# States of Composite Systems, p. 216: Ballentine

1. Jul 6, 2011

### jfy4

Hi everyone,

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 $|a_{m'}b_{n'}\rangle\langle a_{m'}b_{n'}|$ do not form a complete set? Second, why is $\rho$ still in there? Whenever I have dealt with the state operator before, it is represented either by
$$\rho=\sum_n\rho_n|\psi_n\rangle \langle\psi_n| \quad \text{or} \quad \rho=|\psi\rangle\langle\psi|$$
does that mean that the $\rho$ 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,

2. Jul 7, 2011

### strangerep

Sure they do. That expression in the middle with sum over m',n' is just a
resolution of unity. Cf. Ballentine's eq(1.26).

No, $\rho$ is a state operator. The general requirements for an operator to be a state operator are given in eqs 2.6, 2.7, 2.8.

Your $|\psi_n\rangle$ vectors correspond to the $|a_i b_j\rangle$ in the tensor product case.

Try writing your $\rho$ as a double sum, generalizing the single sum you wrote above. Then substitute it into the expression for $Tr(\rho R)$ and see what you get.

3. Jul 7, 2011

### jfy4

Okay, starting with
$$\rho=\sum_m\sum_n\rho_{mn}|a_mb_n\rangle \langle a_mb_n|$$
then
$$\langle R \rangle = \text{Tr}(\rho R)=\text{Tr}\left(\sum_m\sum_n\rho_{mn}|a_mb_n \rangle \langle a_mb_n|R\right)=\sum_m\sum_n\rho_{mn}\langle a_mb_n|R|a_mb_n\rangle$$
Then using the fact that $|a_mb_n\rangle\langle a_mb_n|$ form a complete set
$$\langle R \rangle=\sum_m\sum_n\sum_{m'}\sum_{n'}\langle a_mb_n|\rho_{mn}|a_{m'}b_{n'}\rangle\langle a_{m'}b_{n'}|R|a_mb_n\rangle$$
Are these the steps Ballentine is taking?

Thanks,

4. Jul 8, 2011

### strangerep

I think that's only correct in general if the $|a_mb_n\rangle$ are eigenstates of $\rho$. In a more general basis, the state operator is not diagonal. So my hint about "double-sum" was too easy to misinterpret, sorry.

I don't think you're taking the trace correctly here. But it's probably quickest if I just
show you the following:

For the simpler case of a non-composite system, in an arbitrary basis
$$\def\<{\langle} \def\>{\rangle} \<R\> = Tr(\rho R) = \sum_j \<j| \rho R |j\> = \sum_j \<j| \rho 1 R |j\> = \sum_{jk} \<j| \rho \; |k\>\<k| \; R |j\>$$

Can you generalize that to Ballentine's 2-component composite system where the basis vectors are Ballentine's eq(8.11) ? It should give you eq(8.12) almost trivially.

5. Jul 8, 2011

### jfy4

what you wrote makes it very easy to see the result thanks. My confusion is the above step. I had only ever seen $\rho=|\psi\rangle\langle\psi|$ or the way earlier with the sum. And I thought that when taking the trace, $\rho$ was expressed in terms of it's basis vectors such as in (2.14) and (2.15). I understand that section is on pure states, but rho is expressed in terms of it's basis vectors, not kept as rho. I thought I had to do that every time.

Thanks for your help. I'll try and finish out the section and if I have more questions I'll put them up.

Thanks,