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Quantum Mechanics: total angular momentum of an electron in a hydrogen atom

  1. Mar 25, 2009 #1
    An electron in a hydrogen atom occupies the combined position and spin state.

    [tex]\Psi[/tex][tex]\left(\vec{r},\xi\right)[/tex]=[tex]\left(\sqrt{1/3}Y^{1}_{0}\xi_{+}+\sqrt{2/3}Y^{1}_{1}\xi_{-}\right)[/tex]

    What are the possible measured values of [tex]J^{2}[/tex] (where J is the total angular momentum of the electron L + S) and with what probability will each be found?


    [tex]J^{2}[/tex] = [tex]\hbar^{2}j\left(j+1\right)[/tex]

    [tex]\left|l-s\right|\geq j \geq l+s [/tex] , where [tex]l[/tex] and [tex]s[/tex] are the orbital angular momentum and spin angular momentum quantum numbers, respectively.

    I know that, according to the given position state of the electron and the fact that it is an electron, [tex]l = 1[/tex] and [tex]s = 1/2[/tex].

    I know that [tex]j[/tex] will be [tex]1/2[/tex] or [tex]3/2[/tex]. Therefore, [tex]J^{2}[/tex], when measured, will be either [tex]3/4\hbar^{2}[/tex] or [tex]15/4\hbar^{2}[/tex].

    I am having trouble determining the associated probabilities of the possible measurement values for [tex]J^{2}[/tex]. From what I have read about addition of angular momentum, it would seem that I would need to calculate the Clebsch-Gordon coefficients for the total angular momentum. I am not really sure where to start. I have read Griffiths' explanation for angular momentum and the Clebsch-Gordon coefficients but he doesn't explain how to use them for total angular momentum.

    I feel dumb asking this kind of question but I am having trouble understanding Quantum Mechanics.
     
  2. jcsd
  3. Mar 25, 2009 #2
    The wavefunction should be:

    [tex]\Psi\left(\vec{r},\xi\right)=\left(\sqrt{1/3}Y^{0}_{1}\xi_{+}+\sqrt{2/3}Y^{1}_{1}\xi_{-}\right)[/tex]
     
  4. Nov 23, 2010 #3
    In order to get the probabilities you need to use the Clebsch-Gordan table.

    From the original equation we know that Y01 is the same as |10> and Y11 is the same as |11>. We also know that electrons have an inherent spin of 1/2, so the chi (+) is |1/2 1/2> and the chi (-) is |1/2 -1/2>.

    This is evident because chi + is spin up and chi - is spin down and we are working with an electron.

    So at this point we have:
    Y01 = |1 0>

    Y11 = |1 1>

    Chi (+) = |1/2 1/2>

    Chi (-) = |1/2 -1/2>

    So now we plug this back into the first equation and for now we will ignore the radial part (R21). When we do this we get:

    sqrt(1/3) |10> |1/2 1/2> +sqrt(2/3) |11> |1/2 -1/2>

    This is where the Clebsch-Godradn table comes into play.

    We know that the spin of electron is 1/2, and we also know that l is one (we got that because the lower part of the angular equation is on for both Y01, and Y11, and also in the radial equation Rnl)

    So we go to the 1 X 1/2 clebsch-Gordan table.

    so |10> |1/2 1/2> = sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2>
    This one we got from the line highlighted in yellow
    and |11> |1/2 -1/2> = sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2>
    This one we got from the line highlighted in green

    Now we plug this back in and get
    sqrt(1/3)*(sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2>)+ sqrt(2/3)(sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2>)

    This equals sqrt(2)/3|3/2 1/2>+sqrt(2)/3|3/2 1/2>+(2/3)|1/2 1/2> - (1/3)|1/2 1/2>
    Which simplifies to 2*sqrt(2)/3|3/2 1/2>+(1/3)|1/2 1/2>

    So J2 = (3/2)(3/2+1)hbar2 with a probability of 2*2/9
    = (1/2)(1/2+1)hbar2 with a probability of 1/9

    And Jz = (1/2)hbar with a probability of 1
     

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