# Does the order of operators in the the hamiltonian in second quantised form matter?

For a particles that not interact (free particles) we can write the Hamiltonian in second quantized form as

$$\hat{H} = -\frac{\hbar^2}{2m} \int \psi^{\dagger}(\vec{x}) \nabla^2 \psi(\vec{x}) d^3x \, .$$

Here $$\psi^{\dagger}(\vec{x})$$ and $$\psi(\vec{x})$$ are field operators.

Does the order of the operators, $$\psi^{\dagger}(\vec{x})$$ , $$\psi(\vec{x})$$ and $$\nabla^2$$ matter? Is it alright to change the order of these operators?

• Of course the order of operators matters. Sep 23, 2018 at 17:28

Yes, it matters. $$\nabla^2$$ represents a derivative, so rearranging is as silly as saying something like this is true in general $$f(x) \frac{d}{dx} h(x) = h(x) \frac{d}{dx} f(x) = f(x) h(x) \frac{d}{dx} = \cdots$$
• I don't believe that that is the right analogy; the question was about rearranging operators, not the bringing functions in front of operators. A better analogy might be e.g. $\frac{d}{dx}\frac{d}{dy} \neq \frac{d}{dy}\frac{d}{dx}$ generally, but even that requires some explanation because it does hold true in many cases. Sep 23, 2018 at 21:04