# What state do we have if we don't know if a measurement was performed?

1. Mar 17, 2014

### greypilgrim

Hi,

My question is partially motivated by this discussion. Suppose we have a pure state that is in a superposition for a certain basis, e.g. $\left|\psi\right\rangle=\alpha\left| 0\right\rangle+\beta\left| 1\right\rangle$. Now a colleague of ours measures the state with respect to that basis, but does not tell us the result. From his point of view, he ends up with a pure state (say $\left| 0\right\rangle$), from our point of view we have a mixed state $\rho=\left|\alpha\right|^2 \left| 0\right\rangle\left\langle 0\right|+\left|\beta\right|^2 \left| 1\right\rangle\left\langle 1\right|$.

But what kind of state do we end up with if our colleague doesn't even tell us if he performed the measurement at all?
Or, as a follow-up question, if he doesn't tell us in which basis he measured?

2. Mar 17, 2014

### atyy

Your colleague becomes entangled with the state, but you don't know what state, so you trace him out, and get the reduced density matrix, which will be an improper mixed state.

You have to know that he is interacting with the system, otherwise you don't know the time evolution of the state, since the state is no longer evolving freely.

Last edited: Mar 18, 2014
3. Mar 18, 2014

### Staff: Mentor

This is interpretation dependant. But for most interpretations, including bog standard Copenhagen or ensemble, its |0> once its observed to be that, and is that objectively out there independent of observation for everyone. Remember most interpretations assume a common sense world out there that observations appear in.

Thanks
Bill

4. Mar 18, 2014

### naima

If you have many copies of the qbit, you can know if your colleague have done a measurement.
if there was no measurement there is a direction along which the spin measurement always give the same outcome.
there is a preferred basis where the density matrix has only one non null element.
If yes there is a preferred basis where the density matrix is diagonal with more non null elements.

Decorehence leads to a preferred basis (e g after many scatterings of photons on dust.)
dust is first in a pure superposition of position (in a preferred basis).
Can we think that after the first scattering it is in another basis, in another one after the second and that there is convergence to the preferred basis that emerge from decoherence?

5. Mar 18, 2014

### Demystifier

Let p be the probability that measurement has not been performed. Then the state is
$p|\psi><\psi| + (1-p)\rho$
where $\rho$ is your density above.

More generally, if you know the probabilities $p_i$ for measurements in different bases $i$, then the state is
$\sum_i p_i \rho_i$
where $\rho_i$ is the state when you know that the measurement is performed in the basis $i$.

6. Mar 18, 2014

### Jazzdude

If you know the basis he measured you can even say more than that. The density operator you suggest does not contain the necessary information to deduce the states that make up the ensemble. Like in classical physics, the more accurate description is a simple list of states with their associated probabilities. So simply $\{ \left( \left| 0 \right\rangle , p_0=\left|\alpha\right|^2 \right) , \left( \left| 1 \right\rangle , p_1=\left|\beta\right|^2 \right) \}$ expresses all your information about the resulting state.

Cheers,

Jazz

7. Mar 18, 2014

### greypilgrim

@Jazz:

I see what you mean, but are you sure this list contains more information than the density operator? As far as I can see there is no measurement that could distinguish them.

8. Mar 18, 2014

### Jazzdude

Yes, the list contains more information. Because many different lists can generate the same density operator. What you are asking is rather, if this information is relevant to physics.

It is conceptually a big difference if you have a single pure state after a measurement or just some state that is indistinguishable from an ensemble of pure states by measurement. The fact that we cannot by measurement dissect a single instance of a quantum state doesn't mean we should not have the concept of such a state. If you believe in quantum preparation then you should make that distinction. If you believe that density operators are more fundamental than pure states then don't make it.

Cheers,

Jazz

9. Mar 18, 2014

### greypilgrim

But is the knowledge about the measurement basis really a piece of information that we should relate to the state so closely? It's classical information that our colleague has told us, maybe a long time before he actually performed the measurement.

10. Mar 20, 2014

### naima

Suppose you get two sets of atoms one with knowing the density matrix and the other knowing the list. Can it help you to imagine an experiment with different results. If no what is this information you are talking about?

11. Mar 20, 2014

### Jazzdude

Do you believe that you can prepare a pure state, by post-selecting a measurement for example?

Cheers,

Jazz

12. Mar 21, 2014

### naima

Not absorbed particles which go through the two slits are selected and give a pure state.
You said ther is aloss of information on the density matrix.

