Why can't we use quantum bit entanglement for FTL information?

[ThomasT]

jambaugh, thanks for your clarifications. It's clearly your primary antagonist, unusualname, that has more or less hijacked the thread for personal reasons (he's one of those would be metaphysicians who seems to like to think of reality in 'holographic' terms).

Anyway, the OP's question has been answered.

 

[huge10]

transmit information indirectly.

But basically you're not transmitting any information.
The effects of entanglement are not due to any transfer of information (which would violate Einstein's special relativity).

The information is already contained in the whole system since the beginning.

 

[least_action]

How does one state the original question mathematically? (As one might do before attempting to prove it)

You use the tensor product of dirackets to entangle states but how can you say the information cannot travel faster than light?

 

[unusualname]

How does one state the original question mathematically? (As one might do before attempting to prove it)

You use the tensor product of dirackets to entangle states but how can you say the information cannot travel faster than light?

You essentially argue that a measurement at either end doesn't alter the probability distribution of any observable at the other end, but as mentioned above the argument might be considered tautological.

I can't get tex to work but there should be a derivation on the wikipedia page (if it's not ruined by an edit war)
http://en.wikipedia.org/wiki/No-communication_theorem

For a deeper argument see Eberhard & Ross' 1989 paper Quantum field theory cannot provide faster-than-light communication

 

[mrspeedybob]

But basically you're not transmitting any information.
The effects of entanglement are not due to any transfer of information (which would violate Einstein's special relativity).

The information is already contained in the whole system since the beginning.

How can that be? Suppose two particles are entangled. If I precisely measure the position of one and the momentum of the other then I should know precisly the position and momentum of both. This is impossible.

 

[jambaugh]

How does one state the original question mathematically? (As one might do before attempting to prove it)

You use the tensor product of dirackets to entangle states but how can you say the information cannot travel faster than light?

To begin with, considering your two observers Alice and Bob of two halves of an entangled pair, let us consider whether Alice's actions can affect Bob's observations.

Define the density operator of the entangled pair (or any pair): \rho.

Now consider that Alice's available action is to choose what variable to observe. As unusualname pointed out she can't pick what outcome she is to observe only what variable to look at. To keep it simple let Alice's observation be boolean with a 0 or 1 outcome (eigen-values). Let X be Alice's local observable. (Note the negation is 1-X).

Now consider any variable Bob might look at, call it Y and let it be likewise boolean. You can later generalize these to many variable cases.

A signal could be passed only if the outcome of Bob's measurement depended in some way on the choice of measurement Alice made regardless of the outcome. So we consider the conditional probability:

P(Y=1| X=1 or X = 0) = P(Y=1 and [X = 1 or X = 0]) /P(X = 1 or X = 0).
(The probability Y=1 given either X =1 or X =0, i.e. the probability Y=1 given X was measured.)

Note the denominator is 1, Alice will either see 1 or 0.
P(Y=1 and [X=1 or X= 0]) = P(Y=1 and X=1)+P(Y=1 and X = 0).

Work through the calculations with the density operators and see that P(Y=1 and X = 1) + P(Y=1 and X=0) = P(Y=1). Bob's observation is independent of Alice's choice to measure X though it may correlate with what outcome Alice sees.


P(Y=1 \cap X=1) = Tr( Y\otimes X \circ \rho )
P(Y=1 \cap X=0) = Tr(Y \otimes (1-X)\circ \rho)
Tr(Y\otimes X \circ \rho) +\Tr(Y\otimes (1-X)\circ \rho) = Tr(Y\otimes X\circ \rho + Y\otimes(1-X)\circ \rho)
=Tr([Y\otimes X +Y\otimes(1-X)]\circ\rho) = Tr(Y\otimes 1\circ \rho)
=P(Y=1)

Note that any general observation is a linear combination of boolean observations (the multipliers being the eigen-values and the boolean observations being projectors onto the corresponding eigen-space). Thus this derivation generalizes to any acts of observations.
Alice can't send Bob a message by choosing what to observe.

 

[jambaugh]

How can that be? Suppose two particles are entangled. If I precisely measure the position of one and the momentum of the other then I should know precisly the position and momentum of both. This is impossible.

Not quite and "that's QM". Measuring the momentum of one means you know what you would have measured if you had measured the momentum of the other. Not what it is but what you would have seen. Likewise with the positions. But just as you cannot simultaneously measure the momentum and position of one you cannot simultaneously measure the momentum of one and measure both momentum and position of the other.

Part of the problem is that we speak of "the one" and "the other" as if they were objects but part of how you determine "the one" and "the other" is to label them by position.

Consider an entangled pair first as a whole. Specifically consider one anti-correlated in position and momentum. You can then subdivide the pair by momentum (the one with the +x momentum and the one with the -x momentum) or you can subdivide the pair by position (the one into the left and the one to the right) but these are distinct factorizations of the Hilbert space because the implicit projectors do not commute.

