- #1
Niles
- 1,866
- 0
Hi guys
I am reading a book, where they use vector expressions a lot. In it they write
[tex]
{\bf{M}} = \sum\limits_{\psi ,\psi ',\sigma ,\sigma '} {\left\langle {\psi '} \right|\left\langle {\sigma '} \right|{\bf{m}}\left| \sigma \right\rangle \left| \psi \right\rangle a_{\psi ',\sigma '}^\dag a{}_{\psi ,\sigma }}
[/tex]
where m={t1, t2, t3} is the vector containing the three Pauli spin matrices t1, t2, t3 and a is the annihilation operator. They say this is equal to
[tex]
{\bf{M}} = \frac{\hbar }{2}\sum\limits_{\psi ,\sigma ,\sigma '} {\left\langle {\sigma '} \right|\left( {t_1 ,t_2 ,t_3 } \right)\left| \sigma \right\rangle a_{\psi ,\sigma '}^\dag a{}_{\psi ,\sigma }}
[/tex]
I cannot see why they equal psi = psi', since m is a vector, not a diagonal matrix. What are they doing here?
I am reading a book, where they use vector expressions a lot. In it they write
[tex]
{\bf{M}} = \sum\limits_{\psi ,\psi ',\sigma ,\sigma '} {\left\langle {\psi '} \right|\left\langle {\sigma '} \right|{\bf{m}}\left| \sigma \right\rangle \left| \psi \right\rangle a_{\psi ',\sigma '}^\dag a{}_{\psi ,\sigma }}
[/tex]
where m={t1, t2, t3} is the vector containing the three Pauli spin matrices t1, t2, t3 and a is the annihilation operator. They say this is equal to
[tex]
{\bf{M}} = \frac{\hbar }{2}\sum\limits_{\psi ,\sigma ,\sigma '} {\left\langle {\sigma '} \right|\left( {t_1 ,t_2 ,t_3 } \right)\left| \sigma \right\rangle a_{\psi ,\sigma '}^\dag a{}_{\psi ,\sigma }}
[/tex]
I cannot see why they equal psi = psi', since m is a vector, not a diagonal matrix. What are they doing here?
Last edited: