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Spin matrix

  1. Dec 31, 2008 #1
    Hi everyone,
    I now able to understand spin matrix (if i am correct in other words Pauli matrix).
    For e.g.,
    for S=5/2 systems the spin matrix (say for SX) is given by:

    Sx= 1/2[a 6X6 matrix]

    I hope members will know what is this 6X6 matrix! Since i don't know how to type matrix in this forum]..
    Actually i wanted to know how this 6X6 matrix is obtained??

    thanks and wish u a very happy new yr '09.
    Rajini
     
  2. jcsd
  3. Dec 31, 2008 #2

    malawi_glenn

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    "just" use the definition of the rotation operator:

    [tex] D^{(s)}_{M,m} = <s,M|e^{-i\vec{S}\cdot \vec{n} \phi / \hbar} |s,m> [/tex]

    where [tex] \vec{S} = (S_x, S_y, S_z) [/tex]

    See Sakurai, Modern Quantum mechanics, chapter 3 or similar textbooks.
     
  4. Dec 31, 2008 #3

    malawi_glenn

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    or wait a minute, what i just wrote is crap, it is not what you are looking for.

    You want matrix representation of operators S_x , S_y and S_z

    Just rewrite S_x and S_y in terms of ladder operators and work out each entry in the matrix :-)

    Easy but time-consuming!
     
  5. Dec 31, 2008 #4
    Hi thanks..
    but how?
    Can you please make one calculation..for one (Sx or Sy or Sz? for a S=5/2..
    thanks
     
  6. Dec 31, 2008 #5

    malawi_glenn

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    i mean, this is easy if you are a researcher, the ladder operators are defined as:

    [tex]J_{\pm} \equiv J_x \pm i J_y[/tex]

    And then use:

    [tex] J_{\pm}|j,m> = \hbar \sqrt{j(j+1) - m(m\pm 1 )}|j,m \pm 1 > [/tex]


    That's all you need man

    (Here J is the standard symbol for angular momentum)
     
  7. Jan 3, 2009 #6
    i really dont understand of..how to operate for S=5/2
    I am a expert..but started learning slowly...
    or if possible can u explain each steps found in this page (they did for 1/2)
    http://quantummechanics.ucsd.edu/ph130a/130_notes/node278.html
    i hope if you explain me i can derive for S=5/2,MS> state
    thanks (if u dont find time..i hope i can succeed..but takes at least 1 week)
    rajini
     
  8. Jan 3, 2009 #7

    malawi_glenn

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    1 week?

    Why dont you just try the matrix element

    < S = 5/2, M = 5/2 | S_y | S = 5/2, M = 3/2>

    Write S_y = (S(+) - S(-))/(2i)

    And use the formula in my latest post. It is a really straightforward calculation, good luck
     
  9. Jan 3, 2009 #8

    malawi_glenn

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    a question, do you know how matrix representation of operators work at all? I am having a problem to understand where your lack of understanding is. You only say "explain each step" but it is really hard to know what you know and what you don't know.

    Why don't you tell what part you don't understand?
     
  10. Jan 3, 2009 #9
    really i dont know how matrix rep. of operators work!
    < S = 5/2, M = 5/2 | S_y | S = 5/2, M = 3/2>
    Okay how to rep. the above relation as a matrix.also why m=3/2?
    thanks
     
  11. Jan 3, 2009 #10

    malawi_glenn

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    I just gave a particular matrix element to begin with.

    Now look at the spin 1/2 matrix, lets look at S_z :

    The first entry in the upper left corner (1) is this one: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = 1/2> = 1 (I pull out the factor hbar/2 ...) Make sure you can do this!!

    Then the entry in the upper right corner (0) is: < S = 1/2, M = 1/2 |S_z|S = 1/2, M = -1/2> = 0

    The entry in the lower left is < S = 1/2, M = -1/2 |S_z|S = 1/2, M = 1/2> = 0

    and

    The entry in the lower right corner is < S = 1/2, M = -1/2 |S_z|S = 1/2, M = -1/2> ...

    So what one does is to put the states with decreasing value of M as rows and coloums, then work out each entry in the matrix (that's why one calls <A| H | B> matrix element, it is an element in the matrix representing the operator H)

    Now make sure you can do the S_z for spin 1/2 and also the S_x and S_y for spin 1/2. Then try to do spin 5/2, which is same procedure, but many many more elements to calculate.
     
  12. Jan 3, 2009 #11
    also remember that the s_z matrix is just a diagonal matrix with the eigenvalues going down the diagonal.

    ie for spin 5/2 along the diagonal the values are 1,1,1,-1,-1,-1
     
  13. Jan 3, 2009 #12

    malawi_glenn

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    yes, so in reality, only the S_x and S_y are "tricky" to work out.
    but the diagonal values should be 5/2, 3/2, 1/2, -1/2, -3/2, -5/2 ...
     
