# The meaning of the addition of the spin generator to the Lorentz generators

1. Nov 8, 2012

### liorde

When constructing the Lie algebra of the Lorentz Transformation, the references usually start with an infinitesimal proper-orthochronous transformation, and then find the infinitesimal generators. Let's call the set of these generators L. after finding L, the references usually compute the commutation relations of L, and then add to L the operators S, corresponding to the spin. The reason is that L+S obey the same commutation relations as L, and therefore L+S is more general then L.
My questions are:
a) Is there any physical motivation to add the S (besides "it agrees with the phenomenon of spin")? I'm looking for a symmetric-inspired motivation.
b) Would the physics be any different if we didn't add S?
c) I thought that the commutation relations of the algebra determine the group structure, so why aren't L and L+S equivalent? why do we need the extra S?
d) could we add another term, let's call it T, to L+S such that L+S+T still obey's the algebra?
e) If L represents a transformation which is originally defined as operating on space-time coordinates, what do S and L+S represent in terms of space-time coordinates? if the answer is that S doesn't care about space-time but operates on a different space, can we say that T above (in question d) operates on a third space?
f) I know that in quantum mechanics, angular momentum has integral eigenvalue while spin has half integral values. The reason is that angular momentum is defined via the position operator X and that leads to integral eigenvalue, while spin is defined to operate on "spin space" which is separate from "space space". What does that have to do with my previous discussion?

Thanks

2. Nov 8, 2012

### Fredrik

Staff Emeritus
Can you post a reference to a book or article that does this? If possible, link directly to the page at Google Books. I'm curious, because what you're describing doesn't sound like anything I'm familiar with. It might just be because I've never studied a rigorous presentation of this stuff.

Since I don't understand the approach you're talking about, I'm wondering if what you're talking about is (somehow) that quantum physicists use the universal covering group of the symmetry group instead of the symmetry group itself? (For example, SU(2) instead of SO(3) and SL(2,ℂ) instead of SO(3,1)). If so, I might be able to partially answer some of your questions. (Not today though).

3. Nov 9, 2012

### liorde

Ramond, Field Theory: A Modern Primer, Second Edition, Chapter 1, page 7.

Look at the last paragraph.

I've seen at least two more references using this method (although I might have exaggerated when I said "the references usually start with [...]").

I am not familiar with the concept of "universal covering group". I'll read about it to see if it helps me understand.

Thanks

4. Nov 9, 2012

### Fredrik

Staff Emeritus
Thanks. I'm a bit busy right now, but I will take a look at that in a few hours. I should of course have posted a reference of my own. The approach that I'm somewhat familiar with is the one described in chapter 2, volume 1, of Steven Weinberg's "The quantum theory of fields". Unfortunately, it seems to be impossible to view the relevant pages at Google Books.

Last edited: Nov 9, 2012
5. Nov 9, 2012

### Fredrik

Staff Emeritus
Ramond isn't doing anything to find out what the Lie algebra of SO(3,1) is. He's just saying without proof that it's given by (1.2.27). Now it's easy to see that if two sets of operators $\{L_{\mu\nu}\}$ and $\{S_{\mu\nu}\}$ satisfy those commutation relations, and the two sets commute with each other, then the sums $M_{\mu\nu}=L_{\mu\nu}+S_{\mu\nu}$ satisfy those commutation relations too.

If $\{M_{\mu\nu}\}$ is an arbitrary set of operators that satisfy the commutation relations, we can define three new operators $\{J_i\}$ by (1.2.30). The operator $J^2=J_iJ_i$ commutes with each of the $J_i$, so we can find simultaneous eigenvectors of e.g. $J^2$ and $J_3$. If we write the eigenvalue equations as
\begin{align}
J^2|jm\rangle &=j(j+1)|jm\rangle\\
J_3|jm\rangle &=m|jm\rangle
\end{align} we can prove that 2j must be a non-negative integer and that -j≤m≤j. (Nitpick: We don't actually prove that j is non-negative. j can be ≤-1. But j(j+1) is invariant under the change j→-j-1, so if j≤-1 we can easily redefine j to be non-negative). What's important to note here is that j can be a half-integer. The possible values of j are 0, 1/2, 1, 3/2, 2,... If there's any doubt that there exist irreducible representations with half-integer j, it can be removed by explicitly constructing a j=1/2 representation using the Pauli spin matrices.

