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Angular Momentum Problem in Dirac Notation

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img857.imageshack.us/img857/2079/dirac.png [Broken]
    2. Relevant 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>

    3. 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]>
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 16, 2012 #2
    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'.
     
  4. Jan 17, 2012 #3
    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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