# Marsden/Hoffman Analysis 10.9 ∂psi/dt

1. May 2, 2014

### middleCmusic

[removed post]

Last edited: May 2, 2014
2. May 2, 2014

### 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 $t$.

3. May 2, 2014

### Ray Vickson

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.