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

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)?

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.

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 $$\langle T_{00}\rangle=2m^2$$.

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 $$\langle T_{00} \rangle = \frac{2m^2}{2mV} = \frac{m}{V}$$

• A very good answer. Note however that even though the wavefunction will spread, you should be able to get what OP wants in the classical limit $\hbar\to0$. – Peter Kravchuk Dec 8 '18 at 8:55