1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pauli Spin Matrices - Lowering Operator - Eigenstates

  1. Jan 15, 2017 #1
    This is not part of my coursework but a question from a past paper (that we don't have solutions to).

    1. The problem statement, all variables and given/known data

    Construct the matrix ##\sigma_{-} = \sigma_{x} - i\sigma_{y}## and show that the states resulting from ##\sigma_{-}## acting on the eigenstates of ##\sigma_{z} ## are also eigenstates of ##\sigma_{z} ## and comment on your result.

    2. Relevant equations
    pauli spin matrices

    3. The attempt at a solution
    I need more help with the commenting on the result and the actual physics rather than the maths here,

    I constructed the matrix ##\sigma_{-} =
    \left( \begin{array}{ccc}
    0 & 0 \\
    2 & 0 \end{array}
    \right)
    ##

    and in previous bit of question found the eigenstates of ##\sigma_{z}## to be ##
    \left( \begin{array}{ccc}
    1 \\
    0 \end{array}\right) ## and ##
    \left( \begin{array}{cc}
    0 \\
    1 \end{array}\right)## respectively.

    So therefore ##\sigma_{-}
    \left( \begin{array}{ccc}
    1 \\
    0 \end{array}\right)
    =2
    \left( \begin{array}{ccc}
    0 \\
    1 \end{array}\right)##

    and also ##\sigma_{-}
    \left( \begin{array}{ccc}
    0 \\
    1 \end{array}\right) =
    \left( \begin{array}{ccc}
    0 \\
    0 \end{array}\right)##

    I am pretty sure the math is correct as I ran the math past a few people who agreed but I cant see how/why that that shows they are also eigenstates of ##\sigma_{z}##. I can maybe see it mathematically with the first result, as that is explicitly an eigenstate, but the zero matrix result, I am not sure how in words I can say that it is. And what it means physically. As I said this isnt part of any coursework, just a question from a past exam paper, any help/advice is much appreciated.
     
  2. jcsd
  3. Jan 15, 2017 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    An eigenvector is defined to be a non-zero vector, so the statement that the resulting vector is an eigenstate of ##\sigma_z## as mentioned in the question shall only apply to cases when the resulting vector is non-zero.
    As for the comment, I think it asks you about how ##\sigma_-## changes the eigenvector of ##\sigma_z##, for example how the eigenvalue changes.
     
  4. Jan 15, 2017 #3
    Ah ok, that zero vector/matrix was throwing me. Ok so its just as simple as saying that it lowers the eigenvalue by one? A previous part of the question was to find the eigenvalue of ##\sigma_z## and was found to be 3 and 3.
     
  5. Jan 15, 2017 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    At least that's the most notable property of ##\sigma_-##.
     
  6. Jan 15, 2017 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    3 and 3?
     
  7. Jan 15, 2017 #6
    Sorry, worded that wrong. Meant to say that I found the eigenvalues of the two eigenstates to be 3 and 3.
     
  8. Jan 15, 2017 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's what I thought you meant, but you should have found them to be 1 and -1.
     
  9. Jan 15, 2017 #8

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Not sure why I missed that, but that's clearly not true.
     
  10. Jan 16, 2017 #9
    My bad! I did the previous bit of the question a couple weeks ago, sorry my fault. It seems the 3 and 3 I had in my head was not about that, it was to find the eigenvalues of ##\sigma^2=\sigma_x^2+\sigma_y^2+\sigma_z^2 ## Sorry for the confusion. I just remembered in my head that I found eigenvalues and they were 3, so when I went back to do the rest of the multiple part question, I had those 3's in my head.
     
  11. Jan 16, 2017 #10
    But now that you pointed out the mix up. My original question about commenting on the result ##\sigma_{-}## acting on ##\sigma_z##? as when the lowering operator acts on ##\sigma_z## it results in an eigenstate of ##\sigma_z## but with an eigenvalue of 2, but its original eigenvalues were 1 and -1?
     
  12. Jan 16, 2017 #11

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    No, the constant 2 in ##\sigma_- |+\rangle = 2 |-\rangle## is not the eigenvalue - it should be obvious that this equation is not an eigenvalue equation. What ##\sigma_-## does is that it changes the eigenstate of ##\sigma_z## to another eigenstate whose eigenvalue is one unit smaller than that of the initial eigenstate. Lowering operator, for more general angular momentum operator, obeys the condition ##L_z L_- |n\rangle = (n-1)\hbar L_- |n\rangle## where ##|n\rangle## is an eigenstate of ##L_z##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pauli Spin Matrices - Lowering Operator - Eigenstates
Loading...