4
$\begingroup$

Based on practically proven Quantum-Entanglement principle, any entangled quantum switch in this universe can be flipped instantaneously from elsewhere. Can it be used to send information from far off spaceship to earth instantaneously and vice versa? Is the present quantum technology, with few qubits, sufficient for communication from starship to earth for landing on Mars and mission control? Will radiation play spoil sports to this technology in space?

Chinese quantum entanglement research link: https://www.scientificamerican.com/article/chinese-researchers-achieve-stunning-quantum-entanglement-record/

New contributor
seccpur is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ you're talking about what's described here correct? I think the OP is trying to ask something a little different. QUESS created entangled photon pairs locally in space then sent each one to different locations on Earth. I think an answer to this question will involve entangled states on Earth and Mars, and the optical connection facilitates the entanglement. But I'm not really sure. $\endgroup$ – uhoh Sep 15 at 17:39
  • 6
    $\begingroup$ Your first sentence isn't true. Just because the measurement of one entangled thing instantly collapses the state of its counterpart doesn't mean that anything knows that state has changed. You can't use entanglement to instantly flip switches. $\endgroup$ – Erin Anne Sep 16 at 7:23
  • $\begingroup$ The accepted answer by user1271772 is utterly wrong. Please deaccept it. You should have asked at the physics stack exchange where there's a greater density of people who understand this subject. Of the current answers, Knudsen Number's is the best. It says little but at least what it says is factually accurate. $\endgroup$ – benrg 17 hours ago
  • $\begingroup$ So far the discussions are only on quantum entanglement. None of the answers in this thread till date have not addressed any possible implication of the technology in space application. $\endgroup$ – seccpur 16 hours ago
22
$\begingroup$

No, quantum mechanics cannot be used to transmit information faster than light. This is a common misconception based on misunderstanding how quantum mechanics works. Go here to read more about it. Technologies like quantum communication are valuable for other reasons though, but they still end up transmitting information at the speed of light, but in a cryptographically secure way.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks for this! $\endgroup$ – Organic Marble Sep 15 at 21:43
18
$\begingroup$

There's already two good answers that say a lot of what I wanted to say, and I will refrain from repeating any of their content here. I do think it is useful to add one more item of insight though. You say:

"any quantum switch in this universe can be flipped instantaneously from elsewhere."

When people talk about a "quantum switch flipping instantaneously", they are referring to entangled states of the form:

$$ N\left(|01\rangle + |10\rangle\right),\tag{1} $$

which simply means that that if you measure this system,

  • there is a 50% chance of the measurement telling you that qubit 1 is in state 0 and qubit 2 is in state 1 (the first state in the equation), and
  • there is a 50% chance of the measurement telling you that qubit 1 is in state 1 and qubit 2 is in state 0 (the second state in the equation).

By making the measurement on qubit 1 and getting a 0 or 1, qubit 2 is instantaneously going to become a 1 or 0 (the opposite of the qubit 1), but:

  • We do not know what the other state is without going there and measuring it, because we do not know that the original state is $N(|01\rangle + |10\rangle)$. If we measure qubit 1 and get 0, the original state could also have been $N(|00\rangle + |11\rangle)$ meaning that the other state instantaneously becomes a 0, not a 1. You would either have to spend time traveling to qubit 2 and measure it to find out if the original state was more like $N(|00\rangle + |11\rangle)$ or more like $N(|01\rangle + |10\rangle)$, or you could have someone located near qubit 2 and measuring it at the same time as you measure qubit 1, but then they would have to send a signal to you about the result they measured and it will time time for this signal to reach you. So no information has traveled instantaneously.
  • Nothing is really being "flipped" or "switched". To be flipped or switched implies that something was a 0 and became a 1, or was a 1 and become a 0, but none of that is happening here. We have a qubit that is neither in state 0 nor 1 (it is in a superposition of 0 and 1), and it becomes a 1 or 0 depending on what the other qubit becomes after the measurement.
  • The entire principle does not just apply for any quantum state, it only works for entangled states. If the state was $|00\rangle$, then qubit 1 and qubit 2 are both going to be 0 no matter what, and the measurement outcome for qubit 2 does not depend at all on the measurement outcome of qubit 1.

