- #1

jfy4

- 649

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I am working through Ballentine's Quantum Mechanics books, and I am stuck... I would be appreciative if someone could walk me through this section with some help.

Here is my first confusion. In this case then [itex]|a_{m'}b_{n'}\rangle\langle a_{m'}b_{n'}|[/itex] do not form a complete set? Second, why is [itex]\rho[/itex] still in there? Whenever I have dealt with the state operator before, it is represented either by...we consider a two-component system whose components will be labeled 1 and 2. The state vector space is spanned by a set of product vectors of the form

[tex]

|a_mb_n\rangle = |a_m\rangle\otimes|b_n\rangle,

[/tex]

where [itex]\left\{ |a_m\rangle\right\}[/itex] is a set of basis vectors for component 1 alone, and similarly [itex]\left\{ |b_n\rangle\right\}[/itex] is a basis set for component 2 alone. The average of an arbitrary dynamical variable [itex]R[/itex] is given by

[tex]

\langle R \rangle=\text{Tr}(\rho R)=\sum_{m,n,m',n'}\langle a_mb_n|\rho|a_{m'}b_{n'}\rangle\langle a_{m'}b_{n'}|R|a_,b_n\rangle.

[/tex]

[tex]

\rho=\sum_n\rho_n|\psi_n\rangle \langle\psi_n| \quad \text{or} \quad \rho=|\psi\rangle\langle\psi|

[/tex]

does that mean that the [itex]\rho[/itex] above is (are) the coefficients in the sum I just wrote above?

I have some more questions on the same page just after this part, but perhaps if I can get this, the rest will fall into place.

Thanks,