Angular Momentum Problem in Dirac Notation

  • Thread starter xago
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  • #1
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Homework Statement


http://img857.imageshack.us/img857/2079/dirac.png [Broken]

Homework Equations



H|ψ> = E|ψ>
[itex]L^{2}[/itex]|ψ> = l(l+1)[itex]\hbar^{2}[/itex]|ψ>
[itex]L_{z}[/itex]|ψ> = [itex]m_{l}[/itex][itex]\hbar[/itex]|ψ>

The Attempt at a Solution


I know this problem is very simple since I've seen a very similar problem a while ago but I've completed forgot how to do it over the winter break.
As far as normalization goes its <ψ|ψ> = 1, so I simply multiply the given ket vector by the bra vector of the same state. However I can't for the life of me remember how the bra and ket vectors multiply to an equation is which you just solve for A. Say for the 3rd term, 2|[itex]ψ_{2,1,-1}[/itex]> represents n=1, l=1, [itex]m_{l}[/itex] = -1
So then the Hamiltionian eigenvalue is 1, the [itex]L^{2}[/itex] eigenvalue is 1(1+1)[itex]\hbar^{2}[/itex] = 2[itex]\hbar^{2}[/itex] and the [itex]L_{z}[/itex] eigenvalue is -[itex]\hbar[/itex] but how do I put all this together?

FYI this is for the hydrogen atom |[itex]ψ_{n,l,m_{l}}[/itex]>
 
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Answers and Replies

  • #2
1,083
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This doesn't appear to be an angular momentum problem.

If you have a state c|n>, its corresponding bra is c*<n|, thus the normalization requirement is c*c = <n|n>. And remember orthogonality of eigenstates, <n|n'>= 1 if n=n' and 0 if n=/=n'.
 
  • #3
60
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Thats what I originally thought but I kept thinking i was missing something lol. So is it just
A^2(6+1+4+9+16) = 1
A= 1/6
 

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