Pure and not pure states, an example

In summary: The density matrix for a system in the pure state after a measurement is a_1 if the measurement yields a_1, and a_2 if the measurement yields a_2.
  • #1
jonas_nilsson
29
0
Hi there,

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

[tex] | \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> [/tex]

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
[tex] \rho = |\Psi><\Psi| [/tex]

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
[tex]\rho_1' = p_1 ~\rho ~p_1[/tex]
where p_1 the projection operator
[tex]p_1 = |a_1><a_1|[/tex]
is. After making sure that the norm is OK, I then find that
[tex] Trace(\rho_1^2) = 1 [/tex],
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.
 
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  • #2
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|.

I am not completely sure about this.

Regards,
George
 
  • #3
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:
[tex] |\Psi_1> = c_1 ~(A_{11} |a_1~b_1> + A_{12} |a_1~b_2>) [/tex]
(c for proper norm) with probability
[tex]p_{a1} = A_{11}^2 + A_{12}^2[/tex]
and the analog for measurement yielding a_2:
[tex]| \Psi_2> = c_2 ~(A_{21} |a_2~b_1> + A_{22} |a_2~b_2>) [/tex]
(c for proper norm) with probability
[tex]p_{a1} = A_{21}^2 + A_{12}^2[/tex]

But how about this question: What does the density matrix in the state space for a look like?
Would it simply be
[tex] \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| [/tex]
where [tex]a_{11} = <\Psi | a_1><a_1| \Psi> [/tex] (and so on) ?
 
  • #4
jonas_nilsson said:
But how about this question: What does the density matrix in the state space for a look like?
Would it simply be
[tex] \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| [/tex]
where [tex]a_{11} = <\Psi | a_1><a_1| \Psi> [/tex] (and so on) ?

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

[tex] \rho_a[/tex] is what results when the the [tex]b[/tex] states are "traced out".

Regards,
George
 
  • #5
George Jones said:
[tex] \rho_a[/tex] is what results when the the [tex]b[/tex] states are "traced out".

Regards,
George

Yes, that's correct. In fact, the local density matrix [tex]\rho_a[/tex] is always the initial (pure state) density matrix [tex]\rho[/tex] 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 [tex] Tr(\rho_a A)[/tex].
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.
 

1. What is the difference between a pure state and a not pure state?

A pure state is a quantum state that exists in a single, definite state with 100% certainty. This means that when a measurement is made, the result will always be the same. In contrast, a not pure state, also known as a mixed state, exists in a superposition of possible states and the outcome of a measurement is probabilistic.

2. Can you provide an example of a pure state?

One example of a pure state is a photon with a specific polarization, such as vertical or horizontal. This means that when a measurement is made, the photon will always be found in the same state.

3. How is quantum entanglement related to pure and not pure states?

Quantum entanglement is a phenomenon where two or more particles become connected in such a way that the state of one particle is dependent on the state of the other. In this case, the particles are in a pure state together, but individually they are in not pure states.

4. Can pure and not pure states be changed or altered?

Pure states cannot be changed or altered as they are defined by a single, definite state. However, not pure states can be manipulated through quantum operations, such as measurements or interactions with other particles.

5. How are pure and not pure states used in quantum computing?

In quantum computing, pure states are used to represent qubits, the basic unit of quantum information. These qubits can be in a superposition of states, allowing for multiple computations to be done simultaneously. Not pure states are also used in quantum computing, particularly in quantum error correction, where they are utilized to protect against errors in the system.

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