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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?

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  • $\begingroup$ Of course the order of operators matters. $\endgroup$
    – Qmechanic ♦
    Commented Sep 23, 2018 at 17:28

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Yes, it matters. $\nabla^2$ represents a derivative, so rearranging is as silly as saying something like this is true in general $$f(x) \frac{d}{dx} h(x) = h(x) \frac{d}{dx} f(x) = f(x) h(x) \frac{d}{dx} = \cdots$$

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  • $\begingroup$ 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. $\endgroup$
    – A Nejati
    Commented Sep 23, 2018 at 21:04

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