Infinitesimal conformal transformations

[CAF123]

Homework Statement

Find the infinitesimal dilation and conformal transformations and thereby show they are generated by $D = ix^{\nu}\partial_{\nu}$ and $K_{\mu} = i(2x_{\mu}x^{\nu}\partial_{\nu} - x^2\partial_{\mu})$

The conformal algebra is generated via commutation relations of elements of the set $\left\{P_{\mu}, D, L_{\mu \nu}, K_{\mu}\right\}$. Write down the non vanishing commutation relations for the comformal group given that $P_{\mu} = i\partial_{\mu}$ and $L_{\mu \nu} = i(x_{\nu}\partial_{\mu} - x_{\mu}\partial_{\nu})$, Hence show that $P^2$ is no longer a Casimir operator and thus that only massless theories can be conformally invariant.

2. Homework Equations
$x'^{\mu} = \alpha x^{\mu}$ for finite dilations and $x'^{\mu} = \frac{x^{\mu} - x^2b_{\mu}}{1-2x_{\mu}b^{\mu} +b^2x^2}$ for finite CT's.

The Attempt at a Solution

Write an infinitesimal dilation transformation as $x'^{\mu} = e^{\Lambda}x^{\mu} = (1+i\alpha D)x^{\mu}$ where D is the dilation generator and alpha some infinitesimal scaling. Alternatively, by taylor expanding to first order, $x'^{\nu} = x^{\nu} + x^{\mu}\partial_{\mu}x^{\nu}$. Comparison with the previous result gives $D = ix^{\mu}\partial_{\mu}$. Is that ok?

Similarly, for CT's, $x'^{\mu} = (1+i\alpha^{\rho} K_{\rho})x^{\mu} = x^{\mu} + 2x^{\mu}(b \cdot x) - x^2b^{\mu}$. But if I identify lhs and rhs here I don't get any derivative terms appearing?

And to check my reasoning for the last part: P^2 is not a casimir operator for the conformal algebra anymore because it doesn't commute with D. (A casimir must commute with all the algebra generators). I am not really sure how to show that massless theories must be conformally invariant given the fact that $P^2$ is not a casimir, so a hint here too would be great. Thanks!

 

[strangerep]

$x'^{\mu} = \alpha x^{\mu}$ for finite dilations and $x'^{\mu} = \frac{x^{\mu} - x^2b_{\mu}}{1-2x_{\mu}b^{\mu} +b^2x^2}$ for finite CT's.

You can also use the general formula for a generator, involving differentiation of the new coordinates by the group parameter, and then evaluating at the identity. E.g., for your dilation transformation, it is
$$D ~=~ \left. \frac{\partial x'^\mu}{\partial \alpha} \right|_{\alpha=1} ~ \frac{\partial}{\partial x^\mu}$$ You might want to re-express the dilation as $x'^{\mu} = e^\alpha x^{\mu}$ so that the identity corresponds to $\alpha=0$ rather than $\alpha=1$.

I leave it to you to generalize this to the case of multiple group parameters (which is required to find the SCT generator).

BTW, it should be $b^\mu$ in the SCT numerator.

Re the final part: consider how a representation with $P^2=0$ changes under a dilation...

 

[CAF123]

Re the final part: consider how a representation with $P^2=0$ changes under a dilation...

Hi strangerep, many thanks I see the first part. Hmm if $P^2=0$ and I consider $[P^2, D]$ then this is identically zero. But I can also expand as $P^{\mu}[P_{\mu},D] + [P_{\mu},D]P^{\mu} = 2iP^2 = 0$. But how does this tell us about conformal invariance?

 

[strangerep]

Hmm if $P^2=0$ and I consider $[P^2, D]$ then this is identically zero.

It's only zero if you act with that operator on a state vector in the Hilbert corresponding to a zero-mass field. So you should probably include a state vector in your statements.

Have you studied how (ordinary) QFT is constructed by finding the unitary irreducible representations of the Poincare group? This is done in Weinberg vol1, and also (iirc) Hendrik van Hees's QFT notes. The idea is to construct a Hilbert space which "carries a representation" of the Poincare group (meaning the elements of the group are represented faithfully by operators on that Hilbert space).

