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In Ref. 1 the author states:

The procedure of normal ordering allows us to define the physical Hilbert space. The physical meaning of this approach becomes more transparent in the thermodynamic limit $N\to\infty$ and $V\to\infty$ at constant density $\rho$. In this limit, the Hilbert space is the set of states that is obtained by acting finitely with creation and annihilation operators on the ground state (the vacuum). The spectrum of states that results from this approach consists of the set of states with finite excitation energy. Hilbert spaces that are built on reference states with macroscopically different numbers of particles are effectively disconnected from one another. Thus, the normal ordering of a Hamiltonian of a system with an infinite number of degrees of freedom amounts to a choice of the Hilbert space. This restriction becomes of fundamental importance when interactions are taken into account.

after discussing the particle-hole transformation for $N$ non-interacting, non-relativistic fermions at zero temperature and the normal ordering of the respective Hamiltonian relative to the Fermi sea ground state $|\mathrm{gnd}\rangle$.

A brief and similar discussion of these matters is also given in Ref. 2.

Question: What exactly is meant by the above quote? More specifically, what is a or the physical Hilbert space? And what does it mean for two Hilbert spaces to be effectively disconnected? What is the role of the thermodynamic limit and interactions here?

I always thought that normal ordering is a prescription concerning the order ambiguity in quantization procedures... I guess that the author means that we have "the choice" of two such reference states: The vacuum $|0\rangle$, which is annihilated by all annihilation operators and contains no particles and the filled Fermi sea state $|\mathrm{gnd}\rangle$, which serves as the vacuum state in the particle-hole picture and contains $N$ particles.

All in all, I have the feeling that the above quote concerns the possible unitary inequivalent representations of the canonical anti-commutation relations, but due to my limit background I cannot really make sense of this.


References:

  1. Quantum Field Theory. An integrated approach. Eduardo Fradkin. Princeton University Press. 2021. Chapter 6, p. 151

  2. The Fundamentals of Density Functional Theory (revised and extended version). Helmut Eschrig. Lecture Notes. 2003. Chapter 1, p. 33-34

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  • $\begingroup$ Your question might be a good one for the mathematicians because I can see non-trivial issues cropping up that go beyond the physicality of Hilbert spaces (which I doubt, in the first place... a Hilbert space is a mathematical tool to solve the equations, it's not something we actually observe in nature). See e.g. mathoverflow.net/questions/423471/… (random hit, I don't understand enough of it to even judge the relevance). $\endgroup$ Mar 29, 2023 at 18:28
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    $\begingroup$ @FlatterMann Thanks for your comment. Unfortunately, I don't really it. The question is IMHO perfectly fine for a physics forum; I don't request complete mathematical rigor, to start with. $\endgroup$ Mar 29, 2023 at 18:31
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    $\begingroup$ Sorry for the misunderstanding: I didn't mean to say that it's not fine in the PSE. It absolutely is and I upvoted it myself. I would like to know the answer. My point is that since you are dealing with non-trivial mathematics here, the question you are asking might actually be more pressing to the mathematics community and that you can find more thorough answers there in addition to what somebody might say in this forum. What I should have written is "Your question might ALSO be a good one for the mathematicians.". Fair? $\endgroup$ Mar 29, 2023 at 18:33
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    $\begingroup$ Ah, I see, again thanks for the comment and the link anyway. The point is I don't even know how to phrase the question in a language appropriate for a math forum... And I guess most mathematicians won't know or understand the physical language and the meaning of the quoted paragraph. $\endgroup$ Mar 29, 2023 at 18:36
  • $\begingroup$ I think I would have agreed with you a few decades ago, but I am getting the feeling that some mathematicians did catch on that there is some serious meat on the bones of the QFT constructions of the physicists. In any case, let's see who has a good answer. Looking forward to it. $\endgroup$ Mar 29, 2023 at 18:39

1 Answer 1

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Broadly speaking I'd say your interpretation is basically right.