So these are the points to remember:

  • The state of qubit 2 changing depending on the measurement of qubit 1, does not apply to any state but only entangled states.
  • Nothing is being "switched" or "flipped" like a light switch flipping from off (0) to on (1). A state goes from being in a superposition of 0 and 1, to being in only one of 0 or 1.
  • There is no illusion that information is traveling faster than the speed of light to anybody other than maybe qubit 2 itself. You can think of qubit 2 receiving the signal from qubit 1, that qubit 1 was found in state 0, which leads to qubit 2 settling instantaneously in state 1, but that signalling happens within a single entangled system. Nothing outside of that entangled system is able to witness any super-luminal information transfer. To know that information traveled from qubit 1 to qubit 2 you or someone else would have to measure qubit 2 and the information about the measurement outcome would have to travel from the measurement device to you, which will take time.

Regarding the last point: What if the measurement device measures the states of qubit 1 and qubit 2 at the same time? Does the measuring device witness super-luminal information travel? Well no, because how does the measuring device know that the original state was even entangled? It could have been originally in the state $|01\rangle$ meaning that qubit 1 was in state 0 and qubit 2 was in state 1 the entire time, and no "instantaneous change" occurred.

What about if qubit 2 is a measuring device? The measuring device settles on state 1 immediately when qubit 1 is found to be in state 0, so has the measuring device witnessed super-luminal information travel about the state of qubit 1? Again, this would only be true if the measuring device (qubit 2) knew that it was entangled with qubit 1, and you cannot "know" what state something is in without first measuring it, but measuring this state would mean collapsing its wavefunction into a non-entangled state. So you cannot "know" this state was entangled without making it non-entangled, and if it's non-entangled there is no "instantaneous" information transmission. The measuring device therefore sees the effect of the "instantanous" information transmission but is unable to know whether any information was transmitted at all or if the states were just like that all along. This problem would be the same if the measuring device was both qubits 1 and 2 (the device finds out the states of both qubits at the same time, but it doesn't know whether or not information was transmitted because it cannot know whether or not the qubits were previously entangled, without having un-entangled them).

Then the final question becomes whether or not qubit 2 really changed instantaneously based on the measurement outcome of qubit 1. The theory of how quantum states and measurements work tells us that if the qubits are in the state described by Eq. 1 and qubit 1 is measured to be 0, qubit 2 "instantaneously" settles in state 1, but is there a way to experimentally verify this theory that qubit 2 "instantaneously" settled on state 1? Assume that it takes time, maybe the distance between qubits 1 and 2 divided by the speed of light, for qubit 2 to settle on state 1, then maybe you can come up with some experiment where you do multiple successive measurements and the results would contradict the hypothesis of there being a "delay" in qubit 2 settling on a state? Perhaps that could be the case, but consider for a moment that there is not really any "distance" between qubits 1 and 2, since they are really just one entangled system. If there is zero distance between them, then the speed of information travel does not have to be faster than the speed of light in order for the information to travel 0 metres, so the question now becomes whether or not you can prove that qubits 1 and 2 were more than 0 metres apart at the time of being entangled, and whether or not you can do this without doing any measurements (since measurements un-entangle the qubits) and fast enough to know that the qubits didn't move before your measurement finished.

| improve this answer | |
New contributor
user1271772 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • 2
    $\begingroup$ That is quite an extensive answer for the fact that it is your first one on the site (I saw it from review ^^) ... Thank you for the shared knowledge! $\endgroup$ – finnmglas Sep 16 at 19:39
  • 1
    $\begingroup$ Hey @user1271772 , your intellect is showing... again. +1! $\endgroup$ – Etienne Palos 2 days ago
  • $\begingroup$ Very extensive reply indeed +1. But frankly, quantum system is bit difficult to comprehend. Can these entangled bits be in two different locations? $\endgroup$ – seccpur 2 days ago
  • 2
    $\begingroup$ Thank you for this detailed explanation with easy to follow examples! $\endgroup$ – Knudsen Number 2 days ago
  • 1
    $\begingroup$ I wish I could give this answer more up-votes! $\endgroup$ – Knudsen Number 2 days ago
6
$\begingroup$

Based on practically proven Quantum-Entanglement principle, any quantum switch in this universe can be flipped instantaneously from elsewhere.

