# Various questions

1. Aug 9, 2010

### JK423

I would like some help on the following questions.

1) If we have two interacting particles A and B. Can we still write the state vector of the whole system as the cross product
|AB>=|A>x|B> (1) ?
My professor told me that we can always right the state vector of a bipartite system as in (1) but it doesnt make sense to me if A,B are interacting. Cause eq. (1) indicates that A,B are independent. If we take the position representation of (1) the wavefunctions of A,B will be seperate like ΨA(x)xΨΒ(q) (2).
However if there is interaction its not possible to write the state ΨΑΒ(x,q) as the product (2) with seperate variables.

2) Consider the closed system A+B where A and B are interacting particles as above, whose state's given by the density matrix ρΑΒ. I'm interested in the reduced density matrix of system A, which is given by the partial trace of ρΑΒ over tha variables of B:
ρΑ=TrBΑΒ}
My question is: When we take the trace over B, what states do we actually use??

2. Aug 10, 2010

### JK423

Im writing my undergraduate thesis at the moment and id really want these questions answered!

3. Aug 10, 2010

### meopemuk

ΨA(x)xΨΒ(q) is a perfectly valid wave function for a system of two interacting particles. By looking at system's wave function at a fixed time instant it is impossible to see whether the system is interacting or not. Interaction affects only the wave function's evolution in time.

Eugene.

4. Aug 10, 2010

### strangerep

First of all, you want the tensor product
$$|AB\rangle ~=~ |A\rangle \otimes |B\rangle$$
rather than the ordinary cartesian product. (Check the Wikipedia entry for tensor
product to see the difference.)

Next, it might help if you think in terms of having complete orthonormal bases for
each 1-particle space. I'll call the basis vectors $|a\rangle$ for the "A" space,
and similarly for the "B" space. Any vector $\Phi$ in the tensor product space
can be expressed in the form
$$\Phi ~=~ \sum_{a,b} \, c_{ab} \, (|a\rangle \otimes |b\rangle)$$
for some set of coefficients $c_{ab}$. In your case, the double sum gets
replaced by a double integral since the position variables are continuous.
Powerful math wizards mutter the arcane phrase "Schwartz kernel theorem"

An arbitrary operator on the tensor product space, acting on a particular
vector therein, changes the $c_{ab}$ values.

So like Meopemuk said, you can't tell whether there's interaction or not just
because there's a tensor product. It depends on how the interaction term in
the total Hamiltonian mucks around with all the $c_{ab}$ values.

In my notation above, you'd be tracing over the $|b\rangle$ states.

BTW, if you haven't yet read the early chapters of Ballentine, then go grab a copy
urgently. He presents these sorts of foundational matters much better than many
other textbooks.

Last edited: Aug 10, 2010
5. Aug 10, 2010

### JK423

Your example of expanding the Φ state over the tensor product space basis did help me understand. So i think i'm ok with this!

While thinking on this some basic questions came up..
Suppose that we take the {$$|b\rangle$$} basis of the "B" space. These states should evolve via the Schroedinger equation:
$$i\hbar\frac{\partial|b\rangle}{\partial t}=H|b\rangle$$ which would give for a time-independent hamiltonian $$|b(t)\rangle=\exp( iHt/\hbar )|b\rangle$$.
A question is the following:
a) In general we consider the eigenstates of some observable constant in time not subjected to the Schrodinger's equation.. Why is that?
b) The correct basis to use is $$|b(t)\rangle$$ or $$|b\rangle$$?

Consider the derivative of the density matrix (it has to do with the master equation) of the whole system A+B: $$\frac{d \rho_{A B} }{dt}$$. Literature says that if we take the partial trace over B, its going to be: $$Tr_{B} \{ \frac{d \rho_{A B} }{dt} \} = \frac{d \{Tr_{B} \rho_{A B} \} }{dt} = \frac{d \rho_{A} }{dt}$$
where $$\rho_{A}$$ is the reduced density matrix for the A system.
My question is, how can the trace get inside the derivative? This is related to my previous questions b). If we use the time independent basis for the trace, then yes it can get inside. But if we use $$|b(t)\rangle=\exp( iHt/\hbar )|b\rangle$$ as a basis the it cannot..

