Register to reply

Understanding superposition and entanglement

by martyscholes
Tags: entanglement, superposition
Share this thread:
martyscholes
#1
Sep29-10, 10:23 AM
P: 4
I am a newbie here, just an enthusiast with an above average understanding of math and physics, but exposed to QM after I left college.

I have looked through the posts and did not see a concise summary to the following. Please forgive me if I overlooked some threads.

There are some assertions I have which I want to verify are generally accepted to be true.
A. Measuring entangled particle A only impacts particle B by causing both to decohere where measured and both to go into superposition on a noncommuting measurement.
B. The no communication theorem states that measuring one entangled particle cannot provide measurable information to the other entangled particle.
C. There is no way to measure whether a particle is in superposition for a particular measurement.

Do I understand all of this correctly?

Many thanks,
Marty
Phys.Org News Partner Physics news on Phys.org
'Comb on a chip' powers new atomic clock design
Quantum leap in lasers brightens future for quantum computing
Enhanced NIST instrument enables high-speed chemical imaging of tissues
DrChinese
#2
Sep29-10, 10:43 AM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,290
Quote Quote by martyscholes View Post
Do I understand all of this correctly?

Many thanks,
Marty
Welcome to PhysicsForums, Marty!

Yes, I would say that you have it stated pretty well. I would comment on your A. that a measurement terminates the superposition and sends the particles into a mixed state, at least for non-commuting observables. Occasionally, you can observe entanglement continuing for commuting operators.
martyscholes
#3
Sep29-10, 11:28 AM
P: 4
Quote Quote by DrChinese View Post
Welcome to PhysicsForums, Marty!

Yes, I would say that you have it stated pretty well. I would comment on your A. that a measurement terminates the superposition and sends the particles into a mixed state, at least for non-commuting observables. Occasionally, you can observe entanglement continuing for commuting operators.
Thanks for the fast reply!

Extending the above, if there were a way to determine whether or not an entangled particle was in superposition for a particular observable, then one could violate the no-communication theorem?

Thanks again,
Marty

DrChinese
#4
Sep29-10, 12:27 PM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,290
Understanding superposition and entanglement

Quote Quote by martyscholes View Post
Extending the above, if there were a way to determine whether or not an entangled particle was in superposition for a particular observable, then one could violate the no-communication theorem?

Thanks again,
Marty
That's correct, you cannot determine if the superposition is still effective by a local operation. You must correlate results from both Alice and Bob to learn this after a series of operations, which of course defeats the objective of sending an FTL signal.
martyscholes
#5
Sep29-10, 01:10 PM
P: 4
Quote Quote by DrChinese View Post
That's correct, you cannot determine if the superposition is still effective by a local operation. You must correlate results from both Alice and Bob to learn this after a series of operations, which of course defeats the objective of sending an FTL signal.
Thanks again for staying with me on this and feel free to point out at any time where I am off base.

Suppose I had a device which produced two entangles particles A and B with non-commuting observables X and Y. In addition, suppose that X is not in superposition and is set to a predefined state (stay with me on this fantasy).

Now if we measure observable Y on particle A, then observable X on both particles is in superposition. Once observable X on partible B is measured, it will either have the expected state, which tells us nothing, or it is in the unexpected state, which means that particle A has had observable Y measured.

Where did I go wrong?

Thanks,
Marty
DrChinese
#6
Sep29-10, 01:43 PM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,290
Quote Quote by martyscholes View Post
Thanks again for staying with me on this and feel free to point out at any time where I am off base.

Suppose I had a device which produced two entangles particles A and B with non-commuting observables X and Y. In addition, suppose that X is not in superposition and is set to a predefined state (stay with me on this fantasy).

Now if we measure observable Y on particle A, then observable X on both particles is in superposition. Once observable X on partible B is measured, it will either have the expected state, which tells us nothing, or it is in the unexpected state, which means that particle A has had observable Y measured.

Where did I go wrong?

Thanks,
Marty
If X and Y are non-commuting, then they cannot have a relationship in which X is entangled and Y is not. Either they are both in an entangled state, or neither are.

Also: they cannot be in an entangled state with a known X or Y. X/Y must be unknown prior to the measurement. The result will therefore be random. You cannot send much of a signal using a random sequence.
martyscholes
#7
Sep29-10, 05:20 PM
P: 4
Ok.

Many thanks for the clarification.

Cheers,
Marty


Register to reply

Related Discussions
Superposition problem help Introductory Physics Homework 3
Am I understanding superposition correctly? Is it equivalent to all-potential ? Quantum Physics 4
Superposition representation of particle state in 1-d infitne well (SUPERPOSITION?) Advanced Physics Homework 2
Is understanding one branch of math conducive to understanding another? General Math 4
Proving the superposition of initial conditions gives superposition of motion Introductory Physics Homework 2