Change of basis

  • #1
246
0
Show the matrix representation of [tex] S_z [/tex] using the eigenkets of [tex]S_y[/tex] as base vectors.

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

We get the transformation matrix though:

[tex]U = \sum_k |b^{(k)} \rangle \langle a^{(k)} | [/tex]

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?
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
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Looks fine.

Note: |a> is an eigenket for S_z; <a| would be an eigenbra that is in dual correspondence with |a>.
 
  • #3
246
0
I get:

[tex]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)[/tex]

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

[tex]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)[/tex]

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

looks okay?
 
Last edited:

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