It is easier to see with spin. Consider two spin 1/2 particles and speak of one being "the spin z +1/2" and the other being "the spin z -1/2" one. Or alternatively speak of one being "the spin x +1/2" and the other being "the spin x -1/2". But you can't do both at the same time because Sx and Sz do not commute.

This is what happens when you for example consider the case where "the left one had spin z = +1/2". You have by selection correlated position (usually here treated classically) with spin z (which here is non-classical) and thus you can no longer treat the position of the particle (whether Alice or Bob has it) as a classical variable.

Our implicit instinct to treat the system as classical objects with classical properties leads us to ruin in trying to parse these EPR experiments. You must retrain this instinct and trust the formal language of QM.

 

[DrChinese]

How can that be? Suppose two particles are entangled. If I precisely measure the position of one and the momentum of the other then I should know precisly the position and momentum of both. This is impossible.

jambaugh has provided an excellent response, I just wanted to add a quick comment.

Your last statement is "This is impossible". Not sure which part you are referring to, but of course it IS impossible to know both P and Q. So that means that measuring P of Alice and Q of Bob does NOT yield simultaneous P and Q for either. So that is the source of the contradiction: Once Alice is observed, Bob is no longer entangled with Alice on that basis.

 

[NetMage]

Haha, I had a similar post not too long ago. Essentially you CAN technically have information sent via entanglement. I recently spoke with my math professor about this as well. Theoretically one could set up a system and have a computer understand the bit set up of the entangled particles...if that is one could figure out HOW to set up the bit, this would be meticulous in itself. Next let's say one can set up a bit according to thousands and thousands or whatever arbitrary number one would need to produce the binary for information being sent and understood by x distance. The problem becomes at this point...how would the computer or user know which particles to examine in order to process the information. If you sent the binary for the letter 'g': 01000111 this is the binary for g. But when the entangled particles that you set up the bit for 'g' may send 01010000 which is 'p'. The point being, there is no way for the user or cpu at x distance to know which bits or entangled particles to check to understand the information being sent, it would be all random. Now, this goes to say...that entanglement CAN be used for information security. One could call or text, or w/e method preferred, to let the other user know WHICH particles to check in order to decode a message via entanglement. Better yet, sending encrypted information to let a user know which bits or entangled particles to check...this is highly valuable...there is no way to intercept this information. :D

 

[NetMage]

Remember, entangled particles -- once one particle is observed, it only gives information about the other particle...information is not REALLY sent. Its just redundant information, which is why measurements of information sending via entanglement has really been reduced to limits of security. If you and your friend had a plan A - Z to let's say, rob a bank :D. At some distance you are calling the shots as to which plan shall be executed before the robbery takes place. The feds are hot on your trail, and you call your friend and tell him to check particles 001, 002, 006, and 145. Your friend who already knows what the message means, will then know which plan to execute. This may be an exotic example, but nonetheless, gets the point across as to how it could be used to secure information. The feds may have tapped your phone and heard the whole conversation..but they have no information as to what is about to happen according to your plan that was chosen via entanglement. Now, apply this to I suppose 'better' examples lol.

 

[Gigasoft]

Entanglement is of no consequence in your example. It is no different from any other method involving an one-time pad. It is possible for the police to intercept the particles, measure them, and pass them on, without your friend becoming any wiser.

Now, you might think, what if I tell my friend to measure each particle along a random basis of my choosing? Then, if someone has already measured the particles, my friend will know.

Well, not really. If they are able to measure the particles, they are just as able to store the particles, and replace them with some other particles that are entangled with their own particles. When you call your friend, the feds pick up the phone and pretend to be him. They decrypt the message using the particles they got from you, then they encrypt the message with their own particles, call your friend and pretend to be you.

So, entanglement can only help you if you already have a way of transmitting a message unaltered.

 

[NetMage]

lol, what are you talking about, you cannot 'intercept' the particles...

 

[NetMage]

Gigasoft, you do not seem to understand what the implications of entanglement are...I'm no expert in the field, but I have done enough research to know that what you are talking about is nonsense.

 

[NetMage]

haha, they cannot just randomly and willingly entangle particles from some x distance...lol. They could not entangle their particles with yours...

 

[Gigasoft]

Well, then you obviously can't read.

As I said, they would capture your particles, as you try to send them to your friend. They would then generate new entangled pairs of particles, and send one particle from each pair to your friend. When you think you're talking to your friend on the phone, they decrypt your message using the particles they received from you. They then encrypt the message again, using the particles they kept from the corresponding other pair. Now, they call your friend and deliver the newly encrypted message.

In your scenario, if your agreement is to measure each particle in a particular basis, and the police knows this, the police can just measure each particle in that basis and pass the particles on as they are. When your friend measures the particles, his measurements will be the same as they would have been otherwise, so they don't even have to intercept your phone call.

On the other hand, if you have some means of giving your friend a bunch of particles that can't be intercepted, then you could just as well give him a notebook full of random numbers which can't be intercepted, which would be much cheaper.

