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...

@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).

@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).

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.

And now the important point is that if we build our Fock space from $FS$ in the thermodynamic limit, we, roughly speaking, "decouple" it from the Fock space we obtain by working with $|0\rangle$, correct? And thus we have two theories?! And I guess it is experiment which decides what theory is the correct one?!

@TobiasFünke Yes, that's right. Given a generic Hamiltonian, normal ordering it with respect to a given reference state means expressing it in terms of creation/annihilation operators such that (1) the creation operators are on the left and (2) all of the annihilation operators annihilate the reference state. In my example, $a^\dagger a$ is normal ordered with respect to $|0\rangle$ but not $FS$; to normal order with respect to the latter, we first define our $b$'s (which all annihilate $FS$) and then reorder the original Hamiltonian as necessary.

@TobiasFünke To provide some context as far as interactions go, you might consider $\phi^4$ theory. The free field has the Hamiltonian $\sum_\mathbf k \omega_\mathbf k a^\dagger_\mathbf k a_\mathbf k$ which is normal ordered wrt the "free vacuum" $|0\rangle$, but adding $\lambda \phi^4/4!$ destroys this property. Normal ordering the interaction term implies that we're moving to a different ground state $|\Omega\rangle$ (the so-called physical vacuum or interacting vacuum) which constitutes the foundation for an entirely different Hilbert space carrying a unitarily inequivalent [...]

[...] representation of the CCR/CARs. The fact that the ground state of the interacting theory and the ground state of the free theory actually lie in different Hilbert spaces (suggesting that $|\Omega\rangle$ cannot be expanded in terms of eigenstates of the free theory) is the spiritual content of Haag's theorem and, since that's precisely what we do in the perturbative approach to QFT, constitutes an important unsolved problem in the formal foundations of (interacting) QFT.

these two starting points are inequivalent; but we actually "started" the "FS-representation" from the bare vacuum $|0\rangle$. So my (last) question is, I guess, the following: Starting just from the CAR in the TDL, can we get both representations, and in particular the FS-representation without "knowing" it is constructed from the bare vacuum $|0\rangle$? Do you know a good intro book where I can get a basic understanding of all of this? I guess my thoughts are too confusing rn... Anyway, thank you very much again for your nice answer, time and patience. Always a pleasure to learn from you!

The point about representation theory is that the CARs should be understood to be defining an abstract $*$-algebra which is generated by the elements $\{\mathscr A_\mathbf k\}$ equipped with the product operation $\{\mathscr A_\mathbf k ,\mathscr A^\dagger_{\mathbf q}\} = \mathbf 1 \delta_{\mathbf p \mathbf q}$. These are not operators, just an abstract algebra. A representation of this algebra is a map $\pi$ which takes elements of the algebra to operators on some Hilbert space.

In my first example, we have two representations $\pi_1$ and $\pi_2$ whereby $\pi_1(\mathscr A_\mathbf k) = a_\mathbf k$ and $\pi_2(\mathscr A_\mathbf k) = b_\mathbf k$.

These representations are unitarily equivalent because if you consider the map $U$ (from the Hilbert space to itself) such that $Ua_\mathbf k U^{-1} = b_\mathbf k$, you will find it to be unitary.

Or rather, there exists a unitary map which maps $a_\mathbf k \mapsto Ua_\mathbf kU^{-1} = b_\mathbf k$

Hi. Thanks for your time. Yes, I think I had this in mind already, but your clear words brought it on point. Now in the TDL, it seems, there exists no such U (which I assume can be proven rigorously), and both representations are unitarily inequivalent, correct? So the normal ordering thing which Fradkin discusses, basically picks out the correct representation ( (which has to be proven by experiments, as I think inequivalent representations make different predictions).

Explicitly in this case, such a $U$ is provided by the operator $N^{-1/2} \prod_\mathbf k a^\dagger_\mathbf k$

Which simply maps $|0\rangle$ to $FS$.

Yes, exactly.

Okay so as you say, this operator becomes problematic in the thermodynamic limit

This paper goes through some calculations in detail

link.springer.com/article/10.1007/BF01810425

In any case, yes it can be shown that there is no such $U$

So what happens is that the normal ordering procedure essentially defines a representation. Given a normal-ordered Hamiltonian, we may define the reference state to be annihilated by all of the $a_\mathbf k$'s, and build the corresponding Hilbert space from there.

You can look at the issue forwards and backwards - given a (single particle) Hilbert space, we may define the vacuum state and creation/annihilation operators, then the Fock space (what fradkin calls the physical Hilbert space), then a normal ordered Hamiltonian

or we may start with a normal ordered Hamiltonian (built formally from creation/annihilation operators), define the vacuum as being annihilated by the annihilation operators, and then build up the Fock space from there

1) Thanks for the paper, it looks really interesting and at a first glance I can understand it well (although I find it strange that they compute inner products of vectors of different HIlbert spaces, but I think this is not really important for the following discussion). 2) Ahhh, okay, Now it seems much clearer. I had/have difficulties to connect the different ways I learned about Fock spaces etc. to the more algebraic view...

