1. Limited time only! Sign up for a free 30min personal 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!

Spin operators

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    upload_2014-9-28_15-30-55.png

    2. Relevant equations

    See question

    3. The attempt at a solution

    Ok 1a) is dead simple. But I am struggeling with part b. I plug in all the variables. But cannot seem to get a and b..... I do this:

    (-b)(S_x)(a,b) = E(a,b) and then get

    -b^2h/2= Ea and ah/2=E

    I dont know.. Can you guys help me.
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Note that the b in the Hamiltonian is not the same b as that in the wave function ...
     
  4. Sep 28, 2014 #3
    Well that is a big help thanks. haha. Although I am sure still not sure how to proceed to find the Energy states. Because when I multiply the X matrix with the S_x matrix I get the transpose of X...
     
  5. Sep 28, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Well, it is not really a transpose - you simply switch the elements (a transpose would take a column matrix to a row one). The energy states are simply the states for which ##H\chi = E\chi##, where ##E## is a number, i.e., the eigenstates of the Hamiltonian.
     
  6. Sep 28, 2014 #5
    Ya I understand that. But how can I get back the orginal X to find E. Since, as you say the elements switch places of X.
     
  7. Sep 28, 2014 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So, what did you obtain for the eigenstates? If you have found the energy eigenstates, all you need to do is to check what the constant multiplying ##\chi## is after you apply the Hamiltonian to it.
     
  8. Sep 28, 2014 #7
    [tex] \frac{-b\hbar}{2}\begin{pmatrix}
    0&1\\
    1&0\\
    \end{pmatrix}
    \begin{pmatrix}
    a\\
    b\\
    \end{pmatrix}=
    \frac{-b\hbar}{2}
    \begin{pmatrix}
    b\\
    a\\
    \end{pmatrix}
    [/tex]

    Now what?
     
  9. Sep 28, 2014 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Well, to start I would call the ##b## in the ##\chi## ##c## instead to separate it from the ##b## in the Hamiltonian (or rename the Hamiltonian ##b## to ##B## ...). Now you need to solve the eigenvalue equation that states that this should be equal to ##E \chi##.
     
  10. Sep 28, 2014 #9
    Which is preciously where I'm stuck. I get a relationship for a and b but don't know how to solve for them.
     
  11. Sep 28, 2014 #10

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I suggest starting by solving for the energy eigenvalue E.
     
  12. Sep 28, 2014 #11
    Then I just get what I had over X.... Now sure how that helps. E should be a value and not contain matrices . How can I eliminate them?
     
  13. Sep 28, 2014 #12

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For the matrix equation to hold, it must hold for every element.
     
  14. Sep 28, 2014 #13
    [tex] \frac{-B\hbar}{2}\begin{pmatrix}
    0&1\\
    1&0\\
    \end{pmatrix}
    \begin{pmatrix}
    a\\
    b\\
    \end{pmatrix}=
    \frac{-B\hbar}{2}
    \begin{pmatrix}
    b\\
    a\\
    \end{pmatrix}=E
    \begin{pmatrix}
    a\\
    b\\
    \end{pmatrix}
    [/tex]
    So that implies:

    [tex]\frac{-B\hbar a}{2b} = E [/tex] and [tex]\frac{-B\hbar b}{2a} = E [/tex]

    Equating the:

    [tex]\frac{-B\hbar b}{2a} = \frac{-B\hbar a}{2b} [/tex]

    so
    [tex]a^2=b^2 [/tex]

    Which implies:

    [tex]a=\pm b [/tex]

    But we have the relationship:

    [tex]|a|^2+|b|^2=1 [/tex]

    So plugging it in I get:

    [tex]a=\frac{1}{\sqrt{2}} b=\pm \frac{1}{\sqrt{2}} [/tex]

    So how does this give me the ground state and the excited states? or the value for E.

    Like I said I am not sure how to solve this problem. Please some more guidance.
     
  15. Sep 29, 2014 #14

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You are essentially done. You have two expressions for E in terms of a and b that you now know - so just insert them. The ##\pm## gives you two possible values of E - the lower one corresponding to the ground state and the other the excited.
     
  16. Sep 29, 2014 #15
    so the ground state is:

    [tex]\frac{-B\hbar}{2} = E [/tex]

    and the excited state is:

    [tex]\frac{B\hbar}{2} = E [/tex]
     
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: Spin operators
  1. Spin operators (Replies: 2)

  2. Spin operator (Replies: 5)

Loading...