Conformal group, Conformal algebra and Conformal invariance in field theory

[strangerep]

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?)

 

[samalkhaiat]

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)?

Expectation value of a vector field is not a scalar quantity! Your equations need to be covariant, you can't put vector quantity on one side and scalar quantity on the other side of your equation.
Or, did you mean to ask about the VEV of a scalar field? In this case, Yes, non-vanishing $<0|\phi |0>$ does not violate Poincare invariance (it violates other symmetries ). 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.

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.

Yes, Multiplication by the unit operator does this job.

...
I want to know
whether or not the gauge-transformed fermion field lives in
the same Fock space as the original untransformed field.

Operators don't live in Hilbert space, and multiplying the field by local U(1) phase does not mix the creation and annihilation operators. On the other hand, the operators in Bogolubov transformations are not related by gauge transformations.

E.g: take $\lambda(x) = s t^2$, where $s$ is a real constant.

NO, I don't take this because, it is not compatible with any Lorentz and gauge invariant field theory. Covariant formalism requires:

1) $\partial^2 \lambda =0$ , your choice does not satisfy this equation.
2) time-independent charge, your choice leads to the following "charge"

$$Q = st^{2}\int dx^{3}\psi^{\dagger}\psi$$

[...]
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 $\lambda(x)$...

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:
$$\phi(x) = \int_{p^{3}} a(p)u(x;p) + a^{\dagger}(p)u^{\ast}(x;p) = \int_{p^{3}} b(p)v(x;p) + b^{\dagger}(p)v^{\ast}(x;p) \ 1$$

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

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

[This imposes certain conditions on f & g but I'm not interested in them here]
Put (2) in (1), you find

$$a(p) = \int_{k^{3}} f(p,k)b(k) + g^{\ast}(p,k)b^{\dagger}(k) \ \ 3$$

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

Eq(2), Eq(3) and similar ones for $v^{\ast}, a^{\dagger}$ 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 $(a,a^{\dagger})$ or $(b,b^{\dagger})$ nuchanged.

What are some of these other questions?

In QM, exact symmetries need to be unitarly implimented (on at least a dense subset of the Hilbert space including the vacuum) by operators of the form

$$e^{i \Lambda Q}$$

where $\Lambda$ is a c-number.
If Lambda is an operator;
How can you show that $i \Lambda Q$ is a Lie algebra element? Since we are not dealing with supersymmetry, how can you show that $[\Lambda , Q] = 0$ ?
How can you maintain the Abelian nature of U(1)?
In the case of free EM field, what is Q and how can you construct it?

The above questions represent (more or less) the mathematical side of the trouble. Let me now explain to you the (physical) importance of a c-number gauge function in QFT.

Global gauge invariance of the Lagrangian is an algebraic symmetry in the sense that the conserved (Noether) charge, it generates, commutes with the
S-matrix, i.e., it imposes algebraic condition on it. Local gauge invariance on the other hand is a dynamical symmetry because it does not lead to any new algebraic condition on the S-matrix, i.e., NO new (Noether) charge is obtained from the local gauge symmetry. The proof of the last stetement relies entirely on the behaviour of the (c-number) gauge function;
see the lectures of Weinberg at the 1970 Brandeis Summer School in: "Lectures on Elementary Particles and QFT", edited by S. Deser et al.(MIT press)

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?

QFT Vol I, page 251, the line below Eq(5.9.31).

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 equation is WRONG because, it is not covariant. YOU CAN NOT EQUATE VECTOR WITH SCALAR! To make sense, the {a} (whatever it is) in your equation must carry a spacetime index (i.e., Lorentz vector):

$$a_{\mu}(p) \rightarrow a_{\mu}(p) \pm i p_{\mu} \Lambda (p)$$

Now, this makes a lot of sense because, it describes how the polarization vector changes under local gauge transformations. You need to remember that creation/annihilation operators do not carry spacetime index.

(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?)

This is certainly a good idea

regards

sam

 

[strangerep]

[...]
YOU CAN NOT EQUATE VECTOR WITH SCALAR! [...]

So you're shouting at me for a silly typographical omission? Geez!
Of course I know the difference between scalars and vectors.

