Implicit partial derivative of wave function

Question

I'm working out some QM problems and need to clarify the procedure for calculating the partial of an implicit function. What's needed is to differentiate a wave function twice with respect to t. Here's the function: $φ(x,t) = e^{i(ax − bt)}ψ(x − vt,t)$. My answer differs from the books. I'm messing up the derivative of psi. My question is since t shows up twice in psi how do you handle this? Differentiate each piece seperately then add them? Not sure. Thanks.

Answer 1

Your idea is right about breaking up the problem in parts, but it's a bit more sophisticated than just adding them up. First, note that $\phi$ is a product. The derivative of a product is obtained by Leibniz's rule:

$$\partial_t\phi(x,t) = \partial_t(e^{i(ax − bt)}) \psi(x − vt,t) + e^{i(ax − bt)} \partial_t (\psi(x − vt,t))$$

I shortened the notation for the partial derivative with respect to $t$ as $\partial_t$. This can be further worked out to

$$\partial_t\phi(x,t) = -ibe^{i(ax − bt)} \psi(x − vt,t) + e^{i(ax − bt)} \partial_t (\psi(x − vt,t))$$

Second, there is the last factor in the last term, which is a composite function. Here you should use the chain rule. I'm going to notate the derivative with respect to the first variable of $\psi$ as $\partial_1\psi$, likewise for the derivative w.r.t. the second variable $\partial_2\psi$. Then, we have that

$$\partial_t (\psi(x − vt,t)) = -v\partial_1 \psi(x − vt,t) + \partial_2 \psi(x − vt,t)$$

Combining everything we have

$$\partial_t\phi(x,t) = -ibe^{i(ax − bt)} \psi(x − vt,t) -v e^{i(ax − bt)} \partial_1 \psi(x − vt,t) + e^{i(ax − bt)} \partial_2 \psi(x − vt,t)$$

If you still have trouble with this, I suggest you check up the chain rule for partial derivatives in particular.