Spin 1/2 Basis Change Homework Solution

[beebopbellopu]

Homework Statement

This is from my first-quarter graduate QM course. Part 4 of this problem asks me to compute the unitary operator U which transforms S_n into S_z, where S_n is the spin operator for spin 1/2 quantized along some arbitrary axis n = icos[itex]\phi[/itex]sinθ + jsin[itex]\phi[/itex]sinθ + zcosθ.

Homework Equations

From part 1, I solved S_n = \begin{pmatrix} cosθ&sinθe^{-i[itex]\phi[/itex]}\\ sinθe^{i[itex]\phi[/itex]}&-cosθ \end{pmatrix}

|n;+> = cos(θ/2)|+> + sin(θ/2)e^{i[itex]\phi[/itex]}|->
|n;-> = sin(θ/2)|+> - cos(θ/2)e^{i[itex]\phi[/itex]}|->

where |+> and |-> are the spin up and down states along the z-axis

Also: U(adjoint)S_nU = S_z

The Attempt at a Solution

U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)e^{i[itex]\phi[/itex]}|->]<+| + [sin(θ/2)|+> - cos(θ/2)e^{i[itex]\phi[/itex]}|->]<-|

I then calculated out <+|U|+>, <-|U|+>, <+|U|->, and <-|U|-> to get the 4 elements of the unitary operator matrix, and end up with:

U = \begin{pmatrix}cos(θ/2)&sin(θ/2)e^{i[itex]\phi[/itex]}\\sin(θ/2)&-cos(θ/2)e^{i[itex]\phi[/itex]}\end{pmatrix}

However I used that to transform S_n and could not recover S_z. U(adjoint) is just the transpose complex conjugate so then I should be using:

U(adjoint) = \begin{pmatrix}cos(θ/2)&sin(θ/2)\\sin(θ/2)e^{-i[itex]\phi[/itex]}&-cos(θ/2)e^{-i[itex]\phi[/itex]}\end{pmatrix}

I'm not quite sure where to go from here.

Edit: I also can't figure out how to post matrices, if someone can help with that also. Thanks!

Edit 2: Thanks vela for the fix.

 

[vela]

Fixed your LaTeX.

Homework Statement

This is from my first-quarter graduate QM course. Part 4 of this problem asks me to compute the unitary operator U which transforms S_n into S_z, where S_n is the spin operator for spin 1/2 quantized along some arbitrary axis n = icos[itex]\phi[/itex]sinθ + jsin[itex]\phi[/itex]sinθ + zcosθ

Homework Equations

From part 1, I solved S_n = [itex]\begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi} \sin \theta & -\cosθ \end{pmatrix}[/itex]


|n;+> = cos(θ/2)|+> + sin(θ/2)e^{i[itex]\phi[/itex]}|->
|n;-> = sin(θ/2)|+> - cos(θ/2)e^{i[itex]\phi[/itex]}|->

where |+> and |-> are the spin up and down states along the z-axis.

Also: U(adjoint)S_nU = S_z

The Attempt at a Solution

U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)e^{i[itex]\phi[/itex]}|->]<+| + [sin(θ/2)|+> - cos(θ/2)e^{i[itex]\phi[/itex]}|->]<-|

I then calculated out <+|U|+>, <-|U|+>, <+|U|->, and <-|U|-> to get the 4 elements of the unitary operator matrix, and end up with:
[tex]U = \begin{pmatrix} \cos(\theta/2) & e^{i\phi}\sin(\theta/2) \\ \sin(\theta/2) & -e^{i\phi}cos(θ/2)\end{pmatrix}[/tex]

However I used that to transform S_n and could not recover S_z. U(adjoint) is just the transpose complex conjugate so then I should be using:

[tex]U^\dagger = \begin{pmatrix} \cos(\theta/2) & \sin(\theta/2) \\ e^{-i\phi}\sin(\theta/2) & -e^{-i\phi}cos(\theta/2)\end{pmatrix}[/tex]

I'm not quite sure where to go from here.

Edit: I also can't figure out how to post matrices, if someone can help with that also. Thanks!

 

[beebopbellopu]

Okay so I did out the matrix math a little further, found some more double angle identities and so forth, and if I set [itex]\phi[/itex] to 0 after multiplying through it comes out to σ₃ which is what I want for S_z.

I think that this is because the unitary operator has to keep the inner product and its therefore an orthogonal transformation, meaning U must be orthogonal. If that's the case then it has to have real elements and the only way for that is if phi = 0, which makes sense in that the the azimuthal angle needs to be 0 in order to be orthogonal to the z-axis.

Is that correct?

 

[vela]

Your matrix for S_n is missing an overall factor of 1/2.

If you multiply the second eigenvector by e^-iΦ, you'll have
\begin{align*}
\vert + \rangle_n &= \begin{pmatrix} \cos(\theta/2) \\ e^{i\phi}\sin(\theta/2) \end{pmatrix} \\
\vert - \rangle_n &= \begin{pmatrix} e^{-i\phi}\sin(\theta/2) \\ -\cos(\theta/2)\end{pmatrix}
\end{align*}(It's not really necessary to do this. I'm just doing it for aesthetic reasons.)

If you recall your basic linear algebra, you know that the matrix U that diagonalizes S_n has the eigenvectors of S_n as its columns, so you have[tex]U=\begin{pmatrix}
\cos(\theta/2) & e^{-i\phi}\sin(\theta/2) \\
e^{i\phi}\sin(\theta/2) & -\cos(\theta/2)
\end{pmatrix}[/tex]
You can verify that this matrix is in fact unitary for any value of the angles and that it will diagonalize S_n. Requiring Φ=0 isn't acceptable since n can point in any direction.

For some reason, you ended up with the transpose of this matrix.

 

[paranormal]

I would say that your attempt at a solution is on the right track. However, it seems like there may be a mistake in your calculation of U(adjoint). The correct matrix should be:

U(adjoint) = \begin{pmatrix}cos(θ/2)&sin(θ/2)\\-sin(θ/2)e-i\phi&cos(θ/2)e-i\phi\end{pmatrix}

Notice the negative sign in the second row. This should fix the issue of not being able to recover Sz. Also, I would suggest double checking your calculations for the elements of the unitary operator matrix, just to make sure there are no other errors.

As for posting matrices, you can use LaTeX code to format them. For example, the matrix:

\begin{pmatrix} a&b\\ c&d \end{pmatrix}

would be written as:

\begin{pmatrix} a&b\\ c&d \end{pmatrix}

in your post. Just make sure to use backslashes before special characters like \, &, and <, and to enclose the code in dollar signs ($$) to format it as a math equation. I hope this helps!