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Two quick ones

  1. Apr 12, 2005 #1
    Two quick ones :)

    Hi, two questions:

    1) How can I find the expectation value of the x-component of the angular momentum, [tex]\langle L_x \rangle[/tex], when I know [tex]\langle L^2 \rangle[/tex] and [tex]\langle L_z \rangle[/tex]?

    2) Say, I have a state [tex]|\Psi \rangle[/tex] and two operators A and B represented as matrices. Now, [tex]|\Psi \rangle[/tex] is given as a linear combination of the eigenstates of A and I want to express them as a linear combination of the eigenstates of the operator B instead. How do I do that?

    Thanks :)
     
  2. jcsd
  3. Apr 12, 2005 #2

    vanesch

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    I don't think you can ! I think you can if you are in an EIGENSTATE of L^2 and Lz, but not if you only know the expectation values, because I think I can make up two different wave functions with same expectation for L^2 and Lz and different expectation for Lx...

    cheers,
    Patrick.
     
  4. Apr 12, 2005 #3
    Oh, they ARE eigenstates of both L^2 and Lz. Sorry. What do I do? :)
     
  5. Apr 12, 2005 #4

    vanesch

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    I think you will always find 0, no ? Because of the cylindrical symmetry of these states ?

    The reason is that you can write Jx as 1/2 (J_+ + J_-) and these, acting on a state |j,m> will give you |j,m+1> and |j,m-1>, so
    <j,m| Jx |j,m> = 0, no ?

    cheers,
    patrick.
     
  6. Apr 12, 2005 #5
    I don't quite get it. What does the |J>'s represent?
     
  7. Apr 12, 2005 #6

    vanesch

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    Sorry, I should have written L. J stands in general for an angular momentum, L for an orbital angular momentum and S for a spin angular momentum. One usually uses the notation J if it doesn't matter (as is the case here) whether it is orbital or spin angular momentum one talks about.

    I would think you are aware of J+ and J- (or L+ and L-), the ladder operators of angular momentum ?

    cheers,
    Patrick.
     
  8. Apr 12, 2005 #7

    vanesch

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    I guess you mean that you work in a certain basis, in which you get to know the components of A (matrix) and B (matrix).
    However, there is an ambiguity in the way you know psi: if you know its projections on the eigenvectors of A, you don't necessarily know its components in the basis in which A is given, because eigenvectors are only defined upto a complex multiplication factor ! So depending on how these eigenvectors were fixed your components will have an ambiguity (each of them individually) of a complex factor: just as well say that you don't know them (except if they happen to be 0).

    Happily, you can hold the same reasoning for B :-) so I would say that a possible answer to the question will always be, that in a suitably scaled set of eigenvectors of B, psi will always take on the components {1,1,1,1,...1}
    :redface:

    cheers,
    Patrick.
     
  9. Apr 12, 2005 #8
    Sorry, I'm wasting your time. The situation in 1) is this:

    An electron moving in a Coulomb-field from a proton, is in the following state (at time t=0)

    [tex]|\phi,t=0> = \tfrac4{5}|100> + \tfrac{3i}5|211>[/tex]​

    where [tex]|nlm>[/tex] is the usual energy eigenstates of the hydrogenatom. They are also eigenstates of angularmomentum:


    [tex]L^2|nlm>=l(l+1)\hslash^2|nlm>[/tex]

    [tex]L_z|nlm>=m\hslash|nlm>[/tex]

    The questions are now:

    a) Calculate the expectation value <E> in the state [tex]|\phi,t=0>[/tex].

    b) Calculate the expectation values for L^2 and L_z in the state [tex]|\phi,t=0>[/tex].

    c) Calculate the expectation value for L_x in the state [tex]|\phi,t=0>[/tex].

    It's c) I'm having trouble with. Hope this helps.
     
    Last edited: Apr 12, 2005
  10. Apr 12, 2005 #9

    dextercioby

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    [tex] \langle \hat{L}_{x}\rangle_{|\psi\rangle} =:\langle\psi|\hat{L}_{x}|\psi\rangle [/tex]

    Okay?

    Now,use the fact that

    [tex] \hat{L}_{x}=\frac{1}{2}\left(\hat{L}_{+}+\hat{L}_{-}\right) [/tex]

    and the action of [itex] \hat{\mbox{L}}_{+} [/itex] and [itex]\hat{\mbox{L}}_{-} [/itex] on an arbitrary hydrogenoid wavefunction [itex] |n,l,m\rangle [/itex] (see textbook) and the orthonormalization of the states

    [tex] \langle n,l,m|n',l',m'\rangle =\delta_{nn'}\delta_{ll'}\delta_{mm'}[/tex]

    Daniel.
     
  11. Apr 13, 2005 #10
    Yep, I with you now :) Thank you both.
     
  12. Apr 14, 2005 #11
    I assume that eigenvectors of both A and B provide the complete basis. (?)
    well, about matrices

    you have
    |c>=|A><A|c>

    you need to find

    |c>=|B><B|c>

    where |A> and |B> - eigenvectors

    then

    <b|c>=<B|A><A|c>;
     
  13. Apr 14, 2005 #12

    dextercioby

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    I don't follow.What and who are those A,B,a,b,,blah,blah,blah...?

    Daniel.
     
  14. Apr 14, 2005 #13
    that is an old linear algebra in QM notation. I assume that eigenvectors of each matrix forms an ortogonal basis.

    You have vector in the basis of the A matrix

    [tex]\vec\psi=a_1 \vec{a_1}+a_2 \vec{a_2}+....[/tex] (1)

    you need to find

    [tex]\vec\psi=b_1 \vec{ b_1}+b_2 \vec{ b_2}+....[/tex] (2)

    so multiply (1) by [tex]\vec{b_1}, \vec{b_2}...[/tex] and so on.

    [tex]b_1 =a_1 (\vec {a_1}\vec{ b_1})+a_2 (\vec {a_1}\vec{ b_1})+...[/tex]
     
  15. Apr 14, 2005 #14

    dextercioby

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    Okay.What relevance does it have here?

    Daniel.
     
  16. Apr 14, 2005 #15
    well,

    that was (1)


    that was (2)
     
  17. Apr 14, 2005 #16

    dextercioby

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    Are u hijacking the thread?This was the context

     
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