4
$\begingroup$

I am thinking about symmetries and that their "quantum" consequences are Ward identities of the form $$<\beta|[Q,S]|\alpha>=0,$$ where $Q$ is the conserved charge associated with the symmetry at hand, $S$ is the $S$-matrix and $|\beta>$ and $|\alpha>$ denote out and in states. How do I prove that $$<\beta|[Q,S]|\alpha>=0$$ given that $Q$ is the charge generated by this symmetry? My thinking is that if $$U=e^{iQ\epsilon}$$ is an exponential operator, corresponding to the (finite) symmetry transformation, then I have to expand the transformed $S$-matrix elements and this should give me the commutator. However, isn't $U$ supposed to be transforming the $S$-matrix as well and cancelling, thus, the effects of the state transformation?

I hope the point of my confusion is clear.

$\endgroup$

1 Answer 1

3
$\begingroup$

$U = e^{i\epsilon Q}$ is a symmetry if : $$\langle\alpha | e^{-i\epsilon Q}Se^{i\epsilon Q}|\beta\rangle = \langle \alpha|S|\beta\rangle \tag{1}$$

You can see this as saying that the $S$-matrix is invariant ie $S = e^{-i\epsilon Q}Se^{i\epsilon Q}$ or, equivalently, as saying that if you transform the in and out states as $|\alpha\rangle,|\beta\rangle \to e^{i\epsilon Q}|\alpha\rangle,e^{i\epsilon Q}|\beta\rangle$, then the scattering amplitude is unchanged. The two statements are equivalent.

Then, taking $(1)$ at first order in $\epsilon\to 0$, we get : $$\langle\alpha|[Q,S]|\beta\rangle= 0 \tag 2$$

Edit 1 : why do we not transform both the states and $S$ ?

This is just like the difference between the Schrödinger and Heisenberg pictures of quantum mechanics, or between passive and active transformations. If you transform both the states and operators, everything is trivially preserved. On the other hand, the statement that a given transformation is a symmetry is non trivial (some transformations are not symmetries and others are not).

For example, if you want to test the Lorentz invariance of your theory, you could let your coordinates fixed (and therefore let $S$ unchanged) but rotate your in- and out-states. The amplitudes before and after the transformations are physically distinct. The statement that they are the same is the fact that the theory is Lorentz invariant.

$\endgroup$
3
  • $\begingroup$ Thanks Soluble fish. The derivation is okay. However, I find it difficult to convince myself why the S-matrix and the states are not simultaneously transformed?? But instead, we transform the one or the other. I know it makes more sense to transform one or the other, but I seem to not be able to come up with a convincing reason. Any help? $\endgroup$
    – schris38
    May 12 at 8:32
  • $\begingroup$ Also, since we keep only up to first order terms in both the expansions, does this imply that the Ward identity is somehow related to infinitesimally small transformations? $\endgroup$
    – schris38
    May 12 at 8:33
  • 1
    $\begingroup$ I answered your first question in an edit. For your second question, yes, the Ward identity is the statement that the theory is invariant under infinitesimal transformation. Just like in ordinary quantum mechanics, a hermitian operator $Q$ generates a group of transformation $e^{itQ}$, which acts on operators as $M \mapsto e^{-itQ}Me^{itQ}$ and infinitesimally as $\delta M= -i\epsilon [Q,M]$ $\endgroup$ May 12 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.