent to destroying a particle of momentum $$\vec{p}$$
and creating one of momentum $$e^{w} \vec{p}$$.
I.e: we require that:

$$e^{i w D} |p> = |e^{w} p> \ \ (D1)$$

or:

$$e^{i w D} |p> = a^{\dagger}(e^{w} p) \ a(p) \ |p>$$

This must hold for any $$\vec{p}$$, so we make the ansatz:

$$e^{i w D} = \int d^{3}p \ a^{\dagger}(e^{w} p) \ a(p) \ \ (D2)$$

A short calculation verifies that (D2) indeed satisfies (D1).

We can then find $$D$$ as follows:

$$D = i \lim_{w \rightarrow 0} \frac{d}{dw} e^{i w D} = i \lim_{w \rightarrow 0} \frac{d}{dw} \int d^{3}p \ a^{\dagger}(e^{w} p) \ a(p)$$

from which we find (after a few lines):

$$D = i \int d^{3}p \ p_{i} \ \frac{\partial a^{\dagger}(p)}{\partial p_{i}} \ a(p) \ \ \ (D3)$$

(implicit sum over i in the above).

I've "verified" this technique by doing a similar thing for
ordinary rotation generators. An explicit expression for
these is given in Greiner & Reinhardt (via the longer
Lagrangian/expansion approach). It matches what
one obtains using the (much faster) Stefanovich method.

Most notably, the dilation generator given by (D3) above
annihilates the vacuum. I.e: dilation transformations
do not take us to an inequivalent representation(!).
This was unexpected for me, since your earlier posts
suggested that a different vacuum would arise.
Or have I misunderstood something?

[...]If you are not (already) familiar with the axiomatic formalism of
QFT, I recommend the following book for you

J. Lopuszanski
"An Introduction to Symmetry and Supersymmetry in Quantum Field Theory"
World Scientific, 1991.

I have ordered same from Amazon.

Cheers,

- sr.

 

[samalkhaiat]

I've done a few of those. In this case, I'm more interested in the
destination than the journey

Good luck.

In Fock space, applying such a dilation to a state of momentum
$$\vec{p}$$ is equivalent to destroying a particle of momentum $$\vec{p}$$
and creating one of momentum $$e^{w} \vec{p}$$.
I.e: we require that:

We conclude this from Eq(8.8) which is a direct result of the assumption that scale invariance is not spontaneously broken [see Eq(8.7)], i.e., $$D|0 \rangle = 0$$ :

$$e^{-i \alpha D} |P \rangle = e^{-i \alpha D} a^{\dagger}(p) |0 \rangle = \left [ e^{-i \alpha D}a^{\dagger}(p) e^{i \alpha D} \right ] e^{-i \alpha D} |0 \rangle$$

Now, if the vacuum is invariant, i.e., scale symmetry is not broken, then we can write

$$e^{-i \alpha D}|P \rangle \propto \int \frac{d^{3}\vec{p}}{2 p^{0}} a^{\dagger}(e^{\alpha}p) a(p) |P \rangle$$

$$D = i \int d^{3}p \ p_{i} \ \frac{\partial a^{\dagger}(p)}{\partial p_{i}} \ a(p) \ \ \ (D3)$$

Yes, this is how it is done in the constructive (not Lagrangian!) formalism. But, howfar we can go without a Lagrangian? Not too far, I am afraid. Wienberg starts his (3-vol) book with the constructive approach, only to find himself forced to introduce the Lagrangian formalism in chapter 7 (of vol I)!
By the way, Stefanovich (whoever he is) is not the one who invented the method you described.

Most notably, the dilation generator given by (D3) above
annihilates the vacuum.

This should not surprise you, because you started with such an assumption!

I.e: dilation transformations
do not take us to an inequivalent representation(!).
This was unexpected for me, since your earlier posts
suggested that a different vacuum would arise.
Or have I misunderstood something?

