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Homework Help: S_z matrix in s_y basis: three methods and three results !

  1. Feb 12, 2014 #1
    Hello guys. when i write s_z matrix in s_y basis with unitary transformation i get s_y but in second method when i rotate s_z about x axis ( - pi/2 ) to get s_z in s_y basis , i get -s_y ! and if we consider cyclic permutation, s_z becomes s_x. i am totally confused! three methods and three results !

    thank you .
  2. jcsd
  3. Feb 12, 2014 #2
    In your second method, use the right hand rule to see if you need a rotation of ( - pi/2 ) or ( pi/2 )?
    In the third method, see that for the 3 coordinates you have 3 cyclic permutations: one of them is the one you mention (x–>y, y–>z, z–>x), another one gives you the identity, then what's the third one?
  4. Feb 12, 2014 #3
  5. Feb 12, 2014 #4
    It's the coordinates system you're rotating here (passive transformation), so the angle is +pi/2.
    For third method: yes, this is the other permutation: (x–>z, y–>x, z–>y).
  6. Feb 12, 2014 #5
    but in this sulotion set i attached , he consider -pi/2 . and in the third method , we get s_x , so finally which is correct ? s_x or s_y ?

    Thank you

    Attached Files:

    • sss.jpg
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  7. Feb 12, 2014 #6
    Well it looks to me that he's getting –s_y which is not what you want to get, so i'll keep saying that since we're rotating a coordinates system pi/2 makes sense to me..
    cyclic permutation (x–>z, y–>x, z–>y) gets you from s_z to s_y, and (x–>y, y–>z, z–>x) from s_y to s_z; pick the one you need!
  8. Feb 12, 2014 #7
    the problem is exactly here, with permutation : s_z in s_y basis : s_x but with unitary transformation or rotating method s_z in s_y basis : s_y . i mean why we have 2 results ? at last which of the results are correct ?
  9. Feb 12, 2014 #8
    Put it this way: if you have a vector in real space pointing along the +y-direction with a set of equations involving positions, and want to call this direction your new z+ while keeping all equations you may have intact, it's fine as long as you perform the cyclic permutation (x–>y, y–>z, z–>x) in your coordinate system and in the equations. The same is true with spin matrices in the sense that it leaves the commutation relations unchanged... the other cyclic permutation would make you relabel the y-direction as x and apart from the identity transformation that's all the cyclic permutations you have.
  10. Feb 13, 2014 #9


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    There is some freedom in choosing the ##s_y## eigenstates. For example, in the z-basis, the ##s_y## eigenstates are often taken to be $$\frac{1}{\sqrt{2}} \binom {1} {i},\;\;\frac{1}{\sqrt{2}} \binom {1} {-i}$$

    Or, you can choose $$\frac{1}{\sqrt{2}} \binom {1} {i},\;\;\frac{1}{\sqrt{2}} \binom {i} {1}$$

    See what you get if you take the first set of y-eigenstates as a basis for expressing the ##s_z## matrix.

    Repeat for the second set of y-eigenstates.
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