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Problem with complex conjugates and the Time Independent Schrodinger equation

  1. Apr 6, 2010 #1
    I can understand that if z= a + ib then z*=a - ib, where the definition of the complex conjugate is reversing the sign in front of the imaginary part.

    I'm now confused about how the complex conjugate works in the TISE. in a stationary state,

    i[tex]\hbar[/tex][tex]\frac{\partial\Psi}{\partial\\t}[/tex]= E[tex]\Psi[/tex]

    this can be arranged to give the time derivative as

    [tex]\frac{\partial\Psi}{\partial\\t}[/tex]= -[tex]\frac{i}{\hbar}[/tex] E[tex]\Psi[/tex]

    Now my text book says that the complex conjugate of this derivative is

    [tex]\frac{\partial\Psi\\^{*}}{\partial\\t}[/tex]= [tex]\frac{i}{\hbar}[/tex] E[tex]\Psi\\^{*}[/tex]

    This is where I'm confused. I understand that in order to take the conjugate of the derivative of the wave function, we change the sign in front of the i, but then it also changes the wave function on the right-hand-side to its complex conjugate. There must be something implicit that I'm missing. I'm sure its something very trivial, I just don't understand why the wave function on the right hand has also changed to the complex conjugate.
     
  2. jcsd
  3. Apr 6, 2010 #2

    SpectraCat

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    The time-dependent wavefunction is itself complex, so the complex conjugation operation also changes both [tex]\psi[/tex] and its time derivative to their complex conjugates. Note that the Hamiltonian is Hermitian, so it is its own complex conjugate.
     
  4. Apr 6, 2010 #3

    Cyosis

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    To add to what Spectracat said. Remember that [itex](zw)^*=z^*w^*[/itex] and [itex](z+w)^*=z^*+w^*[/itex] with z and w complex numbers. Therefore [itex]z^*=(a+ib)^*=a^*+i^*b^*=a-ib[/itex], because a and b are real numbers.
     
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