The 2 or 3 lines after Eq(8.8) says:

In the mathematical-world (not the real one) of "exact" scale symmetry, |P> and |P>' belonge to the same Hilbert space.
But such an "exact" symmetry led to unrealistic mass spectrum. So, in the real world, Dilation symmetry must be broken, i.e., Eq(8.8) and therefore your Eq(D3), are not true. This is why I said;

Notice though, if the vacuum was not unique then the state |P>' would belong to a different Hilbert space than the state |P> and our conclusion (about the mass spectrum) would be avoided.

regards

sam

 

[strangerep]

[...] Most notably, the dilation generator given by (D3) above
annihilates the vacuum.

This should not surprise you, because you started with such an assumption!

Did I? Hmmm. Oh, you mean where I said:

In Fock space, applying such a dilation to a state of momentum
$$\vec{p}$$ is equivalent to destroying a particle of momentum
$$\vec{p}$$ and creating one of momentum $$e^{w} \vec{p}$$

and implicit in that is the fact that the Fock space is constructed
from irreps of the Poincare group only, the vacuum being the
state with lowest energy eigenvalue. But we know that dilations
don't play nice with energy-momentum, unlike rotations and
boosts which can be defined on a common dense domain with
the translation generators (as I've just been reading in
Lopuszanski).

OK, that point is now clearer. So onto other things... which
I should probably put in a separate thread.

Thanks again for your efforts.

- strangerep.

 

[meopemuk]

sam. Thanks for great posts. I would like to ask what is the physical meaning of this Poincare group extension? Are you going to extract any new physics from this, or it is just a purely mathematical exercise?
I am asking because Poincare transformations have clear physical meaning, but I haven't experienced dilations in real life, except probably in "Mother, we shrunk ourselves!" movie.

But, howfar we can go without a Lagrangian? Not too far, I am afraid. Wienberg starts his (3-vol) book with the constructive approach, only to find himself forced to introduce the Lagrangian formalism in chapter 7 (of vol I)!

I am a great fan of the Weinberg's book, and I agree that "constructive" or particle-based approach, though being physically transparent, is helpless when it comes to writing down interacting relativistically invariant Hamiltonians. On the other hand, I can't make sense of field-theoretical derivations, canonical quantization, gauge invariance, etc. Their mathematics looks great, but when they start to talk about fields as of physical entities, I don't believe. So, my idea is that all this field theory stuff is just a nice mathematical trick that somewhat mysteriously gives us correct Hamiltonians. Then we can remain within the constructive or particle-based approach and simply borrow Hamiltonians from field theory, without asking how they were obtained. That's how I read Weinberg. I have a feeling that you strongly disagree.

Regards.
Eugene.

 

[samalkhaiat]

sam. Thanks for great posts. I would like to ask what is the physical meaning of this Poincare group extension? Are you going to extract any new physics from this, or it is just a purely mathematical exercise?

A detailed account of the application of conformal group in 4-dimentional QFT and an extensive bibliography of early work on the subject are found in

I.T. Todorov et al., "Conformal invariance in quantum field theory", Scuola Normale Superiore, Pisa, 1978.

In 2D spacetime, conformal invariance of the world-sheet theories is essential for preventing the appearance of ghosts, and provides a perspective on the variety of consistent string theories that can be constructed.
Heisenberg spin chain and fractional quantum Hall effect are examples of scale-invariant quantum systems. See;
J. Cardy, "conformal field theory comes of age", Physics World,(June 1993),P.29.

On the other hand, I can't make sense of field-theoretical derivations, canonical quantization, gauge invariance, etc.

It should make sense! One needs to apply the quantum laws to all possible classical systems (particle/field).

but when they start to talk about fields as of physical entities, I don't believe.

The physical entities in QFT are "field quanta", i.e., fermions & bosons. We are talking about scale of order $$10^{-13}cm$$ .So, what is field and what is particle on such length scale? I believe [field vs particle]-argument only leads to a metaphysical garbage.

my idea is that all this field theory stuff is just a nice mathematical trick that somewhat mysteriously gives us correct Hamiltonians.

There is no mystry involved! The Hamiltonian is just one of many Norther numbers which can be derived within the Lagrangian formalism.[see post #12]

Then we can remain within the constructive or particle-based approach and simply borrow Hamiltonians from field theory, without asking how they were obtained.

