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Question about spin operators and eigenvalues

  1. Dec 31, 2014 #1
    I've been watching Leonard Susskind's videos on quantum entanglements. Naturally, one of the things that he has been discussing is spin and its various operator Hermitian matrices and eigenvalues. Now I have two main questions about this:

    1. I know that if you apply a spin operator σ (which is a matrix) to an eigenvector <a> (I would type it in the form of a ket vector, but I don't see that option in the latex), then you will get an eigenvalue λ multiplied by the same ket vector <a>. In the case of spin, those eigenvalues can be either +1 or -1. Furthermore, a spin can be measured as either up or down. My question to you guys is:

    Does +1 correlate to measuring spin up and -1 to spin down?

    I am asking this because I know that particles can actually be said to be in a linear superposition of states. For example: Spin up (which is represented by <1, 0 > ) can be written as 1 <up> + 0 <d>.
    Now this one is obviously spin up, however:

    Spin when measured against a certain axis has an eigenvectors < 1/squrt(2) , 1/squrt(2) > which can be written as:

    1/squrt(2) <up> + 1/squrt(2) <down>

    In this case, the 1/squrt(2) terms are ,probability amplitudes. When you square the magnitude of these probability amplitudes, you get the probability that the particle will be measured to have that probability amplitude's respective state. This means that in the case of the equation above, the particle would have a 50% chance of being measured as spin up and a 50% chance of being measured as spin down.

    If you apply the spin operator that goes with this particular state vector to said vector however, you get an eigenvalue of +1. Proof:

    σ = a 2 by 2 matrix. In this case the matrix is all 0 on the diagonal and it has the number 1 as the other two entries. When you multiply this matrix by the vector < 1/squrt(2) , 1/squrt(2) > , your result is just that same vector (which means that the eigenvalue is +1).

    Now if +1 implies up and - 1 implies down, then getting this eigenvalue of +1 would seem to guarantee that you would measure the particle to be spin up if you have the eigenvector < 1/squrt(2) , 1/squrt(2) > . This guarantee however, is a direct contradiction to the notion that you'd have a 50% chance of measuring up and a 50% chance of measuring down.

    That is why I came here to verify whether or not +1 and -1 implied up and down respectively, or if these eigenvalues represent something else. If it is something else, what exactly do these eigenvalues tell you? If I was right in thinking that +1 was up and -1 was down, then what becomes of the notion of the linear superposition of states?

    2. Leonard Susskind used 3 different 2 by 2 σ matrices. σ1 had 0's on the diagonal and 1's as the other entries. σ2 had 0's on the diagonal, - i as the top right entry and i as the bottom left entry. σ3 had 1 as the top left entry, -1 as the bottom right entry and 0's as the other entries. Now this may seem like a rather obvious question, but I just want to verify this:

    The 3 different matrices simply correspond to the 3 different Cartesian axes correct? In other words,
    σ1 corresponds to if you're measuring spin against the x- axis while σ2 corresponds to the y - axis and σ3 is the z - axis correct?
  2. jcsd
  3. Dec 31, 2014 #2

    You certainly do not, the superposed state you wrote is not even an eigenstate so you won't get a constant multiple of the state back.
  4. Dec 31, 2014 #3


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    Staff: Mentor

    +1 and -1 imply spin-up and spin-down along the measurement axis, so the eigenstates of ##\sigma_x## with eigenvalues +1 and -1 correspond to spin in the positive and negative directions on the x axis, and likewise for ##\sigma_y## and ##sigma_z## on the y and z axes. By convention, we generally choose the z axis to be up so ##|U\rangle## (spin-up) and ##|D\rangle## (spin-down) are the eigenstates of ##\sigma_z## with eigenvalues +1 and -1 respectively.

    When you write the wave function of a spin-up particle as ##1|U\rangle + 0|D\rangle## the two coefficients are still perfectly good probability amplitudes; it's just that in this particular state, and writing the wave function in terms of the eigenstates of ##\sigma_z##, the probability amplitude for finding the particle in the spin-down state is zero so this isn't a particularly interesting superposition. The exact same state could be written as a more interesting superposition in terms of the eigenfunctions of ##\sigma_x## as ##\frac{\sqrt{2}}{2}(|L\rangle + |R\rangle)## where ##|L\rangle## and ##|R\rangle## are the two eigenfuctions of ##\sigma_x##; which form I choose depends on whether I'm interested in the result of a spin measurement along the z axis or the z axis.

    This whole thing really isn't much different than choosing whether we should treat a vector pointing to the northwest as a superposition of basis vectors pointing to the north and to the west, or whether we should choose our basis vectors to point NW and SE. One way a vector pointing NW isn't a superposition and vectors pointing north and west are; the other way the vector pointing NW is a superposition and the vectors pointing north and west are not.
  5. Dec 31, 2014 #4
    (0 * 1/squrt(2) ) + (1 * 1/squrt(2)) = 1/squrt(2)

    (1* 1/squrt(2) ) + (0 * 1/squrt(2) ) = 1/squrt(2)

    The vector is < 1/squrt(2) , 1/squrt(2) > . Therefore the eigenvalue should be 1. How is it not? Can you show me mathematically what you mean?
  6. Dec 31, 2014 #5
    About the part where you have ##\frac{\sqrt{2}}{2}(|L\rangle + |R\rangle)## : You said that an eigenvalue of +1 would imply spin in the positive direction on the x axis and -1 would imply spin in the negative direction on the x - axis. However, if you multiply σx by this vector, then you get the same vector back. That would mean that you'd always get +1 as your eigenvalue, which in turn implies a 100% chance of getting a spin in the positive direction. The equation ##\frac{\sqrt{2}}{2}(|L\rangle + |R\rangle)## however, implies a 50% chance of both states when you square the probability amplitudes. It is this contradiction that is the source of my confusion. How is this contradiction rectified?
  7. Dec 31, 2014 #6


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    Staff: Mentor

    You do not. We have ##\sigma_x|L\rangle = -|L\rangle## and ##\sigma_x|R\rangle = |R\rangle## because ##|L\rangle## and ##|R\rangle## are the eigenvectors of ##\sigma_x## with eigenvalues -1 and +1 respectively. Therefore:
    ##\sigma_x \frac{\sqrt{2}}{2}(|L\rangle + |R\rangle) = \frac{\sqrt{2}}{2}(\sigma_x|L\rangle + \sigma_x|R\rangle) = \frac{\sqrt{2}}{2}(|R\rangle - |L\rangle)##
    This is a different vector, so ##\frac{\sqrt{2}}{2}(|L\rangle + |R\rangle)## is not an eigenvector of ##\sigma_x##.

    When you measure the spin along the x axis of a particle in the state ##\alpha|L\rangle + \beta|R\rangle## you have a probability ##|\alpha|^2## of getting the eigenvalue of ##|L\rangle## as a result and a probability ##|\beta|^2## of getting the eigenvalue of ##|R\rangle## as a result.
    Last edited: Dec 31, 2014
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