1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, and no!

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.