I am trying to understand the process of Quantum Entanglement for use in Quantum computers.

The problem I have is this: Suppose some nuclear process emits an electron-positron pair. Now after adequate separation, I measure the electron's position and the positron's momentum at time $t$ simultaneously, both with sufficiently high degrees of accuracy.

By momentum conservation, I should be able to tell both the position and momentum of the electron (or positron) at the measuring time $t$, thus violating Heisenberg's Principle.

What is wrong with this logic?

Specifically, I cannot understand how will entanglement affect momentum or position?

I am an Electronics student, not Physics so I apologize if this is too simple.

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    $\begingroup$ It seems to me that your reasoning is supposing the position and momentum of the (center of mass of) the electron-positron pair are both known initially, since uncertainty in say the momentum would hamper the possibility to relate the momentum measured for the positron to the momentum deduced for the electron (and similarly for position). But that supposition already violates the Heisenberg principle. $\endgroup$ – Marc van Leeuwen Feb 18 '15 at 5:08

Since you're an electronics student, I'll speak your language. Think of momentum and position as parameters in time and frequency domain of a signal rather than classical observable that are well defined. If you do so, you can easily realize that your frequency isn't well defined if you don't do an infinitely long measurement. This is simply due to the wave nature of variables that come from a Fourier transform.

It is not an issue of tricking the system to read a position, but rather it is more about whether what you read from an experiment makes sense as a real, reproducible and reliable physical observable.

  • $\begingroup$ Combining with @pwf 's answer, the best understanding I could get is that to get arbitrary small error product (dxdp), I cannot measure wave properties instantaneously, and as there is finite experiment duration, I determined position and momentum not at the same exact time t, and rather dx1dp1 and dx2dp2 would satisfy the uncertainty principle, which makes sense somewhat. That frequency analogy really put it into perspective. Thanks! Now why EPR is rephrased as spin makes more sense as well. $\endgroup$ – ps95 Feb 17 '15 at 21:39
  • $\begingroup$ You got it right now with two momenta and two positions. Cheers! :) $\endgroup$ – The Quantum Physicist Feb 17 '15 at 21:46

What is wrong with this logic is that you are supposing a particle has simultaneously well-defined position and momentum. This is not true - a state localized in real space is delocalized in momentum space, and vice versa. The classical conservation laws hold on the quantum level as operator laws, not as laws on the states.

  • $\begingroup$ As far as I can tell delocalisation as you put it is another way to express the Hisenberg principle. I am supposing a free electron-positron pair and if I perform the above experiment, can you explain how the momentum space of the positron is delocalized if the position space of the elcectron is localized by the experiment? What result will I get? $\endgroup$ – ps95 Feb 17 '15 at 20:57
  • $\begingroup$ @prakharsingh95: What is there to explain? If you measure something's momentum, you force it into an momentum eigenstate, which is not a position eigenstate. (This is indeed the essence of the Heisenberg uncertainty principle) $\endgroup$ – ACuriousMind Feb 17 '15 at 21:01
  • $\begingroup$ indeed that's my problem. I want to get this result, ie I cannot specify p and x simultaneously for either electron or positron. But I am measuring electron's p and positron's x, and this cross measurement is why I cannot satisfy myself with this explanation. $\endgroup$ – ps95 Feb 17 '15 at 21:08
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    $\begingroup$ @prakharsingh95: The problem is indeed that you are using classical conservation laws for a quantum situation. You are trying to force classical logic upon quantum objects - this never works well, and is not justified. You have to use the quantum version of the conservation, which is only a statement about expectation values, and it does not imply that you know the momentum of the electron exactly if you know the momentum of the position exactly. $\endgroup$ – ACuriousMind Feb 17 '15 at 21:11

Suppose the particles are initially in the (entangled) state $$A\otimes B+C\otimes D$$ where $A$ and $C$ are position eigenstates for particle 1 and $B$ and $D$ are position eigenstates for particle 2.

Note that this state is the same as $$X\otimes Y+Z\otimes W$$ where $X=(1/2)(A+C)$ and $Y=(1/2)(A-C)$ are momentum eigenstates for particle 1 and $Z=(1/2)(B+D)$ and $W=(1/2)(B-D)$ are momentum eigenstates for particle 2.

(I am using "position" and "momentum" here for arbitrary observables with eigenstates related as above.)

Now observe the first particle's position. Without loss of generality, you get $A$. Therefore the pair is now in state $$A\otimes B=A\otimes Z+A\otimes W$$

Now observe the second particle's momentum. It is either $Z$ or $W$, equiprobably. The pair is now in either state $A\otimes Z$ or $A\otimes W$.