The BORROWING business is a DEFICIENCY there is no need for it.

That's how I read Weinberg. I have a feeling that you strongly disagree.

Your guessing is right! I believe in the saying:

"Give me the Lagrangian, I give you everything."



Regards.

sam

 

[meopemuk]

I believe in the saying:

"Give me the Lagrangian, I give you everything."

I have a problem with this saying, when it is applied to relativistic Lagrangian quantum field theories (such as QED). All QED textbooks do a good job in presenting calculations of S-matrix elements and physical properties that are related to the S-matrix. These properties include scattering cross-sections, energies of bound states, and decay rates. In order to get these nice predictions one needs to introduce renormalization, which is equivalent to adding infinite (mass and chage renormalization) counterterms to the Hamiltonian. This means that the Hamiltonian becomes ill-defined, strictly speaking. Surely, this infinite Hamiltonian can be used to calculate the S-matrix, because contributions from counterterms cancel in each perturbation order of the Feynman-Dyson perturbation theory. However, I am afraid this Hamiltonian cannot be used for anything else.

For example, we cannot use this Hamiltonian H to construct the time evolution operator $U(t,t') = \exp(\frac{i}{\hbar}H(t-t'))$ and study how states and observables in an interacting system evolve in time. Of course, one might say that this is irrelevant, because in high energy physics particle collisions occur so fast, that there is no chance to measure what is going on during these collisions. However, in my opinion a theory cannot be complete without a well-defined finite Hamiltonian.

I guess, my question is this: am I right that time-dependent calculations cannot be rigorously performed within renormalized QED? Or there is a substantial gap in my education?

Thank you.
Eugene.

 

[samalkhaiat]

I guess, my question is this: am I right that time-dependent calculations cannot be rigorously performed within renormalized QED? Or there is a substantial gap in my education?

Peopel would benefit more from this tutorial if we restrict our discussion to the main subject which is conformal invariance.
However, if you put the above question in a separate thread, I will be happy to participate in the discussion

regards

sam

 

[samalkhaiat]

Finite SCT

I expected someone to ask me about the finite version of special conformal transformations which has not been mentioned in this tutorial! But for some reason no body raised this question, maybe, you all busy doing the exercises , Nevermind because I will in this post derive the finite version of SCT's and explain my excuse for ignoring it.

The finite form of SCT is nothing but a translation T(-c), preceded and followed by an inversion
$$I: \ x^{a} \rightarrow \frac{x^{a}}{x^{2}}$$

To see this, consider the following sequence of transformations;

$$x^{a} \rightarrow x_{1}^{a} = Ix^{a} = \frac{x^{a}}{x^{2}} \ \ 28.1a$$

$$x_{1}^{a}\rightarrow x_{2}^{a} = T(-c)x_{1}^{a} = x_{1}^{a} - c^{a} \ \ 28.1b$$

$$x_{2}^{a}\rightarrow \bar{x}^{a} = Ix_{2}^{a} = \frac{x_{2}^{a}}{x_{2}^{2}} \ \ 28.1c$$

From these it follows that

$$\bar{x}^{a} =IT(-c)Ix^{a} = \frac{x^{a}-c^{a}x^{2}}{1 - 2c.x + c^{2}x^{2}} \ \ 28.2a$$

or

$$\frac{\bar{x}^{a}}{\bar{x}^{2}} = \frac{x^{a}}{x^{2}} - c^{a} \ \ 28.2b$$

But, why should IT(-c)I be regarded as a finite SCT's? Well, you can convince yourself by observing that
i) c = 0 corresponds to the identity transformation;

$$IT(0)I = EI^{2} = E.E = E$$

ii) it is trivial to verify that the infinitesimal version of (28.2)[1st order in c] is indeed Eq(3.2d) [the 4th term in Eq(1.12)];

$$\delta x^{a} = \bar{x}^{a} - x^{a} = 2(c.x)x^{a} - c^{a}x^{2} \ \ 28.3$$

iii) it is an easy exercise to see that the inversion locally scales the metric

$$ds^{2} \rightarrow d\bar{s}^{2} = \frac{ds^{2}}{(x^{2})^{2}} \ \ 28.4$$

Hence, the inversion is a discrete conformal transformation.

