States of Composite Systems, p. 216: Ballentine

[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.

...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
<br /> |a_mb_n\rangle = |a_m\rangle\otimes|b_n\rangle,<br />
where \left\{ |a_m\rangle\right\} is a set of basis vectors for component 1 alone, and similarly \left\{ |b_n\rangle\right\} is a basis set for component 2 alone. The average of an arbitrary dynamical variable R is given by
<br /> \langle R \rangle=\text{Tr}(\rho R)=\sum_{m,n,m&#039;,n&#039;}\langle a_mb_n|\rho|a_{m&#039;}b_{n&#039;}\rangle\langle a_{m&#039;}b_{n&#039;}|R|a_,b_n\rangle.<br />

Here is my first confusion. In this case then |a_{m&#039;}b_{n&#039;}\rangle\langle a_{m&#039;}b_{n&#039;}| 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
<br /> \rho=\sum_n\rho_n|\psi_n\rangle \langle\psi_n| \quad \text{or} \quad \rho=|\psi\rangle\langle\psi|<br />
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,

 

[strangerep]

Here is my first confusion. In this case then |a_{m&#039;}b_{n&#039;}\rangle\langle a_{m&#039;}b_{n&#039;}| do not form a complete set?

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).

Second, why is \rho still in there? Whenever I have dealt with the state operator before, it is represented either by
<br /> \rho=\sum_n\rho_n|\psi_n\rangle \langle\psi_n| \quad \text{or} \quad \rho=|\psi\rangle\langle\psi|<br />
does that mean that the \rho above is (are) the coefficients in the sum I just wrote above?

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.

 

[jfy4]

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.

Okay, starting with
\rho=\sum_m\sum_n\rho_{mn}|a_mb_n\rangle \langle a_mb_n|<br />
then
<br /> \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<br />
Then using the fact that |a_mb_n\rangle\langle a_mb_n| form a complete set
<br /> \langle R \rangle=\sum_m\sum_n\sum_{m&#039;}\sum_{n&#039;}\langle a_mb_n|\rho_{mn}|a_{m&#039;}b_{n&#039;}\rangle\langle a_{m&#039;}b_{n&#039;}|R|a_mb_n\rangle<br />
Are these the steps Ballentine is taking?

Thanks,

 

[strangerep]

Okay, starting with
\rho=\sum_m\sum_n\rho_{mn}|a_mb_n\rangle \langle a_mb_n|<br />

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.

<br /> \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<br />

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
<br /> \def\&lt;{\langle}<br /> \def\&gt;{\rangle}<br /> \&lt;R\&gt; = Tr(\rho R) = \sum_j \&lt;j| \rho R |j\&gt; = \sum_j \&lt;j| \rho 1 R |j\&gt;<br /> = \sum_{jk} \&lt;j| \rho \; |k\&gt;\&lt;k| \; R |j\&gt;<br />

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.

 

[jfy4]

<br /> Tr(\rho R) = \sum_j \langle j| \rho R |j \rangle<br />

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,