According to Weinberg's book, each particle species is associated with an irreducible representation of the Poincaré group. I think that this makes this j (which can be a half-integer) one of the numbers that identify a particle species. We therefore take the particle's Hilbert space to be a direct sum of $L^2(\mathbb R^3)$ and $\mathbb C^{2s+1}$ where s is the particle's spin. Now we need a representation of the Lie algebra on each of these spaces. The L operators defined by (1.2.24) give us a representation on $L^2(\mathbb R^3)$, so now we just need a representation on $\mathbb C^{2j+1}$. This is what the spin operators are for.

Note since the L+S operators are the Lorentz generators on the relevant Hilbert space, we're not adding the spin operators to the Lorentz generators. The spin operators are an essential part of the Lorentz generators, not something that we add to them.

6. Nov 10, 2012

### liorde

You say that

But $\left\{ {{L_{\mu \nu }}} \right\}$, defined in (1.2.24), is an "arbitrary" set of operators that satisfy the commutation relations, so it by itself should give rise to all the different representations (corresponding to the different eigenvalues of ${J^2}$) with no need to add $\left\{ {{S_{\mu \nu }}} \right\}$. Thus, $\left\{ {{L_{\mu \nu }}} \right\}$ by itself will be enough to determine the transformation law of the state/field, the form of the action integral and hence the physics.

As an analogy, I feel that adding ${{S_{\mu \nu }}}$ to ${{L_{\mu \nu }}}$ is like multiplying a state vector (in quantum mechanics) by a phase. It doesn't change any matrix element just like in our case it doesn't change the commutation relations. I guess my mistake here is that I think that the commutation relations determine everything, but that is not true.

Also, I'm used to thinking of a group or algebra as an abstract structure, which has realizations (i.e representations). Equation (1.2.24) is an example of a realization (of the group's infinitesimal generators) which operates on space-time coordinates, while (1.2.27) is a more general-abstract relation that defines how the group behaves in any realizations. So the specific realization (1.2.24) operates on ${L^2}({ℝ^3})$, but that doesn't require that other representations will also operate on space-time. They can just as well operate on spin space.

I realize that there is some fundamental gap in my understanding of all of this. I hope my questions are understandable.

Thanks

7. Nov 10, 2012

### Fredrik

Staff Emeritus
I guess I wasn't clear enough. Given a set of operators that satisfy the commutation relations, and a (re)definition of j that ensures that j≥0, we can prove that j is a member of the set {0,1/2,1,3/2,...} and m is a member of the set {-j,-j+1,...,j-1,j}. This part of the proof doesn't guarantee that there's a common eigenvector |jm> for each of these values of j and m. It only guarantees that there's no |jm> with j or m not in these sets.

To prove existence of a |jm> with specific values of j and m, we need to define a specific representation. The definition of the L operators is the start of an existence proof for integer values of j. The definition of the Pauli spin matrices and the operators $S_i=\frac 1 2 \sigma_i$ is the start of the existence proof for j=1/2. For a more general existence proof, see e.g. theorem 4.9, page 102-106 of "Lie grups, Lie algebras, and representations: an elementary introduction" by Brian Hall.

Note that the |jm> stuff is just what we do to represent the sub-algebra su(2), but any representation of sl(2,ℂ) defines a representation of su(2) as well, so the |jm> stuff will be part of what we consider when we define a representation of sl(2,ℂ).

Last edited: Nov 10, 2012
8. Nov 10, 2012

### liorde

I'm still a little bit confused...

So you're saying that the commutation relations of the algebra (= the structure constants) do not include all the information about the group? Two algebras (e.g L and L+S) can have the exact same commutation relations but correspond to different groups? (in our example, L doesn't have a j=1/2 representation while L+S does, hence L and L+S are not isomorphic)? What additional structure do we need to define on the algebra so that it lets us tell apart two seemingly equivalent algebras? I mean, where is the information about the difference between the groups encoded when looking at the algebra? I thought that there is a one to one correspondence between the algebra and the group via exponentiation, but I guess this is not true since two equivalent algebras give rise to different groups.

