# Spin 1/2 Basis Change

## 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 Sn into Sz, where Sn is the spin operator for spin 1/2 quantized along some arbitrary axis n = icos$\phi$sinθ + jsin$\phi$sinθ + zcosθ.

## Homework Equations

From part 1, I solved Sn = \begin{pmatrix} cosθ&sinθe-i$\phi$\\ sinθei$\phi$&-cosθ \end{pmatrix}

|n;+> = cos(θ/2)|+> + sin(θ/2)ei$\phi$|->
|n;-> = sin(θ/2)|+> - cos(θ/2)ei$\phi$|->

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

## The Attempt at a Solution

U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)ei$\phi$|->]<+| + [sin(θ/2)|+> - cos(θ/2)ei$\phi$|->]<-|

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)ei$\phi$\\sin(θ/2)&-cos(θ/2)ei$\phi$\end{pmatrix}

However I used that to transform Sn and could not recover Sz. 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.

Last edited:

vela
Staff Emeritus
Homework Helper

## 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 Sn into Sz, where Sn is the spin operator for spin 1/2 quantized along some arbitrary axis n = icos$\phi$sinθ + jsin$\phi$sinθ + zcosθ

## Homework Equations

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

|n;+> = cos(θ/2)|+> + sin(θ/2)ei$\phi$|->
|n;-> = sin(θ/2)|+> - cos(θ/2)ei$\phi$|->

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

## The Attempt at a Solution

U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)ei$\phi$|->]<+| + [sin(θ/2)|+> - cos(θ/2)ei$\phi$|->]<-|

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

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

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

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!

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 Sz.

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
Staff Emeritus
Homework Helper
If you recall your basic linear algebra, you know that the matrix U that diagonalizes Sn has the eigenvectors of Sn as its columns, so you have$$U=\begin{pmatrix} \cos(\theta/2) & e^{-i\phi}\sin(\theta/2) \\ e^{i\phi}\sin(\theta/2) & -\cos(\theta/2) \end{pmatrix}$$