# If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states?

+ 8 like - 0 dislike
2120 views

In the paper "Supersymmetry and Morse Theory", on the third page (p. 663 in the journal version), Witten says:

"Now in any quantum field theory if a symmetry operator (an operator which commutes with the Hamiltonian) annihilates the vacuum state, then the one particle states furnish a representation of the symmetry."

Why is this true? Is there a simple explanation or computation that doesn't go too far afield of Witten's relatively informal discussion in the introduction of this paper, or is it more complicated?

This post imported from StackExchange Physics at 2015-05-09 14:34 (UTC), posted by SE-user physics_acolyte
Well, from classical quantum mechanics it is somehow self-evident why are excitations of the vacuum meaningfully organized by representations of the respective symmetries - all the classical problems of QM from the infinite potential well to the hydrogen atom can be understood as representation-theory problems. However, the hard question of QFT (as far as my limited knowledge reaches, unanswered) is why are particles "renormalized irreducible quadratic excitations". So far, I have understood this as a postulate of QFT rather than some kind of mathematical necessity.

This post imported from StackExchange Physics at 2015-05-09 14:34 (UTC), posted by SE-user Void
What does he mean by "the one particle states furnish a representation of the symmetry." Is he basically saying that the symmetry maps one particle states into one particle states?

This post imported from StackExchange Physics at 2015-05-09 14:34 (UTC), posted by SE-user NowIGetToLearnWhatAHeadIs
The case $\hat{H}\propto\hat{n}$ is trivial. $\hat{H}\hat{S}|n\rangle = \hat{S}\hat{H}|n\rangle = n\hat{S}|n\rangle$, so $\hat{S}|n\rangle$ has the same number of particles as $|n\rangle$. In other words, $\hat{S}$ preserves particle number. A "free" field theory is precisely one in which $\hat{H}$ is a sum of $\hat{n}$ operators for every possible mode, so we've shown that the statement is true for a free theory. For an interacting theory I don't know.

This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user DanielSank
More answers (among them mine) together with an extensive discussion can be found at physicsoverflow.org/30822

This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user Arnold Neumaier

+ 3 like - 0 dislike

This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization.

A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the momentum operator, they can have further quantum numbers. These states should exist even in an interactiong field theory, describing a single particle away from any interaction. In a local quantum field theory, these states are associated with local field operators: $$| p, \sigma \rangle = \int e^{ipx} \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x$$ Where $\psi$ is the field corresponding to the particle and $\sigma$ describes the set of other quantum numbers additional to the momentum. A symmetry generator $Q$ being the integral of a charge density according to the Noether's theorem $$Q = \int j_0(x') d^3x'$$ should generate a local field when it acts on a local field: $[Q, \psi_1(x)] = \psi_2(x)$ (In the case of internal symmetries $\psi_2$ depends linearly on the components of $\psi_1(x)$, in the case of space time symmetries it depends on the derivatives of the components of $\psi_1(x)$)

Thus in general:

$$[Q, \psi_{\sigma}(x)] = \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x)])$$

Where the dependence of the coefficients $C_{\sigma\sigma'}$ on the momentum operator $\nabla$ is due to the possibility that $Q$ contains a space-time symmetry. Thus for an operator $Q$ satisfying $Q|0\rangle = 0$, we have $$Q | p, \sigma \rangle = \int e^{ipx} Q \psi_{\sigma}^{\dagger}(x) |0\rangle d^4x = \int e^{ipx} [Q , \psi_{\sigma}^{\dagger}(x)] |0\rangle d^4x = \int e^{ipx} \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) \int e^{ipx} \psi_{\sigma'}^{\dagger}(x) |0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) | p, \sigma' \rangle$$ Thus the action of the operator $Q$ is a representation in the one particle states. The fact that $Q$ commutes with the Hamiltonian is responsible for the energy degeneracy of its action, i.e., the states $| p, \sigma \rangle$ and $Q| p, \sigma \rangle$ have the same energy.

This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user David Bar Moshe
answered May 14, 2015 by (4,095 points)
1) Why does $[Q,\psi]$ have to produce a strictly local field? I guess that the symmetry of the Hamiltonian may lead to the requirement that it does but I do not see it. 2) Do indeed the statements hold in interacting field theory? Could you maybe point to a reference where it is shown that the fullblown self-interacting one-particle states also furnish a representation of the symmetry?