Im really confused about the above

6. Aug 12, 2010

### JK423

7. Aug 12, 2010

### alxm

Are you saying that the time independent wave function of an interacting system is the same as the one for a non-interacting system? That's not true at all.

One can, as strangerep said, use non-interacting states as a basis for the interacting system. But the non-interacting states are not eigenstates of the interacting hamiltonian, and a single non-interacting particle state does not describe the state of a single particle.

(For instance, in the case of fermions a single Slater determinant does not accurately describe an interacting system. E.g. the best wave function written as a simple product of single-particle wave functions for the Helium atom will result in a ground-state energy of -2.86 a.u. when it's actually -2.90)

8. Aug 12, 2010

### meopemuk

No, I didn't say that. "Time independent wave function" is the same as "stationary state" or "eigenfunction of the Hamiltonian". Obviously, eigenfunctions of the free and interacting Hamiltonians are different. For example, eigenfunctions of the free Hamiltonian are plane waves, while eigenfunctions of the Coulomb Hamiltonian are localized. However, nobody can forbid you to prepare a plane-wave state at time t=0 in the Coulomb field. This state will not be stationary, of course, but it is still a valid state.

A single Slater determinant describes SOME state of the interacting system. However, this state is not a good approximation for the energy eigenstate. You are right about that.

Eugene.

9. Aug 13, 2010

### strangerep

Let's take a step back. Any given class of physical system is described by a
set of observable quantitie. For a specific prepared system, we can assign numbers
by measuring these observables. In general, the numbers we get for two
different quantities depends on the order in which we measure the two
quantities. This leads us to infer that the set of observable
quantities in fact forms a Lie algebra. A "state" is then really a
mapping from the set of these quantities to ordinary numbers.
Since these numbers (for many repeated measurements) must form a
probability distribution, that leads to the technique of representing
the states as vectors in a Hilbert space, with the observable
quantities being correspondingly represented as operators on that
Hilbert space. One can then exploit mathematical properties of Hilbert
space vectors: eigenvalues of Hermitian operators are real (hence
physical observables should correspond to Hermitian operators),
and the set of eigenstates of _any_ Hermitian operator spans
the Hilbert space: _any_ state can be expressed as a superposition
of such eigenstates.

This might seem backwards from how some QM courses proceed. They
seem to start from states and then introduce observables, etc. But
what I described in the previous paragraph is the more powerful
and general way of thinking about the whole subject.

For nonrelativistic QM, the Lie algebra of observable quantities
is the Galilean algebra (which we deduce physically by examining
the group of symmetries in our nonrelativistic world. Ballentine
develops QM really well from this perspective.

One of the elements in the Galilean algebra is the Hamiltonian.
The Schrodinger equation is nothing more a statement of what
the Hamiltonian *is* -- the generator of time evolution.

Neither. It's better to think in terms of eigenstates of the
Hamiltonian:

$$H \, |E\rangle ~=~ E \, |E\rangle$$

where E is a real number (energy). The set of all possible values
of E for a given Hamiltonian is called its "spectrum".
Let's denote the spectrum set as

$$E ~\in~ \{ E_1, E_2, E_3, \dots \}$$

Alternative, you could pick any other Hermitian operator, find
its eigenvalue spectrum and associated eigenstates, and expand
any arbitrary state in terms of these different eigenstates.

Any arbitrary state (vector) in the Hilbert space can be expressed
as a linear combination of these eigenstates.

For a system composed of two copies "A" and "B" of the above,
we form a tensor product space. Then we must keep track of which
eigenvector is from each space, and we might write one particular
basis vector of the tensor product space as

$$|E_n\rangle_{(A)} ~\otimes~ |E_m\rangle_{(B)}$$

Sometimes people abbreviate all this and write simply

$$|E_n ,\, E_m \rangle ~,$$

intending that it be understood that a tensor product vector is
being used, with the first arg pertaining to the "A" subsystem,
and the second arg pertaining to the "B" subsystem.

