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Homework Help: Stupid easy question about spin.

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Okay, this would be easy if it hadn't been 15 years since undergrad quantum. Here goes.

    I'm finding the energy spectrum of a Heisenberg "two-electron ferromagnet", if you will, with a Hamiltonian described by


    2. Relevant equations

    3. The attempt at a solution

    Well, after a while of dusting off my brain and groveling to fellow students, I figured out that

    [tex](\hat{S_1}+\hat{S_2})^2 = \hat{S_1}^2 + \hat{S_2}^2 + 2\hat{S_1}\cdot\hat{S_2} \rightarrow \hat{S_1}\cdot\hat{S_2} = \frac{1}{2}( (\hat{S_1}+\hat{S_2})^2 -\hat{S_1}^2 -\hat{S_2}^2 ) [/tex]

    So my Hamiltonian is now

    [tex]H = -\frac{1}{2}J((\hat{S_1}+\hat{S_2})^2 - \hat{S_1}^2 -\hat{S_2}^2 ) - h(\hat{S_{z1}}+\hat{S_{z2}})[/tex]

    Okay. Now, the eigenvalues of [tex]\hat{S}^2[/tex] are [tex]s(s+1)[/tex] (we're doing the usual [tex]\hbar=1[/tex] trick). And the eigenvalues of [tex]\hat{S_z}[/tex] are [tex]m[/tex]. And I know that electrons have [tex]s=\frac{1}{2}[/tex] and [tex]m=-s...s[/tex] in integer steps.

    So... it should just be a matter of plugging in possible values for, er, s&m, so to speak. But the [tex](\hat{S_1}+\hat{S_2})^2[/tex] term confuses me. My gut feeling is to treat that as an [tex]\hat{S}^2[/tex] term but use values [tex]-1,0,1[/tex] as possible values of [tex]\hat{S_1}+\hat{S_2}[/tex]. Is that the right way to handle it?
  2. jcsd
  3. Nov 18, 2007 #2


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    Gold Member

    yes, that's roughly right.

    Your H may be written as [itex] -J/2 ( S_{tot}^2 - S_1^2 - S_2^2 -h S_{tot,z}) [/itex]
    This is diagonal if you use for basis the three spin 1 states. Applying [itex]S_{tot}^2 [/itex]will give [itex] 1 \times (1+1) \hbar^2 = 2 \hbar^2 [/itex] for any of the spin 1 states. Applying S_1^2 or S_2^2 will give [itex]1/2(1/2+1) hbar^2 = 3/4 \hbar^2[/itex]. The only term that will distinguish between the three S=1 states is the [itex]S_{tot,z}[/itex] operator that gives [itex] m_{tot} \hbar [/itex].
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