# Quantum Entanglement: what if each particle assumes a \*different state\* simultaneously?
**URL:** https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435
**Category:** Factual Questions
**Created:** [December 14, 2015, 6:52pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435 "2015-12-14T18:52:33Z")
**Posts on this page:** 6
**Page:** 2
### Author:  [@rat\_avatar](https://boards.straightdope.com/u/rat_avatar)
#### Post date: [July 15, 2016, 9:46pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/21 "2016-07-15T21:46:07Z")
> [@Half\_Man\_Half\_Wit](#):
>
> Ah, I hadn’t interpreted the OP to intend to make different measurements on both particles (nor had I noticed that this is a zombie thread). However, even in that case, there’s no problem, since the two measurements are carried out on different subspaces of the joint Hilbert space, and thus, commute trivially, and the joint measurement operator is just the tensor product of the two local measurements.
I would argue that it is not just different subspaces of the joint Hilbert space in a GR context.
Q: How does a physicist milk a cow?
A: Well, first let us consider a spherical cow…
Minkowski space or joint Hilbert space or other Euclidean (analog) models are nice in areas like special relativity because you can have nice things like invariant Lorentz transformations to preserve the spacetime interval between any two events. But they only work in physical systems over finite distances in systems without significant gravitation.
In general relativity Minkowski space is still a good description in a local reference frame, assuming you are not next to a singularity but this is only true small regions of a gravitational field. In general if you are using anything Euclidian it will only apply to very small areas or to nonexistent fictional conditions which are simplified for a thought experiment. Even if you try to expand this to London and San Francisco the tidal forces will become noticeably non-Euclidean and will not be invariant transformations.
A practical example of this effect is that Under special relativity GPS satellite fall behind ground clocks 7 microseconds per day.
Under general relativity GPS satellites gain about 45 microseconds per day due to their altitude. But even that is a simplification where GPS time is synchronized appropriately for a frame anchored to the center of the earth.
They really don’t make efforts to correct for the Sagnac effect or frame dragging. They chose to observe from the one reference frame where they can ensure simultaneous. As a ground based observer your clock would need significant corrections to see the same but you could not correct for simultaneous with all GPS satellites at once with one correction.
Here is a paper that will explain how the Galilei group (nonrelativistic quantum mechanics), or Lorentz group (nongravitational) are the only ones that work in 3+1 and greater dimensional space.
> **[Page Not Found](https://projecteuclid.org/x429.xml?aspxerrorpath=%2FjournalArticle%2FDownload)**
>
> Page Not Found
> [@](#):
>
> We prove in this paper that in an (n + l)-dimensional space-time
> (n ^ 3) the only non trivial linear kinematical groups which are compatible
> with the isotropy of space are the Lorentz and Galilei groups.
The error introduced by using spherical cows is small enough on localized experiments and you may be able to construct a test well enough to account for wayward udders in some cases. But the horns and legs start poke out and accumulate errors pretty easily as the transforms would become less and less invariant.
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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [July 15, 2016, 10:46pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/22 "2016-07-15T22:46:13Z")
> [@Leo\_Bloom](#):
>
> I’m dipping into this, yet again, but leaving lurking:
>
> You have said it’s a green card-red card world or worlds. Lots of probabilities unknown until game time.
>
> But how does probability of those two states suggest pink and violet?
>
> A 40% chance of red doesn’t mean, at one state, a 100% chance of 2/5 red, does it?
>
> Isn’t 2/5 red disallowed by the original (metaphoric) rules?
I’m honestly not sure what you’re asking, but I’ll try to answer as best I can. At each measurement, you’ll observe a definite property; however, there are different measurements you can do. One measurement has either a red card, or a green card as outcome. Another has either a yellow or blue card. You can only do one of those measurements; in the end, you’re left with a card that’s definitely the color that the measurement revealed—i.e., if you measured red/green, and got red, then any subsequent red/green measurement will likewise yield red.
If we have an equal superposition, that is, there’s a 50/50 chance of observing red versus green, then there’s also a 50/50 chance of observing blue versus yellow, or pink versus violet. However, if there is an unequal chance of observing either property, things are more complicated—besides there being some chance of observing pink/violet or violet/pink, there is also a chance of observing pink/pink or violet/violet, as the correlation will no longer be perfect.
> [@rat\_avatar](#):
>
> I would argue that it is not just different subspaces of the joint Hilbert space in a GR context.
Well, nobody really knows what happens when you combine QM and GR. However, even in quantum field theory in curved backgrounds it is the case that equal-time commutators of operators at different points in space must vanish—that’s just the criterion of relativistic causality.
There might be some nonlocality in a final theory of quantum gravity, as seems to be suggested by things like the holographic principle, but the OP’s question really does not depend on these details—the problem, such as it might be, exists even in flat space, absent any GR corrections.