I'll put my question in another way:
Suppose I define a new algebra thus:
The basis operators are $\left\{ {{Q_{\mu \nu }}} \right\}$ where $\mu ,\nu \in \left\{ {0\;,\;1\;,\;2\;,\;4} \right\}$ and where the the set of operators $\left\{ {{Q_{\mu \nu }}} \right\}$ is antisymmetric in $\mu ,\nu$ (so there are only 6 different basis operators).
The commutation relations are:
$\left[ {{Q_{\mu \nu }}\;,\;{Q_{\rho \sigma }}} \right] = i{\eta _{\nu \rho }}{Q_{\mu \sigma }} - i{\eta _{\mu \rho }}{Q_{\nu \sigma }} - i{\eta _{\nu \sigma }}{Q_{\mu \rho }} + i{\eta _{\mu \sigma }}{Q_{\nu \rho }}$
where
${\eta _{\nu \rho }} = {\rm{diag}}\left( {1\;,\; - 1\;,\; - 1\;,\; - 1} \right)$
and summation notation is used.
Now define the group $G$ as the set of elements ${e^{i{t_{\mu \nu }}{Q_{\mu \nu }}}}$ for real parameters ${t_{\mu \nu }}$.
My set of questions is:
Does the following question has a definite answer: "what are precisely the irreducible representations of the group $G$ ?" ? Can we know, without defining additional structure, if $G$ has irreducible representations in vector spaces of all finite dimensions? Or is there missing information and hence we can't determine which representations exist? Is $G$ isomorphic to the proper orthochronous Lorentz transformations?
Since both ${{L_{\mu \nu }}}$ and ${{L_{\mu \nu }} + {S_{\mu \nu }}}$ obey the same commutation relations that ${{Q_{\mu \nu }}}$ does, but ${{L_{\mu \nu }}}$ doesn't have representations that ${{L_{\mu \nu }} + {S_{\mu \nu }}}$ does, I'm wondering where is the missing information.

Thanks

9. Nov 10, 2012

### Fredrik

Staff Emeritus
Unfortunately so am I. I don't know this subject as well as I'd want to, but I'll try to answer anyway. I'm going to have to repeat some of what I said in my previous posts, but this is at least getting a little clearer in my head each time I improve the explanation.

The commutation relations define the algebra completely. This is one of three important parts of the answer to your question.

For example, if $\big\{J_i\big|i\in\{1,2,3\}\big\}$ is the basis of a 3-dimensional Lie algebra, and $[J_i,J_j]=i\varepsilon_{ijk}J_k$, then that Lie algebra is (isomorphic to) su(2).

Suppose that R is a representation of su(2). This means that R is a Lie algebra homomorphism into a Lie algebra whose members are linear operators on a vector space, and whose Lie bracket is the commutator. So the definitions of "representation" and "su(2)" imply that $[R(J_i),R(J_j)]=i\varepsilon_{ijk}R(J_k)$. Let's simplify the notation by defining $K_i=R(J_i)$. The $K_i$ satisfy commutation relations that are identical to the Lie bracket relations of the $J_i$. Since $[K^2,K_3]=0$, these two operators may have simultaneous eigenvectors. The commutation relations for the $K_i$ imply that if there's a simultaneous eigenvector $\mathbf{v}$, then there exist numbers (j,m) such that $j\in\{0,1/2,1,3/2,\dots\}$, $m\in\{-j,-j+1,\dots,j-1,j\}$, and
\begin{align}
K^2\mathbf{v} &=j(j+1)\mathbf{v}\\
K_3\mathbf{v} &=m\mathbf{v}.
\end{align} It's convenient to denote this $\mathbf{v}$ by $|jm\rangle$.

The commutation relations do not imply that there exists such $|jm\rangle$ for each (j,m) such that $j\in\{0,1/2,1,3/2,\dots\}$ and $m\in\{-j,-j+1,\dots,j-1,j\}$. Such things are determined by the function R, not by the commutation relations. That's the second important part of the answer to your question.

For example, if
\begin{align}
\sigma_1 &=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\qquad
\sigma_2 =\begin{pmatrix}0 & -i\\ i & 0\end{pmatrix},\qquad
\sigma_3 =\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\\
K_i &=R(J_i)=\frac{i}{2}\sigma_i,
\end{align} then the $K_i$ satisfy the commutation relations, but every simultaneous eigenvector $|jm\rangle$ has j=1/2 and m=±-1/2. So there are only two of them. This was to be expected, because the vector space on which the linear operators (here represented by 2×2 matrices) are defined is 2-dimensional.