Indeed, quantum entanglement can connect (alias entangle) two particles in a way that the change in one particle results in an immediate change in state for the other one.

Theories about transmitting information faster than light are exciting, however, (as far as I know) there haven't been any actual cases where humans managed to actually flip the quantum switches in a way that allowed for faster-than-light transmissions of data.

Is the present quantum technology, with few qubits, sufficient for communication from starship to earth for landing on Mars and mission control?

The present quantum technology is not only incapable of but also rarely built for purposes of communication. Like with every exciting technology, scientist will run experiments and work out thories about this though. It just doesn't seem acheivable in the near future.

Current quantum technology that is actually on the market, like the quantum computers from D-Wave Systems (BC, Canada) claim to use quantum technology, however, there is a controversial debate about their Qubits. Those technically aren't actual Qubits. They are small rings of niobium, that have similar properties to actual quantum objects when they are cooled to very low temperatures (15 millikelvin, "which is about 180 times colder than interstellar space", as stated in this document).

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ While most of this answer is good, it is false to say that "those technically aren't qubits". You have not provided any evidence for why they are "not qubits" and you have also not provided any citation about there being a "controversial debate about their qubits". There was indeed controversy when D-Wave machines were first announced, but it's all settled now. D-Wave's qubits are real qubits. $\endgroup$ – user1271772 Sep 16 at 18:26
  • 2
    $\begingroup$ I see. No problem at all. It's good to see you're learning! I don't know of any SE thread, but D-Wave showed that their qubits do have entanglement with each other: journals.aps.org/prx/abstract/10.1103/PhysRevX.4.021041. If they are entangled, then they cannot be classical bits, they must be qubits. $\endgroup$ – user1271772 Sep 16 at 20:15
  • 1
    $\begingroup$ @user1271772 but it's not really a binary choice, is it? Couldn't company X's bits be slightly entangled for marketing purposes yet the mechanism by which things are computed not be true quantum computing? I don't mean to say that's the case for D-wave's computers (I'm thinking of their early ones in particular), but we can't just say "there's some entanglement, so ipso facto this is a true quantum computer" can we? $\endgroup$ – uhoh 2 days ago
  • 1
    $\begingroup$ @uhoh We could get into a long debate about what D-Wave does, but for now I'm just refuting the statement in this answer that "Those technically aren't actual Qubits". If there is even slight entanglement, they must be qubits, because classical bits cannot even have the slightest of entanglement. $\endgroup$ – user1271772 2 days ago
  • 1
    $\begingroup$ @WaterMolecule It's true that classical computers can be considered "quantum technology" but the author of the answer means "quantum-information based technology". All those devices that you mentioned, use classical and quantum mechanics to manipulate classical information (i.e. 1s and 0s), whereas "quantum technology" usually refers to technology that uses quantum information (e.g. entanglement) to enhance some functionality. $\endgroup$ – user1271772 2 days ago
2
$\begingroup$

While the given answers are very complete, I feel like they are also very complex to understand. So here's my two cents:

When two qubits are entangled, they have to be in a "superposition". Practically, this means that they both have a 50/50% chance of being either 1 or 0.

The only things we can do with qubits is reset them to zero (which breaks entanglement), or flip them in certain ways. However, given that we start with an unknown value, there is no way for us to arrive at a known value, that would not break entanglement.

As we can not "set" a qubit to a known value (apart from 0 which breaks the entanglement) we can not send data using qubits.

| improve this answer | |
New contributor
Daniël van den Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ Super +1. Thanks for simplifying. $\endgroup$ – seccpur yesterday

Your Answer

seccpur is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.