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Measuring the moment in some direction

  1. Jan 3, 2006 #1
    ok, now im lost... my base is the eigenvectors of [tex]L_z[/tex] how do i measure the values of moment in some direction [tex]n=(a,b,c)[/tex]?
     
  2. jcsd
  3. Jan 4, 2006 #2
    ok, its [tex](J_x,J_y,J_z)(a,b,c)[/tex]
    when n is normalized ofcourse...
    but i cant obtain [tex]J_x[/tex] values at the same base as [tex]J_z[/tex]... i hope it doesnt matter, the measured quantities are allways the eigen values right?
     
    Last edited: Jan 4, 2006
  4. Jan 5, 2006 #3

    CarlB

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    I'm not quite certain what you're saying here. If what you mean is "but i cant obtain [tex]J_x[/tex] values in the [tex]J_z[/tex] basis", then this is not true. You can use any basis you wish to write J_x.

    In the J_z basis, the eigenvectors for J_z are particularly simple, namely (0,1) and (1,0) transpose. I seem to recall that with the usual Pauli matrices, the eigenvectors for J_x, in the J_z basis, are (1,i) and (1,-i), but my memory is subject to error. In any case one can certainly choose ones basis completely independently of how one chooses the eigenvectors and eigenvalues.

    And you're right that the way operator for spin in the (a,b,c) direction is simply [tex]a\sigma_x + b\sigma_y + c\sigma_z[/tex].

    Sigh. It turns out that there is a much simpler way of doing this that uses (pure) density matrix formalism. In that formalism, the operators and the wave functions are all 2x2 matrices and the student doesn't need to solve eigenvector equations to find the eigenvectors. For example,

    [tex]\sigma_z (1 \pm \sigma_z) = \pm (1 \pm \sigma_z)[/tex]

    so in the pure density matrix formalism, the eigenvector of [tex]S_z = \sigma_z/2[/tex] with eigenvalue +1/2 is obviously just [tex]1 + \sigma_z[/tex]. Thus the eigenvector for the [tex](a\sigma_x + b\sigma_y + c\sigma_z)/2[/tex] operator with eigenvalue -1/2 is simply [tex]1 - a\sigma_x - b\sigma_y - c\sigma_z[/tex]. The disadvantage of working in this formalism is that you lose the ability to add two solutions to a wave equation.

    To get back to the usual spinor formalism from pure density matrix formalism, simply right multiply by a constant spinor. For example, if you want to convert the spin -1/2 eigenvector for the S(a,b,c) operator into sigma_z form, the answer is simply:

    [tex]|(a,b,c)> = (1 - a\sigma_x - b\sigma_y - c\sigma_z) \left( \begin{array}{c}1\\0\end{array}\right)[/tex]

    [tex] = \left(\begin{array}{c}1-c\\-a-ib\end{array}\right)[/tex].

    Note that instead of using (1,0) in the above, you could equally well have used (0,1). The resulting vector will be different, but will still be an eigenvector of S_(a,b,c). In fact, it will be:

    [tex]\left(\begin{array}{c}-a+ib\\1-c\end{array}\right)[/tex]

    but these two vectors are proportional (and therefore correspond to the same state) as (a,b,c) has unit length (try the arithmetic and see).

    Also note that if the eigenvector you're looking for is an eigenvector of S_z itself, then one of the two choices, that is, (1,0) or (0,1), will be annihilated by the 2x2 matrix that corresponds to the eigenstate.

    Carl
     
    Last edited: Jan 5, 2006
  5. Jan 7, 2006 #4
    thanks for the reply, i dont know yet density matrix formalism... we learn it only on QM3... (i mean we had a lecture about it that lasted 2 houres, but its not for the exam and im all booked with homework so i havent tried to sit down and make sense of it all...)

    but you made me see that the eigenvector for the J_x operator can be any combination of the base vectors, it just means they wont be the eigenvectors of the hamiltonian if H is applied in the Z direction, which means the moment would rotate in the X-Y plane...

