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I Angular momentum operator commutation relation

  1. May 10, 2017 #1
    I am reading a proof of why

    [tex] \left[ \hat{L}_x, \hat{L}_y \right ] = i \hbar \hat{L}_z [/tex]

    Given a wavefunction [itex]\psi[/itex],
    [tex] \hat{L}_x, \hat{L}_y \psi = \left( -i\hbar \right)^2 \left( y \frac{\partial}{\partial z} - z \frac {\partial}{\partial y} \right ) \left (z \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial z } \right )[/tex]

    [tex]= -\hbar ^2 \left ( yz \frac{\partial ^2 \psi}{\partial z \partial x} + y \frac{\partial \psi}{\partial x} - yx \frac{\partial ^2 \psi}{\partial z^2} - z^2 \frac{\partial ^2 \psi}{\partial y \partial x} + zx \frac{\partial ^2 \psi}{\partial y \partial z} \right )[/tex]

    This is a simple expansion of the brackets. I don't understand however, where the second term in the brackets of the last equation comes from?
     
    Last edited: May 10, 2017
  2. jcsd
  3. May 10, 2017 #2

    blue_leaf77

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    Use the product rule to calculate ##y \frac{\partial}{\partial z} \left(z\frac{\partial \psi}{\partial x}\right)## because both ##z## and ##\frac{\partial \psi}{\partial x}## are a function of ##z##.
     
  4. May 10, 2017 #3

    fresh_42

    Staff: Mentor

    The product rule:
    $$
    y\cdot \frac{\partial}{\partial z}(z \cdot \frac{\partial \psi}{\partial x})= y\cdot \frac{\partial}{\partial z}(z)\cdot \frac{\partial \psi}{\partial x} + y\cdot z\cdot \frac{\partial}{\partial z}(\frac{\partial \psi}{\partial x})=y \cdot 1 \cdot \frac{\partial \psi}{\partial x} + y\cdot z \cdot \frac{\partial^2 \psi}{\partial z \partial x}
    $$
     
  5. May 10, 2017 #4
    yes I understand that but why not, by the same reasoning, this is not applied on the other terms? (or maybe it is and im not seeing it) i.e. why not

    [tex]y \frac{\partial}{\partial z} \left( -x \frac {\partial \psi}{\partial z} \right ) = y \frac{\partial(-x)}{\partial z} \frac{\partial \psi}{\partial z} - xy \frac{ \partial ^2 \psi}{\partial z^2} [/tex]

    but just [tex]y \frac{\partial}{\partial z} \left( -x \frac {\partial \psi}{\partial z} \right ) = - xy \frac{ \partial ^2 \psi}{\partial z^2} [/tex]
     
  6. May 10, 2017 #5

    blue_leaf77

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    What is ##y \frac{\partial(-x)}{\partial z}## equal to? Recall that ##x## and ##z## are independent variables.
     
  7. May 10, 2017 #6
    yep that is zero. it just slipped my mind. Thanks
     
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