# Does Clairaut's Theorem apply to the Wave Function?

In Griffiths Intro to Quantum Mechanics, I came across a problem that asks the student to prove one of the consequences of the Ehrenfest theorem:

$$\frac{d \langle p \rangle}{dt} = \left\langle - \frac{\partial V}{\partial x} \right\rangle$$

The intended solution begins,

$$\langle p \rangle = -i\hbar \int \left( \Psi^*\frac{\partial \Psi}{\partial x}\right) \, dx$$

Thus,

$$\frac{d\langle p \rangle}{dt} = -i\hbar \int \frac{\partial }{\partial t}\left( \Psi^*\frac{\partial \Psi}{\partial x}\right) \, dx = -i\hbar \int \frac{\partial \Psi^*}{\partial t}\frac{\partial^2 \Psi}{\partial t \partial x} \, dx$$

The next step relies on the fact that the mixed partials here commute and we can write $\frac{\partial^2 \Psi}{\partial t \partial x} = \frac{\partial^2 \Psi}{\partial x \partial t}$. The rest of the proof follows fairly simply after you assume this. However, I was wondering what exactly allows this. Some threads online state that Clairaut's theorem allows us to switch the order of differentiation. However, the information online seems to say that Clairaut's theorem applies only to a real function, and the wave function is not. What's the formal reason that the mixed partials apply here?

Thanks!

• Apart from an error in the last equality (you end up with two terms, each one with a single derivation in time...), there are a couple of mathematical caveats to do that calculation. First of all, you should be worried by the possibility of exchanging the derivative in time with the integration in x... – yuggib Jan 26 '15 at 9:16
• To answer the question in your title, if both the real and imaginary parts of $\Psi$ separately fulfill the continuity requirements for symmetry of second derivatives, then you're good to go with Monsieur Clairaut! – WetSavannaAnimal Jan 26 '15 at 12:21
• @WetSavannaAnimalakaRodVance Thanks! That seems like the right way to go. However, is there some way to prove the continuity of the second derivatives of those parts? Perhaps using the Schrodinger equation, though I haven't studied PDEs yet, so I'm not sure how. – nphirning Jan 28 '15 at 6:44