Matrix Representation of S_z Using Eigenkets of S_y

[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?

 

[Gokul43201]

Looks fine.

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

 

[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&amp;1\\i&amp;-i\end{array}\right)

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

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

looks okay?