samalkhaiat said:
Could you elaborate on the last statement about why the vev of A_{\mu} must be 0 by Poincare invariance? [...]
First method:Domb math!
[...]
OK, thanks. I figured it must be something like that. It seems quite a
lot of similar things are proven that way. I.e: use Lorentz transf formula
to move a c-number J^{ab} outside the vev, then use
vacuum translation invariance to get the final result.
Second method:
[...]
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![...]
Diverting on another tangent briefly... I presume that problem doesn't
arise if we're dealing with a non-vanishing fixed scalar (rather than vector)?
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.
Yes, I've thought about that in the past. Quantization amounts to constructing
a mapping from c-number classical functions f to operators \Phi(f) on a Hilbert space.
But \partial_{a}\lambda(x) originally came from classical EM, so it seems reasonable
to expect that it should be mapped to a Hilbert space operator.
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?
I wrote (clumsily) about a related question over on sci.physics.research some
time ago in a thread called "Gauge Transformations in Momentum Space".
Here's a (revised) version:
------- s.p.r. post -------------------
I know that we need the EM potential to get a gauge-covariant
derivative, and all that. But I want to focus just on U(1)
acting on the fermion field since I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.
I'll write \psi(x) for the position-space field operator, and
\Psi(p) for the corresponding momentum-space field operator.
In this notation, a U(1) transformation acting on \psi(x) is written:
\psi(x)\ \rightarrow \ \psi'(x) \ = U[\lambda] \psi(x) U^\dagger[\lambda] \ = \ e^{i \lambda(x)} \ \psi(x)
I claim that a time-dependent gauge transformation, where \lambda(x)
depends on time, not merely space, will (in general) mix the annihilation and
creation operators from which \Psi(p) is built. Hence it must be handled
as a Bogoliubov transformation between unitarily inequivalent representations.
E.g: take \lambda(x) = s t^2, where s is a real constant.
Suppressing the 3-space coordinates for brevity, we have:
\psi'(t) \ = \ e^{i s t^2} \ \psi(t)
Taking Fourier transforms...
\Psi'(p) = F(e^{i s t^2}) \ast \Psi(p)
where "F" denotes Fourier transform, and "\ast" denotes convolution.
This gives:
[...] = \int dq \ \sqrt{\frac{2i\pi}{s}} \ exp(-\frac{iq^2}{4s}) \ \Psi(p-q)
For \psi(t) = e^{iEt}, its Fourier transform is a delta fn, so:
\Psi'(E) \propto exp(-\frac{iq^2}{4s})
which has support from E = [-\infty , +\infty].
I.e: this gauge transformation maps a +ve energy delta fn \delta(E)
into a function with support from E = [-\infty , +\infty]. The
new field is thus an infinite superposition of both +ve and -ve frequencies.
In terms of annihilation/creation operators, this means that an
annihilation operator corresponding to energy E gets mapped into a
complicated infinite combination of both annihilation
and creation
operators. That's usually the signal that we're dealing with a Bogoliubov
transformation between disjoint Fock spaces, and that's what I'm trying
to prove (or disprove) definitively, for arbitrary \lambda(x).
---- (End s.p.r. post) ----------------------------------
If what I've written above is on the right track, it would mean that
local gauge transformations correspond to Bogoliubov transformations
between disjoint Fock spaces, and questions about the meaning of
|\lambda(x)| \ll 1 must be addressed in that context.
See below for related stuff about the action of gauge transformations
on A_{\mu}.
and tons of other questions!
What are some of these 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!
Yes, I vaguely remembering reading that - but now I cannot find it. Could you please give me a
more specific reference to the place where Weinberg says that?
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.
It occurs to me that if \lambda (x) is a c-number function, then a gauge
transformation of A_{\mu} (expressed in momentum space) induces a
transformation on the annihilation/creation operators like this:
a(p) \rightarrow a'(p) \ = \ a(p) \ + \ p_{\mu} \Lambda(p)
This has the form of a "field displacement" transformation, which in general maps
between unitarily inequivalent representations. So it seems the same thing is
happening for the EM field as happened above for the fermion field under local U(1).
But of course, this doesn't yet resolve the puzzle - since \Lambda(p) is a
c-number here. So maybe it does need to be an operator after all. But I'll wait
until you tell me what the "tons of other questions" are before I ramble on further.
(BTW, as this is drifting away from conformal group matters, would you prefer
I move this sub-conversation back over into the Quantum Physics forum?)
But, you were the only one who had the courage to ask and this is good!
There are, though, two ways of avoiding the troubles:
). Indeed, if you repeat the domb math method for scalar field (instead of vector field), you will see that Poincare invariance does not force <0|\phi |0> to vanish. This is why Goldstone was able to prove his theorem.