This post imported from StackExchange Physics at 2015-06-16 14:51 (UTC), posted by SE-user Void
@Void 1) Please think for a moment of $Q$ as the electric charge operator, you can write it uzing the Gauss' law as a surface integral of the electric field over a very large sphere at infinity. Due to the large distance these fields do not produce singularities when multiplied by other fields, thus the only singualrities coming from the commutator are those due to the local field $\psi$, thus the commutator itself is also a local field at $x$.

This post imported from StackExchange Physics at 2015-06-16 14:51 (UTC), posted by SE-user David Bar Moshe
@Void 2) Assuming Lorentz symmetry, the renormalization factors Z in ψ R =Z(ψ)ψ are Lorentz scalars, thus nothing essential changes in the analysis when the true renormalized field is used: Its transformation properties remain the same.

This post imported from StackExchange Physics at 2015-06-16 14:51 (UTC), posted by SE-user David Bar Moshe
+ 2 like - 0 dislike

This is a conceptual issue; only trivial computations are needed. Note that for an interacting theory, the 1-particle states are the (renormalized) states of the interacting theory and not the(bare) states of the free theory in terms of which the bare Hamiltonian is formulated!

The vacuum states of a Poincare covariant quantum field theory are defined as the states of the trivial (1-dimensional) representation of the Poincare group in the decomposition of the unitary representation of the Poincare group on the vacuum sector of the observable algebra of the theory. Similarly, the 1-particle states of the theory are defined as the states of the irreducible representations of the Poincare group in this decomposition.

If a self-adjoint operator $Q$ generates a 1-parameter group of symmetries $U(s):=e^{isQ}$ of the theory, it commutes with all generators of the Poincare group, i.e., if for each Poincare transformation $\Lambda$ and all real $s$,, the transform $U(s)\Lambda U(-s)$ is another Poincare transformation. This implies that the image of a representation space of the Poincare group under $U(s)$ is another representation space of the Poincare group. In particular, since symmetries are by definition invertible, the image of a vacuum state under a symmetry $U(s)$ has dimension $\le 1$, hence must be another vacuum state, and the image of an irreducible 1-particle space under a symmetry $U(s)$ is again irreducible, hence another irreducible 1-particle space. In particular, the space spanned by all vacuum states and the space spanned by all 1-particle states are both invariant under the symmetry, and hence furnishes a representation of it. (In particular, the assumption that $Q$ annihilates a vacuum state is not needed for this argument.)

answered May 9, 2015 by (14,019 points)
edited May 20, 2015

A self-adjoint operator Q generates a symmetry of the theory if it commutes with all generators of the Poincare group.

This can't be true, even the momentum operator won't commute with all generators of the Poincare group. for example with the angular momentum operators.

@JiaYiyang: Corrected, thanks!

if exp(Q) maps each Poincare transformations to another Poincare transformations.

Emm, what exactly does it mean? Will it cover internal transformations?

@JiaYiyang: I updated my answer again to make very explicit what I had meant.

if for each Poincare transformation Λ and all real s,, the transform U(s)ΛU(−s) is another Poincare transformation.

I still suspect a loophole can be found in this definition, but I can't construct a counterexample now. I think a minimal requirement for a "symmetry" would be that its generator is a conserved quantity(this would include boost operators even though they don't commute with Hamiltonian), but this is not evident from your definition.

since symmetries are by definition invertible, the image of a vacuum state under a symmetry U(s) is another vacuum state

I fail to see the connection between your first half of the sentence and the second half.

@JiaYiyang: 2nd part: Invertibility forces the dimensions to be equal, hence the image representation is trivial.

''this would include boost operators even though they don't commute with Hamiltonian'' - in which sense can they then be called conserved?

There are different levels of generality of the term symmetry. What they have in common is that solutions of the field equations are mapped to solutions of the field equations. If the field equations are covariant, this seems to me to require that a symmetry commutes with the Poincare group in the sense I stated, though I didn't check it. But I know of no counterexample.

