# Change of basis

1. Sep 30, 2007

### indigojoker

Show the matrix representation of $$S_z$$ using the eigenkets of $$S_y$$ as base vectors.

I'm not quite sure on the entire process but here's what i think:

We get the transformation matrix though:

$$U = \sum_k |b^{(k)} \rangle \langle a^{(k)} |$$

where |b> is the eigenket for S_y and <a| is the eigenket for S_z

this will give me a change of basis operator that i can operate on the S_z operator to get it into the S_y basis.

would this be the correct though process?

2. Oct 1, 2007

### Gokul43201

Staff Emeritus
Looks fine.

Note: |a> is an eigenket for S_z; <a| would be an eigenbra that is in dual correspondence with |a>.

3. Oct 2, 2007

### indigojoker

I get:

$$U =\left( \frac{1}{\sqrt{2}} |+ \rangle \langle +| + \frac{i}{\sqrt{2}} |- \rangle \langle +| \right) + \left( \frac{1}{\sqrt{2}} |+ \rangle \langle -| - \frac{i}{\sqrt{2}} |- \rangle \langle -| \right)$$

$$U= \frac{1}{\sqrt{2}} \left(\begin{array}{cc}1&1\\i&-i\end{array}\right)$$

$$US_z=\frac{1}{\sqrt{2}}\frac{\hbar}{2}\left(\begin{array}{cc}1&1\\i&-i\end{array}\right)\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$$

thus, S-z in S_y basis is:
$$S_z=\frac{\hbar\sqrt{2}}{4}\left(\begin{array}{cc}1&-1\\i&i\end{array}\right)$$

looks okay?

Last edited: Oct 2, 2007