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Math Help - Quantum Mechanics - Angular Momentum (Eigen Functions) part 4

  1. #1
    Senior Member bugatti79's Avatar
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    Quantum Mechanics - Angular Momentum (Eigen Functions) part 4

    Hi Folks,

    I am having difficulty expanding this exercise.

     <br />
\displaystyle L_+ L_- f = - \hbar ^{2} e^{i \phi} \left (\frac{\partial}{\partial \theta}+i cot \theta \frac{\partial}{\partial \phi} \right ) \left [e^{-i \phi} \left (\frac{\partial f}{\partial \theta}-i cot \theta \frac{\partial}{\partial \phi} \right )\right ]<br />

    My attempt involved as attached but the next line of derivation has 3 more terms...I thought it was just a matter of multiplying out terms and simplifying...?

    The answer is

     <br />
\displaystyle L_+ L_= - \hbar ^{2}\left (\frac{\partial^{2}}{\partial \theta^{2}}+cot \theta \frac{\partial}{\partial \theta} + cotan^{2} \theta \frac{\partial^{2}}{\partial \phi^{2}} +i \frac{\partial}{\partial \phi}\right )<br />

    Thanks
    Attached Thumbnails Attached Thumbnails Quantum Mechanics - Angular Momentum (Eigen Functions) part 4-dsc_0819.jpg  
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  2. #2
    A Plied Mathematician
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    I think you're forgetting the product rule in a place or two. Also, your \phi's and \theta's are too much alike. Try this:

    \displaystyle-\hbar^{2}e^{i\phi}\left(\frac{\partial}{\partial\t  heta}+i\cot(\theta)\,\frac{\partial}{\partial\phi}  \right)\left[e^{-i\phi}\left(\frac{\partial f}{\partial\theta}-i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right)\right]


    \displaystyle=-\hbar^{2}e^{i\phi}\left(\frac{\partial}{\partial\t  heta}+i\cot(\theta)\,\frac{\partial}{\partial\phi}  \right)\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]

    \displaystyle=-\hbar^{2}e^{i\phi}\left[\frac{\partial}{\partial\theta}\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]+i\cot(\theta)\,\frac{\partial}{\partial\phi}\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]\right]

    \displaystyle=-\hbar^{2}e^{i\phi}\left[e^{-i\phi}\,\frac{\partial^{2}f}{\partial\theta^{2}}-e^{-i\phi}i\,\underbrace{\frac{\partial}{\partial\thet  a}\left(\cot(\theta)\,\frac{\partial f}{\partial\phi}\right)}_{\text{Product Rule}}+i\cot(\theta)\,\underbrace{\frac{\partial}{  \partial\phi}\left(e^{-i\phi}\,\frac{\partial f}{\partial\theta}\right)}_{\text{Product Rule}}+\cot^{2}(\theta)\underbrace{\frac{\partial}  {\partial\phi}\left(\,e^{-i\phi}\frac{\partial f}{\partial\phi}\right)}_{\text{Product Rule}}\right].

    See what that does for you.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I think you're forgetting the product rule in a place or two. Also, your \phi's and \theta's are too much alike. Try this:

    \displaystyle-\hbar^{2}e^{i\phi}\left(\frac{\partial}{\partial\t  heta}+i\cot(\theta)\,\frac{\partial}{\partial\phi}  \right)\left[e^{-i\phi}\left(\frac{\partial f}{\partial\theta}-i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right)\right]


    \displaystyle=-\hbar^{2}e^{i\phi}\left(\frac{\partial}{\partial\t  heta}+i\cot(\theta)\,\frac{\partial}{\partial\phi}  \right)\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]

    \displaystyle=-\hbar^{2}e^{i\phi}\left[\frac{\partial}{\partial\theta}\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]+i\cot(\theta)\,\frac{\partial}{\partial\phi}\left[e^{-i\phi}\,\frac{\partial f}{\partial\theta}-e^{-i\phi}i\cot(\theta)\,\frac{\partial f}{\partial\phi}\right]\right]

