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In my book "Quantum Theory for Mathematians" By B. Hall there is a discussion about the derivative of the inner product of a time-dependent wave functions $\psi(t)$ (note: no position dependence is indicated) (p. 72).

In the proof of a proposition in section 3.7 dealing with the time-dependent Schrödinger eqaution, stating $$\frac{d}{dt}\langle A\rangle_{\psi(t)} =\langle\frac{1}{i\hbar}[A,\hat{H}]\rangle_{\psi(t)}$$ where $\langle A\rangle_{\psi(t)}=\langle \psi(t), A\psi(t)\rangle$ and $A$ is an operator, there is a calculation showing that $$\frac{d}{dt}\langle\psi(t),A\psi(t)\rangle=\langle\frac{d\psi}{dt},A\psi\rangle+\langle\psi,A\frac{d\psi}{dt}\rangle.$$ Now, the inner product is defined to be the usual inner product for $L^2(\mathbb{R})$ with respect to position, but since $\psi$ is not dependent on position how can this inner product even be defined? And if the author meant for the wave function to be $\psi(t,x)$, should the (total) derivatives with respect to $t$ in the inner product not be partial derivatives with respect to $t$?

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There are at least two plausible interpretations for the equation $$ \frac{d}{dt} \langle\psi(t), A\psi(t)\rangle = \left\langle\frac{d\psi}{dt}, A\psi\right \rangle + \left\langle\psi, A\frac{d\psi}{dt}\right\rangle; \tag{1} $$ not sure which, if either, the author intended.

  1. The time-dependent wave function $\psi$ is taken to be $L^{2}$-valued, not complex-valued; the time derivatives are also $L^{2}$-valued, and (1) is a formal consequence of bilinearity of the inner product.

  2. Equation (1) should be taken as holding at each position $x$, and the $L^{2}$ inner product found by integrating each side.

Technically, an element of $L^{2}$ isn't a complex-valued function, but an equivalence class of functions with any two identically equal except on a set of measure zero. Consequently, any equation involving pointwise (spatial) values of $\psi$ has to be interpreted carefully. It's possible this explains your author's notation.

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  • $\begingroup$ Hi Andrew, and thanks for your answer. Do you think it would be appropriate to contact the author about this directly? $\endgroup$ – msx Jul 11 '16 at 19:02
  • $\begingroup$ The primary question may be: Does not knowing this detail prevent you from reading onward? If not, I'd just keep going. Context may clarify the technicalities as you keep reading. If this Really Matters, on the other hand, there's probably no harm in writing to the author, but do understand that your letter may well go unanswered. $\endgroup$ – Andrew D. Hwang Jul 11 '16 at 19:33

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