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Question:

What is the general definition of one particle states $|\vec p\rangle$ in an interacting QFT? By general I mean non-perturbative and non-asymptotic.


Context. 1)

For example, in Weigand's notes, page 42, we can read (edited by me):

As $[H,\vec P]=0$, these operators can be diagonalised simultaneously. Write the eigenstates as $$\begin{aligned} H|\lambda(\vec p)\rangle&=E(\lambda)|\lambda(\vec p)\rangle\\ \vec P|\lambda(\vec p)\rangle&=\vec p|\lambda(\vec p)\rangle \end{aligned}\tag{1}$$ One particle states are those with $E(\vec p)=\sqrt{\vec p^2+m^2}$.

For me, this is on the right track as a general definition, because we are defining $|\vec p\rangle$ as the eigenstates of operators that commute, which is fine. But there are some issues with this definition:

  • Why is $E$ a function of $\vec p$? why is $E$ not an independent label on $|\vec p\rangle$? Just as $p_x,p_y,p_z$ are independent, why is $p_0$ not independent as well$^{[1]}$?

  • Once we know that $E=E(\vec p)$, how do we know that $E^2-\vec p^2$ is a constant? (we define mass as that constant, but only once we know that it is actually a constant). By covariance, I can see that $p^2$ has to be a scalar, but I'm not sure what's the reason that this scalar has to be independent of $p$ (for example, in principle we could have $p^2=m^2+p^4/m^2$).

  • How do we know that these $|\vec p\rangle$ are non-degenerate? Why couldn't we have, for example, $H|\vec p,n\rangle=E^n(\vec p)|\vec p,n\rangle$, with $E^n\sim \alpha/n^2$, as in the hydrogen atom?

These properties are essential to prove, for example, the Källén-Lehmann spectral representation, which is itself a non-perturbative, non-asymptotic result.

Context. 2)

In Srednicki's book, page 53, we can read (edited by me):

Let us now consider $\langle p|\phi(x)|0\rangle$, where $|p\rangle$ is a one-particle state with four-momentum $p$. Using $\phi(x)=\mathrm e^{iPx}\phi(0)\mathrm e^{-iPx}$, we can see that $$ \langle p|\phi(x)|0\rangle=\mathrm e^{ipx} \tag{2} $$

This only works if $P^\mu|p\rangle=p^\mu|p\rangle$, which we could take as the definition of $|p\rangle$. Yet again, I don't know why we would expect $p^2$ to be a constant (that we could identify with mass). Or why are the eigenstates of $P^\mu$ non-degenerate.

The expression $(2)$ is essential to scattering theory, but the result itself is supposedly non-asymptotic (right?). Also, it fixes the residue of the two-point function, which is one of the renormalisation conditions on the self-energy ($\Pi'(m^2)=0$). This means that $(2)$ is a very important result, so I'd like to have a formal definition of $|p\rangle$.


$^{[1]}$ in standard non-relativistic QM, $E$ is a function of $\vec p$ because $H$ is a function of $\vec P$. But in QFT, $H$ is not a function of $\vec P$, it is a different operator.

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    $\begingroup$ Any theory that is Lorentz invariant has a Hilbert space that forms a representation of the Lorentz algebra. $p^2$ is a Casimir invariant of this algebra meaning that every representation may be labelled by the value of $p^2$. In particular, a particle is described in Hilbert space language as a single irreducible representation of the Lorentz algebra. A one-particle state corresponding to the particle is a state in this representation and therefore has a constant value of $p^2$. $\endgroup$
    – Prahar
    Jun 27, 2016 at 18:07
  • $\begingroup$ You might be looking for something more specific, but I suggest to give a look at Bjorken&Drell, "Relativistic Quantum Fields", §16.2. $\endgroup$
    – pppqqq
    Jul 2, 2016 at 18:19

3 Answers 3

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I'll address your issues with definition (1):

