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\phisinθ + jsin\phisinθ + zcosθ.

Homework Equations

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

|n;+> = cos(θ/2)|+> + sin(θ/2)e^i\phi|->
|n;-> = sin(θ/2)|+> - cos(θ/2)e^i\phi|->

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\phi|->]<+| + [sin(θ/2)|+> - cos(θ/2)e^i\phi|->]<-|

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\phi\\sin(θ/2)&-cos(θ/2)e^i\phi\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\phi&-cos(θ/2)e^-i\phi\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\phisinθ + jsin\phisinθ + zcosθ

Homework Equations

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


|n;+> = cos(θ/2)|+> + sin(θ/2)e^i\phi|->
|n;-> = sin(θ/2)|+> - cos(θ/2)e^i\phi|->

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\phi|->]<+| + [sin(θ/2)|+> - cos(θ/2)e^i\phi|->]<-|

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(\theta/2) &amp; e^{i\phi}\sin(\theta/2) \\ \sin(\theta/2) &amp; -e^{i\phi}cos(θ/2)\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^\dagger = \begin{pmatrix} \cos(\theta/2) &amp; \sin(\theta/2) \\ e^{-i\phi}\sin(\theta/2) &amp; -e^{-i\phi}cos(\theta/2)\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!

 

[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 \phi 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 haveU=\begin{pmatrix} <br /> \cos(\theta/2) &amp; e^{-i\phi}\sin(\theta/2) \\ <br /> e^{i\phi}\sin(\theta/2) &amp; -\cos(\theta/2)<br /> \end{pmatrix}
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.