Schrodinger equation in terms of complex conjugate

  • #1

Main Question or Discussion Point

I know there's a similar post, but i didn't understand it. Why the derivative respect to t in terms of the complex conjugate of ψ is:
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instead of being the original S.E in terms of ψ*
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or the equation in terms of ψ with the signs swapped
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Answers and Replies

  • #2
Paul Colby
Gold Member
1,034
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One is simply applying conjugation to the equation as a whole. This yields an equation which is valid provided the original is valid. Some c-number conjugation rules that may be helpful are,

  1. ##(a b)^\ast = a^\ast b^\ast##
  2. ##(a + b)^\ast = a^\ast + b^\ast##
  3. ##(i)^\ast = -i##
  4. ##(\frac{\partial \psi}{\partial t})^\ast = \frac{\partial \bar\psi}{\partial t}##
 
  • #3
I got it!. I always forget to use the product rule, Thanks
 

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