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Marsden/Hoffman Analysis 10.9 ∂psi/dt

  1. May 2, 2014 #1
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    Last edited: May 2, 2014
  2. jcsd
  3. May 2, 2014 #2

    micromass

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    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##.
     
  4. May 2, 2014 #3

    Ray Vickson

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    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.
     
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