quantum mechanics - One particle states in an interacting theory

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.

Answer

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 $\lambda\neq\lambda'$.