So, what was my excuse for ignoring the finite SCT's?
The inversion turns out to be indetermnate for the poins of the light cone sueface $$x^{2}=0$$ in Minkowski space. This leads to the fact that SCT's (28.2) are not defined globally on spacetime. Indeed, they mix up in a very complicated way the topology of spacetime.
For instance, if you take
$$x^{a}= (t,0,0,0), \ c^{a} = (c,0,0,0)$$

you get

$$\bar{t} = \frac{-1}{c} \ . \frac{t}{t - \frac{1}{c}}$$

For c>0, we see that, as

$$t\rightarrow \pm \infty , \ \bat{t}\rightarrow ( \frac{-1}{c})_{\mp}$$

and for

$$t \rightarrow ( \frac{1}{c})_{\mp}, \ \bar{t}\rightarrow \pm \infty$$

This shows that it is not useful to interpret the finite conformal transformations as mappings of spacetime. Thus, the conformal group is not a true transformation group of spacetime; its global action can be considered only on compactified version of spacetime. So, when we talk about conformal transformations on spacetime, you should keep in mind that only infinitesimal conformal transformations are well defined on Minkowski space;
From the infinitesimal scale transformations

$$\delta x^{a} = - \alpha x^{a}$$

we find that

$$\frac{1}{2} \delta x^{2} = - \alpha x^{2} \ \ 28.5$$

and, from the infinitesimal SCT's (28.3), we see that

$$\frac{1}{2} \delta x^{2} = x^{2}(c.x) \ \ 28.6$$

Observe that in either case $$\delta x^{2}$$ vanishes if $$x^{2}$$ does. This means that a point on the light cone is transformed into a point on the light cone.

regards

sam

 

[strangerep]

I expected someone to ask me about the finite version of special conformal transformations [...]

For my part, that's only because I was already aware of it.

The finite form of SCT is nothing but a translation T(-c), preceded and followed by an inversion

Curiously, some authors also talk a lot about "reversions", i.e: TIT instead of
SCT's ITI. I guess this doesn't matter in the end.

The inversion turns out to be indetermnate for the poins of the light cone surface $$x^{2}=0$$ in Minkowski space. This leads to the fact that SCT's (28.2) are not defined globally on spacetime. Indeed, they mix up in a very complicated way the topology of spacetime.[...]

Is this related to how SCTs correspond to uniform accelerations, which
"split" Minkowski spacetime into causally-disconnected Rindler wedges?

- - - - -

On a separate matter (while I've got your attention), there's another bigger
question which has always bugged me when people talk about the
conformal group. In GR, we have a transformation group larger than
Poincare, and Einstein's solution was to invent a general-covariant
definition of mass using a non-Lorentzian metric. So why do people
insist on retaining the usual (+,-,-,-) metric when (eg) dilations are
contemplated? I would have thought that, physically, it's obviously the
wrong thing to do, in light of the immense successes of GR and its
general covariance?

Cheers.

 

[samalkhaiat]

Curiously, some authors also talk a lot about "reversions", i.e: TIT instead of
SCT's ITI.

Do they?

I guess this doesn't matter in the end.

Well,

$$T(0)IT(0) = EIE = I \neq E$$

is not the identity transformation! Furthermore, from Eq(28.2a), we see;

$$I\bar{x} = T(-c)Ix = Ix - c$$

which is the Eq(28.2b). This can be written as;

$$I \bar{x} = \{T(-c)IT(-c)\}T(c)x$$

this shows that TIT is not a SCT.

Is this related to how SCTs correspond to uniform accelerations, which
"split" Minkowski spacetime into causally-disconnected Rindler wedges?

