# 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
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
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? Aug 3, 2021 at 20:46

In short, the answer to the last question is yes. Let $$|p_0\rangle$$ be a single-particle state with definite spatial momentum $$\begin{equation} \hat P |p_0\rangle = p_0 |p_0\rangle. \end{equation}$$ Ignoring spin labels, this is the same as what Weinberg calls $$\Psi_{p,\sigma}$$. Then, it is true that $$\begin{equation} \int dp |p\rangle \langle p| = 1_{\rm single-particle} \end{equation}$$ 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).