# Measuring one qubit in an entangled pair in another basis?

Qubits are usually measured in the computational basis, but we can change the basis by a unitary $$U$$ to measure in the basis formed by the columns of $$U$$.

For example, if $$| \psi \rangle = | 0 \rangle$$, we can convert $$| \psi \rangle$$ to the basis $$U$$:

$$U | \psi \rangle = \frac{1}{\sqrt{2}} | a \rangle + \frac{1}{\sqrt{2}} | a_\perp \rangle$$.

In this case, our basis is $$\{ | + \rangle, | - \rangle \}$$. But what if we entangle two qubits:

$$| \psi \rangle = | + \rangle$$, $$| \phi \rangle = | 0 \rangle$$

$$\texttt{CNOT} | \psi \rangle | \phi \rangle = \frac{1}{\sqrt{2}}\texttt{CNOT} | 00 \rangle + \frac{1}{\sqrt{2}}\texttt{CNOT} | 10 \rangle = \frac{1}{\sqrt{2}} | 00 \rangle + \frac{1}{\sqrt{2}} | 11 \rangle$$

And of course, this state is entangled. Now, we send the first qubit (originally storing the state $$| \psi \rangle$$) to another party. This party wants to measure $$\psi$$ in a different basis, $$X$$. How can we represent this basis change mathematically? We can't write an ket for the first qubit anymore, as the state is non-separable.

My assumption is that we can represent the other party measuring through the basis $$X$$ as the unitary transformation: $$X \otimes I$$, where $$I$$ is the identity matrix. But I would like some confirmation if this is correct; and if it is, is there some physical interpretation?

Thanks.

If you have some basis change unitary $$U$$ for a single qubit, then the person holding the first qubit can perform that unitary on their qubit, using the mathematical description $$U\otimes I$$. If the person were holding the second qubit, they would describe it as $$I\otimes U$$. So, that's the "yes" part!
The "no" part is what the unitary looks like. If you want to measure in the $$X$$ basis, you need a unitary that transforms $$X$$ into $$Z$$ (the standard measurement basis), i.e. $$U^\dagger XU=Z$$. These things are easy enough to write down: you have $$U^\dagger |+\rangle=|0\rangle$$ and $$U^\dagger|-\rangle=|1\rangle$$, but you might happen to already know that Hadamard does this job.