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?
  • $\begingroup$ Can you spell out what $|\phi\rangle$ is? $\endgroup$ May 27, 2020 at 19:29

1 Answer 1


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 $U$.

Note that a similar relation holds for any maximally entangled state.

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.

  • $\begingroup$ 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 $I$)? $\endgroup$ May 28, 2020 at 7:55
  • $\begingroup$ @PeeyushKushwaha I recommend learning some basics of quantum information, like the first one or two chapters of a basic textbook. $\endgroup$ May 28, 2020 at 11:13
  • $\begingroup$ Alright, I'll try to get some more practice. $\endgroup$ May 28, 2020 at 12:16

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