I've been discussing some things with Samalkhaiat over in the conformal(adsbygoogle = window.adsbygoogle || []).push({});

field theory tutorial. A part of that conversation (indicated by the new

title) was drifting away from CFT matters, so we both thought it was better

to move it into the Quantum Physics forum, to minimize pollution of the CFT

tutorial.

For the benefit of other readers, I'll first summarize some background...

The sub-conversation arose from Sam's post from about a year ago:

https://www.physicsforums.com/showpos...58&postcount=5 [Broken]

Specifically, Samalkhaiat wrote:

Among other things, I asked Sam whether he still regarded this [...EM gauge transformations...]

[itex]A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda[/itex]

Everything in nature indicates that this is an exact symmetry.

So we expect to find a unitary operator U , such that

[itex]U|0>=|0>[/itex]

and,

[itex]A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda[/itex]

are satisfied in the quantum theory of free EM field.

But this leads to the contradictory statement;

[itex]<0|A'_{\mu}|0> = <0|A_{\mu}|0> = <0|A_{\mu}|0> + \partial_{\mu}\Lambda[/itex]

or,

[itex]\partial_{\mu}\Lambda = 0[/itex]

(this can not be right because [itex]\Lambda[/itex] is an arbitrary function).

So, one is led to believe that the EM vacuum is not invariant

under the gauge transformation. i.e

[itex]U|0> \neq |0>[/itex]

or in terms of Q

[itex]Q_{\Lambda}|0> \neq 0[/itex]

This means that the gauge symmetry is spontaneously broken!

i.e

[itex]<0|[iQ_{\Lambda},A_{\mu}]|0> = \partial_{\mu}\Lambda \neq 0[/itex]

but this is equivalent to the statement that the field operator has

non-vanishing vacuum expectation value;

[itex]<0|A_{\mu}|0> \neq 0[/itex]

which is wrong because of Poincare' invariance.

So, I am baffled!! [...]

as a puzzle, and whether he had since resolved it. He said:

Intrigued by the above, I (perhaps foolishly) recalled an old spr NO, I have not There are, though, two ways of avoiding the troubles:

1) CHEATING: If we "say" that [itex]\lambda[/itex] is an operator (-valued

distribution), i.e., if we regard the gauge function as a q-number instead

of being a c-number, then

[tex]\langle 0|\partial_{a}\lambda |0\rangle = 0[/tex]

and the paradox does not arise.

This is, however, a plain cheating, because [itex]\lambda[/itex] is the

parameter of the (infinite-dimensional) Lie group U(1) . If one takes it to

be an operator, then one should explain what [tex]|\lambda(x)| \ll 1[/tex]

means? and tons of other questions! Of course, in the existing literature,

people never practice what they preach! When the paradox hit them in the

face, they claim that [itex]\lambda[/itex] is operator function, everywhere

else, they treat it as an arbitrary c-number function! For example; Weinberg says;

"[itex]\lambda[/itex] is linear combination of [itex]a[/itex] and [itex]a^{\dagger}[/itex] whose precise form

will not concern us .." but everywhere, in his book and papers, [itex]\lambda(x)[/itex] is used as

c-number function!

(2) [...]

post about "Gauge Transformations in Momentum Space" where I was

trying to figure out whether there was some relationship between

EM gauge transformations, and (generalized) Bogoliubov transformations.

(I should probably add "generalized field displacement transformations"

to the latter.)

Sam scolded me and wrote:

I want to focus (for now) on the last 2 sentences, in particular the If you have two Fock spaces based on two different (complete orthonormal)

sets of modes u(x;p) and v(x;p), then you can expand the same field in each

set:

[tex]\phi(x) = \int_{p^{3}} a(p)u(x;p) + a^{\dagger}(p)u^{\ast}(x;p) =

\int_{p^{3}} b(p)v(x;p) + b^{\dagger}(p)v^{\ast}(x;p) \ \ (1)[/tex]

Completeness allows you to expand the bases in terms of each other

[tex]v(x;p) = \int_{k^{3}} f(p,k)u(x;k) + g(p,k)u^{\ast}(x;k) \ \ (2)[/tex]

[This imposes certain conditions on f & g but I'm not interested in them

here] Put (2) in (1), you find

[tex]a(p) = \int_{k^{3}} f(p,k)b(k) + g^{\ast}(p,k)b^{\dagger}(k) \ \ (3)[/tex]

[when g = 0, the two Fock spaces coincide]

Eq(2), Eq(3) and similar ones for [itex]v^{\ast}, a^{\dagger}[/itex]

define the B.T's. Now, I know of no gauge transformation that can produces

these B.T's let alone multiplying by U(1) phase! Such phase can always be

absorbed by the base functions u(x;p) or v(x;p), leaving

[itex](a,a^{\dagger})[/itex] or [itex](b,b^{\dagger})[/itex] unchanged.

bit about multiplying by a (local) U(1) phase and absorbing it into

the base functions u(x;p) or v(x;p) ...

Since we're talking about local U(1) gauge transformations on charged

fermions in EM, I need to clarify that (at least in my understanding)

[itex]\phi(x)[/itex] is really represents a Dirac spinor field, but the spin-related

indices have been suppressed in the above. In the Hilbert space the

transformation needs to be unitarily implemented, i.e:

[tex]\phi(x)\ \rightarrow \ \phi'(x) = U[\lambda] \phi(x) U^\dagger[\lambda]

= \ e^{i \lambda(x)} \ \phi(x) \ \ \ (sr1) [/tex]

Deferring (for now) the issue of what [itex]U[\lambda][/itex] looks like, I just

want to focus on the far-right side my eq(sr1) above.

Using Sam's eq(1), this becomes

[tex]\phi'(x) = \ e^{i \lambda(x)} \ \phi(x)

= \ \ e^{i \lambda(x)} \int_{p^{3}} ( a(p)u(x;p) + a^{\dagger}(p)u^{\ast}(x;p) )

\ \ \ (sr2)[/tex]

As stated, I don't understand how the [itex]e^{i \lambda(x)}[/itex] can be

sensibly and consistently absorbed into both [itex]u(x;p)[/itex] and

[itex]u^{\ast}(x;p)[/itex] simultaneously, leaving the creation/annihilation

operators unchanged. (Though maybe that's not what Sam meant?)

[There are some other matters arising from our discussion in the

other thread that I also want to pursue here, but it's probably better

to clarify one point at a time.]

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# Gauge Transformations and (Generalized) Bogoliubov Transformations.

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