2
$\begingroup$

Let $|\psi\rangle = \alpha|0\rangle + \beta |1\rangle$. In Bloch sphere representation, this is $\cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}e^{i\phi}|1\rangle$.

In matrix representation:

$|\psi\rangle\langle\psi| = \cos^{2}\frac{\theta}{2}|0\rangle\langle0| + \cos\frac{\theta}{2}\sin\frac{\theta}{2}e^{i\phi}|0\rangle\langle 1| + \sin\frac{\theta}{2}\cos\frac{\theta}{2}e^{i\phi}|1\rangle\langle0| + \sin^{2}\frac{\theta}{2}e^{2i\phi}|1\rangle\langle 1|\,.$

However, I am told it is also equivalent to restate the above as

$|\psi\rangle\langle\psi| = \cos^{2}\frac{\theta}{2}|0\rangle\langle0| + \cos\frac{\theta}{2}\sin\frac{\theta}{2}e^{-i\phi}|0\rangle\langle 1| + \sin\frac{\theta}{2}cos\frac{\theta}{2}e^{i\phi}|1\rangle\langle0| + \sin^{2}\frac{\theta}{2}|1\rangle\langle 1|\,.$

Note the difference in the (0,1) and (1,1) entry.

Why is this so?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

You are missing the fact that $$ \langle \psi | = \bigg( |\psi\rangle \bigg)^{\dagger} \,.$$

"$\langle \psi|$" is conjugate transpose of "$|\psi\rangle$".

So, if

$$|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}e^{i\phi}|1\rangle\,,$$

then

$$\langle \psi| = \cos\frac{\theta}{2}\langle 0| + \sin\frac{\theta}{2}e^{-i\phi}\langle 1 |\,.$$

Since $e^{i \phi}$ is a complex number, it will gain a negative sign in the exponent.

So now, if you calculate $|\psi\rangle\langle\psi|$, you will get the second equation, which is the correct one.

$\endgroup$
1
  • 2
    $\begingroup$ thank you for the spot $\endgroup$
    – Physkid
    Oct 31, 2023 at 2:35

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.