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In principle, it is possible to entangle any property of two particles, including speed and momentum. Surely then, this could be used to defy the Uncertainty Principle, which states that the momentum and position of a particle cannot be together measured with accuracy, but the more accurate one gets, the less accurate the other becomes:

$\Delta x\Delta p=\dfrac{\hbar}{2}$.

If one was to set up an experiment where two particles are entangled, one could then measure the position of one particle with great accurary at the same time as measuring the momentum of the other particle with great accuracy, which would mean that, as the two particles are entangled, one would then know the momentum of the particle they measured the position of.

Therefore, the Uncertainty Principle would have been defied, as both momentum and position have been measured with a high accuracy.

However, this can't be true, as someone would have thought of this before.

So why wouldn't this work?

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6 Answers 6

It wouldn't work simply because there doesn't exist any entangled – or unentangled – state that would violate the uncertainty principle. The inequality $$\Delta x_1\cdot \Delta p_1 \geq \frac\hbar 2$$ is a universal law of physics (which may be easily proved as a mathematical theorem if one assumes that the quantities are described by the maths of quantum mechanics) and it holds regardless of values or uncertainty or properties of $x_2,p_2$ (properties of another particle) or their entanglement with $x_1,p_1$ (subscripts $1,2$ label the two particles here).

So you can prepare a state in which $x_1$ is sharply determined and $p_2$ is sharply determined – they commute with each other – but because $x_1$ is sharply determined, it follows that $p_1$ is completely uncertain and similarly $x_2$ is completely uncertain, so the two particles just can't have the same positions or the same momenta. They're not entangled in the usual sense.

Alternatively, you may prepare entangled states in which the two particles always have the same position $$ |\psi\rangle = \int_{-\infty}^{+\infty} dx \, f(x)\ |x\rangle \otimes |x+a\rangle $$ where the first tensor factor describes the first particle and the second tensor factor describes the other particle (I added a variable shift $a$ so that the particles don't overlap if you don't want to). But for a particle in this state, we can prove that the uncertainty $\Delta p_2$ is equal to the uncertainty $\Delta p_1$ simply because the state is totally symmetric with respect to the exchange of the two particles so the calculations of $\Delta p_2$ and $\Delta p_1$ are manifestly the same and yield the same results.

Because $\Delta p_1\geq \hbar / 2(\Delta x_1)$ by the uncertainty principle, it follows that – in this state – $$\Delta p_2 \geq \frac{\hbar}{2\cdot \Delta x_1} $$ as well. The momentum of the second particle is greater than the usual multiple of the inverse uncertainty of the position of the first particle. This inequality doesn't hold for a general state because the two particles have nothing to do with each other. But if you impose an additional condition that the particles are entangled, then their properties are really the same and one may prove additional inequalities.

Your problem is that you seem to think that you are free to impose an arbitrary collection of conditions on the state vector and it still exists. But you're not free to do so. In other words, if you do so, your condition have no solution. Such states can't exist in Nature. They're mathematically impossible.

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"[S]o the two particles just can't have the same positions or the same locations." Is that what you intended to write? –  Glen The Udderboat Mar 13 '13 at 21:10
    
Perhaps you did. I didn't read into the question that entanglement was understood to be "stuck together" (spatially). –  Glen The Udderboat Mar 13 '13 at 21:38
    
Gugg, nope, you caught a typo. One of the words should have been momenta. Fixed. Otherwise entanglement may correlate any quantities but the OP used the entanglement of particular quantities such as positions and momenta and momenta that identifies the corresponding properties of both particles - and this entanglement does stick the particles together. –  Luboš Motl Mar 14 '13 at 6:23

This has been the main point of argument between Bohr and Einstein until the end of the lives of these great men.

I am pleased to see this question has being asked here. It sounds like a well planned conspiracy, which I have though myself a few decades ago, and I am sure many other people must have wondered about this. The answer I have given to this question goes something like this.

The whole notion of entanglement is based on the idea (observed fact) that the two entangled particles, A and B say, are subjected to quantum mechanical behaviour which is determined by a single wave function, the “wave function of the entanglement.” This means that any two mechanically conjugate quantities $Q_A$ and $P_B$ belonging to the particle A and particle B respectively are subjected to the uncertainty principle.

