Marsden/Hoffman Analysis 10.9 ∂psi/dt

[middleCmusic]

[removed post]

 

[micromass]

You are correct, it makes little sense.

I think you can correct the problem statement as follows. Consider $\psi:\mathbb{R}^4\rightarrow \mathbb{C}:(x,t)\rightarrow \psi(x,t)$ such that for each fixed $t$, we have that $\psi(\cdot,t)$ is a quantum mechanical state, meaning that $\psi(\cdot,t)\in \mathcal{V}$ and $\|\psi(\cdot,t)\|=1$. Suppose that it satisfies the Schrödinger equation (we should probably demand that $\psi(x,\cdot)$ is differentiable), then $<\psi(\cdot,t),H\psi(\cdot,t)>$ is independent of $t$.

 

[Ray Vickson]

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.



Then, on pg. 614, it says


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.



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.

Some books are a lot more careful---especially (I think) the classics that were written in the early 20th century. The ones I have employed use something like $\psi(x)$ in the steady-state case and $\Psi(x,t)$ (with a capital $\Psi$) in the time-dependent case.

However, I think you understand perfectly well what the author intends, and his sloppiness should not cause problems, at least at the start. Maybe later it will cause real confusion---let's hope not.