In our QFT course we show that for a free scalar field:

$$(\square + m^2) \phi(t,x) = 0\tag{1}$$ i.e. that the field operator satisfies the Klein-Gordon equation (as we expect).

But also that:

$$(\square + m^2) \pi(t,x) = 0\tag{2}$$ i.e. that the momentum density operator satisfies the Klein-Gordon equation too.

What is the significance of the momentun density operator also satisfying the Klein-Gordon equation. Is this a general result for all quantum fields, or is it special here? Is it to be expected for relativistic reasons?


2 Answers 2


This is a fact about the harmonic oscillator where the field degrees of freedom are going along for the ride. Consider the Hamiltonian $$H = \frac{1}{2} p^2 + \frac{1}{2} q^2.$$ Because this is symmetric in $p$ and $q$, we can regard either one as the configuration variable. But now let's add another term involving only one of them. $$H = \frac{1}{2}p^2 + \frac{1}{2}q^2 + g q^4.$$ In this case, $Q = p, P = -q$ is still a canonical transformation so $Q$ and $P$ will satisfy Hamilton's equations. But they will do so for a new Hamiltonian, namely $$K = \frac{1}{2}Q^2 + \frac{1}{2}P^2 + g P^4.$$ It is now easy to show that, for $g \neq 0$, the Euler-Lagrange equation satisfied by $Q$ is more complicated.


Well, for the Klein-Gordon field eq. (2) holds because momentum $\pi\approx\dot{\phi}$ is just the velocity on-shell, and we can differentiate eq. (1) with time $t$. This is not true for generic fields.


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