Why is psi equal to psi' in vector expressions in QM?

[Niles]

Hi guys

I am reading a book, where they use vector expressions a lot. In it they write

$${\bf{M}} = \sum\limits_{\psi ,\psi ',\sigma ,\sigma '} {\left\langle {\psi '} \right|\left\langle {\sigma '} \right|{\bf{m}}\left| \sigma \right\rangle \left| \psi \right\rangle a_{\psi ',\sigma '}^\dag a{}_{\psi ,\sigma }}$$

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

$${\bf{M}} = \frac{\hbar }{2}\sum\limits_{\psi ,\sigma ,\sigma '} {\left\langle {\sigma '} \right|\left( {t_1 ,t_2 ,t_3 } \right)\left| \sigma \right\rangle a_{\psi ,\sigma '}^\dag a{}_{\psi ,\sigma }}$$

I cannot see why they equal psi = psi', since m is a vector, not a diagonal matrix. What are they doing here?

 

[fzero]

Without having more details about what definitions the book is using, it appears that in the expression

$${\left\langle {\psi'} \right|\left\langle {\sigma'} \right|{\mathbf{m}}\left| \sigma \right\rangle \left| \psi \right\rangle$$

$$|\psi\rangle$$ is the state describing the spatial physics, while $$|\sigma\rangle$$ is an internal spin state. If this is true, these states lie in different Hilbert spaces. Furthermore, the operator $$\mathbf{m}$$ acts only on the spin states, not $$|\psi\rangle$$. So we can rearrange

$${\left\langle {\psi'} | \psi \right\rangle\left\langle {\sigma'} \right|{\mathbf{m}}\left| \sigma \right\rangle .$$

 

[Niles]

Ah, of course... I should have known that. I will not forget it. Thanks, you have helped me a lot today. I really appreciate it.

Best wishes,
Niles.