Imagine as you suggest that you measured the position $x_a$ of particle A. This measurement alone introduces an uncertainty, $\Delta p_a$, to the momentum of particle A, nothing strange about this. Whether you measure the momentum of A or not, this uncertainty is there. This means that the momentum of particle B, $p_b=p_T-p_a$ is totally undetermined because $p_a$ is, and this will reflect in your measurement of the momentum of particle B.

CONCLUSSION:

As long as the two particles are entangled and described by a single wave function, it is impossible to do an experiment on two pieces of the system through which we can violate the uncertainty principle. It is one thing to come up with various scenarios, but is another to actually do the experiment to find out what is actually happening.

NATURE TELLS US OTHERWISE!!

I must say I was disappointed when I reached this conclusion!! I hope I am wrong!

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Historically, I have my doubts with regards to the first sentence. First of all, Einstein died in 1955 and Bohr died in 1962. But leaving that detail aside, I haven't found out whether the argument between them continued until 1955. I actually thought their argument ended (in non-agreement) in 1935 or 1936 with Bohr's response to the EPR paper. –  Glen The Udderboat Mar 13 '13 at 22:32
    
@Gugg Thanks for your kind comment. I did not say they died at the same time. The sentense was aimed at simply stretching the depth of their beliefs and their differences. It is not related to the chronology and outcome of their debate, but thanks for bringing this into our attention. –  JKL Mar 13 '13 at 22:51

Actually, you can use entanglement between two detectors to violate the uncertainty principle in the noise-disturbance formulation, see my contribution Phys. Rev. Lett. 110, 120403 (2013)/ http://arxiv.org/abs/1212.2815v3 . You cannot violate the Kennard inequality, however.

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Hi Antonio, And welcome to Physics SE. Would you mind elaborating more information about how the violation works? –  Ali Aug 26 '13 at 8:44

Position and momentum are canonically conjugate variables, you either be in the momentum space or in the position space. In order to do what you want I have to do a direct product of hilbert space of two particles one being in position space and other in momentum space. And not only that you want entanglement. So for example you could at most have states like,

$|\psi(x_1,k_1,x_2,k_2)\rangle=|x_1\rangle\otimes|k_1\rangle+|x_2\rangle\otimes|k_2\rangle$ $(x_1,x_2)$(position's of 1st particle) and $(k_1,k_2)$(momentum's of 2nd particle)

Which can make sense since you are entangling the space degree of freedom of one particle with momentum degree of freedom of another particle. So by doing a position measurement(x) on 1st particle you will have exact momentum information(k) of 2nd particle. And now if your system of two particles have momentum conservation i.e. ($\vec{k}+\vec{k'}$=conserved) it seems I should be able to figure out momentum of first particle. And then my friend we would be boiling down to the EPR paradox simultaneous measurement of non-commuting variables. That is measurement of position & momentum simultaneously. Check this links out http://www.drchinese.com/David/EPR.pdf http://en.wikipedia.org/wiki/EPR_paradox

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Consider two entangled quantum systems which are described by three properties, A, B and C each which can take a value of up or down.

The two systems can be entangled such that if you measure the same property on both systems then you will get the same result 100% of the time, and each case occurs 50% of the time. That is to say both systems give up 50% of the time or both systems give down 50% of the time, but you will never get a result where one systems gives up and the other down. Just to reiterate is is only when you measure the same property, ie: both A, both B or both C.

But here is the problem if you measure different properties then you will get the same direction 25% of the time and opposite directions 75% of the time. For example if I measure A on one system and it gives a result of up, then when I measure B on the other system I will get up 25% of the time and down 75% of the time.

So there are two ways to make sense of this, either all properties have values and the first measurement sends an instantaneous signal to the other system to randomly change its value accordingly or you have to accept that the properties don't have values until you make the measurement.

So basicly when you are measuring two different properties of two different (albeit entangled) systems you are not in general allowed to infer that the systems also had the same value of the corresponding property.

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Lubos's answer is excellent, but a simpler answer is that after you measure the particles, they're not entangled anymore. If the first is known (after a measurement) to be in some state A and the second is known to be in some state B, then (quite regardless of whether you measured the same or different observables), the pair is in state $A\otimes B$, which is not entangled.

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