Srednicki's "Quantum Field Theory", an electronic copy of which is freely available here, seems to state on p 205 that the states eq. (32.3) which differ by a phase factor that can range through [0,2$\pi$) are mutually orthogonal. But if the underlying Hilbert space is separable this does not seem possible. Who can enlighten me?
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2$\begingroup$ Could you explain a little bit more why you think the mutually orthogonal vacuum states are not compatible with the Hilbert space being separable? $\endgroup$– HeteroticCommented Nov 1, 2013 at 12:47
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$\begingroup$ This is not the problem, they will be orthogonal. But in a separable Hilbert space there can at most be a countable number and the points in the interval [0,2$\pi$) are not. $\endgroup$– UrgjeCommented Nov 1, 2013 at 15:48
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$\begingroup$ There are two different concepts : the space of parameters defining the real vaccum (the moduli space) which is $U(1)$, so is continuous and not countable, and the usual space of Fock states. For the Fock states, this is math, while, because the momentum space is continous, I would say, that, naturally, the Hilbert space has not a countable basis... $\endgroup$– TrimokCommented Nov 1, 2013 at 18:48
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$\begingroup$ There are aspects of QFT practice (the way physicists use it) which have never found mathematically rigorous formulation, I wonder if this is another example? $\endgroup$– Mitchell PorterCommented Nov 1, 2013 at 21:13
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$\begingroup$ Somehow I missed the closely related question physics.stackexchange.com/q/56520 where in the answers a reference was made to a paper by Brauner, "arxiv.org/abs/1001.5212". This work indicates that for the case at hand the overall Hilbert space is not separable (although the Fock spaces built on the different vacua may) and that different vacuum states are indeed orthogonal. $\endgroup$– UrgjeCommented Nov 4, 2013 at 11:21
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2 Answers
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For the benefit of others who read this, note that if $\mathcal H$ is a separable Hilbert space, then there exists a countable, orthonormal basis for $\mathcal H$. Notice that this does not immediately imply that there cannot exist an uncountable basis for $\mathcal H$, but this nonetheless turns out to be true as a consequence of the dimension theorem;
Let $V$ be a vector space, then any two bases for $V$ have the same cardinality.
See also the following math.SE posts:
Now, for your question. Let's suppose that there are an uncountable number of orthogonal vacuua $|\theta\rangle$ in the Hilbert space where $\theta\in [0,2\pi)$, then we have the following possibilities
The Hilbert space of the theory is not separable. In this case, there is no contradiction.
The Hilbert space of the theory is separable. In this case, there is a contradiction, and we need a resolution.
As far as I am aware, most axiomatizations of QFT assume that the Hilbert space of the theory is separable, but there is discussion in the literature about relaxing this assumption. I'll attempt to dig up some references.
Let's therefore assume separability and look for a resolution. The standard resolution is that when constructing the Hilbert space of the theory, one chooses only one of these (physically equivalent) vacuua to be the vacuum of the Hilbert space, then one constructs the rest of the physical Hilbert space about this vacuum. The rest of the vacua are not elements of the Hilbert space of the theory.
There is another perspective on this which is interesting. Let's suppose that there is some larger, non-separable Hilbert space $\mathcal H_\mathrm{big}$ containing all of the vacua $|\theta\rangle$ and which is an orthogonal direct sum all of the Hilbert spaces $\mathcal H_\theta$ that could have been generated from each of the possible vacua and used as the physical Hilbert space of the theory.
\begin{align}
\mathcal H_\mathrm{big} = \bigoplus_{\theta\in[0,2\pi)} \mathcal H_\theta
\end{align}
Then we view each of the Hilbert spaces $\mathcal H_\theta$ as a superselection sector of the larger Hilbert space $\mathcal H_\mathrm{big}$. In this case, if the physical system occupies a state $|\psi\rangle$ in a given superselection sector $\mathcal H_\theta$, then the state of the system will remain in the sector for all times under the Hamiltonian evolution, so we may as well view "the" Hilbert space of the system as simply the superselection sector it started in. In a sense, this is essentially the same as originally having picked a vacuum on which to build the Hilbert space because the different superselection sectors don't "talk" to each other.
The following physics.SE post is useful for understanding superselection sectors:
What really are superselection sectors and what are they used for?
I also found the following nLab page on superselection theory to be illuminating:
answered Nov 1, 2013 at 18:56
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$\begingroup$ 1) The Wightman axioms demand the Hilbert space to be separable. But see the section about separable spaces on Wikipedia (Hilbert space). 2) I understand your elegant argument about H_big. In the separable case the |theta>'s are not linearly independent and there is some "multiple counting". 3) In writing down the Higgs Lagrangian what is assumed about the underlying Hilbert space? I never found this explicitly mentioned. Conventionally it would be separable. It would be nice to see a thorough discussion of this matter. $\endgroup$– UrgjeCommented Nov 1, 2013 at 21:42
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$\begingroup$ @Urgje 1) Yes that's what the Wightman axioms demand; that doesn't mean there isn't another (perhaps undiscovered) set of axioms that don't assume this. 2) I'm not saying that the $|\theta\rangle$'s are not linearly independent. The $\mathcal H_\mathrm{big}$ construction assumes that they each belong to a different, orthogonal subspace in the direct sum, so they are, in particular, linearly independent. 3) I don't think there is currently a sufficiently satisfactory axiomatization of standard model quantum field theory for such a thorough discussion to exist, but I may be wrong. $\endgroup$ Commented Nov 1, 2013 at 21:50
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$\begingroup$ 1) Indeed, these axioms also require the vacuum state to be unique and it seems this is not always the case. 2) Point taken. In this way the dimensionality issue is avoided. However, if the Hilbert space is separable, then there is a countable basis and all |theta>'s can be expressed in terms of them. Then it seems that the direct integral H_big can be replaced by an infinite direct sum. 3) This is really a pity. Already a discussion of the isolated Higgs model would be enlightening. $\endgroup$– UrgjeCommented Nov 2, 2013 at 19:57
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Not 100% sure but here is an attempt to a solution:
1)First of all, it's not clear to me why the underlying Hilbert space needs to be separable. Is this mentioned somewhere in the book and is there a physical reason for this requirement?
2)Nevertheless, I will assume here that the Hilbert space of physical states needs indeed to be separable. A possible way out of the contradiction is the following: There is an uncountable number of solutions that minimize the potential but all of them are physically equivalent. You can imagine that these solutions live in some kind of space if you want to, but this is not the Hilbert space of physical states. Out of these solutions we arbitrarily pick one and define it as the true (physical) vacuum. But any choice is physically equivalent and all of them lead to the same (single) physical state.
