States of Composite Systems, p. 216: Ballentine

In summary, the confused person is working through Ballentine's Quantum Mechanics book and is stuck on a section on the state operator. They ask for help with understanding the state operator and are given a quick summary of the state operator and how it works in terms of a product of basis vectors. They are then asked to generalized the double sum to a more general case of a system with more than two components. Once they understand the generalized double sum, they are asked to apply it to a more complicated situation involving a two component system. They are then shown the trace of the state operator and are able to understand it easily.
  • #1
jfy4
649
3
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
[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]
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
[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,
 
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  • #2
jfy4 said:
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?
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 [itex]\rho[/itex] still in there? Whenever I have dealt with the state operator before, it is represented either by
[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?

No, [itex]\rho[/itex] 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 [itex]|\psi_n\rangle[/itex] vectors correspond to the [itex]|a_i b_j\rangle[/itex] in the tensor product case.

Try writing your [itex]\rho[/itex] as a double sum, generalizing the single sum you wrote above. Then substitute it into the expression for [itex] Tr(\rho R) [/itex] and see what you get.
 
  • #3
strangerep said:
No, [itex]\rho[/itex] 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 [itex]|\psi_n\rangle[/itex] vectors correspond to the [itex]|a_i b_j\rangle[/itex] in the tensor product case.

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

Okay, starting with
[tex]\rho=\sum_m\sum_n\rho_{mn}|a_mb_n\rangle \langle a_mb_n|
[/tex]
then
[tex]
\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
[/tex]
Then using the fact that [itex]|a_mb_n\rangle\langle a_mb_n|[/itex] form a complete set
[tex]
\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
[/tex]
Are these the steps Ballentine is taking?

Thanks,
 
  • #4
jfy4 said:
Okay, starting with
[tex]\rho=\sum_m\sum_n\rho_{mn}|a_mb_n\rangle \langle a_mb_n|
[/tex]
I think that's only correct in general if the [itex]|a_mb_n\rangle[/itex] are eigenstates of [itex]\rho[/itex]. In a more general basis, the state operator is not diagonal. So my hint about "double-sum" was too easy to misinterpret, sorry.

[tex]
\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
[/tex]
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
[tex]
\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\>
[/tex]

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
strangerep said:
[tex]
Tr(\rho R) = \sum_j \langle j| \rho R |j \rangle
[/tex]

what you wrote makes it very easy to see the result thanks. My confusion is the above step. I had only ever seen [itex]\rho=|\psi\rangle\langle\psi|[/itex] or the way earlier with the sum. And I thought that when taking the trace, [itex]\rho[/itex] 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,
 

Related to States of Composite Systems, p. 216: Ballentine

1. What is a composite system in the context of quantum mechanics?

A composite system in quantum mechanics refers to a system that is made up of multiple subsystems. These subsystems can be individual particles, such as atoms or electrons, or they can be larger systems, such as molecules or crystals. The behavior of a composite system is described by the laws of quantum mechanics, which govern the behavior of particles at the microscopic level.

2. How do the states of composite systems differ from the states of individual particles?

The states of composite systems are described by a mathematical combination of the states of the individual subsystems. This is known as the tensor product, and it allows for a more complex description of the system as a whole. In contrast, the states of individual particles are described by a single state vector.

3. Can composite systems exhibit quantum entanglement?

Yes, composite systems can exhibit quantum entanglement, which is a phenomenon where the states of two or more particles become correlated in such a way that the measurement of one particle affects the state of the other. This is a key aspect of quantum mechanics and has been demonstrated in various experiments.

4. What is the significance of the density operator in describing the states of composite systems?

The density operator is a mathematical tool used to describe the states of composite systems. It allows us to calculate the probability of finding a system in a particular state, as well as the average values of physical observables. This is important because it allows us to make predictions about the behavior of composite systems and compare them to experimental results.

5. Are there any real-world applications of the understanding of states of composite systems?

Yes, there are many real-world applications of the understanding of states of composite systems in quantum mechanics. For example, it is important in the development of quantum computers, which use the principles of quantum mechanics to perform calculations much faster than classical computers. It is also essential in the study of advanced materials, such as superconductors, which exhibit quantum behavior at the macroscopic level.

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