0
$\begingroup$

If a Hadamard gate is applied to a qubit that is in the state $|0\rangle$, then it becomes the state $$|+\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)$$ In density matrix form, this state is:

$$\rho=|+\rangle\langle+|=\frac{1}{2}\left(|0\rangle+|1\rangle\right)\left(\langle0|+\langle1|\right)$$

But, if we start with the state $|0\rangle$ in density matrix form, $|0\rangle\langle0|$, and then apply a Hadamard gate, we get:

$$\rho=|+\rangle\langle0|=\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)\langle0|$$

Why is there a discrepancy between these two results? Which is correct?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

Applying a unitary $U$ to a state $\psi$ is expressed by $$U\psi.$$

Applying a unitary to a density matrix $\rho$ is expressed by

$$U\rho U^\dagger.$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ to clarify why this is the case: If $\vert\psi\;^\prime\rangle = U\vert\psi\rangle$, then $\rho^\prime = \vert\psi\;^\prime\rangle\langle\psi\;^\prime\vert = U\vert\psi\rangle\langle\psi\vert U^\dagger$ $\endgroup$
    – paulina
    Commented Apr 26 at 15:01
  • 1
    $\begingroup$ Yes absolutely. The same (or rather its "dual") is also encoded in the Heisenberg equation of motion. $\endgroup$
    – lcv
    Commented Apr 26 at 17:23

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.