13. Mar 21, 2014

### Jazzdude

I mean in general, not just for this experiment. And I had to ask if you believe that, because the answer and argument depends a lot on it. There are so many ways to think about quantum theory as you know, and the way you argue reflects in that.

So if you believe that a pure state can be prepared, i.e. you know that the system is in a pure state after preparation, then you must accept that as a logical consequence that you know more than what the density matrix encodes about an ensemble that you prepare by randomly selecting from pure states.

It's true that you cannot measure the difference with a Born-measurement. But that may just be a limitation of measurement and not one of the system you describe. Quantum measurement is notoriously difficult to grasp, and we surely don't make our live easier by being imprecise about something we can be very precise about.

If you believe that density states are more fundamental than pure states then this may all seem unnecessary to you. But it doesn't seem so to me. I know what I prepared.

Here's an example how this can be of relevance: I prepare an ensemble by selecting two different pure states, throwing a die and picking one of the two based on the result. Two of my friends now receive this state together with a description. I tell one that the state is a density matrix and the other I give the exact state list with probabilities.

At some point I realise that the die has not been perfectly fair and I tell them that the real probabilities are actually a bit different. The friend with the list can fix that easily and arrives at a new correct state description. The other friend looks at his density matrix and has no clue what he should fix, because he doesn't know how the density matrix decomposes. He doesn't even know if the two states in the ensemble are orthogonal or not.

Cheers,

Jazz

14. Mar 21, 2014

### Staff: Mentor

It is a well known property of mixed states that they cant be uniquely decomposed into convex sums of pure states.

Of course if you tell the colleague the way it was prepared he has more information than the state by itself tells us.

I am scratching my head what the issue is here.

States simply do not tell us everything about its preparation. As Ballentine says they are the equivalence classes of preparation procedures that are observationally equivalent. Randomly selected pure states give mixed states that do not necessarily tell us the states it was randomly selected from.

I think that's the point Jazzdude was, correctly, trying to make.

Thanks
Bill

Last edited: Mar 21, 2014
15. Mar 22, 2014

### naima

I agree with this relation of equivalence.
Relative numbers can be defined as equivalent sets of ordered integers:
-1 = (0,1) = (1,1) = (2,3) ...
There is not more information in (2,3) than in -1. This is useless information.
with a Jazzdude list you have a useless information about a phase. Density matrix is the good tool to erase this useless information.
I was interested in what Jazzdude said about the possiblenon orthogonality of the states in his list.
look at this paper
there are all the ingredients on this topic.

16. Mar 22, 2014

### greypilgrim

Well then you could attribute a lot more information about the preparation procedure to the state, like time and place of its creation or the experimental setup that was used. In fact all states would be different even if you used the exactly same preparation procedure simply because you cannot create two states at the exact same point in time.

Or you could say an electron extracted from a hydrogen atom was different from an electron extracted from a helium atom. However, as is confirmed by experiments, they are in fact indistinguishable.

The density operator is the simplest COMPLETE description of the state, two states are experimentally (in general statistically) distinguishable if and only if their density operator is different. I'm not saying the information about the basis that was used for its creation is worth nothing, I'm just saying that this information is not more important to the state than time an place of its creation, the experimental setup, the name of the preparator, his favorite football team or the weather outside at the day of creation. Hence I see no reason why we should attribute it to the state as {density operator,creation basis}=your list.

17. Mar 22, 2014

### Staff: Mentor

By definition a state is a positive operator of unit trace.

Thanks
Bill

18. Mar 22, 2014

### greypilgrim

I think you're splitting hairs here, but okay, a quantum state is not described by a density operator, a quantum state IS a density operator. But this only reinforces my notion that all other information we have about the preparation of the state, be it the specific basis used, the experimental setup, name or hair color of the experimentalist or phase of the moon at the time of preparation is additional, classical information that's nice to know but doesn't relate directly to the state and has no influence on whatever experiment we perform on the state.

19. Mar 22, 2014

### naima

We saw that density matrix is the generalization of pure states. They are peculiar ones.
Decoherence tells us that a preferred basis emerges (a set of orthogonal pure vector states).
Is there a reason why nature would prefer these pure states?

20. Mar 22, 2014

### atyy

It's a matter of taste whether the density matrix is a generalization of pure states. In dBB, quantum mechanics is only an effective theory, so it is natural that mixed states and POVMs have a fundamental status. In many-worlds, the pure state is fundamental, and mixed states are derived. As long as there is a chance that quantum mechanics is complete, then pure states can be considered fundamental, and mixed states to be derived.