The next step is to see how a density operator (aka state operator)
can be expressed in terms of these eigenstates, but I'll leave
it at this for now.

[BTW, are you able to get a copy of Ballentine? It would be orders
of magnitude easier if I could just refer you to particular chapters
therein, where these things are explained much better than I can ever
do here.]

10. Aug 21, 2010

### JK423

Strangerep,
First of all, thanks again for the detailed answer! I'm sorry for the late response but i didn't have an internet connection these days. Btw, yes i do have a copy of Balentine so you can refer me there.
The first part of your post, i must read it a couple of times in order to get it.. :-)
According to the second part,
You say 'eigenvectors' from each space. But, eigenvectors of what Hamiltonian operator??
Since A, B are coupled, only the tensor product is an eigenvector of the total Hamiltonian! And not each one separately! So, as i undertstand it, you cannot have an energy basis for each space separately.. And if thats correct, when we trace the density matrix over the B variables (say energy states), what basis in B space do we use since it doesnt have an energy basis...?

I've totally lost it...

John

11. Aug 23, 2010

### strangerep

Let's consider each component space separately at first. There is a "free" Hamiltonian
$H_0$ acting in each space. (Think of it as a Hamiltonian acting only on component A,
and a separate copy of it acting only on component B.)

If we start with an uncorrelated tensor product state, then evolution under the
free Hamiltonian does not produce any correlation. But if we now add an interaction
term in the 2-particle Hamiltonian which mixes up the components, then correlation
develops over time.

I don't know how much of Ballentine you've studied, but it's probably useful to
(re-)study section 8.3 "States of Composite Systems" until you understand in detail
what the terms "correlation" and "reduced state operator" mean, and the conditions
under which a tensor product state can be said to be "correlated"or not. (I need to be
sure you understand that aspect of QM theory before I try to go further.)

12. Aug 23, 2010

### JK423

I just read Ballentine's chapter. Its the first time i read this book and realized that he analyzes things in great detail, comparing to other textbooks!
What ive understood so far is this.
At first we suppose that systems A and B are not interacting, so they are described by their free Hamiltonians $$H_{A}$$ and $$H_{B}$$ whose corresponding eigenvectors are $$| a_{m} >$$ and $$| b_{n} >$$. So, an arbitrary uncorrelated state of the bipartite system will be

$$| \phi > = \sum_{mn} c_{mn} | a_{m} > \otimes | b_{n} >$$

When the interaction term appears, the total Hamiltonian will be

$$H_{total} = H_{A} + H_{B} + V_{AB}$$

and state |φ> will evolve

$$| \Psi > = \exp (i H_{total} t / \hbar) | \phi >$$

in a correlated state |Ψ>, which is expressed in terms of the basis $$| a_{m} >$$ and $$| b_{n} >$$.

In other words, a state of two interacting subsystems can always be expressed in the basis of the free Hamiltonians. So, in my original question on partial tracing the total density matrix, the answer is that i will trace over the states of the free Hamiltonian $$H_{B}$$ of the B system in order to get the recuced density matrix of A.

Am i correct?

13. Aug 25, 2010

### strangerep

Yes, it's great to have a text that develops the subject well from a modern perspective.
Every serious student of QM should buy a copy. :-)

Right.

Yes -- for the cases you're considering involving a finite number of degrees of freedom
(i.e., a finite-dimensional Hilbert space) and reasonably well-behaved interaction terms.

For the infinite-dimensional case, and/or with less well-behaved interactions, it turns
out there's other subtleties broadly known as "Haag's theorem" where the above does
not apply, or does not apply without extra conditions. But for your current purposes I don't
think you need to worry about this. I mention it only to avoid getting my butt kicked if
I fail to mention this caveat. :-(