To perform such a construction, one first finds the Casimirs of the group. (For Poincare, these are the mass and total angular momentum.) The Casimirs are important because if we choose a field with specific values of those Casimirs, then other Poincare group elements won't change those values.

But this is a rather long (but very important) story. You might want to start a thread about it in the main QM forum.

But I can also expand as $P^{\mu}[P_{\mu},D] + [P_{\mu},D]P^{\mu} = 2iP^2 = 0$. But how does this tell us about conformal invariance?

Consider a Hilbert space consisting of states of a massless field. Dilations will simply map these states among themselves. In this sense, that Hilbert space "carries a representation" of dilations.

In contrast, for the Hilbert space of a massive field, those dilations will map out of the Hilbert space (since they change the mass). Therefore, that Hilbert space does not carry a representation of dilations.

 

[CAF123]

It's only zero if you act with that operator on a state vector in the Hilbert corresponding to a zero-mass field. So you should probably include a state vector in your statements.

Ok, so I would write $[P^2, D]|P^2, W^2\rangle = 2iP^2 |P^2, W^2\rangle$ where $P^2$ and $W^2$ are the Casimir operators of the Poincare group and as such as can label states. Extending to dilations, we see that the r.h.s of the above is not identically zero so $P^2$ is no longer a viable Casimir operator for the conformal group (which contains dilations). Is this better?

Consider a Hilbert space consisting of states of a massless field. Dilations will simply map these states among themselves. In this sense, that Hilbert space "carries a representation" of dilations.

In contrast, for the Hilbert space of a massive field, those dilations will map out of the Hilbert space (since they change the mass). Therefore, that Hilbert space does not carry a representation of dilations.

If the states are all massless then $P^2=0$ and so $[P^2,D]|P^2=0, W^2\rangle = 0 \cdot |P^2=0, W^2\rangle$. All we know at this stage is that the state we get back is a massless state but to characterise what massless state it actually is we require knowledge of $W^2$ obtained by the action of $W^2$ on the state. Is this what you meant by saying the massless states transform amongst themselves? Thanks!

 

[strangerep]

Ok, so I would write $[P^2, D]|P^2, W^2\rangle = 2iP^2 |P^2, W^2\rangle$ where $P^2$ and $W^2$ are the Casimir operators of the Poincare group and as such as can label states. Extending to dilations, we see that the r.h.s of the above is not identically zero so $P^2$ is no longer a viable Casimir operator for the conformal group (which contains dilations). Is this better?

I would have used different symbols to label the states, e.g., $|m,j\rangle$. In that notation, $$P^2|m,j\rangle = m^2 |m,j\rangle ~.$$ ($m,j,$ are numbers, whereas $P^2, W^2$ are operators.)

If the states are all massless then $P^2=0$ and so $$[P^2,D]|P^2=0, W^2\rangle = 0 \cdot |P^2=0, W^2\rangle$$. All we know at this stage is that the state we get back is a massless state but to characterise what massless state it actually is we require knowledge of $W^2$ obtained by the action of $W^2$ on the state. Is this what you meant by saying the massless states transform amongst themselves?

I meant the following: let $H_m$ be a Hilbert space of states carrying a representation of the Poincare group corresponding to a field of mass $m$. Then (by definition), for every $\psi\in H_m$, we have $$P_\mu\psi\in H_m ~,~~~~ J_{\mu\nu}\psi\in H_m ~,$$ and, of course, $$P^2 \psi\in H_m ~.$$ However, for $m\ne 0$, then $$\phi ~:=~ D\psi \notin H_m ~,$$ but for $m=0$ we have $$\chi ~:=~ D\psi \in H_0 ~.$$

 

[CAF123]

I would have used different symbols to label the states, e.g., $|m,j\rangle$. In that notation, $$P^2|m,j\rangle = m^2 |m,j\rangle ~.$$ ($m,j,$ are numbers, whereas $P^2, W^2$ are operators.)

Ok, I see, so you just used the eigenvalues of the operators to label the states like it is done in quantum mechanics.

However, for $m\ne 0$, then $$\phi ~:=~ D\psi \notin H_m ~,$$ but for $m=0$ we have $$\chi ~:=~ D\psi \in H_0 ~.$$

So this is the statement I need to prove to answer the question, yes?

