# Physical mass in a quantum field theory

1. Apr 15, 2013

### center o bass

In special relativity we have the relation that for a free particle

$$E^2 = \vec p^2 + m_0^2$$

and that also hold in relativistic free field theories (free Klein-Gordon etc) where one can show that we have a completeness relation

$$1 = \int \frac{d^3 \vec p}{(2\pi)^3} \frac{1}{2E_{\vec p}} |\vec p\rangle \langle \vec p|.$$

Now in proving the so called 'Kallen-Lehman representation' one starts from a completeness relation of the form

$$1 = |\Omega\rangle \langle \Omega| + \int \frac{d^3 \vec p}{(2\pi)^3} \frac{1}{2E_{\vec p}} |\vec p\rangle \langle \vec p|_{\text{one particle}} + \text{two particle} + \ldots$$

where it is stated that

$$E^2 = \vec p^2 + m^2$$

where m is the physical mass of the particle. This leads to the result that the two point correlation function (as a function of the momentum) has a pole at the physical mass of the particle.

But isn't this really just a definition of the physical mass of the particle? Do we have any justification to say that E in the coefficient $1/2E$ for the completeness relation above
actually obeys

$$E^2 = \vec p^2 + m^2$$

or do we just postulate it? If so what are the empirical reasons that we call this the physical mass; how do we know that we actually measure m and not m_0?

Thanks to anyone that can shed some light on this.

2. Apr 15, 2013

### DarMM

The reason is quite simple. The Hamiltonian will have an isolated second lowest eigenvalue, the lowest being the vacuum. This state will have some four-momentum

$$p_{*} = (m,0,0,0,)$$

All other one-particle states will be boosts of this eigenvalue, so to integrate over them you need a Lorentz invariant measure on the mass-shell corresponding to boosts of

$$p_{*} = (m,0,0,0,)$$

The measure

$$\int{d^4 p \delta(p^2 - m^2)\theta(p^{0})}$$

is the Lorentz invariant measure in momentum space, which is concentrated purely
on that mass-shell. It allows you to integrate over all one particle states.

You can then easily derive:

$$\int{d^4 p \delta(p^2 - m^2)\theta(p^{0})} = \int{d^3 p \frac{1}{2E_{p}}}$$

So this is derived, not assumed. $$m$$ is the value of the second lowest eigenvalue of the Hamiltonian. The normal Källén–Lehmann derivation then shows that the two-point function has a pole at $$p_{*}$$ and also at any momenta which are a boost of it. There is no reason that this should be identical to the coefficient in front of the quadractic term in the Hamiltonian. For free theories it turns out they are, but for interacting theories they're not.