Distinguishing between superposition and hidden information

[michael879]

ok I have this question and I am having some trouble putting it into words. Heres an example.

Lets say there are particles X and Y that can be in states A or B. X is in a superposition between state A and state B. Y has a 50% chance of being in state A and a 50% chance of being in state B. However Y is NOT in a superposition. Assuming you can manipulate the particles in any way possible in the future, and you can create as many X and Y like particles as you want, is there anyway to tell the difference between an X and Y particle?

 

[jostpuur]

A quick guess from me is that you should be able to tell the difference by using some interference phenomenon. For example, if A and B are eigenstates of the energy first, but you can change the system so that the hamiltonian operator changes, it will cause some oscillation between A and B. The probabilities for particles to be in A or B will depend on interferences, that will instead depend on whether there is superposition or not in the first place.

 

[michael879]

ok, then answer this. If its possible to tell the difference between particles X and Y, then isn't it possible to communicate faster than the speed of light using entanglement?? Looking at X causes it to collapse into Y. Therefore if you can distinguish between X and Y you can tell when someone far away has looked at an entangled particle.

 

[meopemuk]

Lets say there are particles X and Y... Looking at X causes it to collapse into Y.

X and Y are two different particles. They cannot "collapse" into each other. You need to formulate your question more precisely. At this point I don't understand what you are asking.

 

[michael879]

X and Y are two different particles. They cannot "collapse" into each other. You need to formulate your question more precisely. At this point I don't understand what you are asking.

X and Y are the same particle. That was a unclear in my original post. X and Y are two different states of the same particle. X particles are in a superposition between A and B, while Y particles are either A or B. Y is what you get if you observe X.

|X> = 1/sqrt(2) (|A> + |B>)

|Y> = |A>
or
|Y> = |B>

 

[meopemuk]

X and Y are the same particle. That was a unclear in my original post. X and Y are two different states of the same particle. X particles are in a superposition between A and B, while Y particles are either A or B. Y is what you get if you observe X.

|X> = 1/sqrt(2) (|A> + |B>)

|Y> = |A>
or
|Y> = |B>

In quantum mechanics, your state Y is called "mixed state", while the state X is called "pure state". In order to explain the difference, it is convenient to use the "density matrix" representation of states which has the advantage of being general enough to cover both pure and mixed states. The more usual wavefunction representation of states is applicable only to pure states.

In the "density matrix" representation one associates with each state a positive definite Hermitean matrix $\rho$, such that

$$Tr(\rho) = 1$$

Then the expectation value of each operator $F$ in this state is given by formula

$$\langle F \rangle = Tr(F \rho)$$...(1)

Assuming that vectors $| A \rangle$ and $| B \rangle$ are orthogonal, the density matrices corresponding to your two states can be written as

$$\rho_X = \left[ \begin{array}{cc} 1/2 & 1/2 \\ 1/2 & 1/2 \\ \end{array} \right]$$...(2)

and

$$\rho_Y = \left[ \begin{array}{cc} 1/2 & 0 \\ 0 & 1/2 \\ \end{array} \right]$$...(3)

respectively.

Note that the density matrix corresponding to the pure state vector $|X \rangle = 1/ \sqrt{2} (|A \rangle + |B \rangle)$ must be the projection operator on this vector. It is easy to see that eq. (2) satisfies this requirement. In particular, the characteristic property of projection $\rho_X^2 = \rho_X$ is valid.

Note also that diagonal elements 1/2 in the mixed state matrix (3) are simply probabilities of finding the pure states $| A \rangle$ and $| B \rangle$ in this mixture.

So, your two states are quite different, and one can find an observable, which has different expectation values (see formula (1)) in these states. Apparently, the operator of this observable should be non-diagonal in the basis $| A \rangle , | B \rangle$ in order to see the difference.

Eugene.

 

[michael879]

I know these states are different, that wasnt my question... I was asking if its possible to tell the difference between these two states physically. Because if it is, it would seem like information COULD travel faster than light through entanglement. If you entangle two X state particles and then view one of them, the other one moves to the Y state. If it is possible to distinguish between X and Y, two entangled X particles moved far enough apart could send information at faster than c. right?

 

[meopemuk]

I know these states are different, that wasnt my question... I was asking if its possible to tell the difference between these two states physically.

Pick any observable which is not diagonal in the basis $|A \rangle, |B \rangle$ (e.g., the observable of position). The expectation values of this observable are different for states X and Y. So, by measuring these expectation values you will be able to distinguish the two states physically.

Eugene.

 

[michael879]

ok, so if you can distinguish between X and Y, how can it not be possible to send information instantly with entanglement?? You have two particles in state X, and you measure one. The other particle instantly switches to state Y, and you can tell its switched to state Y. You could create a simple faster than light morse code with this couldn't u?

 

[meopemuk]

ok, so if you can distinguish between X and Y, how can it not be possible to send information instantly with entanglement?? You have two particles in state X, and you measure one. The other particle instantly switches to state Y, and you can tell its switched to state Y. You could create a simple faster than light morse code with this couldn't u?

There are tons of information and disinformation about entanglement are related superluminality in physics journals and Internet. I prefer to stay away from this crowd as it makes me dizzy. Perhaps, somebody else can help you here.

Eugene.

 

[michael879]

ok I am just drawing a conclusion from what you told me is true.. The way I see it faster than light communication with entanglement being impossible contradicts what you said about being able to distinguish X from Y.

 

[JesseM]

If you entangle two X state particles and then view one of them, the other one moves to the Y state.

Isn't X a single-particle state? I think if you're dealing with entangled particles, then by definition you have a two-particle state, which is different than just a pair of particles in the X state. In your notation the two-particle state might be something like |Z> = 1/sqrt(2) (|A>|B> - |B>|A>)

 

[michael879]

well yea I was expanding the scenario to make entanglement possible.

 

[JesseM]

well yea I was expanding the scenario to make entanglement possible.

But my point is, for a two-particle state, where a superposition would be something like |Z> = 1/sqrt(2) (|A>|B> - |B>|A>) and a non-superposition would be something like |W> = |A>|B> or |W> = |B>|A> (measuring one of the particles with respect to properties A or B would collapse the |Z> state to the |W> state), I believe the only way to determine which state the two-particle system was in would be to perform a measurement on both particles and pool the information, so this won't be of any use for FTL communication.