Or, if you insist on treating the two measurements as "simultaneous", note that the initial state is also equal to $$A\otimes Y+A\otimes W+C\otimes Y+C\otimes W$$ so that a measurement of "particle 1's position and particle 2's momentum" returns $(A,Y)$, $(A,W)$, $(C,Y)$ or $(C,W)$ equiprobably.

Where's the problem?

  • $\begingroup$ thank you. I think I understand now. you seem to be good at it. can you help me please to understand this: en.wikipedia.org/wiki/EPR_paradox#Mathematical_formulation where it says "It remains only to show that Sx and Sz cannot simultaneously possess definite values in quantum mechanics." and this: " One may show in a straightforward manner that no possible vector can be an eigenvector of both matrices." and this: " More generally, one may use the fact that the operators do not commute," $\endgroup$ – Árpád Szendrei Dec 5 '16 at 18:00

I think the key point that you're missing is that as soon as you make a measurement, the entanglement between the two particles is broken. It should also be noted that the original particle also obeyed the uncertainty principle, and that at a quantum level there is no direct relationship between position, momentum and time.

Another confusing factor is that you haven't specified what is actually causing the entanglement. One possibility is that it is because the particles can be shot off in any direction, but one particle must go in the opposite direction to the other; that's both boring and difficult to think about, because the entangled part of the uncertainty in the position is perpendicular to the direction of travel. So I'm going to examine the one-dimensional case; we can create entanglement by making the energy of the original particle uncertain. In this case, we don't know what momentum the particles have relative to the center of mass, but we do know that they are equal.

If you make the position measurement (on the electron) first:

  • the uncertainty about the position of the electron becomes arbitrarily small;

  • the momentum of the electron changes, with the uncertainty becomes correspondingly large;

  • the uncertainty about the position of the positron becomes smaller, but not arbitrarily so;

  • the uncertainty about the momentum of the positron also becomes smaller (because it is correlated with the energy of the original particle, and hence with the position of the electron) but not arbitrarily small.

It may be helpful to imagine that the remaining uncertainty in the position of the positron is because of the uncertainty of the position of the original particle, though this is not entirely accurate - the uncertainty of the position of the original particle does affect the uncertainty of the position of the positron after measuring the electron position, but the relationship isn't as straightforward as that.

However, even though we haven't actually calculated exactly what the wavefunction looks like, we can guarantee that the uncertainty principle holds, simply because it is true for any wavefunction, no matter how constructed.

Also, since the entanglement between the particles is broken by the position measurement, when we then measure the momentum of the positron:

  • nothing happens to the electron or our knowledge of it;

  • our uncertainty about the momentum of the positron becomes arbitrary low;

  • the position of the positron is changed, and the uncertainty becomes correspondingly high.

If, on the other hand, we made the momentum measurement (on the positron) first:

  • the uncertainty about the momentum of the positron becomes arbitrarily small;

  • the position of the positron changes, with the uncertainty becoming correspondingly large;

  • the uncertainty about the momentum of the electron becomes smaller, but not arbitrarily so;

  • the uncertainty about the position of the electron becomes smaller (because it is correlated with the energy of the original particle, and hence with the momentum of the positron) but not arbitrarily so.

And, as before, the entanglement is broken, so when we then measure the position of the electron nothing happens to the positron or our knowledge of it.

What if you make the measurements at the same time? Well, that would be complicated to analyze, but we can cheat by calculating the results in a different frame of reference, one in which the measurements did not occur at the same time. It so happens that the results of QM never depend on the reference frame, so we can be sure that this produces the right result.

(OK, the fact that in the real world the measurements take a finite time messes this up, unless you do them sufficiently far apart. At that point you would really have to model the exact behaviour of both measurement devices to figure out exactly what happens. But the end result will be the same: the particles are no longer entangled, and the wavefunction always obeys the uncertainty principle.)

  • $\begingroup$ Initially, I had assumed that it is indeed possible to measure momentum and position to an arbitrarily small degree of precision simultaneously. Even then, I don't understand how an experiment on electron's position changes the positron's momentum. I can digest the corresponding example in the case of spin as the wave vectors are only sharing information and the wave is just collapsing to one state upon observation. $\endgroup$ – ps95 Feb 18 '15 at 4:27
  • $\begingroup$ OK, I believe this update addresses your question. You're quite right; measuring the electron's position doesn't change the positron's momentum, other than in the sense that it reduces the uncertainty by collapsing part of the wavefunction. $\endgroup$ – Harry Johnston Feb 18 '15 at 10:36
  • $\begingroup$ Suppose you use totally independent (wavefunction wise) electron and positron, you make a momentum measurement on the positron, then intuitively you should be able to exactly specify the momentum of the electron (LM conservation, with exact, not expected values) withe the same degree of error as you have in the positron measurement. As a consequence, you need to calculate the position of the electron at the exact same time (hence independent of inertial frames -relativity), to ensure that disentanglement doesn't occur (which is an instantaneous event, hence again independent of frames. $\endgroup$ – ps95 Feb 18 '15 at 13:09
  • $\begingroup$ Contd... But simultaneous p and s measurements don't make sense in any frame (you would need infinite time dilation). In the limiting case when you see the experiment from a speed of light frame, you will approach the h/4pi limit. $\endgroup$ – ps95 Feb 18 '15 at 13:11
  • $\begingroup$ Firstly, note that if the two particles have independent wavefunctions then the momentum measurement on the positron does not change the wavefunction for the electron or give us any new information about it. (This is actually what it means for the wavefunctions to be independent.) In the experiment you're talking about, even if there is no "interesting" entanglement there are still correlations between the particles, and mathematically it's the same thing. $\endgroup$ – Harry Johnston Feb 18 '15 at 18:23