@ArnoldNeumaier; Your argument really only works when Q annihilates the vacuum, because only in this case is exp(iQ) well defined. The assumption of an invariant vacuum is needed because when the symmetry is spontaneously broken, exp(iQ) on a single particle state symmetry rotates both the particle and the whole vacuum, so in addition to transforming the particle, it produces infinitely many 0-momentum Goldstone bosons, so it isn't even defined as an operator in the usual Hilbert space.

If you formally expand the Hilbert space in a nonseparable way to include all the different vacua in a single Hibert space which is not physical (nobody does this in a canonical formalism, but this is what the path integral does in practice), then it is true that the image of exp(iQ) where Q is the broken symmetry, is still a single particle state, for the reason you say, but it's a single particle state in a completely different vacuum.

@JiaYiyang: The argument is talking about an internal symmetry, just ignore supersymmetries and the Poincare group generators.

$e^{iQ}$ is defined if $Q$ is self-adjoint, which I assumed. In interacting field theories you often need to use a nonseparable Hilbert space (the direct sum of all superselection sectors)  to get a unitary representation of the Poincare group.

In QED one has a unique vacuum sector but the charged sectors are not Lorentz invariant. In QCD one has a circle of vacuum states parameterized by the theta angle.

In the canonical formalism one must (by definition) work in Fock space, which of course doesn't allow all this. This lack of flexibility makes the canonical formalism awkward for describing physical but noncanonical features of topological origin.

Algebraic QFT (though you seem to dislike it) nicely transcends the limitations of the canonical framework.

@ArnoldNeumaier: Ok, so you did mean to speak about the extended Hilbert space, with all the superselection vacuum sectors included. But this is unusual, so you need to say that you are thinking about a vastly expanded Hilbert space in the answer, because your argument doesn't work if you use the ordinary physical Hilbert space. This one-vacuum plus Fock space Hilbert space is what the student who asked the question certainly had in mind.

It's not that I don't like the way algebraic QFT "transcends" this limitation of a single vacuum, it's just that AQFT is not transcending anything, as I see it. It is simply defining properties of quantum fields formally, without any construction. The result is a formal and involved way to avoid speaking about path integration. See the discussion here.

@RonMaimon: The question posed discusses a quote from Witten. The context of Witten's statement clearly talks about the possible presence of many equivalent vacua. Thus one cannot restrict discussion to Fock space.

Concerning the path integral, see the discussion here.

t's not that I don't like the way algebraic QFT "transcends" this limitation of a single vacuum, it's just that AQFT is not transcending anything, as I see it. It is simply defining properties of quantum fields formally, without any construction.

In 2D there are lots of rigorous constructions, some of them very explicit, where one can understand superselection sectors, theta-vacua, charged states, etc. in a very detailed way. It is only in 4D where there are no constructions apart from quasifree fields - but this is because of the difficulty of the subject, not something intrinsic to AQFT.

@ArnoldNeumaier,

Invertibility forces the dimensions to be equal, hence the image representation is trivial.

$\exp(ia^{\dagger})$ is also invertible, but it connects 1-particle state with many-particle state.

''this would include boost operators even though they don't commute with Hamiltonian'' - in which sense can they then be called conserved?

Just in the usual sense： the boost generator $K$ satisfies $\frac{dK}{dt}=0$. Since $K$ explicitly depends on time, $\frac{dK}{dt}=i[H,K]+\frac{\partial K}{\partial t}$, this does not contradict with the fact that $H$ and $K$ do not commute.

What they have in common is that solutions of the field equations are mapped to solutions of the field equations.

I totally agree, but we seemed to be trying to give a definition without directly referring to field equations or Lagrangian.

If the field equations are covariant, this seems to me to require that a symmetry commutes with the Poincare group in the sense I stated, though I didn't check it.

Well, I may be missing something simple, but I haven't been able to show myself that "if $f$ is a solution of the EOM then $U(s)f$ must also be a solution of the EOM", given your condition on $U(s)$.

equal dimensions: The symmetry is already known to map a representation space into another one. The dimension of the spaces being 1, they must correspond to thee trivial representation.

Whether $K$ depends explictly on time depends on how the representation is given. The argument should be independent of the form of the representation.

