# A basic question

#### cdevarennes

I have a question that is probably very basic, but I simply don't get it yet.

Quantum entanglement, simply put, is a correspondence between particles or systems of particles such that taking a measurement of one will predict the outcome of a similar measurement of the other.

The Heisenberg Uncertainty Principle states that there are variables associated with particles or systems of particles such that certain knowledge of one variable precludes the same degree of knowledge of the other - if you know the spin of a particle along the Z axis, you can't then know the spin of that same particle along the X axis.

What I was wondering is this - given a pair of entangled particles, measuring an arbitrary variable "A" on particle number 1 sets the value for that variable on particle number 2...so what happens when you then attempt to measure variable "B" on particle number 2 when variables "A" and "B" are incompatible observables? What result would such a measurement produce?

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#### DrChinese

Gold Member
cdevarennes said:
I have a question that is probably very basic, but I simply don't get it yet.

Quantum entanglement, simply put, is a correspondence between particles or systems of particles such that taking a measurement of one will predict the outcome of a similar measurement of the other.

The Heisenberg Uncertainty Principle states that there are variables associated with particles or systems of particles such that certain knowledge of one variable precludes the same degree of knowledge of the other - if you know the spin of a particle along the Z axis, you can't then know the spin of that same particle along the X axis.

What I was wondering is this - given a pair of entangled particles, measuring an arbitrary variable "A" on particle number 1 sets the value for that variable on particle number 2...so what happens when you then attempt to measure variable "B" on particle number 2 when variables "A" and "B" are incompatible observables? What result would such a measurement produce?
The answer is surprisingly simple. It is no different than the situation in which you look at both observables on the same particle. If you look at A on particle 1, and then look at B on particle 1, what do you see? Each time you perform an "incompatible" observation, you lose the previous knowledge you gained. Particle 1 no longer acts as if it is in state A.

Thus, when you perform observation B on particle 2 (after performing A on particle 1), particle 2 no longer acts as if it is in state A (relative to particle 1).

Once the observation of A on particle 1 is performed, the entangled state is collapsed and the pair is no longer entangled. So don't expect subsequent B observations on particles 1 and 2 to yield the same value any more often than would be expected by chance alone.

#### cdevarennes

Hmm...and there is, axiomatically, no way to observe a single particle and determine whether or not it is entangled, without also observing the suspected object of that entanglement...crap. I'll probably have some questions on Planck's quantization of time later, but I need to chew on this a bit. Thanks for your help.

#### DrChinese

Gold Member
cdevarennes said:
Hmm...and there is, axiomatically, no way to observe a single particle and determine whether or not it is entangled, without also observing the suspected object of that entanglement...
That is an interesting point, but I am not certain if it is entirely correct. I guess part of the question is: determined relative to what?

If I recall correctly (questionable), Vanesch and/or several others convinced me that an entangled photon will not exhibit double slit interference effects. An otherwise identical photon which is not currently entangled will show such effects. I might even be able to cite a reference.

#### ZapperZ

Staff Emeritus
2018 Award
DrChinese said:
If I recall correctly (questionable), Vanesch and/or several others convinced me that an entangled photon will not exhibit double slit interference effects. An otherwise identical photon which is not currently entangled will show such effects. I might even be able to cite a reference.
I'm not sure if this is true. I believe that several (at least 2) experimental results have shown such effects, mainly to show that entanglement photons beats the diffraction limit since the sum of their energies (wavelength) is larger (smaller) than for the individual photon.[1,2] This clearly show that these entangled photons (however many that in entangled) are really just "one" object that are connected.

[1] P. Walther et al., Nature v.429, p.158 (2004).
[2] M.W. Mitchell et al., Nature v.429, p.161 (2004).
or at http://physicsweb.org/article/news/8/5/6

Zz.

#### cdevarennes

Hmm...if, in fact, either of you are correct, then an interesting situation arises. In an interpretation of the EPR paper (on Wikipedia, I think), one of the ways in which General Relativity is reconciled with the idea of quantum entanglement is that no "useful information" can actually be communicated using entanglement - any such communication of "useful information" (and don't ask about the definition of useful information - they didn't give one) would violate (or possibly violate) general relativity by exceeding the speed of light. Thing is, if the observation of one member of an entangled pair causes the entangled state to collapse, then imagine a set of 10 individually entangled pairs of particles, seperated by an arbitrary distance. Alice takes measurements on particles 1, 3, and 4 - but leaves 2, and 5-10, alone. If Bob can then examine the second set of 10 individually and determine that numbers 2 and 5-10 are all still entangled, then pass those results through a simple NOT gate, he gets the binary number 1011, or 13...useful information if a context is defined.

Hmmm....then it simply becomes a matter of maintaining the entanglement of particles for long enough periods of time to be useful, and Alice can communicate a message to Bob in binary.