    so [tex]J_n|j,m> = (\frac{a}{2}(\sqrt{(j-m)(j+m+1)}|j,m+1>+\sqrt{(j-m)(j+m-1)}|j,m-1>), [/tex]
    [tex]\frac{-ib}{2}(\sqrt{(j-m)(j+m+1)}|j,m+1>-\sqrt{(j-m)(j+m-1)}|j,m-1>),[/tex]
    [tex] cm|j,m>)[/tex]
    am i right?

    by the way... is this equivalent: [tex]J_+|j,m>=\sqrt{(j-m)(j+m+1)}|j,m+1>=\sqrt{j^2+j-m^2-m}=\sqrt{j(j+1)-m(m+1)}[/tex] ?
    i found the first form and the last one used in different places, but in the first one if j+m=-1 or j-m=0 it equels 0, and in the last one the condition is different: j=-1 or m=-1... so it seems that the answer will change for these values...
     
    Last edited: Jan 7, 2006
  6. Jan 8, 2006 #5
    okay, nevermind that last bit about [tex]J_+|j,m>=\sqrt{(j-m)(j+m+1)}|j,m+1>=\sqrt{j^2+j-m^2-m}=\sqrt{j(j+1)-m(m+1)}[/tex]
    ... i sometimes get confused :blushing:
    anyway, after thinking over it again.. the vector i wrote down there isnt the eigenvector, i need to make a matrix with these vectors and diagonize it to find the eigenvalues and then the eigenvectors...
    i guess that the eigenvalues in any direction sould be integers and between j and -j... thats my intuision.
     
  7. Jan 8, 2006 #6

    CarlB

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    I think what you've done here is very cool. But I think that it's slightly wrong. I would replace the "," with "+" on the RHS, and I think that the "cm" coefficient needs to be something like "(1+cm)". Let me take another look at it and comment in the next day or two.

    What you're doing, I think, is writing the j-spin spinor that has spin +m in the (a,b,c) direction in terms of the j-spinors for j-spin in the usual (0,0,1) direction. To do this, you are writing the j-spin operator in the (a,b,c) direction by taking the vector product of the j-spin operators in the x, y and z directions with the vector (a,b,c), then writing the operator that is the obvious "eigenoperator" for this operator (i.e. [tex](1+J_{(a,b,c)})[/tex] ), and then converting the eigenoperator to an eigenvector by right multiplying by the (0,0,1) j-spinor.

    I think that the method is correct, but this is outside of what I usually spend my time doing. I mess around with spin-1/2 only because I'm interested in the elementary fermions all of which are spin-1/2. You can get the higher spin particles by combining spin-1/2 particles, but its messy, at least in the usual method of doing it.

    When you take a spinor |x> and convert it to a density matrix by |x><x|, the result is sometimes called by the mathematicians a "square spinor". Square spinors contain all the information about a state that spinors do. They do not do some of the embarassing things that spinors do. For example, they do not multiply by -1 when you rotate them 360 degrees. And they don't have arbitrary non physical complex phases like spinors do. So I think that they're more fundamental.

    Standard QM uses density matrices to model classical mixtures of quantum states. I think that density matrices should be used to model QM at the fundamental level, but that's pretty much just me.

    By the way, were you assigned the problem of computing j-spin in the (a,b,c) direction from j-spin in the usual (0,0,1) direction as an assignment?

    Carl
     
    Last edited: Jan 8, 2006
  8. Jan 9, 2006 #7

    CarlB

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    I still haven't looked at this, but will. I'm surprised no one else is commenting already.

    Carl
     
  9. Jan 12, 2006 #8

    CarlB

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    I finally got around to applying a pencil to this and I couldn't get it to work. The problem basically boils down to the fact that the spin-1/2 rep of SU(2) is unique in that, for example:

    [tex]\sigma_x \sigma_y = - \sigma_y \sigma_x[/tex]

    This means that with spin-1/2 reps, the Lie algebra relation:

    [tex][S_x,S_y] = iS_z[/tex]

    can be rewritten as

    [tex]S_x S_y = \frac{i}{2}S_z[/tex]

    and this means that one can apply Clifford algebraic notation to this representation. For spin 1, for example, the above is no longer true. Thus you can't convert a square 1-spinor to a standard spinor by simply multiplying by a spinor as is possible in the spin-1/2 case.

    Anyway, thanks for getting me to think about it.

    Carl
     
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