# What is the expression for $|\psi\rangle\!\langle\psi|$ if $|\psi\rangle=\cos(\theta/2)|0\rangle+\sin(\theta/2)e^{i\phi}|1\rangle$?

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?

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.

• thank you for the spot Oct 31, 2023 at 2:35