Hilbert space and wave functions of single-particle states in QFT (Weinberg)

So, following Weinberg (chapter 2), he derives all the transformation properties of the states $$\Psi_{p,\sigma}$$. These are eigenstates of the four-momentum (and some other observable with a discrete spectrum), which obey the orthonormality relation $$(\Psi_{p',\sigma'}, \Psi_{p,\sigma}) = \delta(p'-p) \delta_{\sigma' \sigma} \tag{2.5.19}.$$ Also, I assume (but am uncertain) that they form a complete basis for the single-particle Hilbert space. If so, then the most general single-particle state can be expanded in terms of these eigenstates as

$$\Psi = \sum_\sigma \int \frac{d^3p}{2p_0} c_\sigma(p) \Psi_{p,\sigma}$$

where $$\frac{d^3p}{2p_0}$$ is the Lorentz invariant integration measure over the mass-shell and $$c_\sigma(p)$$ are simply the expansion coefficients. Is this representation of a general state correct? I ask because Weinberg never writes anything of the sort. He only deals with momentum eigenstates.

Also, from my prior knowledge of quantum mechanics, the expansion coefficients $$c_\sigma(p)$$ are usually interpreted as a wave function in momentum representation. As such, the Fourier transform of $$c_\sigma(p)$$ gives the corresponding wave-function in position representation. Now, I know that QFT doesn't use wave functions. But still, what prevents me from defining wave functions in this way?

From writting this question, I suspect that the problem is with the definition of the single-particle Hilbert space. Weinberg never properly defines what is the Hilbert space of these one particle states, much less state that the $$\Psi_{p,\sigma}$$ states form a complete basis. Can a one-particle Hilbert space be consistently defined in this way? Or does it only make sense as a part of Fock space?

Edit: in short, can someone confirm if the following resolution of the identity holds for the one-particle Hilbert space?

$$\sum_\sigma \int d^3p \Psi_{p,\sigma} \Psi^\dagger_{p,\sigma} = 1\!\!1$$

This is what I understand as a "completeness" relation for the $$\Psi_{p,\sigma}$$ basis, and the expansion of a general state $$\Psi$$ in terms of the states $$\Psi_{p,\sigma}$$ depends on the validity of this relation.

• The momentum operators are hermitian and commute with one another. This implies that there is one (improper) basis of eigenstates of these operators which is what we denote as $\Psi_{p,\sigma}$ the second label just indexing possible degeneracies. The rigorous version of this is the so-called SNAG theorem. In short, that is the reason why these states fo form a basis in the one-particle Hilbert space.
– Gold
Commented Aug 3, 2021 at 1:05
• Also, regarding what should be a one-particle Hilbert space, that lies in the requirement that the representation of the Poincare group be irreducible. In particular this demands $P^2 = -m^2$ over the states in the one-particle Hilbert space.
– Gold
Commented Aug 3, 2021 at 1:07
• So they DO form a basis of the one-particle Hilbert space? This means that the resolution of the identity (added to the question as an edit) holds? Can I interpret the coefficients $c_\sigma(p)$ as a wave function in momentum representation? Commented Aug 3, 2021 at 20:46

Nothing prevents you from writing a state like that! Always remember, ordinary QM is a type of QFT, a 1-dimensional one to be exact. In QM, the relevant "fields" are $$X$$ and $$P$$, and they are parametrized by time alone instead of space and time.

So now you see QFT and QM are not that different, BUT you have to be careful, and remember that in QFT the position is just a parameter the same way time is in QM, it is NOT an operator, the fields are the operators. So you can't have a "wavefunction" in the sense of a complex function that depends on position the same you have it in QM. Well, to be extra precise, there is something "like" a position operator that you can define in QFT, it's called Newton-Wigner operator, but it's not really the same thing as in QM, and it has some problems. There is another SE post about it here: What exactly is the difference between the position operator in non-relativistic QM and the Newton-Wigner operators in QFT?

The same is NOT true for momentum, momentum is fine as an operator. Remember how in QM time is not an operator, but energy is. It's more or less the same thing. Your resolution of unity is fine, everything is fine. There is just one problem: since you have no position operator you also have no complete set of position eigenstates. Since you have no complete set of position eigenstates, you don't have a resolution of unity in terms of positions, and so whatever you do, including taking the Fourier transform of your $$c_{\sigma}(p)$$, it won't be a position space wavefunction.

The one-particle Hilbert space is a separate matter, I don't really see how it is related.

EDIT: I realize now what you were saying about the one-particle Hilbert space, and I have to say you are right, you can come up with a "position basis" for one particle following means similar to what you described. In fact that's more or less what Schwartz does in page 23 of his QFT book to extract the wave function and the Schrodinger equation for the large mass (low energy) limit. But it course it's not some kind of wave function for the whole QFT, it simply describes a subspace, and if you take it too seriously you run into troubles with antiparticles etc.

Additionally, the Fourier-transformed momentum eigenstates won't truly represent localized particles, as you won't get a delta function, again exactly because you don't have a position operator.

In short, the answer to the last question is yes. Let $$|p_0\rangle$$ be a single-particle state with definite spatial momentum $$$$\hat P |p_0\rangle = p_0 |p_0\rangle.$$$$ Ignoring spin labels, this is the same as what Weinberg calls $$\Psi_{p,\sigma}$$. Then, it is true that $$$$\int dp |p\rangle \langle p| = 1_{\rm single-particle}$$$$ is the unity operator in the single-particle Hilbert space. (Above I was sloppy with the integration measure: there can be a $$\sqrt{p^2+m^2}$$-dependent factor, which depends on how $$|p\rangle$$ are normalised).