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I Schrodinger equation in terms of complex conjugate

  1. Jun 22, 2016 #1
    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:
    instead of being the original S.E in terms of ψ*
    or the equation in terms of ψ with the signs swapped
  2. jcsd
  3. Jun 23, 2016 #2

    Paul Colby

    User Avatar
    Gold Member

    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}##
  4. Jun 23, 2016 #3
    I got it!. I always forget to use the product rule, Thanks
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