Rotation Operator Matrix Representation using |+z> and |-z> Basis

[cragar]

Homework Statement

Determine the matrix representation of the rotation operator
$R(\phi k)$ using the states |+z> and |-z> as a basis. Using your matrix representation verify that $R^{\dagger}R=1$

The Attempt at a Solution

Do I need to write $R| \psi>$ in terms of a matrix.
If I have $|\psi>=a|+z>+b|-z>$
Then do I just operate R on $\psi$ and then write this in terms of a matrix.
are these related to the Pauli spin matrices

 

[vela]

Homework Statement

Determine the matrix representation of the rotation operator
$R(\phi k)$ using the states |+z> and |-z> as a basis. Using your matrix representation verify that $R^{\dagger}R=1$

The Attempt at a Solution

Do I need to write $R| \psi>$ in terms of a matrix.
If I have $|\psi>=a|+z>+b|-z>$
Then do I just operate R on $\psi$ and then write this in terms of a matrix.

I'm not sure exactly what you had in mind, but it's probably not the most straightforward way to solve this problem.

are these related to the Pauli spin matrices

Yes. Remember that the spin operators $\hat{S}_x$, $\hat{S}_y$, and $\hat{S}_z$ are generators of rotations. (This is definitely covered in your textbook.) Use that fact to calculate R.

 

[cragar]

ok thanks for your help. My book gives the matrix for R and it is in the z basis.
And I took $R^{\dagger}R$ and it equaled one. But If the matrix wasn't in the
z basis would I use the roatation matrix to get the answer.
I would take $S^{\dagger}RS$ and this would give the correct R for the problem.

 

[vela]

I don't understand what you're asking. Well, I sort of do, but I'd like you to clarify your question. What is S? How did you find S?

 

[cragar]

$s= \[left (begin{array}{cc}<+z|+x>& <+z|-x>\\ <-z|+x>& <-z|-x> \end{array}\right)\]$
where the bras are what basis I am going to and the kets are the basis that I was in. I don't really know how S is derived though

 

[vela]

Yes, that's the regular method you use to change bases. You're doing the same thing you learned in linear algebra. It's just the notation that's different.