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Pulling out partial derivatives?

  1. May 18, 2013 #1
    I'm reading through the book Quantum Mechanics (Second Edition) by David J. Griffiths and it got to the part about proving that if you normalise a wave function, it stays normalised (Page 13).

    That part that I don't get is how they say:

    ## \dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial^2 \Psi}{\partial x^2} - \dfrac{\partial^2 \Psi^*}{\partial x^2} \Psi \right) = \dfrac{\partial}{\partial x} \left[\dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \Psi \right) \right] ##

    How can they just pull out a partial operator like that?
    Because if you expand it out again it would give you:

    ## \dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right) ##

    The operator will be applied to the wrong ## \Psi ## and also won't you need to apply the product rule to is as well?
  2. jcsd
  3. May 18, 2013 #2
    If you do just what you said, you'll see that the extra terms cancel and you'll be back at the original expression. Take out a scrap of paper.
  4. May 18, 2013 #3
    If you apply the product rule to the first term of ## \dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right) ##, you'll get a term ## \dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}##, which cancels out with a term ##- \dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}## you'll get if you apply the product rule to the second term.
  5. May 18, 2013 #4
    Haha right, thank you. I thought there was a special property of partial derivatives that I didn't know.
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