Anyway, there are quite a number of things in what you wrote
that I'd like to pursue, so I've begun a new thread at:

https://www.physicsforums.com/showpost.php?p=1406589&postcount=1

to continue the discussion without polluting this CFT thread further.

 

[Jim Kata]

quick question Sam
does the algebra for the conformal group have a central extension?

I"m looking at Fayet and Ferrara and they're mentioning this thing called the chiral charge that looks like a central extension. If it does have one in which commutator relation does it occur? I don't feel like doing jacobi identities all day.

 

[samalkhaiat]

quick question Sam
does the algebra for the conformal group have a central extension?

On space-time of dimension greater than two, the conformal group C(1,n-1) has no
non-trivial central extensions,i.e., it cannot be extended by U(1) in a non-trivial way.
Or, in the terminology of mathematical literature, the 2nd cohomology group of C(1,n-1) is trivial;

$$H^{2}_{0}(C(1,3),U(1)) = 0$$

Due to this trivial cohomology, the fields carry ordinary (non-projective) representations
of the conformal group.
The fact that there are no non-trivial central charges (in the FINITE-DIMENSIONAL conformal algebra) can also be deduced from the fact that C(1,n-1) is isomorphic to the SEMI-SIMPLE "Lorentz" group
SO(2,n)[ see post#6]. It was proved by Bargmann that every local exponent of a semi-simple group is equivalent to zero. This means that any central charges in semi-eimple Lie algebras (such as the conformal algebra so(2,n)) can be eliminated by a redefinition of the generators. This implies that every representation of the conformal group SO(2,n) can be reduced to ordinary (non-projective) representation.
Try to prove that, there is only ONE Jacobi identity in so(2,n)!

I"m looking at Fayet and Ferrara and they're mentioning this thing called the
chiral charge that looks like a central extension.

I think they are talking about the chiral symmetry algebra, also known as the observable algebra or simply the W-algebra, which contains 2 commuting copies (left and right) of the Virasoro algebra. Their symmetry group is the LOCAL (infinite-dimensional) conformal group in two dimensions, $D^{\infty}_{loc}(2)$,i.e., the set of all (anti)holomorphic mappings of $\mathbb{C}^{2}$. Essencially, it is the direct product group

$$D^{\infty}_{loc}(2) = Diff(S^{1}) \times Diff(S^{1})$$

where $Diff(S^{1})$ is the infinite-dimensional group of orientation preserving diffeomorphisms of a circle.
On the Hilbert space of QFT, the representations of this local group are intrinsically projective, i.e., the central charge in the Lie algebra of $D^{\infty}_{loc}(2)$ cannot be eliminated.
Indeed, one can show that the (infinite-dimensional) Lie algebra of
$Diff(S^{1})$ (i.e., the algebra of vector fields on a circle; the Witt algebra) allows for non-trivial central extension, the result is the Virasoro algebra.

In general, it is difficult to determined wether a given Lie algebra admits
non-trivial central extensions or not. But, it is easy to show that FINITE-DIMENSIONAL (SEMI-)SIMPLE Lie algrbras do not possesses non-trivial central extensions, whereas their (INFINITE-DIMENSIONAL) LOOP algebras possesses a unique non-trivial central extension by a single generator. The resulting algebras are called the Kac-Moody (or untwisted affine) algebras.
(loop algebra $\mathcal{L}(S^{1}) = Map(S^{1}, \mathcal{L})$ is the space of smooth mappings from a circle $S^{1}$ to some (compact) finite-dimensional Lie algebra $\mathcal{L}$).
The Kac-Moody algebras are really a special cases of the good old current algebras,$\mathcal{L}(M)= Map(M, \mathcal{L})$, of QFT; they are obtained from current algebras in the case in which space-time is 2-dimensional and space itself is compact, i.e., a circle.
The Virasoro and Kac-Moody algebras form the corner stone of 2D conformal physics. They are not unrelated structures. Indeed, with each Kac-Moody algebra there is a Virasoro algebra associated in a natural way so that together they form a semi-direct product. For example, in the WZW models, the currents realize 2 Kac-Moody algebras, and the components of the stress-tensor close into 2 Virasoro algebras; this is true in any 2D conformal field theory. The relation between them corresponds to the construction of the stress-tensor in terms of (normal ordered) bilinears of the currents, as originall proposed by Sugawara.

regards

sam