Marsden/Hoffman Analysis 10.9 ∂psi/dt

My question has been posted on stackexchange here: http://ift.tt/1iRMGZl



But I wanted to ask you all since it's really a better fit for physicsforums' format. The question is copied below.



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This is a question about Chapter 10.9 of Elementary Classical Analysis by Marsden and Hoffman.



The following passages are quoted from page 612 of the second edition.

Quote:

[Suppose] [itex]\mathcal{V} = L^2[/itex] is the space of functions [itex]\psi: \mathbb{R}^3 \to \mathbb{C}[/itex] that are square integrable: [tex] \| \psi \|^2 = \langle \psi, \psi \rangle = \int \psi(x) \overline{\psi(x)} dx < \infty. [/tex] A quantum mechanical state is (by definition) a function [itex]\psi \in \mathcal{V}[/itex] such that [itex]\| \psi \| = 1[/itex]; that is, [itex]\psi[/itex] is normalized.

Then, on pg. 614, it says

Quote:

[This] operator governs the time dependence of [itex]\psi[/itex] by means of the celebrated Schrödinger equation, which reads [tex] i\hbar \frac{\partial \psi}{\partial t} = H\psi. [/tex]

So we have the partial derivative with respect to time of a function defined on a space of *positions*. What is going on here? How can one take the derivative with respect to time of a function defined on 3-space? It seems to me to be at worst, totally illogical, and at best, extremely careless.



The reason I ask these questions (besides just wanting to understand) is because I have to do the following problem, which makes use of the second concept.

Quote:

Suppose [itex]i\hbar(\partial \psi/\partial t) = H\psi[/itex]. Prove that [itex]\langle \psi, H\psi\rangle[/itex] is constant in time. This result is called conservation of energy.

As the above is a homework problem, I don't want any help on it. But if the problem is inherently contradictory, please let me know. Thanks in advance.