$P^2|m,j\rangle = m^2|m,j\rangle$ and $[P^2, D]|m,j\rangle = 2iP^2|m,j\rangle=2im^2|m,j\rangle$. So if $m=0$ then $P^2$ is still a Casimir of the conformal group (provided it also commutes with SCT generator). So its eigenvalues (=0) can be used to label states living in a Hilbert space carrying a representation of the conformal group and these massless states within a representation are distinguished by the action of another Casimir operator onto the state which determines it uniquely. Is this right and enough to answer the question?

If it is right, then that would mean $D|m=0,j\rangle = |m=0,j'\rangle$ (i.e action of D produces a massless state but this state need not be the same from the one denoted by j).
Then if also, $[D,W^2]=0$ then $D|m=0, j \rangle = |m=0,j\rangle$ i.e state is invariant.

Is any of that ok? Hope I'm getting there :)

 

[strangerep]

So this is the statement I need to prove to answer the question, yes?

I hope so. (Tbh, I'm not entirely sure how much the question expects from you.)

$P^2|m,j\rangle = m^2|m,j\rangle$ and $[P^2, D]|m,j\rangle = 2iP^2|m,j\rangle=2im^2|m,j\rangle$. So if $m=0$ then $P^2$ is still a Casimir of the conformal group [...]

Your terminology is not quite right. You'd have to say a bit more, like: "$P^2$ is still a Casimir of the conformal group if restricted to the massless sector".

(provided it also commutes with SCT generator). So its eigenvalues (=0) can be used to label states living in a Hilbert space carrying a representation of the conformal group and these massless states within a representation are distinguished by the action of another Casimir operator onto the state which determines it uniquely. Is this right and enough to answer the question?

As I read the question, it's not really asking you about the other eigenvalues, but only about the mass eigenvalue.

If it is right, then that would mean $D|m=0,j\rangle = |m=0,j'\rangle$ (i.e action of D produces a massless state but this state need not be the same from the one denoted by j). Then if also, $[D,W^2]=0$ then $D|m=0, j \rangle = |m=0,j\rangle$ i.e state is invariant

Again, I'm not sure you need to be specific about the other eigenvalues. You could just write $|m=0,\; \dots\rangle$

 

[CAF123]

Again, I'm not sure you need to be specific about the other eigenvalues. You could just write $|m=0,\; \dots\rangle$

Ok, many thanks. Does this imply then that a massless theory need not be conformally invariant only that for a theory to be conformally invariant it must be a massless theory? What we've shown above is that massless theories are invariant under dilations (the action of dilations is just to mix between different states of the representation) and since dilations are part of the conformal group, conformal invariance in massless sector would need dilation invariance in the massless sector.

Thanks!

 

[strangerep]

Ok, many thanks. Does this imply then that a massless theory need not be conformally invariant

I guess so. What really matters is the maximal dynamical group for the (class of) system(s) being considered.

only that for a theory to be conformally invariant it must be a massless theory?

Or we're dealing with a rather weird case that admits a continuous spectrum of nonzero masses. But I don't know any of those.

What we've shown above is that massless theories are invariant under dilations (the action of dilations is just to mix between different states of the representation) and since dilations are part of the conformal group, conformal invariance in massless sector would need dilation invariance in the massless sector.

Yes, that's the idea. BTW, to complete the picture you might want to check what happens with SCTs.

[Aside:] I'm not very happy with the term "conformally invariant theory", since it's not always clear what is meant thereby. I prefer to think about the maximal dynamical group for the system -- by which I mean the maximal group which preserves the equations of motion, i.e., which maps solutions into solutions. This is easy enough to understand in the classical case. In the quantum case, it means we require a Hilbert space carrying a representation of the group.

But this perspective gets lost in the voluminous reams of material about Lagrangians, etc. It doesn't hurt to be aware that the Lagrangian+Noether approach might not always tell the whole story -- at least, not without some extra work. E.g., for the ordinary classical Kepler problem there's an extra time-dependent dynamical symmetry of the equations of motion which does not correspond to a conserved quantity. Nevertheless, that symmetry is the underlying reason for Kepler's 3rd law, and hence is reasonably important .