# B How does Quantum Teleportation Work?

1. Sep 26, 2016

### Anderson Yuan

I am in high school trying to create a video about Quantum Teleportation. Can someone please verify this is what happens during Quantum Teleportation?
1. An entangled pair of qubits are generated. Qubit A is is sent to one location, Qubit B to another.

2. The qubit to be teleported interacts with qubit A, and the relationship between the quantum states of qubit A and the qubit being teleported can be measured. (Does the Heisenberg Uncertainty Principle tie into this at all?)

3. After measuring the relationship between quantum states with qubit A and the qubit being teleported, the information is sent to qubit B. (This is something that I'm struggling with. How is the information being sent?)

4. Because qubits A and B are entangled, the information about how the quantum state of the qubit being teleported interacts with the quantum state of qubit A will also reveal how the quantum state of the qubit being teleported interacts with the quantum state of qubit B.

5. Qubit B uses this information to replicate the quantum state of the qubit being teleported. (How does qubit B change like this?)

What happens to the original qubit being teleported?

2. Sep 26, 2016

### Simon Bridge

If I am understanding you, 2 is not correct. The relationship between the states is determined by the entanglement.
Measurement is only applied to one or the other particle, locally, and only on the local particle state.
Try to distinguish information from "a message".

Also see entanglement, quantum nonlocality and decoherence.

3. Sep 26, 2016

### Staff: Mentor

It does not. The uncertainty principle says that you cannot measure two incompatible observables with unlimited precision at the same time, but here we're only measuring one observable. We can do that with as much precision as our measuring instrument allows, and we can always find a better instrument if we need more precision.

We write it down on a piece of paper, roll the paper up into a little ball and tie it to the leg of a carrier pigeon, and then set the pigeon loose to fly to the distant laboratory that we're teleporting our qubit to. Or we could put it in an envelope, write the address of the distant laboratory on the envelope, attach a first-class postage stamp to the envelope, and give the envelope to the postal service. There are many ways of sending information.

This is the key to the teleportation protocol (which would be better called the "copying protocol"). When you interact with a quantum system you change its state. Qubits A and B are entangled, meaning that they are a single quantum system with a single state. Thus, the teleported qubit is not interacting with just A at step 2, it is interacting with the entangled A-B pair, and the measurement affects the entire A-B system. If the people with particle B know what that measurement was (this is the thing that we have to send by carrier pigeon, or snail mail, or just by shouting to our lab partner on the other side of the room) they can perform whatever measurement is needed to unwind the A measurement and change the state of their particle to match the state of the "teleported" particle.
It's gone. It was lost in the interaction with particle A of our entangled pair.

4. Sep 27, 2016

### Staff: Mentor

Let me bring in a bit of notation, which I hope you will be able to understand, but should clear things up. Let say that the state to be teleported is
$$|\psi_1\rangle = a |+\mathbf{z}\rangle_1 + b |-\mathbf{z}\rangle_1$$
That means that we want another particle, particle 3, to end up in the same state,
$$|\psi_3\rangle = a |+\mathbf{z}\rangle_3 + b |-\mathbf{z}\rangle_3$$
that is, with the same coefficients a and b in from of the +z and -z components.

Now, by performing a special measurement on particles 1 and 2 (the latter you called A) together, these two particles can end up in one of four states $|\Psi^{(+)}\rangle$, $|\Psi^{(-)}\rangle$, $|\Phi^{(+)}\rangle$, $|\Phi^{(-)}\rangle$, with equal probability. Just before that measurement, the three particles, taken together, were in the state
\begin{align*} |\psi_{123}\rangle &= \frac{1}{2} |\Psi_{12}^{(-)}\rangle \left( -a |+\mathbf{z}\rangle_3 - b |-\mathbf{z}\rangle_3 \right) + \frac{1}{2} |\Psi_{12}^{(+)}\rangle \left( -a |+\mathbf{z}\rangle_3 + b |-\mathbf{z}\rangle_3 \right) \\ &\quad + \frac{1}{2} |\Phi_{12}^{(-)}\rangle \left( b |+\mathbf{z}\rangle_3 +a |-\mathbf{z}\rangle_3 \right) + \frac{1}{2} |\Phi_{12}^{(+)}\rangle \left( -b |+\mathbf{z}\rangle_3 + a |-\mathbf{z}\rangle_3 \right) \end{align*}
After the measurement, the state for particle 3 can be found in the parenthesis next to the state for particle 1 and 2: if the result was $|\Psi_{12}^{(+)}\rangle$, then
$$|\psi_3\rangle = -a |+\mathbf{z}\rangle_3 + b |-\mathbf{z}\rangle_3$$
As you see, the coefficients a and b are still there, but not necessarily at the right place, or with the correct relative sign (as in the example above, where a is negative while b is positive). So after the measurement, information, meaning "what was the result of the measurement of particles 1 and 2," must be given to whoever has particle 3 so they can make the proper modification (called a rotation) to get the a and b in front of the correct $|\pm\mathbf{z}\rangle_3$ with the right sign.

5. Sep 27, 2016

### Strilanc

Alice and Bob share a pair of qubits A and B. The qubits are entangled so that they will agree if measured along the X axis (their X parity is "SAME") and also they will agree if measured along the Z axis (their Z parity is "SAME").

Alice wants to send a qubit Q to Bob. She does a measurement that compares Q with A along the X axis, finding out if they agree or disagree X-wise. If they're different she yells out "HEY BOB! The X axis of Q was DIFFERENT from A! So it's also different from B! Flip B over to fix that!". Otherwise she yells out "X AXIS IS FINE!".

Then she does the same thing again, but with the Z axis. She compares A and Q along the Z axis, finding out their Z-parity. If they differ, then "HEY BOB! Z axis is backwards! Flip it over!". Otherwise "Z AXIS IS FINE!".

And that's it. After Bob does the two corrections (or lack-of-corrections) that Alice asked for, B is definitely in the state that Q was in at the start. Alice managed to send that state, but at the cost of totally ruining her own two qubits in the process (because of the measurements).

There's three main surprising things about this protocol, I think. 1) measuring X-parity doesn't mess up Z-parity. 2) The state space is continuous, but the corrections are all-or-nothing: SAME or DIFFERENT. 3) This is literally the quantum equivalent of a one-time pad cipher.

6. Oct 2, 2016

### Anderson Yuan

I am still having trouble understanding what you mean. Can you make it clearer?

7. Oct 2, 2016

### PeroK

How can you make a video about something you don't even begin to understand?

8. Oct 2, 2016

Staff Emeritus
Youtube is filled with such things.

9. Oct 2, 2016

### Staff: Mentor

Yes, but they are made by people who misunderstand, as opposed to someone who has the good sense to know that he doesn't understand.

@Anderson Yuan, you will want to try reviewing @Strilanc's #5 above. See how far into it you get before you find something you don't understand and then post a more specific question about exactly where you got stuck, and we may be able to help you through the hard spot.
(And I do have to caution you that the best understanding you can get as a high school student is going to be incomplete and oversimplified in many ways. As you may have figured out from reading @DrClaude's answer, you need a fair amount of math beyond high school to do quantum mechanics for real; even if you do AP calculus in high school you won't be ready for QM until your second year of college).

10. Oct 2, 2016

### Simon Bridge

This sounds like one of the more open-ended tasks a high school student may be set to do.
The important part is how you go about assembling the final project, and it's presentation, in a scientific way and not so much to learn the subject well.
This also makes it trickier to avoid the whole "doing someone's homework for them" thing.