    \displaystyle=-\hbar^{2}e^{i\phi}\left[e^{-i\phi}\,\frac{\partial^{2}f}{\partial\theta^{2}}-e^{-i\phi}i\,\underbrace{\frac{\partial}{\partial\thet  a}\left(\cot(\theta)\,\frac{\partial f}{\partial\phi}\right)}_{\text{Product Rule}}+i\cot(\theta)\,\underbrace{\frac{\partial}{  \partial\phi}\left(e^{-i\phi}\,\frac{\partial f}{\partial\theta}\right)}_{\text{Product Rule}}+\cot^{2}(\theta)\underbrace{\frac{\partial}  {\partial\phi}\left(\,e^{-i\phi}\frac{\partial f}{\partial\phi}\right)}_{\text{Product Rule}}\right].

    See what that does for you.
    Oh I see...My mistake was to assume that the test function f is just f and not a function of  \phi and \theta and hence just multiply out the brackets. That wouldnt make sense of course to have a test function without in dependant variables.

    Just one question. On the last line, could'nt we change the partial to ordinary derivatives because we have pulled out all the constants etc.

    Ok, I will reply with results when I can... Thanks Ackbeet!!
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  4. #4
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    Quote Originally Posted by bugatti79 View Post
    Oh I see...My mistake was to assume that the test function f is just f and not a function of  \phi and \theta and hence just multiply out the brackets. That wouldnt make sense of course to have a test function without in dependant variables.

    Just one question. On the last line, could'nt we change the partial to ordinary derivatives because we have pulled out all the constants etc.
    No, because whatever the operator acts on (test function or not) is going to be a function of both independent variables.

    Ok, I will reply with results when I can... Thanks Ackbeet!!
    You're welcome, as always!
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Hi Folks,

    I am having difficulty expanding this exercise.

     <br />
\displaystyle L_+ L_- f = - \hbar ^{2} e^{i \phi} \left (\frac{\partial}{\partial \theta}+i cot \theta \frac{\partial}{\partial \phi} \right ) \left [e^{-i \phi} \left (\frac{\partial f}{\partial \theta}-i cot \theta \frac{\partial}{\partial \phi} \right )\right ]<br />

    My attempt involved as attached but the next line of derivation has 3 more terms...I thought it was just a matter of multiplying out terms and simplifying...?

    The answer is

     <br />
\displaystyle L_+ L_= - \hbar ^{2}\left (\frac{\partial^{2}}{\partial \theta^{2}}+cot \theta \frac{\partial}{\partial \theta} + cotan^{2} \theta \frac{\partial^{2}}{\partial \phi^{2}} +i \frac{\partial}{\partial \phi}\right )<br />

    Thanks
    Actually just prior to deriving the very first equation in #1, we have this equation which is

    \hat L^2= \hat L_+ \hat L_- + \hat L^2_z- \hbar \hat L_z

    I believe the last 2 terms are suppose to cancel each other but I dont see how, see my attempt below

    by definition \displaystyle \hat L_z = \frac{\hbar}{i} \frac{\partial}{\partial \phi} therefore \displaystyle \hat L^2_z = -\hbar^2 \frac{\partial^2}{\partial \phi^2} and \hbar \hat L_z = -\hbar^2 \frac{\partial}{\partial \phi}???
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Actually just prior to deriving the very first equation in #1, we have this equation which is

    \hat L^2= \hat L_+ \hat L_- + \hat L^2_z- \hbar \hat L_z

    I believe the last 2 terms are suppose to cancel each other but I dont see how, see my attempt below

    by definition \displaystyle \hat L_z = \frac{\hbar}{i} \frac{\partial}{\partial \phi} therefore \displaystyle \hat L^2_z = -\hbar^2 \frac{\partial^2}{\partial \phi^2} and \hbar \hat L_z = -\hbar^2 \frac{\partial}{\partial \phi}???
    I have sorted the problem. I work out what the first term is as in this post then, i can combine all know terms to get L hat squared. Thanks ackbeet
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  7. #7
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    Ok, that's fine. I was definitely going to need to do some research to answer that problem!
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