YES, as I said before; the finite SCT's are well defined only on some compact domains in Minkowski space. Penrose diagram and Rindler space are such domains.
In the penrose diagram, the compact region $$- \pi \leq ( \bar{u},\bar{v})\leq \pi$$ is conformal to the whole of Minkowski space $$-\infty \leq (u,v) \leq \infty$$

i.e., the boundary of Penrose diagram represens $$\pm \infty$$ of Minkowski space.

In Rindler space, the coordinates cover only a quadrant of Minkowski space (i.e., the wedge x>|t|), eventhough the metric is conformal to the whole of Minkowski space.

... So why do people insist on retaining the usual (+,-,-,-) metric

The idea that spacetime can be modeled by a differentiable manifold (M) is implicit in the requirement that metric on M should have signature (+,-,-,-). This is true with or without the conformal group. As you might know, in order to be able to solve field equations, that is to be able to pose a Cauchy problem, spacetime must be globally hyperbolic and time orientable. That is to say that the manifold must admit a direction vector field. This can be shown to follow if and only if M admits a Lorentzian metric.
R.Geroch and G.T.Horowitz wrote a very nice article about the "Global structure of spacetimes" in
General Relativity;An Einstien Centenary Survey, Edited by Hawking & Israel, Cambridge, 1979.
Have a look at it.

when (eg) dilations are
contemplated? I would have thought that, physically, it's obviously the
wrong thing to do, ...?

NO, you see, Weyl tensor is equal to the Riemann tensor formed by

replacing $$g_{ab}$$ by its conformal transform;

$$\bar{g}_{ab} = (-g)^{-1/4}g_{ab}$$

Consequently, the vanishing of the Weyl tensor implies the vanishing of $$R_{abcd}(\bar{g})$$ , which in turn implies that there exists a mapping such that $$\bar{g}_{ab}$$ is everywhere diagonal, with $$(\pm 1)$$ appearing along the diagonal,
and therefore $$\pm(-g)^{1/4}$$ appears along the diagonal of $$g_{ab}$$ . I hope this made sense to you.

regards

sam

 

[strangerep]

Curiously, some authors also talk a lot about "reversions", i.e: TIT instead of SCT's ITI.

Do they?

Well, no, not exactly. I was mis-remembering. They never said T(-c)IT(c) is
the same as SCT, though they used it for other things. But let's not pursue that.

The idea that spacetime can be modeled by a differentiable manifold (M) is implicit in the requirement that metric on M should have signature (+,-,-,-). This is true with or without the conformal group. As you might know, in order to be able to solve field equations, that is to be able to pose a Cauchy problem, spacetime must be globally hyperbolic and time orientable. That is to say that the manifold must admit a direction vector field. This can be shown to follow if and only if M admits a Lorentzian metric.
R.Geroch and G.T.Horowitz wrote a very nice article about the "Global structure of spacetimes" in
General Relativity;An Einstien Centenary Survey, Edited by Hawking & Israel, Cambridge, 1979.

[...]
Weyl tensor is equal to the Riemann tensor formed by
replacing $$g_{ab}$$ by its conformal transform;

$$\bar{g}_{ab} = (-g)^{-1/4}g_{ab}$$

Consequently, the vanishing of the Weyl tensor implies the vanishing of $$R_{abcd}(\bar{g})$$ , which in turn implies that there exists a mapping such that $$\bar{g}_{ab}$$ is everywhere diagonal, with $$(\pm 1)$$ appearing along the diagonal, and therefore $$\pm(-g)^{1/4}$$ appears along the diagonal of $$g_{ab}$$ . I hope this made sense to you.

Unfortunately, I've been unable to find a freely downloadable version
of the Geroch-Horowitz paper. But I understand the essential point you're
making. And thanks, your answer clarified some things that were
previously hazy. Let me now check my understanding...

To be able to pose a Cauchy problem in GR, we can effectively start with
a metric which is diag(+,-,-,-), multiplied by an arbitrary position-dependent
dilation function. [Such position-dependent dilations correspond to SCTs.]
I presume that one must then still solve the Einstein gravitational eqns to
narrow down the form of these functions applicable to different physical
situations?

 

[samalkhaiat]

Let me now check my understanding...

