# Pure and not pure states, an example

1. Jun 23, 2005

### jonas_nilsson

Hi there,

I was looking at a problem, where the state (a pure state) of a two-particle system was given as:

$$| \Psi> = A_{11}| a_1 b_1 > + A_{12} | a_1 b_2> + A_{21} | a_2 b_1> + A_{22} | a_2 b_2>$$

In the system particle a can only be in the states a_1 or a_2 and particle b in b_1 or b_2.

The density operator should be
$$\rho = |\Psi><\Psi|$$

The question is, what happens if we make a measurement that determines the state of particle a? Will the collapsed state still be pure, or can it somehow fall into a mixed state? The second alternative sounds unintiutive to me, but I have been surprised many times enough...

Assuming that the measurement yielded the result a_1, then I think the density operator should collaps into
$$\rho_1' = p_1 ~\rho ~p_1$$
where p_1 the projection operator
$$p_1 = |a_1><a_1|$$
is. After making sure that the norm is OK, I then find that
$$Trace(\rho_1^2) = 1$$,
which means that the state after the measurement is still pure. Is this correct? I don't feel confident with this, and I sense that I might be missing something.

2. Jun 23, 2005

### George Jones

Staff Emeritus
For simplicity, consider a 1-particle system in the state

|phi> = c1|a1> + c2|a2>,

with |phi>, |a1>, and |c1> normalized. The density, as you say is,

rho = |phi><phi|.

Now perform a measurement.

It seems to me that there is a difference between the following 2 situations:

1) a measurement is known to be made, and the outcome is known;

2) a measurement is known to be made, and the outcome is unknown.

You have considered case 1), i.e., suppose the result of the measurement is known to be a1. Then the nortmailzed state now is

|phi'> = |a1>

= (p1 |phi>)/|p1 |phi>|,

where p1 = |a1><a1|.

The new density matrix is

rho' = |phi'><phi'|

= (p1 rho p1)/|p1 |phi>|^2

Now, case 2). With the probabilty |c1|^2, the state of the system is |a1>. With probability |c2|^2, the state of the system is |a2>. Thus, this is a mixed state with density matrix

rho'' = |c1|^2 |a1><a1| + |c2|^2 |a2><a2|.

Regards,
George

3. Jun 23, 2005

### jonas_nilsson

Aaaah, thanks George, now I sort of get the question I was trying to answer. Didn't make any sense to put question of type 1 in a test (which I am looking at).

For case 2 I certainly get a mixed state. I did this approach:
In case we measure a_1 the wave function collapses into:
$$|\Psi_1> = c_1 ~(A_{11} |a_1~b_1> + A_{12} |a_1~b_2>)$$
(c for proper norm) with probability
$$p_{a1} = A_{11}^2 + A_{12}^2$$
and the analog for measurement yielding a_2:
$$| \Psi_2> = c_2 ~(A_{21} |a_2~b_1> + A_{22} |a_2~b_2>)$$
(c for proper norm) with probability
$$p_{a1} = A_{21}^2 + A_{12}^2$$

But how about this question: What does the density matrix in the state space for a look like?
Would it simply be
$$\rho_a = |a_1>a_{11}<a_1| ~+~ |a_1>a_{12}<a_2| ~+~ |a_2>a_{21}<a_1| ~+~|a_2>a_{22}<a_2|$$
where $$a_{11} = <\Psi | a_1><a_1| \Psi>$$ (and so on) ?

4. Jun 23, 2005

### George Jones

Staff Emeritus
Yes, I think so. A little care is needed with the notation, though. For example, $$<a_1| \Psi>$$ is not a scalar, it is ket in the state space for the second particle. Consequently, $$<\Psi | a_1>$$ is a bra for the second particle, and $$<\Psi | a_1><a_1| \Psi>$$, as the inner product of a bra and a ket, is a scalar.

$$\rho_a$$ is what results when the the $$b$$ states are "traced out".

Regards,
George

5. Jun 23, 2005

### vanesch

Staff Emeritus
Yes, that's correct. In fact, the local density matrix $$\rho_a$$ is always the initial (pure state) density matrix $$\rho$$ with b traced out, whether or not a measurement on b is performed. But if no measurement is performed, it is called an "improper" density matrix, while if a measurement is performed (Born rule and projection), it is a "proper" density matrix.
All observables only related to a have expectation value $$Tr(\rho_a A)$$.
So for an observable only related to a, the expectation value is independent of whether, or what, measurement has been performed on b.

cheers,
Patrick.