The third important part of the answer to your question is that even though the commutation relations define the Lie algebra, they do not determine the Lie group, because two Lie groups can have the same Lie algebra. For example SO(3) and SU(2) have the same Lie algebra, su(2). I'm not sure, but it may be the case that for each finite-dimensional Lie algebra g there's exactly one simply connected Lie group with Lie algebra g. SO(3) isn't simply connected, but SU(2) is. SU(2) is homeomorphic to a 3-sphere, while SO(3) is homeomorphic to a 3-sphere with opposite points identified. In other words, SO(3) can be thought of as the collection of 1-dimensional subspaces of $\mathbb R^4$.

No, I don't think so. If they have the same commutation relations, they define the same algebra, but then each algebra corresponds to several groups.

The algebras are isomorphic, but the representations aren't equivalent.

Here you seem to be talking about different representations of the same Lie algebra, rather than different Lie algebras. They are distinguished by the dimension of the vector space that's the domain of the operator, and by the possible values of j and m for |jm> vectors in that space.

I don't know the theorems about reconstruction of the group from the algebra well, but I think the answer is that yes, G is isomorphic to the proper orthochronous Lorentz group, and therefore its Lie algebra has representations in which "eigenvalue" variables like j and m have other values that are consistent with the commutation relations, and not just the values they could have in |jm> vectors in the domain of the original Q operators.

I think this is the proper way to view the L+S thing: Let $J_{\mu\nu}$ denote the components of an antisymmetric matrix whose 6 independent components are the generators of the abstractly defined proper orthochronous Lorentz group. The map $J_{\mu\nu}\mapsto L_{\mu\nu}$ defines a representation of the Lie algebra on $L^2(\mathbb R^3)$. The map $J_{\mu\nu}\mapsto S_{\mu\nu}$ defines a representation of the Lie algebra on $\mathbb C^{2j+1}$. The map $J_{\mu\nu}\mapsto L_{\mu\nu}+S_{\mu\nu}$ defines a representation on the orthogonal direct sum $L^2(\mathbb R^3)\oplus \mathbb C^{2j+1}$. This means that we can also think of $L_{\mu\nu}$ as the restriction of $L_{\mu\nu}+S_{\mu\nu}$ to a proper subspace, and $S_{\mu\nu}$ as the restriction of $L_{\mu\nu}+S_{\mu\nu}$ to the orthogonal complement of that proper subspace. The j and m values of the |jm> vectors in these two mutually orthogonal subspaces are determined by the representations, not by the algebras, so differences are to be expected.

10. Nov 11, 2012

### liorde

Hi,

Your last paragraph raises some of my original questions again:

Thanks

11. Nov 11, 2012

### samalkhaiat

We do not “add” the S, it shows up even in the Lorentz transformation of the coordinates. Indeed, the infinitesimal Lorentz transformation can be written as,
$$\delta x^{\alpha} = (1/2) \omega^{\mu \nu} ( S_{\mu \nu})^{\alpha}{}_{\beta} x^{\beta}, \ \ (1)$$
where $S_{\mu \nu}$ are six 4 by 4 matrices with elements given by
$$( S_{\mu \nu})^{\alpha}{}_{\beta} = \eta_{\mu \beta}\delta^{\alpha}_{\nu} - \eta_{\nu \beta} \delta^{\alpha}_{\nu}.$$
You can easily see that these matrices do form a representation of the Lorentz algebra. Indeed, they are the spin one matrices associated with Lorentz 4-vector fields. They show up here (in that particular form) because the coordinates $x^{\mu}$ carry vector indices. In general, they act on abstract index space (we will come to that in a minute). We can also write the infinitesimal Lorentz transformation in terms of the generalized orbital angular momentum
$$L_{\mu \nu} = x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu},$$
as
$$\delta x^{\alpha} = (1/2) \omega^{\mu \nu} L_{\mu \nu}x^{\alpha}. \ \ (2)$$
Using (1) and (2), we can write the infinitesimal Lorentz transformation as
$$\delta x^{\alpha} = (1/4) \omega^{\mu \nu}\{ L_{\mu \nu} \delta^{\alpha}_{\beta} + ( S_{\mu \nu} )^{\alpha}{}_{\beta} \} x^{\beta}. \ \ (3)$$
Notice that L acts on the coordinates whereas S acts on the INDICES. So, to go over to abstract indices, only S will change. Now, at each space-time point x, we define some generic FINITE-DIMENTIONAL index space with elements
$$\Phi^{A}(x) = \phi , V^{\mu}, \psi_{r}, F^{[\mu \nu ]}, G^{(\mu \nu)},\psi_{r}^{\mu}, \ … \$$
(to be exact A = 1,4,4,6,10,16,…)
The question about the transformation law of $\Phi^{A}$ can be answered in many different ways, but now we can just generalize eq(3) to
$$\delta \Phi^{A}(x) = \frac{1}{2}\omega^{\mu \nu}\{ L_{\mu \nu} \delta^{A}_{B} + ( S_{\mu \nu} )^{A}{}_{B} \} \Phi^{B}(x).$$
This is exactly what you get when expanding the finite transformation
$$\bar{\Phi}^{A}(x) = U^{-1}(\Lambda ) \Phi^{A}(x) U(\Lambda ) = D^{A}{}_{B}( \Lambda ) \Phi^{B}( \Lambda^{-1} x)$$
where
$$D = \exp (\frac{1}{2} \omega \cdot S ), \ \ U = \exp ( \frac{1}{2} \omega \cdot J )$$
The $D$ forms a FINITE-DIMENSIONAL NON-UNITARY REPRESENTATION of the Lorentz group $SO(1,3)$ while the $U$ forms an INFINITE-DIMENTIONAL UNIRARY REPRESENTATION.

