# If I start with an EPR pair and evolve one qubit through a unitary operation, what's really happening?

Suppose I have the EPR pair (maximally entangled state) as $$|\phi\rangle$$ and I apply an arbitrary matrix $$U_3$$ on the first qubit.

$$U_3 = \displaystyle \left[\begin{matrix}\cos{\left(\frac{\theta}{2} \right)} & - e^{i \lambda} \sin{\left(\frac{\theta}{2} \right)}\\e^{i \phi} \sin{\left(\frac{\theta}{2} \right)} & e^{i \left(\lambda + \phi\right)} \cos{\left(\frac{\theta}{2} \right)}\end{matrix}\right]$$

This is what I get after applying it to the first qubit:

$$|\phi_1\rangle = \dfrac{1}{\sqrt{2}} \left(e^{i \lambda} e^{i \phi} \cos{\left(\frac{\theta}{2} \right)}{\left|11\right\rangle} - e^{i \lambda} \sin{\left(\frac{\theta}{2} \right)}{\left|10\right\rangle } + e^{i \phi} \sin{\left(\frac{\theta}{2} \right)} {\left|01\right\rangle } + \cos{\left(\frac{\theta}{2} \right)}{\left|00\right\rangle }\right)$$

I was expecting it to be similar to what would happen if I applied $$U_3$$ to $$|+\rangle$$, but it's vastly different.

Curiously, if I measure either of the qubits, then I get half probability of measuring either $$|0\rangle$$ or $$|1\rangle$$ (though the remaining qubit remains in different states depending on if you measure first qubit or the second).

1. Why does application of $$U_3$$ not affect the probability with which I get result of the partial measurement?
2. Is such a thing (applying just a single-qubit operation on EPR Pair) a part of some phenomena / algorithm which is more easily understood?
• Can you spell out what $|\phi\rangle$ is? May 27, 2020 at 19:29

The maximally entangled state $$|\omega\rangle=|00\rangle+|11\rangle$$ has the property $$(U\otimes I)|\omega\rangle = (I\otimes U^T)|\omega\rangle\ .$$ That is, applying a unitary to the left qubit can be replaced by applying the different unitary $$U^T$$ to the right qubit.
However, applying a unitary to the 2nd qubit does not change the reduced density matrix of the first qubit. Thus, your result does not depend on whether you apply $$U$$.
A different way of seeing the result is to consider the reduced density matrix of the first qubit: It is $$\rho_1 = \tfrac12 I\ ,$$ the maximally mixed state. This is invariant under $$U$$, $$U\rho_1 U^\dagger = \rho_1\ ,$$ and thus, measurement outcomes are the same (and completely random in any basis!) regardless whether you apply $$U$$ or not.
• It's wild how applying a unitary to one of the qubits doesn't change the reduced density matrix of the second. I think it sort of makes sense, given the no-communication theorem. I still don't understand the general procedure involved in doing the calculation of these (e.g. how did you find that reduced density matrix of first qubit is half $I$)? May 28, 2020 at 7:55