The field equations are still arbitrary in Witten's context.

"if f is a solution of the EOM then U(s)f must also be a solution of the EOM", given your condition on U(s).

What I meant was the following; I corrected my answer accordingly.

Let the field equations be $F(\phi)=0$. In order to be a symmetry, $U(s)\phi$ must also be a solution. Since the Poincare group is a symmetry, $\Lambda\phi$ is a solution for every very Poincare transformation $\Lambda$. Thus $U(s)\Lambda U(-s)\psi$ must also be a solution, and since the extra symmetry is undone, it should be the solution $\Lambda'\phi$ for another Poincare transformation $\Lambda'$ - there are no other ''canonical'' solutions one can construct easily. Thus I conclude that we should have $U(s)\Lambda U(-s)=\Lambda'$.

A valuable discussion about the question "Is the path integral in 4 dimensions still only at the level of useful heuristic?" that is however off-topic to the question at hand, has been moved here to chat.

In particular, since symmetries are by definition invertible, the image of a vacuum state under a symmetry U(s) is another vacuum state, and the image of an irreducible 1-particle space under a symmetry U(s) is another irreducible 1-particle space.

I don't quite mind your conclusion on the vacuum state, since it's implied by the assumption of $Q$ annihilating the vacuum anyway. However, it still seems like a non sequitur to me that an invertible $U$ should map 1-particle states to 1-particle states, or in other words preserving the irreduciblity.

@JiaYiYang: I added a few words to my answer. The point is that the image is a representation, and it can be neither the vacuum representation nor a reducible one, hence is irreducible, hence belongs to a 1-particle space.

I just don't see how invertiblity implies preservation of irreduciblity. So is "symmetry must preserve irreduciblity" part of your definition of a symmetry?

@JiaYiYang: If the image representation were reducible you could split it into at least to, and take the preimages to get reducibility of the original representation.

@ArnoldNeumaier,

Yeah you are right, +1 to your answer. But I still wonder if you adopt Witten's definition of symmetry (an operator which commutes with the Hamiltonian), will you still be able to get the same result?

@JiaYiYang: I don't think so. His definition is in any case flawed; I don't think that he meant it literally.

+ 0 like - 0 dislike

What does he mean by "the one particle states furnish a representation of the symmetry." Is he basically saying that the symmetry maps one particle states into one particle states?

I think this is basically it. To prove it, you need to show the Hermitian generator $Q$ of your symmetry has the properties:

(1)$\langle 0|Q|1\rangle=0$,

(2)$\langle n|Q|1\rangle=0$ for $n>1$.

(1) has to be true since $Q$ is Hermitian and annihilates vacuum, and you just let $Q$ act to the left. For (2) I can't say I have a completely rigorous proof, but here's my heuristics: For $\langle n|Q|1\rangle$ to be nonzero, $Q$ must increase the particle number when acting on $|1\rangle$, and such an operator must also increase the particle number when acting on vacuum(this is where I don't I have a rigorous proof), and if so we arrive at a contradiction, and we conclude (2) must be true.

Note that reasoning for (2) is far from complete.

answered May 9, 2015 by (2,640 points)
edited May 26, 2015

$Q=a^*a^*a$ annihilates the vacuum but creates a 2-particle state from a 1-particle state; so the argument must be more involved.

@ArnoldNeumaier, yes, I'm suspecting Hermiticity might play a role.

Yes. To turn this into a complete argument, you'd have to assume that you can expand $Q$ into a series in normally ordered c/a operators, then a hermiticity argument applies and gives after some more work the desired contradiction. But the whole argument works only under the assumption that the interacting theory is in a Fock representation, and even then it is not fully rigorous.

@JiaYiyang; The way you prove the result is to fold in the proof of Goldstone's theorem. You need to know that $Q = \int j_0(x) d^3x$, there there is a local current for the symmetry, and then consider smoothed out finite volume versions of Q $\int j_0(x) \rho(x) d^3x$. When the symmetry is spontaneously broken, the current produces a Goldstone boson, otherwise it just rotates the one particle states into symmetry related one particle states. This is standard current algebra, Gell-Mann's paper on this is really good.

@RonMaimon, thanks, let me think about it.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.