To be able to pose a Cauchy problem in GR, we can effectively start with
a metric which is diag(+,-,-,-), multiplied by an arbitrary position-dependent
dilation function. [Such position-dependent dilations correspond to SCTs.]
I presume that one must then still solve the Einstein gravitational eqns to
narrow down the form of these functions applicable to different physical
situations?

What you are saying is true in 2 and 3-dimensional gravity where the Weyl tensor vanishes everywhere, i.e., where the metric is conformally flat.
In 4-dimensional gravity, Weyl tensor does not need to vanish (except, maybe, at infinity, as per Penrose's suggestion) but we still need the Lorentz signature, i.e., "spacetime" needs to have a direction field. For example;

$$ds^2 = f(r) dt^{2} - g(r) dr^{2}$$

In principle, f and g can be found by solving Einstein's equations.

regards

sam

 

[samalkhaiat]

If people are interested, I could extend this tutorial to include some advanced topics such as:
"approximate" spontaneously broken scale symmetry,
From Einstein and Weyl invariant action to conformally invariant action, and possibly, chiral and conformal invariance in SU(n)XSU(n) model.

sam

 

[strangerep]

If people are interested, I could extend this tutorial to include some advanced topics such as:
"approximate" spontaneously broken scale symmetry,
From Einstein and Weyl invariant action to conformally invariant action, and possibly, chiral and conformal invariance in SU(n)XSU(n) model.

Coincidentally, I was just about to ask about the models you alluded to
earlier in the tutorial where dilations don't preserve the vacuum.

But first, I feel a need for some clarification about a related matter...

About a year ago, over on the Quantum Physics forum there was a thread
titled "QFT and unitary Lorentz representation":

https://www.physicsforums.com/showthread.php?t=126310

In that thread, you wrote about a puzzle arising in connection with
EM gauge transformations:

https://www.physicsforums.com/showpost.php?p=1050058&postcount=5

Specifically, you wrote:

$A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda$

Everything in nature indicates that this is an exact symmetry. So we expect to
find a unitary operator $U$, such that
$U|0>=|0>$
and,
$A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda$

are satisfied in the quantum theory of free EM field. But this leads to the contradictory statement;
$<0|A'_{\mu}|0> = <0|A_{\mu}|0> = <0|A_{\mu}|0> + \partial_{\mu}\Lambda$
or,
$\partial_{\mu}\Lambda = 0$
(this can not be right because $\Lambda$ is an arbitrary function).

So, one is led to believe that the EM vacuum is not invariant under the gauge transformation. i.e
$U|0> \neq |0>$
or in terms of Q
$Q_{\Lambda}|0> \neq 0$
This means that the gauge symmetry is spontaneously broken!
i.e
$<0|[iQ_{\Lambda},A_{\mu}]|0> = \partial_{\mu}\Lambda \neq 0$

but this is equivalent to the statement that the field operator has non-vanishing vacuum expectation value;
$<0|A_{\mu}|0> \neq 0$
which is wrong because of Poincare' invariance.
So, I am baffled! [...]

I have two questions:

1) Could you elaborate on the last statement about why the vev of
$A_{\mu}$ must be 0 by Poincare invariance? (I suspect this is
something I ought to understand already.) :uhh:

2) Do you still regard this whole thing as a puzzle, or have you since
resolved it?

 

[samalkhaiat]

Coincidentally, I was just about to ask about the models you alluded to
earlier in the tutorial where dilations don't preserve the vacuum.

Unfortunatly, very little space is usually devoted in the textbooks and graduate courses to the treatment of these models. The reason for this, I believe,
1) the pressure to cover aspects of QFT that are necessary for its important applications.
2) the absence of "simple, easy to follow" mathematical formulation.
So, I will be facing the difficult task of;
i) explaining in simple terms what are at times rather difficult ideas![by "simple terms" I meant the level of mathematics already used in this tutorial]
ii) keeping the promise (I made in post #1) of "self-contained work".
So, I don't know howlong it will take me to prepare the notes!

1) Could you elaborate on the last statement about why the vev of
$A_{\mu}$ must be 0 by Poincare invariance?

