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Math Help - Quantum Mechanics - Angular Momentum

  1. #1
    Senior Member bugatti79's Avatar
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    Quantum Mechanics - Angular Momentum

    Determine  [\hat L_x, \hat L_y]f(x,y,z)

    where \displaystyle \hat L_x=\frac{\hbar}{i}(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}) and

    \displaystyle \hat L_y=\frac{\hbar}{i}(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})

     <br />
\displaystyle [\hat L_x, \hat L_y]f(x,y,z)=\left(\frac{\hbar}{i}\right)^2 \left[ y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right) -y \frac{\partial}{\partial z} \left( x \frac{\partial f}{\partial z} \right)<br />

     <br />
\displaystyle -z \frac{\partial}{\partial y} \left( z \frac{\partial f}{\partial x} \right)+z \frac{\partial}{\partial y} \left( x \frac{\partial f}{\partial z} \right)-z \frac{\partial}{\partial x} \left( y \frac{\partial f}{\partial z} \right) \right)<br />

     <br />
\displaystyle +z \frac{\partial}{\partial x} \left( z\frac{\partial f}{\partial y} \right)+x \frac{\partial}{\partial z} \left( y \frac{\partial f}{\partial z} \right)-x \frac{\partial}{\partial z} \left( z \frac{\partial f}{\partial y} \right) \right]

    What I dont understand is what's done next... ie if you look at the first term

     <br />
\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)<br />
. This expands out to  <br />
\displaystyle y\frac{\partial f}{\partial x} +yz\frac{\partial^{2} f}{\partial z \partial x}<br />

    What is happening here?..
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  2. #2
    A Plied Mathematician
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    That's just the product rule, isn't it?
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  3. #3
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    TheEmptySet's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Determine  [\hat L_x, \hat L_y]f(x,y,z)

    where \displaystyle \hat L_x=\frac{\hbar}{i}(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}) and

    \displaystyle \hat L_y=\frac{\hbar}{i}(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})

     <br />
\displaystyle [\hat L_x, \hat L_y]f(x,y,z)=\left(\frac{\hbar}{i}\right)^2 \left[ y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right) -y \frac{\partial}{\partial z} \left( x \frac{\partial f}{\partial z} \right)<br />

     <br />
\displaystyle -z \frac{\partial}{\partial y} \left( z \frac{\partial f}{\partial x} \right)+z \frac{\partial}{\partial y} \left( x \frac{\partial f}{\partial z} \right)-z \frac{\partial}{\partial x} \left( y \frac{\partial f}{\partial z} \right) \right)<br />

     <br />
\displaystyle +z \frac{\partial}{\partial x} \left( z\frac{\partial f}{\partial y} \right)+x \frac{\partial}{\partial z} \left( y \frac{\partial f}{\partial z} \right)-x \frac{\partial}{\partial z} \left( z \frac{\partial f}{\partial y} \right) \right]

    What I dont understand is what's done next... ie if you look at the first term

     <br />
\displaystyle y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)<br />
. This expands out to  <br />
\displaystyle y\frac{\partial f}{\partial x} +yz\frac{\partial^{2} f}{\partial z \partial x}<br />

    What is happening here?..
    This is just the product rule

    \displaystyle   y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)=y\left[\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x}\right) \right]=y\left[ \frac{\partial f}{\partial x}\frac{\partial }{\partial z}z+z\frac{\partial }{\partial z} \frac{\partial f}{\partial x}\right]=y\left[ \frac{\partial f}{\partial x}(1)+\frac{\partial^2 f}{\partial z\partial x}\right]
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    This is just the product rule

    \displaystyle   y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)=y\left[\frac{\partial}{\partial z}\left( z \frac{\partial f}{\partial x}\right) \right]=y\left[ \frac{\partial f}{\partial x}\frac{\partial }{\partial z}z+z\frac{\partial }{\partial z} \frac{\partial f}{\partial x}\right]=y\left[ \frac{\partial f}{\partial x}(1)+\frac{\partial^2 f}{\partial z\partial x}\right]
    ok, what if for arguments sake that the function f only depend on x and y then

    \displaystyle   y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)= y \frac{\partial f}{\partial x}?

    Any how, the original reduces down to

    \displaystyle \left(\frac{\hbar}{i}\right)^2 \left(y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right) f(x,y,z) = \left( \frac{\hbar}{i} \right) \left( \frac{\hbar}{i} y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right) f (x,y,z)= i \hbar \hat L_z f (x,y,z)

    What is happening in the middle terms with the hbar over i moved inside the brackets? It may be a typo as I was writing furiously trying to keep up with the lecturer!

    Thanks
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  5. #5
    A Plied Mathematician
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    Quote Originally Posted by bugatti79 View Post
    ok, what if for arguments sake that the function f only depend on x and y then

    \displaystyle   y\frac{\partial}{\partial z}\left(z \frac{\partial f}{\partial x}\right)= y \frac{\partial f}{\partial x}?
    Yes.


    Any how, the original reduces down to

    \displaystyle \left(\frac{\hbar}{i}\right)^2 \left(y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right) f(x,y,z) = \left( \frac{\hbar}{i} \right) \left( \frac{\hbar}{i} y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y} \right) f (x,y,z)= i \hbar \hat L_z f (x,y,z)

    What is happening in the middle terms with the hbar over i moved inside the brackets? It may be a typo as I was writing furiously trying to keep up with the lecturer!

    Thanks
    The hbar over i should distribute and multiply both terms. Your

    \hat{L}_{z}=-i\hbar\left[x\,\dfrac{\partial}{\partial y}-y\,\dfrac{\partial}{\partial x}\right].

    Computing i\hbar\hat{L}_{z} should get you the far left term (minus the test function f).
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