Spin 1/2 Basis Change

  • #1

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[itex]\phi[/itex]sinθ + jsin[itex]\phi[/itex]sinθ + zcosθ.


Homework Equations



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

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

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

Also: U(adjoint)SnU = Sz

The Attempt at a Solution



U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)ei[itex]\phi[/itex]|->]<+| + [sin(θ/2)|+> - cos(θ/2)ei[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)ei[itex]\phi[/itex]\\sin(θ/2)&-cos(θ/2)ei[itex]\phi[/itex]\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[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.
 
Last edited:

Answers and Replies

  • #2
vela
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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 Sn into Sz, where Sn 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 Sn = [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)ei[itex]\phi[/itex]|->
|n;-> = sin(θ/2)|+> - cos(θ/2)ei[itex]\phi[/itex]|->

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

Also: U(adjoint)SnU = Sz

The Attempt at a Solution



U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)ei[itex]\phi[/itex]|->]<+| + [sin(θ/2)|+> - cos(θ/2)ei[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 Sn and could not recover Sz. 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!
 
  • #3
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 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?
 
  • #4
vela
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Your matrix for Sn 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 Sn has the eigenvectors of Sn 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 Sn. 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.
 

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