But do you know how to demonstrate that  "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
and 
are the eigenvalues for 
and 
?
for Stache: 
Sunday, June 27, 2010
You know nothing
It is common for physics grad students to remember the angular momentum operator . Many will even remember that it commutes with and we can therefore form a set of simultaneous eigenfunctions to describe a particle in angular momentum space. Some may even recall that the standard conventions yield the following eigenequations.
\vert%20j%20,m%20\rangle "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
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I must know nothing. I thought that m was the eigenvalue of Lz by definition. But in my memory, the eigenvalue of L squared follows from the commutation relations. Right?
ReplyDeleteYou can of course always renormalize the quantum numbers to make one of them whatever you wish. However you still must show that m ranges from -j to j and moves in integer steps. Of course you begin with J_z|j,m> = \mu |j,m> and J^2|j,m>=\lambda|j,m>. Then later you relate \lambda to \mu and then decide to take \mu = m which in turn forces \lambda = j(j+1)
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