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.
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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.
Raskolnikov
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