The canonical quantization recipe for basic QFT is to write down a classical Lagrangian, transform to a classical Hamiltonian, Fourier transform to momentum space, and then put hats on the $a$'s (the coefficients in the Fourier transform) and interpret this as a Hamiltonian operator.

So what we end up with at the end of the day is a Hamiltonian which has a bunch of creation and annihilation operators in it

The important point for high energy QFT is that the free Hamiltonian has one reference state and the Hamiltonian with interaction terms has a different reference state

Two distinct vacuums, which in the thermodynamic limit do not belong to the same Hilbert space

To perhaps phrase it differently, $:H_0 + H_{int}: \neq :H_0: + :H_{int}:$

@J.Murray By that <you mean that the respective representations are unitarily inequivalent, no?

The representations of the CAR on the respective Hilbert spaces are unitarily inequivalent, yes

But what I mean with that last line on normal ordering is the following

It's a big of a pain in the ass to carry the calculation out (though it is a useful exercise) but notice that if we have something of the form $a^\dagger a + \lambda (a+a^\dagger)(a+a^\dagger)(a+a^\dagger)(a+a^\dagger)/4!$

The first term is the free Hamiltonian $H_0$, and the latter term is the interaction Hamiltonian

@J.Murray This really helped a lot. So to summarize (let me know whether or not this is roughly correct): We can start with a single-particle Hilbert space and from that create our Fock space with the bare vacuum |0\rangle, but then, through normal ordering, can also define a new vacuum, here FS.

For finite systems, these are unitarily equivalent. But in the TDL, these are unequivalent. What I missed before is that the representation "induced" by the normal ordering/def. of FS is independent of the way we "discovered" it. So from the starting point of the abstract algebra, we have, in the TDL, (in the scope of our discussion) two inequivalent representations.

Yes, that looks right to me

With respect to your earlier reservation about computing inner products between vectors of different Hilbert spaces, perhaps you could think of it as assuming the vectors belong to the Hilbert space, computing their inner product under that assumption, and obtaining a nonsensical/contradictory result.

@J.Murray You write: "but notice that if we have something of the form" - is there something missing? Sorry, but I don't understand what you mean with this paragraph.

@J.Murray Ah, this makes sense indeed!

It's a bit like checking to see if a function belongs to the Hilbert space by checking to see that its norm is finite. Objects which don't belong to the Hilbert space don't even have norms, but we still go through the motions.

Yes, one second I need to compose my thoughts on that final point

Sure, take your time

Okay, let me rephrase, what I wrote doesn't really make sense and isn't what I meant

What I mean is that if you take a term like $\lambda \phi^4 \sim \lambda (a+a^\dagger)(a+a^\dagger)(a+a^\dagger)(a+a^\dagger)$

and then examine the difference between that and its normal-ordered counterpart $:\lambda \phi^4:$

Ah, okay

Then you find that $:\lambda \phi^4: = aaaa + 4a^\dagger a a a + 6a^\dagger a^\dagger a a + 4 a^\dagger a^\dagger a^\dagger a + a^\dagger a^\dagger a^\dagger a^\dagger$

However, $\lambda \phi^4$ (no normal order) also contains additional terms which arise from the commutation relations, i.e. $a^\dagger a a^\dagger a = a^\dagger a^\dagger a a + a^\dagger a$

So at the end of the day we find that $\lambda \phi^4 = :\lambda \phi^4: + $ terms like $aa, a^\dagger a^\dagger,$ and $a^\dagger a$

as well as constants $(a a a^\dagger a^\dagger = a^\dagger a^\dagger a a + a^\dagger a + 1)$

I see...and this, of course, describes different physics (which I guess is the general problem in quantization procedures)!

Right.

so the fact that we perturbatively expand the interacting vacuum in terms of the free particle eigenstates is ... problematic

this also ties in with renormalization, adding counterterms to the Hamiltonian to cancel infinities, etc

The recognition that the infinities which arise in QFT are, in a large part, due to these quantization ambiguities is the key insight which takes renormalization from a dirty trick to a brilliant procedure for overcoming our fundamental ignorance of high energies and short distances.

hmh, I only have a rough understanding of all of this, but I think I can follow your line of thoughts

I'm not a high energy guy, so I'm probably not the best one to talk to for an intuitive understanding of the issues at play. Things make more sense in condensed matter - to me, at least. But I'm glad I could help a bit.

Anyway, I have to go to bed now... But let me thank you once more. I learned a lot through our discussion!

Likewise, always good to straighten out my own thoughts by talking to other people. Cheers.

See you around!