# Predicting rotations with many EPR pairs

I was explaining to a colleague that you can't use EPR pairs to communicate information, as it violates the no-communication theorem. This lead me to thinking... If I have let's say 1,000,000 EPR pairs shared with someone far away. We agree ahead of time that if I perform a rotation $$\theta$$ on my qubit, then I mean to communicate the bit $$0$$, and when I perform the rotation $$\phi$$, then I mean to communicate the bit 1.

We agree that at some time in the future (assume we have synchronized clocks), that I will perform the same rotation on all 1,000,000 of my EPR pair halves. After some agreed upon time, the holder of the other 1,000,000 pair halves measures his qubits to predict which rotation I made on my qubits. Then with high probability they can guess the message I wanted to send.

I know there is something wrong in this configuration, but I can't point it down. If this would be possible, then I could communicate faster than light...

Can anyone point out the error in my reasoning?

Thanks.

• You haven't stated how the measurement statistics will differ in the two cases, making them distinguishable. What is the specific condition that the receiver is checking? In both cases they're going to be getting measurements that look exactly like random coin flips. You'll basically find that this reduces to "Well, in case A there's a 50/50 chance that it's ON whereas in case B it's ON half of the time...". – Craig Gidney Jun 29 at 18:30

Well, the thing that is wrong in this configuration is that the holder of the other halves of the EPR pairs will not get any information about the rotation you did.

Let's simplify this protocol to be more specific: let's say you'll apply X gate to communicate 1 and I gate (i.e., do nothing) to communicate 0. These two actions are Ry rotations around by $$0$$ and by $$\pi$$, so they are a lot easier to distinguish than arbitrary rotations by $$\theta$$ and $$\phi$$.

You start with the EPR state $$\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$$. If you apply an X gate to the first qubit, you'll get state $$\frac{1}{\sqrt2}(|10\rangle + |01\rangle)$$. Now you have to distinguish these two states by doing an experiment on the second qubit (and you can repeat that experiment 1kk times).

Since you only have access to 1 qubit, your experiment is limited to performing some rotations and a single-qubit measurement. Regardless of the rotation you perform, the measurement probabilities are going to be the same for those two states, so you won't be able to distinguish them.

• I see now. I was experimenting with some simulators and as you said, there no change in the statistics as I (wrongly) thought. Thank you. – Stephen Diadamo Jun 29 at 18:41

Whatever rotation you do on one half of the state does not affect the reduced state on the other half,

\begin{align*} \mathrm{Tr}_A[(U \otimes \mathbb{I}) |\psi \rangle\langle \psi | ( U^* \otimes \mathbb{I}) ] &= \mathrm{Tr}_A[(U^* U \otimes \mathbb{I}) |\psi \rangle\langle \psi | ] \\ &= \mathrm{Tr}_A[(\mathbb{I} \otimes \mathbb{I}) |\psi \rangle\langle \psi | ] \\ &= \mathrm{Tr}_A[ |\psi \rangle\langle \psi | ] . \end{align*}

So the other party will never be able to distinguish between the two rotations you do and hence no communication.

• Ahh ok, very clear now. Thanks! – Stephen Diadamo Jun 29 at 18:46