The problem is that you propose to make the two measurements "at time $t$ simultaneously". Measuring the particle's momentum cannot be done instantaneously; the more precisely you want to measure it, the longer the minimum required observation time becomes. (Rougly speaking that's because knowing the particle's momentum is equivalent to measuring its frequency, but to know its frequency precisely you have to count for a significant fraction of its period. That's an effect you should be familiar with from electronics, where it shows up in signal processing. The more cycles you count for, the more precisely you'll know its frequency and hence its momentum.) But then your knowledge of the other particle's position becomes fuzzy because you're now asking for its location not at a single point in time, but over some small but nonzero time interval.

So, the longer the measurement takes, the better you may be able to know the one particle's momentum, but the worse you'll know the other's position.

  • $\begingroup$ From QM dx*dp >= h/(4*pi). So I can choose dx arbitrarily small for the positron, dp arbitrarily small for the electron, so as to violate above. For one particle, this I can understand your explanation, but I don't get this analogy for two particles. $\endgroup$ – ps95 Feb 17 '15 at 21:05
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    $\begingroup$ To choose $\Delta p$ arbitrarily small for the electron you must take some time to make the measurement. When are you then going to measure $x$ for the positron? If you measure it at the beginning of the $p$ measurement, I can say, "But that's not exactly the position I wanted - the positron must have been a little farther on when the electron had momentum $p$." If you measure $x$ at the end of the $p$ measurement, I'll say, "No, it must have been a little closer than that." So you won't be able to make $\Delta x$ arbitrarily small; it is limited by how small you want to make $\Delta p$. $\endgroup$ – pwf Feb 17 '15 at 21:15
  • $\begingroup$ So you're saying that it's impossible to simultaneously measure (experiment) both p and x (at the exact same time t because I can never synchronize)? If I repeated this experiment with all possible` t1 (start to dp measure) < dt (time after x is measure after t1) < t2 (end of dp measure)`, I would get a result close to h/4pi but never equal? $\endgroup$ – ps95 Feb 17 '15 at 21:32
  • $\begingroup$ I don't understand that last question. But the issue is not trying to synchronize the $p$ and $x$ measurements; it's trying to identify the time that the $p$ measurement belongs to. It's necessarily spread out in time, so that means the corresponding $x$ measurement is imprecise in time no matter how well synchronized to the $p$ measurement, and that means that $x$ itself is imprecise. $\endgroup$ – pwf Feb 17 '15 at 22:51

Ok. Let's suppose that the initial state of the two particle are an eigenstate of the momentum operator (momentum is well defined). Quantum mechanics tell us that the position of the center of mass is not well defined. If we measure the position of the particle 1 (electron), then we do two thing in the system:

  1. We apply a measurement in a part of the whole system (trace in the hilbert space of the second particle), this means that we don't know precisely what state actually describing the particle, we can only determine the probability of each state be the right state if we know the state of whole system (two particle).
  2. We measure the position of the particle: we "collapse" our possible momentum state in a position state.

Now let's see the another particle:

  1. We apply a measurement in a part of the whole system (trace in the hilbert space of the second particle), this means that we don't know precisely what state actually describing the particle, we can only determine the probability of each state be the right state if we know the state of whole system (two particle).
  2. We measure what momentum state is right for the second particle (positron), and consequently the first particle by conservation law.

The key point is actually in the first procedure made by both measurements. We are ignoring a part of the system. If I exchange some informations after or before the measurements we need to a better description of what actually happens. We need to look for the total state of the particles, and apply the measurements in whole state. Then we see that when we measure the position of some of the particle and the momentum of another particle the total momentum turn to be ill defined (we collapse the total state). Is not so certainly that, when we measure what momentum state is right for the second particle (positron), the momentum of the first particle is achieved by conservation law, because we don't know what happens in the first particle. We can't make assumption of the total state if we are working in a part of this state.


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