Consider a system of spinless, non-interacting fermions in a box with volume $V$. The vacuum state $|0\rangle$ is annihilated by all of the annihilation operators $a_{\mathbf k}$. In order to construct a Hilbert space for our theory, we first consider the vector space of objects $\mathscr V$ which can be obtained by applying a finite number of creation operators to $|0\rangle$, and linear combinations thereof. We can equip this vector space with an inner product which is defined via the canonical anticommutation relations: $$\psi = a_{\mathbf k_1}^\dagger a_{\mathbf k_2}^\dagger |0\rangle \qquad \phi = a^\dagger_{\mathbf q_1} a^\dagger_{\mathbf q_2} |0\rangle \qquad \{a_\mathbf k,a_\mathbf q\}=\{a^\dagger_\mathbf k,a^\dagger_\mathbf q\}=0 \qquad \{a_\mathbf k,a^\dagger_{\mathbf q}\} = \delta_{\mathbf k\mathbf q}$$ $$\implies \langle \psi,\phi\rangle = \langle 0|a_{\mathbf k_2} a_{\mathbf k_1} a^\dagger_{\mathbf q_1} a^\dagger_{\mathbf q_2}|0\rangle = \delta_{\mathbf k_1 \mathbf q_1}\delta_{\mathbf k_2 \mathbf q_2}-\delta_{\mathbf k_1 \mathbf q_2}\delta_{\mathbf k_2 \mathbf q_1}$$

$V$ is not a Hilbert space yet because it is not toplogically complete, so our final step is to take the Hilbert space completion of it by "adding" the Cauchy sequences of elements of $V$. This space $\overline V$ is what Fradkin calls the physical Hilbert space built from $|0\rangle$.

If we equip this space with a (non-interacting) Hamiltonian $H = \sum_\mathbf k \epsilon(\mathbf k) a^\dagger_\mathbf k a_\mathbf k$, we may talk about the ground state of an $N$-particle system, which is the filled Fermi sea. Clearly this state can be obtained by acting on $|0\rangle$ with a finite number of creation operators; explicitly, it is given by $$FS = \frac{1}{\sqrt{N}}\prod_{|\mathbf k|<k_F} a^\dagger_\mathbf k |0\rangle$$

If we'd like, we may start with $FS$ and try to build a Hilbert space over that instead. We define $$b_\mathbf k := \begin{cases} a_\mathbf k & |\mathbf k|>k_F \\ a^\dagger_{-\mathbf k} & |\mathbf k|<k_F\end{cases}$$ which all annihilate $FS$. The original Hamiltonian can be re-expressed as $$H = \sum_{\mathbf k} \epsilon(\mathbf k) a^\dagger_\mathbf k a_\mathbf k = \sum_{|\mathbf k|> k_F}\epsilon(\mathbf k)b^\dagger_\mathbf k b_\mathbf k - \sum_{|\mathbf k|<k_F}\epsilon(-\mathbf k)b^\dagger_\mathbf k b_\mathbf k + E_0$$

where $E_0 \equiv \sum_{|\mathbf k|<k_F} \epsilon(\mathbf k)$ is the ground state energy of the filled Fermi sea. The (positive-energy) excitations created above the Fermi surface are "genuine" particles, while the (negative-energy) excitations created below the Fermi surfaces are "holes."


It's fairly easy to see that the physical Hilbert space constructed from $|0\rangle$ is the same as the physical Hilbert space constructed from $FS$. However, we now consider the thermodynamic limit $N,V\rightarrow \infty$ such that $n\equiv N/V$ remains fixed. The problems with that arise immediately.

The Fermi momentum is implicitly defined such that $N = \sum_{|\mathbf k|<k_F}$. In the thermodynamic limit of large $V$ the sum can be converted to an integral, where we have $$N = \sum_{|\mathbf k|<k_F}\approx \frac{V}{(2\pi)^3}\int_{|\mathbf k|<k_F} \mathrm d^3 k = \frac{V}{(2\pi)^3}\frac{4}{3}\pi k_F^2 $$ $$\implies k_F \propto \sqrt{N/V}$$ As a result, the Fermi momentum and energy are functions only of $n$, and don't change as we take the thermodynamic limit. On the other hand, $E_0 = \frac{V}{(2\pi)^3} \int_{|\mathbf k|<k_F} \epsilon(\mathbf k) \mathrm d^3k\rightarrow \infty$, making the re-expressed Hamiltonian ill-defined.

Additionally, $FS$ can no longer be obtained by acting on $|0\rangle$ with a finite number of creation operators, nor can it be written as the limit of a Cauchy sequence of such vectors. As a result, taking the thermodynamic limit decouples the physical Hilbert space over $|0\rangle$ from the physical Hilbert space over $FS$.


