Angular Momentum Question

1. Jun 21, 2008

Domnu

In http://web.utk.edu/~tbarnes/website/qm1/qm1hw_2006/hw1_2006/hw1_2006.pdf [Broken] , on page 82, where it converts the momentum operators from Cartesian to spherical coordinates, I am a bit confused on how they get from 5.74-5.76 to 5.80-5.82 using the given relations. Could someone help me? Thanks a bunch

Last edited by a moderator: May 3, 2017
2. Jun 21, 2008

Fredrik

Staff Emeritus
That pdf has 3 pages.

3. Jun 21, 2008

Domnu

Last edited by a moderator: May 3, 2017
4. Jun 21, 2008

Fredrik

Staff Emeritus
That's a long and tedious calculation that starts by noting that if f depends on x,y,z and eacho of those depend on the spherical coordinates, we can write down the derivatives with respect to the spherical coordinates...

$$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta} +\frac{\partial f}{\partial z}\frac{\partial z}{\partial \theta}$$

...and so on. This shows that the derivative operators in spherical coordinates are linear combinations of the derivative operators in cartesian coordinates, so you get a system of equations which is possible to solve by the method of of adding a (well-chosen) multiple of one equation to another (lots of times).

5. Jun 22, 2008

Domnu

Hmm... could an easier way be

$$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial z} +\frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial z} +\frac{\partial f}{\partial r}\frac{\partial r}{\partial z}$$ ?

You can compute $$\partial \theta / \partial z,$$ for example, by differentiating implicitly with $$z = r \cos \theta$$... would this work?

6. Jun 22, 2008

Fredrik

Staff Emeritus
Unfortunately no. There is a formula for the derivative of the inverse of the map $(r,\phi,\theta)\mapsto (x,y,z)$, but figuring out what to put into it involves a procedure that's equivalent to the one I described. However, I forgot that it's actually quite easy to express $(r,\phi,\theta)$ as a function of (x,y,z) instead of the other way round. Maybe it's easier to start out that way. http://en.wikipedia.org/wiki/Spherical_coordinates

7. Jun 22, 2008

reilly

Yes, this will work. Look at the last term, with df/dr.
The term dr/dz = z/ sqrt(x*x+Y*y+z*z), or $$\cos \theta$$.

The theta term requires $$\frac{\partial \theta}{\partial z}$$
with z= r $$\cos \theta$$, from which you can compute the desired formula in spherical coords(hold r constant), and so on This is pretty much the standard way to do the problem -- grind it out.
Regards,
Reilly Atkinson

Last edited: Jun 22, 2008