 

[NetMage]

capture your particles? you are assuming that they are able to find you in a hypothetical situation that i created...wow

 

[NetMage]

not to mention the police don't know what any of the particles mean...so you are still wrong..lmao only you and your friend know haha

 

[NetMage]

I refuse to engage in a battle of witts with an unarmed individual...I'm done having this conversation lol...wow

 

[Gigasoft]

Oh great, so Mr. junior undergraduate's "cryptography" basically consists of assuming that no one knows where you or your friend are, or how you're communicating. Congratulations on completely missing the point of quantum cryptography.

Good luck at university. You'll really impress your professor with your "lmao"'s and your "wow"'s, I'm sure.

 

[NetMage]

Well naturally I assume they don't know where we are...it is the principal of the argument rather than the realism as stated...and quantum cryptography typically relates the release of a photon assigned to a data packet that when disturved notifies the network security. Entanglement is not REALLY cryptography. Not to mention, EVEN IF the feds were to find out where you were, they could not just send 'information' to your friend...the feds do not know what any of the entangled particles mean...only you and your friend do...they would have no method of communication with him. You have no clue as to what it is you're talking about. I'm done hijacking this thread with your nonsense. Good day.

P.S. I am doing just fine with my studies thank you. More than you know.

 

[Gigasoft]

Well, you really don't get it, do you. The entanglement of the particles has no effect on your scenario. Nothing in your scenario requires that the particles are in an entangled state. You could send him a copy of the script to a Days of our Lives episode, and it would serve the exact same purpose as your particles. Or, you could send him a diskette with a photograph of the president as a JPEG image, and use the image bits as the decryption key. Most importantly, you could send him particles prepared in a known state, without ever producing any entangled pairs. Your friend won't know the difference, and the police won't know the difference either.

 

[NetMage]

"Lets pretend that you're going back in time and for some crazy reason you want to make sure that the Nazis win WWII. You remember that one of the biggest coups for the Allied forces was intercepting the code device on a U-Boat. You don't care about faster than light comms, but you want it to be instant, and SECURE. So, you bring the technology to use entangled photons to send information, but to keep it secure you can't verify by radio. You make sure that the captain and first officer of each U-Boat knows what "signal 1" "signal 2" and so forth means with their entangled photons. It doesn't matter if they get a 0 or 1, because each time there is a spin flip, they call to memory a conversation they had with their commanding officers at base. They now have completely secure signaling devices, even if they can't send so much as the message "hello" via direct means.

This is the simplest use of a single entangled pair, but there is such a thing as using quantum means of communication at less than c, which would be altered by any spying (or any perturbation). In this case, it's not the speed, but the security that's so desirable. In essence, the benefits of QM comms are many, but FTL isn't one of those benefits."

Maybe you can wrap your brain around this...it is the same concept. Stated by another person from this forum.

 

[unusualname]

@NetMage , the thread is titled "why can't we use quantum bit entanglement for FTL information?", I think it's well-known that there may be applications in subluminal communication scenarios, in fact, as well as security it can be used to improve efficiency of communication channels (dense coding) and in pooled-data situations eg see the article by W Wooters "Quantum entanglement as a resource for communication" (from this book, or do a google search)

but FTL communication is ruled out, (jambaugh posted a detailed argument above for those who don't like the simplistic "cos it's random" type reply)

 

[Gigasoft]

No, what you quoted does not constitute a secure means of communication using quantum entanglement. With only 1 bit measured, it is of course impossible to detect that the message has been tampered with or spied on. In fact, it is not even communication, since no information is transferred from the first submarine to the second (assuming that I interpreted his post correctly).

I recommend that you read up on http://en.wikipedia.org/wiki/Quantum_key_distribution" . What is essential in secure communication using quantum entanglement is that enough bits are transferred that it can be detected with high probability that an attempt at eavesdropping has taken place, and that an eavesdropper doesn't know how the particles will be measured, until the recipient is already in possession of the particles.

 

[NetMage]

Yes I realize all of this, in my original post on here I spoke of entanglement not being a tool for FTL comm. But instead stated as to what it could be used for ie security purposes etc. Then I got tangled up in a ridiculous hijacking of the thread thanks to gigasofts absurd posts.

 

[NetMage]

Gigasoft...you cannot eavesdrop on entangled particles...EVEN IF someone was able to find out measurements of the particles...it does not matter, because the REAL information was already pre determined, the entanglement only gives commands in order to execute the information that was exchanged b4 hand. The only people that even know what the corresponding entangled particles mean, is the people who are using them and understand what information they represent. No one else COULD know.

 

[NetMage]

Basically the purpose is that no one could intercept the information.

 

[Gigasoft]

"The only people that even know what the corresponding entangled particles mean, is the people who are using them and understand what information they represent. No one else COULD know."

"The only people that even know what the corresponding letters in the script to Days of our Lives mean, is the people who are using them and understand what information they represent. No one else COULD know."

Do you believe that there is a conceptual difference between these two lines of reasoning?

 

[NetMage]

I believe you sound ridiculous.