We would be living in a world with no electrons, no light, no strong or weak forces and no gravity.

The algebra determines the LOCAL group structure.

They are the same for scalar field where S = 0.

Try to find it. What do you have other than a coordinate space and abstract index space? Nothing, all geometrical objects are included in the set $\Phi_{A}(x)$.

See eq(1), eq(2) and eq(3).

It does, S acts on all possible index spaces including the index space of space-time; $A = \mu , \mu \nu , \mu \nu \rho , …$

Proper treatment can be given in terms of the universal covering $SL(2,C) = SU(2) \times SU^{*}(2)$

Sam

12. Nov 12, 2012

### Fredrik

Staff Emeritus
I don't understand this. If $x'=\Lambda(\theta)x$ where $\theta$ is a 6-tuple of parameters, then to first order in the parameters,
$$\Lambda(\theta) =I+\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0\Lambda(\theta).$$ This first-order term is what I'd denote by $\omega$ (suppressing $\theta$ from the notation). To first order in the parameters, $x'=(I+\omega)x=x+\omega x$. In component form,
$$x'^\mu =x^\mu+\omega^\mu{}_\nu x^\nu.$$ Is the second term on the right what you would denote by $\delta x^\mu$? In that case, how do you get the S to appear, and why do we want it?

13. Nov 12, 2012

### samalkhaiat

Yes it is.

play with the indices:
$$\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu}= \eta_{\nu \alpha}\delta^{\mu}_{\beta}\omega^{\beta \alpha}x^{\nu}$$
Now, interchange the dummy indices,
$$\delta x^{\mu}= \{ \frac{1}{2} \eta_{\nu \alpha}\delta^{\mu}_{\beta}\omega^{\beta \alpha} + \frac{1}{2} \eta_{\nu \beta}\delta^{\mu}_{\alpha}\omega^{\alpha \beta}\}x^{\nu}$$
Now use $\omega^{\alpha \beta} = - \omega^{\beta \alpha}$, you find
$$\delta x^{\mu} = \frac{1}{2} \omega^{\alpha \beta}( \eta_{\nu \beta}\delta^{\mu}_{\alpha} - \eta_{\nu \alpha} \delta^{\mu}_{\beta}) x^{\nu}$$

These matrices form a 4-dimentional representation of the Lorentz algebra and this is a good enough reason to find them.

Sam

14. Nov 12, 2012

### Fredrik

Staff Emeritus
Aaaah...I've been thinking that there should be a simple way to obtain the Lorentz algebra from the Lorentz group, but I didn't know how to do that until now. Thank you.

15. Nov 12, 2012

### TrickyDicky

But why would you consider only the proper-orthochronous transformation inf. generators?, that is just one component (the identity) of the Lorentz transformations. You are not "adding" anything when considering the general case you call L+S, that's just the general case. Maybe you are just getting confused by the way it is explained in the reference you linked in the other post.

16. Nov 12, 2012

### dextercioby

One needs a full discussion of specially relativistic invariance from the most general point of view, i.e. physics: special relativity (+ quantum mechanics); mathematics: differential geometry (+ harmonic analysis of locally compact groups on rigged Hilbert spaces).

There's no fully rigorous treatment, of course, each book has its own version, some more algebraic, other more analytic. I warmly reccomend the book by Fonda and Ghirardi, the Italian guys have a thorough treatment without using too much fancy mathematics.

Last edited: Nov 12, 2012