Quantum Mechanics - Angular Momentum (Eigen Functions) part 4

Printable View

March 2nd 2011, 12:42 PM
bugatti79

1 Attachment(s)

Quantum Mechanics - Angular Momentum (Eigen Functions) part 4

Hi Folks,

I am having difficulty expanding this exercise.

$ \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 ] $

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

$ \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 ) $

Thanks
March 2nd 2011, 01:04 PM
Ackbeet

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.
March 2nd 2011, 01:34 PM
bugatti79

Quote:

Originally Posted by Ackbeet

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. (Doh)

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!!
March 2nd 2011, 02:09 PM
Ackbeet

Quote:

Originally Posted by bugatti79

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. (Doh)

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.

Quote:

Ok, I will reply with results when I can... Thanks Ackbeet!!

You're welcome, as always!
March 4th 2011, 10:47 AM
bugatti79

Quote:

Originally Posted by bugatti79

Hi Folks,

I am having difficulty expanding this exercise.

$ \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 ] $

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

$ \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 ) $

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 $\hbar \hat L_z = -\hbar^2 \frac{\partial}{\partial \phi}$ ???
March 4th 2011, 01:57 PM
bugatti79

Quote:

Originally Posted by bugatti79

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
March 4th 2011, 03:05 PM
Ackbeet

Ok, that's fine. I was definitely going to need to do some research to answer that problem!