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### Author:  [@WordMan](https://boards.straightdope.com/u/WordMan)
#### Post date: [July 16, 2016, 3:49pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/23 "2016-07-16T15:49:35Z")
Wow, this has stayed interesting - thanks! The fact that “simultaneous” makes no sense in a relativistic world was the flaw in my initial thinking.
Carry on!
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### Author:  [@Derleth](https://boards.straightdope.com/u/Derleth)
#### Post date: [July 16, 2016, 6:04pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/24 "2016-07-16T18:04:28Z")
> [@Leo\_Bloom](#):
>
> I’m dipping into this, yet again, but leaving lurking:
>
> You have said it’s a green card-red card world or worlds. Lots of probabilities unknown until game time.
>
> But how does probability of those two states suggest pink and violet?
>
> A 40% chance of red doesn’t mean, at one state, a 100% chance of 2/5 red, does it?
>
> Isn’t 2/5 red disallowed by the original (metaphoric) rules?
Mathematically, a measurement is represented by what’s called an operator. The state of the world is represented by a vector called a state vector. When you want to know the probability of a measurement turning out a specific way, you take the inner product of the state vector with the corresponding eigenstate of the operator and square it. The result is a real number between zero and one, inclusive; in other words, a probability.
I will now endeavor to teach you enough math to make some sense of what I just said.
First, an inner product, also called a dot product, is a measure of how much two vectors overlap. If they’re the same vector, the inner product is just the square of their length; since quantum vectors are all normalized, that length is 1. If they’re at right angles to each other, they don’t overlap at all, so the inner product is zero. High school stuff, right?
In simple terms, an operator can be written as a matrix. The essential part of matrices, from our standpoint, is that when you multiply a vector by a matrix, you can rotate the vector, resize the vector, or both. All of the matrices we’re interested in have a set of vectors called their eigenvectors which they don’t rotate, but simply resize. (_Eigen_ is a German word meaning ‘specific’, so _Eigenvector_ is just ‘specific vector’, which is a hilariously non-specific piece of terminology.) The important point is, _after you perform a quantum-mechanical measurement, the state vector must be one of the eigenvectors associated with the operator corresponding to the measurement you performed._ What’s more, quantum physicists don’t care how long a state vector is; as far as they’re concerned, all state vectors are normalized, so they have length 1.
In short, if the cards are either red or white, there’s no way to get pink. Pink isn’t an eigenvector of the operator. And, once you measure the state, the state vector is known to be one of the possible eigenvectors of that operator. Measuring it the same way again can’t change it.
What if there are two operators? What if you can measure either red-white or circle-square? Well, this depends: Do the operators commute? That is, when you multiply the matrices representing the operators, let’s say C for red-white and S for circle-square, does CS = SC?
In general, the answer is not always. Some operators commute, such as the positions on the x, y, and z axes. Some operators don’t commute, such as position (any of the position operators) and momentum (momentum in any direction). When two operators commute, they have the same eigenvectors, so measuring the position on the x axis means the quantum state isn’t going to be changed by measuring its position on either of the other axes: It’s already an eigenvector of those operators. They can’t do anything more to it. Going back to the first paragraph, measuring its probability with respect to the eigenvectors of those operators will result in either one (for one specific eigenvector, corresponding to its current state) or zero (for every other eigenvector). Perfect certainty.
If they _don’t_ commute, something happens. Something wonderful. You get uncertainty. Logically, if operators which _do_ commute _must share_ eigenvectors, operators which _don’t_ commute _can’t share_ eigenvectors. Therefore, measuring one aspect means the quantum state vector is now _not_ an eigenvector of the operator corresponding to the other aspect. For example, measuring position means the state vector is not an eigenvector of the momentum operator, and vice-versa.
So what does that give us? Uncertainty. Since the state vector isn’t an eigenvector of the second operator, measuring its probability with respect to any of the eigenvectors of that operator produces a result which isn’t zero and isn’t one. In the limit, if you have absolutely perfect knowledge of position, you have absolutely zero knowledge of momentum, so every possible momentum is equally probable. We know this as the Heisenberg Uncertainty Principle. Note that this falls directly out of the mathematics and has nothing to do with how we observe in any physical sense; the observer effect is not the uncertainty principle, and vice-versa.
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### Author:  [@Derleth](https://boards.straightdope.com/u/Derleth)
#### Post date: [July 16, 2016, 8:37pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/25 "2016-07-16T20:37:16Z")
And I got the first paragraph wrong: You take the _magnitude_ of the inner product and square it.
(I mean, I probably got other stuff somewhat wrong as well, even noting that that wasn’t _intended_ to be a general definition of the inner product, but I noticed that one.)
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### Author:  [@Leo\_Bloom](https://boards.straightdope.com/u/Leo_Bloom)
#### Post date: [July 17, 2016, 2:25pm UTC](https://boards.straightdope.com/t/quantum-entanglement-what-if-each-particle-assumes-a-different-state-simultaneously/740435/26 "2016-07-17T14:25:53Z")
Just saw this now…
Chewing it over. Thanks, and to **Half Man** as well.
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