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.

  • $\begingroup$ 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. $\endgroup$
    – Gold
    Aug 3, 2021 at 1:05
  • $\begingroup$ 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. $\endgroup$
    – Gold
    Aug 3, 2021 at 1:07
  • $\begingroup$ 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? $\endgroup$ Aug 3, 2021 at 20:46

1 Answer 1


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).


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