Is it possible to define an energy momentum tensor for classical point particles from a QFT?

Question

I have a question about the semi-classical limit of a QFT that so far I have never been able to solve. Let's start with a second quantized Klein-Gordon field with Lagrangian $$\mathcal{L}(\phi)=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^2 \phi^2$$

According to Noether's theorem, the following QFT admits an energy momentum tensor which acts as operator on a proper Fock space

$$\hat{T}_{\mu \nu}: F \rightarrow F$$ $$\hat{T}_{\mu \nu}=\partial_{\mu}\hat{\phi}\partial_{\nu}\hat{\phi}-\eta_{\mu \nu}\hat{\mathcal{L}}$$

Now, here is my question:

Does exist an element $|\psi \rangle$ $\in$ $F$ such that $\langle \psi$|$\hat{T}_{\mu \nu}| \psi \rangle = T_{\mu \nu}^{c}$, being $T_{\mu \nu}^{c}$ the energy momentum tensor of a classical point particle (e.g. $T_{\mu \nu}^{c}(x,t) = m \delta^3(x)$ for a static point particle)?

Answer 1

There are a number of problems with localization in space in QFT. But even in ordinary quantum mechanics the wavefunction of a particle at a definite position will spread out as time passes due to the uncertainty principle, so a delta function for all time is definitely not possible.

So I'll answer your question for a momentum eigenstate instead. I won't show all the calculations, but I'll show you how to do them yourself and see that for finite volume V, in the rest frame of the particle, $T_{00}=m/V$ and all other components vanish.

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First of all, since your expression for $T$ involves products of fields at the same spacetime point we need to normal order. So to calculate $\langle q|T_{\mu\nu}|q\rangle$, you will need to find $\langle q|:\partial_\mu\phi(x)\partial_\nu\phi(x):|q\rangle$ and $\langle q|:\phi(x)^2:|q\rangle$

To do this, I'll follow the normalization of e.g. Peskin's textbook. The free scalar field is $$\phi(x)=\int\frac{d^3 p}{(2\pi)^3}\frac {1}{\sqrt{2E_p}}\left(a_p e^{-ip\cdot x}+a^\dagger_p e^{+ip\cdot x}\right)$$ And the annihilation operators acting on the momentum eigenstates give $$a_p|q\rangle=\sqrt{2E_p}(2\pi)^3\delta^{(3)}(p-q)|0\rangle$$ So if you try it yourself, you'll find $$\langle q|:\phi(x)\phi(y):|q\rangle=2\cos q(x-y),\qquad\langle q|:\partial_\mu\phi(x)\partial_\nu\phi(y):|q\rangle=2\cos q(x-y)q_\mu q_\nu$$ So then taking $x=y$,$$\langle q|T_{\mu\nu}|q\rangle =2q_\mu q_\nu,$$ so in the rest frame of the particle $$\langle T_{00}\rangle=2m^2$$.

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Now it's important to realize that the states $|q\rangle$, are not normalized to be unitless. Since, $$\langle p|q\rangle=2E_p(2\pi)^3\delta^{(3)}(p-q),$$ in a box of volume $V$ we have, $$\langle p|p\rangle = 2E_p V.$$ If we want to consider an expectation value it makes sense to normalize these states to have unit norm.

So using normalized momentum states in the rest frame $$\langle T_{00} \rangle = \frac{2m^2}{2mV} = \frac{m}{V}$$