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?


1 Answer 1


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

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

So, if

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


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

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.

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

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