# Preparation of Superposition

1. Mar 23, 2015

### Edward Wij

What is the difference in the preparation of superposition for the 2 cases where:

1. The system is in either one state.. for example.. one electron in spin up or spin down. Do these interfere?
2. The system is in superposition of two states (for example 2 entangled electrons that is spin up and spin down).

Are they both called Superposition? What specific term distinguish them?

2. Mar 23, 2015

### Staff: Mentor

No.

All superposition does is reflect the vector space structure of pure states:
http://www.math.ucla.edu/~tao/resource/general/121.1.00s/vector_axioms.html

Any state is the superposition of tons of other states.

Thanks
Bill

3. Mar 23, 2015

### Edward Wij

Do you generally agree that mixed states occurred because the measuring device is entangled with the system breaking the pure state.. or is it possible for mixed state to occur without entangling with the measurement device or environment? Any example?

4. Mar 23, 2015

### vanhees71

First of all you have to specify, what you mean by "superposition". You have to choose a basis to do so in order to say, a pure state is represented either by a basis vector or is a super position (linear combination) of several such states. Physically a basis in Hilbert space is defined as the complete set of eigenvectors of a self-adjoint operator, representing an observable. More precisely, a pure state (which represents complete knowledge about the system's state within quantum theory) is represented by a ray in Hilbert space (i.e., a non-zero Hilbert-space vector up to a non-zero factor).

Further, a mixture (mixed state) is something different. It represents the description of a system, for which we have not complete knowledge about its state. It is represented by a Statistical operator, which is a positive semidefinite self-adjoint operator with trace 1. Also a pure state can be represented by such an operator, and thus this is the most general (and also most correct) representation of a general state. A state in this sense is pure if and only if the statistical operator is a projection operator, i.e., of the form
$$\hat{R}=\hat{P}_{\psi}=|\psi \rangle \langle \psi|,$$
where $|\psi \rangle$ is a normalized Hilbert-space vector.

5. Mar 23, 2015

### Staff: Mentor

Well its not really on topic but what the heck.

Generally in practice mixed states are from entanglement, but that's not the only way.

You can also do it by presenting states to be randomly observed - such mixed states are called proper - otherwise they are improper.

It is the essence of the modern view of the measurement problem that you cant tell the difference between the two.

Thanks
Bill

6. Mar 23, 2015

### Edward Wij

Can you give an example where in mixed state.. it is there prior to observation and there is no collapse. Looking at your message archives.. I read you kept saying that. I'd like to know if this your only position or others too? In the stern-gerlach spin equipment, either y or z spin can form.. this means it is not in that state before measurement, this seems to conflict with what you said that in mixed state it is in definite state between measurement (no collapse).. can you please elaborate what the distinctions using the stern-gerlach experiments.. thanks in advanced..

7. Mar 23, 2015

### Staff: Mentor

I gave it above.

Thanks
Bill

8. Mar 23, 2015

### vanhees71

The most common mixed state is given by the canonical statistical operator of a many-body system
$$\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
It describes the many-body system at temperature $T=k_{\text{B}}/\beta$.

9. Mar 23, 2015

### Edward Wij

I'd like to know if it is just your belief that in proper mixture.. you present states to be randomly observed - hence it is there prior to observation and no need for collapse. For mainstream physicists.. do they also hold this belief? In pure Copenhagen, the spins in stern-gerlach doesn't have definite state before measurements.. you seemed to believe it is because in the final outcome, it is classical.. so you believe it is classical all the way to the quantum state. (Vanhees71, do you also believe that in proper mixture, there is no collapse and the state is there prior to observations, how about others?)

10. Mar 23, 2015

### vanhees71

In my opinion, there is no such thing as collapse at all. For me, the only consistent interpretation of quantum theory which is compatible with Einstein causality is the minimal statistical interpretation.

11. Mar 23, 2015

### Edward Wij

Oh. So both you and Bhobba follow the minimal statistical interpretation. How about those who do not? Or to be interpretation free. Can you give an actual experiment or example where the mixed state is there prior to observation and no collapse? Because Im not sure what experiment you two are thinking about.

12. Mar 23, 2015

### Staff: Mentor

I gave it bfore. What exactly dont you get about states presented randomly for observation?

And I agree entirely with Vanhees. There is no collapse in the axioms of QM - its simply something some interpretations introduce.

Thanks
Bill

13. Mar 23, 2015

### Edward Wij

Let's use the stern-gerlach experiment, there it is shown that the spins of electron don't have definite values before measurements. So how do you apply the mixed state in the stern-gerlach?

14. Mar 23, 2015

### Staff: Mentor

Its a definition - you cant really argue with a definition.

You are getting a bit confused about the whole issue. Think of it this way - the minimum statistical interpretation is consistent with a number of interpretations - that's because its minimal. Now one interpretation it's consistent with is BM which has no collapse. Another is MW. An interpretation that has explicit collapse is GRW - its consistent with that. Copenhagen takes a different view - the state is entirely subjective - it has explicit collapse but since it is entirely subjective it matters not.

Thanks
Bill

15. Mar 23, 2015

### Staff: Mentor

Its entangled with the apparatus - you yourself mentioned entanglement creating mixed states.

Its not a proper mixed state - or rather the QM formalism doesn't say it must be, but in some interpretations it is.

Thanks
Bill

Last edited: Mar 23, 2015
16. Mar 23, 2015

### Edward Wij

Ok. My primary question is without the system entangling with the apparatus (or environment) and the system is in pure state, would it be possible to have mixed state inside the pure state by measuring the subsystem of the pure state?

17. Mar 23, 2015

### Staff: Mentor

I have no idea what you are trying to say.

Thanks
Bill

18. Mar 23, 2015

### Edward Wij

Let's say particle A is entangled with particle B and particle C. If you just measure particle A and B (subsystem of A, B, C). Can A and B form mixed state?

19. Mar 23, 2015

### Staff: Mentor

Yes. But I fail to see your issue/point etc.

My suspicion is you are angling for some way a purely quantum process to create a proper mixed state. It can't.

Can you please be more explicit about what exactly you are getting at. If you find that difficult can I suggest you think about it more carefully before posting?

Thanks
Bill

20. Mar 23, 2015

### Edward Wij

Oh. I made a mistake. Let me rephrase it. Let's say particle A is entangled with particle B and particle C. If you just consider (Not Measure) particle A and B (subsystem of A, B, C). Can A and B form mixed state? Without measurement, can A and B as subsystem of A,B,C for mixed state? Please reply again on this exact context (to emphasize the truth).