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Normalization of a spin function

  1. Jul 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Given that the antisymmetric spin function for a 2 electron system is N[a1b2-a2b1], find the normalization constant N. (and by a and b I mean the alpha and beta spin states and by 1 and 2, I mean the labels on the two electrons...


    2. Relevant equations
    Normalization: 1=integral over all relevant space of (wavefunction*wavefunction)


    3. The attempt at a solution
    So I tried to square the spin function given, set it equal to 1, and solve for N. However, as the squared value of [a1b2-a2b1]... or what I THINK is the squared value of that, I kept on getting zero... what am I doing wrong in doing:

    square of spin function=(N[a1b2-a2b1])^2
    =(N^2)<a1b2-a2b1|a1b2-a2b1>
    =(N^2)[<a1|a1><b2|b2>-<a1|a2><b2|b1>-<a2|a1><b1|b2>+<a2|a2><b1|b1>]

    and because any sort of <a|a> is 1 and so is <b|b>, all those braket stuff are equal to 1, which overall makes the equation 0... I'm confused :(
     
  2. jcsd
  3. Jul 25, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    no, while <a1 b2|a1 b2> = 1, <a1 b2|a2 b1>=0
    therefore you get
    1= N^2 (1+0+0+1)
    1= 2N^2
    N = 1/sqrt(2) which is what one would expect naively.

    note: the 2e- system has 4 states:
    |a1 a2>,|a1 b2>,|a2 b1>,|a2 b2>
    these are tensor products. eg. |a1 a2> = |a1>|a2>
     
  4. Jul 26, 2007 #3
    how did you get (1+0+0+1)? I kept on getitng something like 1-1+1-1 or something that kept on cancelling all out to 0 :(
     
  5. Jul 26, 2007 #4
    Your state

    [tex]
    |\psi\rangle = N|\alpha_1\beta_2-\alpha_2\beta_1\rangle
    [/tex]

    looks bad. Use this kind of notation instead:

    [tex]
    |\psi\rangle = N\big(|\alpha_1\beta_2\rangle-|\alpha_2\beta_1\rangle\big)
    [/tex]

    When you compute

    [tex]
    \langle\psi|\psi\rangle
    [/tex]

    don't start splitting states [itex]|\alpha_i\beta_j\rangle[/itex] into sums of states [itex]|\alpha_i\rangle[/itex] and [itex]|\beta_j\rangle[/itex], because that is wrong.
     
  6. Jul 26, 2007 #5
    Ok, I got it now : ) but now my question is, I got to the point where N=sqrt(1/2). So can N be both positive AND negative of sqrt(1/2)?
     
  7. Jul 26, 2007 #6
    In fact

    [tex]
    N=e^{i\theta}\sqrt{1/2}
    [/tex]

    are all valid normalization constants, where theta is some arbitrary constant, but the simplest [itex]\sqrt{1/2}[/itex] is usually preferred.
     
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