Rotations of spins and of wavefunctions

wondering12
Messages
17
Reaction score
0
Thread moved from a technical forum, so homework template missing
This is a question regarding the intrinsic angular momentum S of a particle of spin 1.
Assuming S = s(s+1)I = 2I and I is the identity operator. In our case s = 1.
Let |z> be a ket of norm 1 such that Sz |z> = 0, and let |x> and |y> be the ket vectors
obtained from it by rotations of + 1/2 Pi about the y-axis and − 1/2 Pi about the x-axis
respectively. Prove the following relations, as well as those resulting from circular
permutation of x, y, and z:
Sx |x> = 0 , Sx |y> = i|z> , Sxˆ2 |y> = |y> , Sx |z> = −i|y> , Sx2|z> = |z>
Use these to show that |x>, |y>, |y> form an orthonormal basis and that the matrices
representing Sx , Sy and Sz in that basis are:
Sx = \begin{bmatrix}
0 & 0 & 0 \\
0& 0 & -i\\
0&i &0
\end{bmatrix}
Sy = \begin{bmatrix}
0 & 0 & i \\
0& 0 & 0\\
-i&0 &0
\end{bmatrix}
Sz = \begin{bmatrix}
0 & -i & 0 \\
i& 0 & 0\\
0&0 &0
\end{bmatrix}
Show that the eigenvalues of Sz are the same as those of Jz , where Jz is the matrix
representation of the z component of the angular momentum in the |j m> basis, for
j = 1. Then find the most general unitary matrix that transforms
Sz into Jz : U Sz U† = Jz .Choose the arbitrary parameters in U so that it also
transforms Sx into Jx and Sy into Jy .
This problem can be found in Quantum Mechanics by Albert Messiah volume 1 chapter 8 problem 10 part ii and iii.
see attachment.
My humble attempt at the problem,
Used a rotation 3-d matrix for x,y,z or:
Rx = \begin{bmatrix}
1 & 0 & 0 \\
0& cos(a) & -sin(a)\\
0&sin(a) &cos(a)
\end{bmatrix}
Ry = \begin{bmatrix}
cos(a) & 0 & sin(a) \\
0& 1 & 0\\
-sin(a)&0 &cos(a)
\end{bmatrix}
Rz = \begin{bmatrix}
cos(a) & -sin(a & 0 \\
sin(a) &cos(a) & 0\\
0&0 &1
\end{bmatrix}
After that substituting angle for x and y does not give me any meaning to go further. Probably I am starting in a wrong way. Help appreciated.
 

Attachments

  • messiah problem.jpeg
    messiah problem.jpeg
    38.4 KB · Views: 523
Physics news on Phys.org
Those rotation matrices aren't the correct ones for this problem. You need the ones that work on the angular momentum space.
 
vela said:
Those rotation matrices aren't the correct ones for this problem.
Why not? I seem to get the right answer by the unitary transform ##S_x = R_y S_z R_y^{\dagger}##.
 
I think one could also do this problem matrix-less and hence saving lots of space, instead only works with ket notations. For example for the first task to prove ##S_x|x\rangle = 0##, one can use the substitution ##|x\rangle = \exp(-iS_y \pi/2) |z\rangle##. Have tried following this path, unfortunately stuck with the involved algebra. By the way the OP seems to be long absent anyway.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top