All in all, I have the feeling that the above quote concerns the possible unitary inequivalent representations of the canonical anti-commutation relations, but due to my limit background I cannot really make sense of this.

Yes, that's the idea. For finite $N$ and $V$, there are a countable number of allowed momentum modes, each with a countable occupation number (for fermions of course this occupation number is either 0 or 1, but the generalization is painless). Therefore, by enumerating these modes $\mathbf k_1,\mathbf k_2,\ldots$ we may specify states in using the occupation number representation $|n_1,n_2,\ldots\rangle$. The occupation number basis can be associated with the set $\mathbb N^\mathbb N$.

If we consider the linear span of the occupation number basis elements, we obtain a vector space $\mathscr S$ which is a strict superspace of the $\mathscr V$ described at the top of my answer. The reason for this is that $\mathscr V$ does not include basis vectors with an infinite number of nonzero occupation numbers, e.g. the vector $|1,1,1,\ldots\rangle$. In order to make a well-defined (i.e. finite) inner product, we first choose a unit norm "reference state" (such as $|0,0,0,\ldots\rangle$), then consider the vector space generated by vectors which differ from this in a finite number of places (our $\mathscr V$), equip it with an inner product derived from the CCR/CARs, and finally take the Hilbert space completion to define the physical Hilbert space.

When $N$ and $V$ are finite, $FS$ differs from $|0\rangle$ in a finite number of places (by definition) and therefore belongs to the same physical Hilbert space. In the thermodynamic limit, however, there are an uncountable number of momentum modes (so the occupation number basis set becomes isomorphic to $\mathbb R^\mathbb N$) and $FS$ does not differ from $|0\rangle$ in a finite number of places. In this case, the vector space $\mathscr S$ decomposes into an uncountable infinity of subspaces, each of which can be made into a physical Hilbert space in its own right.

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    $\begingroup$ Dear J. Murray, thank you for your very beautiful answer!!! This is very insightful! I have to think about some parts of your answer, however. If you allow, let me ask two questions: a) From what I understand, for finite $N,V$, both approaches yield the same physical predictions - correct? But in the TDL, it makes a difference if we work with FS or $|0\rangle$ as a vacuum state? Is there a way to see this/the consequences more concretely? b) Do you know what happens if we turn on interactions? I have the feeling that all of this is indeed similar to relativistic QFT... $\endgroup$ Mar 30, 2023 at 18:12
  • $\begingroup$ @TobiasFünke a) I'm not sure how convincing or concrete you'll find it, but it's not difficult to show that since $FS$ differs from $|0\rangle$ in an infinite number of places, $\langle FS| \alpha\rangle = 0$ for all $|\alpha\rangle$ in the physical Hilbert space built constructed from $|0\rangle$ (more precisely, assuming that $FS$ lies in that Hilbert space and then computing that inner product yields zero, which is a contradiction since the inner product must be non-degenerate and $FS$ is not the zero vector). $\endgroup$
    – J. Murray
    Mar 30, 2023 at 22:54
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    $\begingroup$ @TobiasFünke b) I'm not entirely certain what specific aspect of interacting QFTs Fradkin is referring to, but my guess would be that normal ordering a kinetic term (e.g. $\sum \epsilon_i a^\dagger_i a_i$) like we did above induces an overall shift in the total energy (the $E_0$ in my answer), which is ultimately not measurable. On the other hand, an interaction term has more creation/annihilation operators in it, and shuffling them around leads to less trivial differences (take my example of a free Hamiltonian re-expressed in terms of the $b$'s and add an interaction term). $\endgroup$
    – J. Murray
    Mar 30, 2023 at 23:02
  • $\begingroup$ Hi, thanks for your reply! I begin to better understand, but I have the feeling that I miss something crucially here...As far as I understand, working in the particle-hole "picture" effectively normal orders the Hamiltonian relative to $FS$, no? So $:H:=H-E_0$; working with this renormalized/ normal ordered Hamiltonian and in the P-H-"picture", we have a finite ground state energy even in the TDL, and moreover the vacuum, now $FS$ is not polarized in the sense that it is the state which is annihilated by all annihilation operators and is the ground state of the Hamiltonian. $\endgroup$ Mar 31, 2023 at 7:11
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – J. Murray
    Mar 31, 2023 at 20:16

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