$E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes the index set from which the $\lambda_0$ are drawn) with $\vec P \lvert \lambda_0\rangle = 0$ and $H\lvert \lambda_0\rangle = E_0(\lambda)\lvert\lambda_0\rangle$ and then you obtain the state $\lambda_{\vec p}$ by applying the Lorentz boost associated to $\vec v = \vec p/E_0(\lambda)$ to $\lvert\lambda_0\rangle$. Since $H$ and $\vec P$ area components of the same four-vector, that means that $E_{\vec p}(\lambda)$ is a function of $E_0(\lambda)$ and $\vec p$ - and $E_0(\lambda)$ is determined by $\lambda$, so $E_{\vec p}(\lambda)$ is really a function of $\vec p$ and $\lambda$.

Conversely, every state of momentum $\vec p$ can be boosted to a state with zero momentum. We examine this zero-momentum state for what $\lambda_0$ it is and then name the state we started with $\lambda_p$, so this really constructs all states.

From the above it also directly follows that $E_{\vec p}(\lambda)^2-\vec p^2 = E_0(\lambda)^2$. I'm not sure what your issue with degeneracy is - degeneracy in $P$ and $H$ is not forbidden - nowhere is imposed that $E_{\vec p}(\lambda) \neq E_{\vec p}(\lambda')$ should hold for $\lambda\neq\lambda'$.

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  • $\begingroup$ Thank you, this really helps to clear up my thoughts :-). So you're basically saying that the definition of $|\vec p\rangle$ is by construction, right? In the case of the photon, where we write $\langle 0|A^\mu(x)|\vec p,\sigma\rangle=\varepsilon^\mu_\sigma(\vec p) \mathrm e^{-ipx}$, the construction is not as straightforward. Any idea how to prove that, in this case, $E_{\vec p}^2-\vec p^2=0$? Anyway, my issue with degeneracies is that in the free case $|\vec p\rangle$ is non-degenerate. I'm having a hard time to see how can interactions produce degeneracy, when they usually break them! $\endgroup$ Jun 27, 2016 at 18:08
  • $\begingroup$ @AccidentalFourierTransform: $\lvert\vec p \rangle$ is not non-degenerate in the free case: The two-particle state $a^\dagger(\vec p /2)a^\dagger(\vec p/2)\lvert 0 \rangle$ is degenerate w.r.t. $H$ and $P$ to $a^\dagger(\vec p )\lvert 0 \rangle$ for a massless field, for example. (I might be misunderstanding what you exactly mean by degeneracy...) The mass of a gauge boson is zero because it is protected by gauge invariance. $\endgroup$
    – ACuriousMind
    Jun 27, 2016 at 18:25
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In completely precise, nonperturbative terms, a one particle state in an interacting QFT is a state that belongs to an irreducible invariant subspace of the Hilbert space of the interacting representation of the Poincare group corresponding to an isolated shell in the spectrum of the translation subgroup.

A one particle momentum state $|p\rangle$ is an eigenstate of the four components of the momentum operator (translation generators in the representation of the Poincare group) that belongs to such a subspace.

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  • $\begingroup$ Sorry for taking so long to comment, I've been out for some time. Anyway, thank you very much, as usual, for your answer! It is actually very helpful to me to precisely define terms, so I appreciate your answer as well as the others :-) $\endgroup$ Aug 9, 2016 at 13:34
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One-particle states are the eigenstates of the mass-squared operator which are orthogonal to the vacuum.

Multi-particle states correspond to the continuous part of the mass spectrum .

Electrons however are infraparticles, see en.wikipedia.org/wiki/Infraparticle .

And for quarks it is not clear what mass means since they are confined.

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  • $\begingroup$ Sorry for taking so long to comment, I've been out for some time. I wanted to thank you for your answer, it is always very helpful to have different answers from different users, so as to get a more complete picture of what's going on. I hope to see more answers from you in stack-exchange in the future :-) $\endgroup$ Aug 9, 2016 at 13:37

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