SU(2) generators in j=1, and pauli uncertainty

  • Thread starter Kontilera
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  • #1
Kontilera
137
2
Hello!
I have two questions regarding QM.

First, I tried to rotate J_z into J_y for the j=1 representation by the transformation rule for matrices.. I took my rotationvector to be (1,0,0) and rotated about 3/4 pi radians which I thought would give me S_y. Instead I got S_z again!
Wolfram alpha says that this transformation always gives S_z, no matter what angle.
My idea was: (exp(-i*pauli_x*x*pi/2)^(-1)) S_z exp(-i*pauli_x*x*pi/2)
and in wolfram alpha I wrote:

(exp(-i*{{0,1,0},{1,0,1},{0,1,0}}*x*pi/2)^(-1)){{1,0,0},{0,0,0},{0,0,-1}} exp(-i*{{0,1,0},{1,0,1},{0,1,0}}*x*pi/2)

and gets {{1,0,0},{0,0,0},{0,0,-1}} with no dependence of x.
Where does my logic break down?


Second, I just want a confirmation that I understood the exlusion principle correct.
For a three (identical)particle system we get a six dimensional subspace of the hilbertspace that corresponds to the same physical system,
|x>|y>|z>, |y>|x>|z>, ... etc.

But nature does only use the subspaces , of this six dimensional subspace, which are totally antisymmetric or symmetric. This means that we have no ambiguity in choosing the state vector, if we know wether our particles are fermions or boson.

Does this seem right? :-)
 

Answers and Replies

  • #2
Bill_K
Science Advisor
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rotated about 3/4 pi radians which I thought would give me S_y
:confused: Don't you want to rotate through 90 degrees = π/2?
My idea was: (exp(-i*pauli_x*x*pi/2)^(-1)) S_z exp(-i*pauli_x*x*pi/2)
The rotation matrix is R(θ) = exp(iθJx) and to get Sy you might want to calculate R(π/2)Sz. If you calculate R(π/2)SzR-1(π/2) instead, you have not rotated, what you have done instead is change the basis. So now you have Sy in the new basis, in which it is again diagonal!
 

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