#### vanesch

Staff Emeritus
Gold Member
DrChinese said:
If I recall correctly (questionable), Vanesch and/or several others convinced me that an entangled photon will not exhibit double slit interference effects. An otherwise identical photon which is not currently entangled will show such effects. I might even be able to cite a reference.
Vanesch will then try to explain why :tongue2:

your single entangled photon appears (locally) to be in a mixture ! So "an otherwise identical photon" would have to appear in the same mixture, which would then also NOT give rise to interference effects, in a trivial way (for instance, it would be white light).
The mixture is described exactly by the reduced density matrix of the single photon you're looking at, and whether this reduced density matrix is an "improper" one (it is a *reduced* density matrix, deduced from a bigger one, of the entangled pair), or a proper density matrix (it is a "true" statistical density matrix) doesn't matter. A photon in a statistical state given by that density matrix doesn't show interference.

cheers,
Patrick.

#### vanesch

Staff Emeritus
Gold Member
ZapperZ said:
mainly to show that entanglement photons beats the diffraction limit since the sum of their energies (wavelength) is larger (smaller) than for the individual photon.[1,2] This clearly show that these entangled photons (however many that in entangled) are really just "one" object that are connected.
But that's using BOTH photons, no ?
DrChinese is talking about using only ONE of a pair of entangled photons, not caring about the second one.

cheers,
patrick.

#### ZapperZ

Staff Emeritus
2018 Award
vanesch said:
But that's using BOTH photons, no ?
DrChinese is talking about using only ONE of a pair of entangled photons, not caring about the second one.

cheers,
patrick.
Oh....

Yes, in the above, they were using ALL the photons (not both since one has 3 entangled photons while the other has 4).

Zz.

#### vanesch

Staff Emeritus
Gold Member
cdevarennes said:
Thing is, if the observation of one member of an entangled pair causes the entangled state to collapse...
Yes, but you cannot detect that. The local reduced density matrix DOESN'T CHANGE whether you performed a measurement or not on the OTHER part of the entangled pair (you can say that the local density matrix is invariant under a collapse induced by a measurement on the OTHER part).
Collapse isn't necessary, btw, otherwise things such as MWI wouldn't make sense, and they do.
I worked this out in detail in a post here some while ago, should do a search on it to find it again...

cheers,
Patrick.

#### cdevarennes

So what you are saying is that there is no way to observationally determine if a particle is entangled without information regarding the particle you suspect it to be entangled with?

#### DrChinese

Gold Member
cdevarennes said:
Thing is, if the observation of one member of an entangled pair causes the entangled state to collapse, then imagine a set of 10 individually entangled pairs of particles, seperated by an arbitrary distance. Alice takes measurements on particles 1, 3, and 4 - but leaves 2, and 5-10, alone. If Bob can then examine the second set of 10 individually and determine that numbers 2 and 5-10 are all still entangled, then pass those results through a simple NOT gate, he gets the binary number 1011, or 13...useful information if a context is defined.

Hmmm....then it simply becomes a matter of maintaining the entanglement of particles for long enough periods of time to be useful, and Alice can communicate a message to Bob in binary.
The method I described allows you to distinguish between an entangled photon (one of a pair) versus an unentangled one (never part of a pair). In my example, Alice does not get a chance to alter the outcome and therefore cannot send a message to Bob.

In considering the entangled pair of particles: as far as anyone knows, there is no such thing as one being unentangled *before* the other. Regardless of whether you observe one at T=1 and the other at T=5000, there is no experimentally observable difference when considering the order of the observations. Eventually you realize that Alice's actions can be no more meaningful than Bob's.

#### DrChinese

Gold Member
cdevarennes said:
So what you are saying is that there is no way to observationally determine if a particle is entangled without information regarding the particle you suspect it to be entangled with?
As I mentioned earlier, you have to ask relative to what, as there can be a difference in some scenarios. If you specify a particular hypothetical setup, I think its ability to send FTL signals can be shown to be non-existent.

Not that you shouldn't try, mind you. I always seem to be right on the edge of figuring FTL signalling out... Oh, and then ZapperZ and Vanesch come along LOL.

#### vanesch

Staff Emeritus
Gold Member
cdevarennes said:
So what you are saying is that there is no way to observationally determine if a particle is entangled without information regarding the particle you suspect it to be entangled with?
Yes, that's what I'm saying.

cheers,
Patrick.

EDIT: more specific: for each thinkable entangled state of 2 systems A and B, there is an EQUIVALENT statistical state for system A alone, which will give IDENTICAL measurement results for measurements only performed on A.

That equivalent statistical state for system A alone is given by the reduced density matrix rho_A calculated from the entangled state of A and B.

What's even more spectacular: this rho_A doesn't change when you perform a measurement on B (even no matter what measurement), so A "doesn't feel" the measurement on B in the case it was part of an entangled pair.

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#### cdevarennes

Vanesh,

That makes sense - for whatever reason (brain fart?) rho_A didn't enter my thought processes, despite the fact that I actually did know that it had to exist. *shrug* I'm going to keep pondering, and teaching myself the math needed to really get into this...and I'll fire any more deep thoughts out this way to see what shakes loose. Thank you (and DrChinese, and ZapperZ) for y'alls input.

Chris
(Professional computer geek, and gifted ameteur at everything else)

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