# Can I really take the classical field equations at face value in QFT?

To be concrete, let's say I have a relativistic $\phi^4$ theory [with Minkowski signature $(+,-,-,-)$]

$$\tag{1} \mathcal{L} ~=~ \frac{1}{2} \left ( \partial_{\mu} \phi \partial^{\mu} \phi - m^2 \phi^2\right ) - \frac{\lambda}{4!} \phi^4.$$

The classical equation of motion for $\phi$ is:

$$\tag{2} (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 ~=~ 0.$$

I knew that canonical quantization is basically replacing all Poisson' brackets with (anti-)commutators. From that point of view, I would expect a classical field equation to remain valid as an operator equation even after quantization. Am I wrong?

If I am indeed correct, then specifically to the $\phi^4$ example, does that mean

$$\tag{3} \left \langle \left [ (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 \right ] \mathcal{O} \right \rangle ~=~ 0$$

for any operator $\mathcal{O}$, in the full interacting theory?

And how do I reconcile this with the path integral picture?

Only the classical paths follow classical equations of motion to the letters. But to quantize a theory, every path is assigned a weight $e^{iS}$, and obviously none of these new inclusions will follow the classical equations. Then, how can the field equations still hold?

## 1 Answer

First of all, OP's eq. (3) needs a quantum correction given by the Schwinger-Dyson equation:

$$\tag{A} \left< T\left\{ F[\varphi]\frac{\delta S[\phi]}{\delta \phi^{\alpha}(x)}\right\}\right>~=~i\hbar\left< T\left\{\frac{\delta F[\phi]}{\delta \phi^{\alpha}(x)} \right\}\right>.$$

Here the Euler-Lagrange expression is $\frac{\delta S[\phi]}{\delta \phi^{\alpha}(x)}.$ However, one should realize that this is just the beginning into a long discussion about operator ordering ambiguities, time ordering, and quantum correction, cf. e.g. this Phys.SE question.

• I thought that $(2)$ and $(3)$ were correct. Where is my mistake ? – Trimok Nov 1 '13 at 19:21