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Homework Help: Simple angular momentum problem. QM

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that Lx is a Hermitian operator

    2. Relevant equations

    Well, since Lx is an observable it must be represented by an Hermitian operator.

    [tex] L = -i\cdot\hbar\cdotr\times\nabla [/tex]

    If an operator is Hermitian, then it's equal to it's Hermitian conjugate

    3. The attempt at a solution

    I do realise what I have to do, however there are holes in my math.

    So in order to show that the Lx operator is Hermitian I could show that:

    [tex] < L_{x} f | f > = < f | L_{x} f > [/tex] is that correct?

    If I assume that's correct and since [tex] < L_{x} f | f > = (< f | L_{x} f >)^* [/tex] I have a complex conjugate missing somewhere. Thus there is an error
     
  2. jcsd
  3. Nov 15, 2008 #2

    CompuChip

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    You don't have a conjugate missing, the identity on the last line is true.
    The point is that, by definition of Hermitian conjugate, if you want,
    [tex]\langle f \mid L_x f \rangle = \langle L_x^\dagger f \mid f \rangle[/tex]
    so if you show that it's also [itex]\langle L_x f \mid f \rangle[/itex] then effectively you are showing that [itex]L_x^\dagger = L_x[/itex].
     
  4. Nov 15, 2008 #3
    Thanks. That I understand but I probably don't understand something else or I don't really understand either, anyway to be more specific here is what I mean

    [tex] \langle f \mid L_{x} f \rangle [/tex] which is

    <f|-ih(y*d/dz - z*d/dy)f>. (I am sorry, Latex doesn't show up for this one. I guess I typed it wrong)

    now if [tex] L_{x} [/tex] is Hermitian then as far as I can understand the above expression must be equal to

    [tex] \langle L_{x} f \mid f \rangle [/tex] which is [tex] \langle f \mid L_{x}f \rangle^* [/tex]

    and here is my problem, I get that [tex] \langle L_{x} f \mid f \rangle [/tex] is equal to [tex] \langle f \mid L_{x} f \rangle [/tex] conjugated and because of the i in the [tex] L_{x} [/tex] expression the two don't really match, do they? :confused: What am I missing?

    Just to be absolutely clear I don't like the fact that [tex]
    \langle L_{x} f \mid f \rangle [/tex] is <-ih(y*d/dz - z*d/dy)f|f> doesn't seem to match with <f|-ih(y*d/dz - z*d/dy)f>
     
  5. Nov 16, 2008 #4

    malawi_glenn

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    you can give a proof of this since the L_x operator is = y*p_z - z*p_y

    now you proove that x_ip_j is hermitian. i.e show that

    [tex] \int ( x_i\hat{p}_j \phi)^*\psi d^3r = \int \phi ^*x_i\hat{p}_j\psi d^3r [/tex]
     
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