First method:Domb math!

Under LT, vector field transforms according to[see Eq(9.7)]

$$\bar{A}_{a}(\bar{x}) = U^{-1}(\omega) A_{a}(\bar{x}) U(\omega) = A_{a}(x) + \omega_{a}{}^{b}A_{b}(x)$$

Take VEV and use Lorentz invariance;
$$U(\omega)|0\rangle = |0\rangle$$
or $$J^{ab}|0\rangle = 0$$ to find

$$\langle 0|A_{a}(\bar{x})|0\rangle = \langle 0|A_{a}(x)|0\rangle + \omega_{a}{}^{b} \langle 0|A_{b}(x)|0\rangle$$

Now, the result follows from Eq(11.2a) and translation invariance;

$$e^{iP.y}|0\rangle = |0\rangle$$

****
Second method :(I discovered this while working on monopole field)

Translation invariance requires

$$\langle 0|A_{a}(x)|0\rangle = \langle 0|A_{a}(0)|0 \rangle$$

So, If VEV is not zero, then it must be a constant Lorentz vector! But we don't have such vector. So, we must start our theory with this vector on borad, i.e., we let our "Lorentz invariance" Lagrangian to depend on this constant vector and write

$$\bar{\mathcal{L}}(\bar{x},\bar{n}) = \mathcal{L}(x,n) \ \ (a)$$

where

$$\bar{\mathcal{L}}(\bar{x},\bar{n}) = \mathcal{L}\left( \bar{A}(\bar{x}), \bar{\partial}\bar{A}(\bar{x}), \bar{n}\right )$$
$$\bar{x} = \Lambda x, \ \ \bar{n} = \Lambda n$$
and
$$\bar{A}_{a}(\bar{x}) = \Lambda_{a}{}^{b}A_{b}(x)$$

Then, from the infinitesimal form of Eq(a) and Eq(13.13), we find

$$\partial_{a}M^{abc} = \left ( n^{b}\frac{\partial}{\partial n_{c}} - n^{c}\frac{\partial}{\partial n_{b}} \right ) \mathcal{L} \neq 0$$

This means: the presence of a non-vanishing fixed vector breaks Lorentz symmetry! One can show that Lorentz invariance can be restored only if the Lagrangian depends on singular field (monopole). Therefore, in the absence of magnetic charge, Lorentz invariance requires

$$n_{a} \equiv \langle 0|A_{a}|0\rangle = 0$$

(I suspect this is something I ought to understand already.) :uhh:

Don't worry about it. Obviously many people in here did not know the answer But, you were the only one who had the courage to ask and this is good!

2) Do you still regard this whole thing as a puzzle, or have you since
resolved it?

NO, I have not There are, though, two ways of avoiding the troubles:
1) CHEATING: If we "say" that $$\lambda$$ is an operator (-valued distribution), i.e., if we regard the gauge function as a q-number instead of being a c-number, then
$$\langle 0|\partial_{a}\lambda |0\rangle = 0$$
and the paradox does not arise.
This is, however, a plain cheating, because $$\lambda$$ is the parameter of the (infinite-dimensional) Lie group U(1) . If one takes it to be an operator, then one should explain what $$|\lambda(x)| \ll 1$$ means? and tons of other questions! Of course, in the existing literature, peopel never practise what they preach! When the paradox hit them in the face, they claim that $$\lambda$$ is operator function, everywhere else, they treat it as an arbitrary c-number function! For example; Weinberg says; $$\lambda$$ "is linear combination of $$a$$ and $$a^{\dagger}$$ whose precise form will not concern us .." but everywhere, in his book and papers, $$\lambda (x)$$ is used as c-number function!

2)Working with the so-called B-field formalism;
in this formalism, $$\lambda (x)$$ is a c-number function,
<0|A|0> = 0, and

$$\langle 0|[Q_{\lambda},A_{a}]|0 \rangle = \partial_{a}\lambda$$

i.e., there is no conflict with Poincare' invariance and Q is always spontaneosly broken (the A field contains massless spectrum) unless $$\lambda$$ is a constant.


regards

sam