    Last edited: Jan 3, 2009
  14. Jan 4, 2009 #13
    I think i have to read a bit!. Since u wrote 'M' decreases (by 1) in rows and columns...Then for S=5/2 (2S+1=6 i.e., a 6X6 matrix) one will not get a diagonal matrix..(but this is not true)
    or i dont understand properly the <A|H|B> methods..
     
  15. Jan 4, 2009 #14

    malawi_glenn

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    Yes you will obtain a diagonal matrix for S_z, it is trivial to see that.
     
  16. Jan 4, 2009 #15
    One question. How you say its 1. There are four 1/2 so which u take? (i understand u take 1/2 outside the matrix so its 1...but which 1/2 u take)
    thanks
     
  17. Jan 4, 2009 #16

    malawi_glenn

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    I said "upper left corner"...
     
  18. Jan 4, 2009 #17
    Then here upper right corner means i assume u r talking upper right corner M=-1/2 (minus 1/2)...so decreasing means it should be -1/4..but u say zero!
     
  19. Jan 4, 2009 #18

    malawi_glenn

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    eh? can you evaluate < S = 1/2, M = 1/2 |S_z|S = 1/2, M = -1/2>
    for me and show how it could be zero...
     
  20. Jan 4, 2009 #19
    actually i dont know so i asked you how it could be zero! but you are asking me the same question..anyway i dont want to disturb you now..normally i like QM but only without lesser than, greater than and pipe symbols (i wont say it bra-ket)..since i dont understand all these properly...also i guess there is lots of books which deals with all these lesser than, greater than and pipe symbols but NOT WITH EXAMPLES (or only for 1/2=S)...Any way i think i can manage to search for some books (tomorrow) and hope to find the solution..
    anyway thanks
    rajini
     
  21. Jan 4, 2009 #20

    malawi_glenn

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    But the bra-ket notation is as simple as wavefunctions. Look:

    wave function:

    [tex] S_z \chi _+ = +\frac{\hbar}{2}\chi _+ [/tex]

    Bra-ket

    [tex] S_z |+> = +\frac{\hbar}{2}|+>[/tex]

    Wave function:

    [tex] \int d\vec{x} \chi _+ ^*\chi _+ = 1 [/tex]
    [tex] \int d\vec{x} \chi _- ^*\chi _+ = 0 [/tex]

    Bra-Ket:
    [tex] <+|+> = 1 [/tex]
    [tex] <-|+> = 0 [/tex]

    I also mean, why do you want matrix representation of operators if you don't even know what it is or how bra-ket works???
     
  22. Jan 4, 2009 #21
    [tex]
    S_z \chi _+ = +\frac{\hbar}{2}\chi _+
    [/tex]
    [tex]
    S_z = +\frac{\hbar}{2}\chi _+ /\chi _+ = ?
    [/tex]
    here if we take X_+ to otherside what will happen..i mean S_z=?
    if the above relation is correct...how to find S_z
    I think i have to read some basic of such notations..can you advice me some books which explains with a example of <A,M|H|B,M> with high order spins (not 1/2)
     
    Last edited: Jan 4, 2009
  23. Jan 4, 2009 #22

    malawi_glenn

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  24. Jan 4, 2009 #23
    woops that was obvious, I guess break has worn on me a bit.
     
  25. Jan 5, 2009 #24
    Now i am somewhat clear! [if something is wrong here...pls. correct me]
    But may be for others (non experts like me) it will be helpful..
    For S=5/2 system:
    the matrix for Sz:
    the symbolic notation is
    <S,M's|Sz|S,Ms>
    Now take S=5/2 (this is fixed)
    and Ms=5/2,3/2,1/2,-1,2-3/2,-5/2 (i.e., 2S+1=2(5/2)+1=six Ms states)
    |S,Ms> = 5/2,3/2,1/2,-1,2-3/2,-5/2 (other way |5/2 5/2>, |5/2 3/2>, ...|5/2 -5/2>) (this is a row)
    and
    <S,M's| = [5/2,3/2,1/2,-1,2-3/2,-5/2]T (other way [|5/2 5/2>, |5/2 3/2>, ...|5/2 -5/2>]T) note: this is a column
    Now make table (matrix) see figure.
    and then fill the matrix elements..
    Here → represents |S=5/2,Ms>
    And ↓ represents <S=5/2,Ms|
    Whenever Ms=M’s => the matrix element is simply Ms or M’s (for Sx).
    If they are not equal => just zero [not same states and they dont exist so zero]
    And finally multiply with [tex]\ 6.62606876(52)/pi\times\ 10^{-34}\ J\ s[/tex]
    Now i will do the same for S_y and S_x (i.e., Ms-1 and Ms+1)
     
    Last edited: Jan 13, 2009
  26. Jan 5, 2009 #25

    malawi_glenn

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    Now we're talking! you got the S_z correct, congrat!

    Now comes to tricky and hard part, to do all S_x and S_y

    You might want to make sure you can reproduce those for the S = 1/2 case, just as a warm up
     
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