Expectation of Complex Operators

Question

Given an operator $\hat{\alpha}$, how do we obtain,

$$ \sqrt{ \left\langle \left( \hat{\alpha} - \left\langle\hat{\alpha}\right\rangle \right)^2 \right\rangle } = \sqrt{ \left\langle\hat{\alpha}^2\right\rangle - \left\langle\hat{\alpha}\right\rangle^2 } $$ My thoughts... To simplify, I define $A$ as,

$$ A\equiv \left( \hat{\alpha} - \left\langle\hat{\alpha}\right\rangle \right)^2 = \hat{\alpha}^2 + \left\langle\hat{\alpha}\right\rangle^2 - 2\hat{\alpha}\left\langle\hat{\alpha}\right\rangle $$ so what do I do with the extra $-2 \left\langle \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle$ in $\langle A\rangle$?

$$ \implies \left\langle \hat{\alpha}^2 \right\rangle + \left\langle \left\langle\hat{\alpha}\right\rangle^2 \right\rangle -2 \left\langle \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle $$

Answer 1

$$ -2 \left\langle \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle = -2 \left\langle \hat{\alpha} \right\rangle \left\langle\hat{\alpha}\right\rangle $$ and that is true because the $\langle \ \ \rangle $ "sees" $\hat{\alpha}$, since $\langle \hat{\alpha} \rangle $ has already been acted upon, and thus, is a number. In general, this is complex and in the case of hermitian $\hat{\alpha}$, real. So, we get,

$$ \left\langle \hat{\alpha}^2 \right\rangle + \left\langle \left\langle\hat{\alpha}\right\rangle^2 \right\rangle -2 \left\langle \hat{\alpha}\left\langle\hat{\alpha}\right\rangle \right\rangle = \left\langle \hat{\alpha}^2 \right\rangle + \left\langle\hat{\alpha}\right\rangle^2 -2 \left\langle \hat{\alpha} \right\rangle^2 = \left\langle \hat{\alpha}^2 \right\rangle - \left\langle\hat{\alpha}\right\rangle^2 $$ and the rest follows.

Answer 2

For the ease of notation I will write $T$ for the operator instead of $\hat \alpha$

Remember that $\langle T \rangle$ is defined as $\langle T \rangle = \langle \psi ,T \psi\rangle$ for some state $\psi$ (in the domain of $T$ with $\langle \psi,\psi \rangle = 1$). Then $\langle T \rangle$ is indeed real if $T$ is hermetian.

In order to calculate $\left\langle (T - \langle T \rangle)^2\right\rangle$ note that $\langle T \rangle$ is a number, but $T$ is an operator, so we're acutally interested in calculating $\left \langle (T - \langle T \rangle I )^2 \right\rangle$, where $I$ is the identity operator.

For $A= (T - \langle T \rangle I )^2$, we have

$$A=(T - \langle T \rangle I )^2 = T^2 - \langle T \rangle T I - \langle T \rangle I T + \langle T \rangle^2 = T^2 - (2\langle T \rangle) T + \langle T \rangle ^2$$

Thus, $$\langle A \rangle = \left \langle (T - \langle T \rangle I )^2 \right\rangle = \left\langle T^2 + (-2\langle T \rangle) T + \langle T \rangle ^2 \right\rangle= \langle T^2 \rangle + \left\langle \left(-2\langle T \rangle \right) T \right\rangle + \left\langle \langle T \rangle ^2 \right\rangle.$$

Now, since $\langle T \rangle$ is a number, we get $\left\langle\left(-2\langle T \rangle \right) T \right\rangle = -2 \langle T\rangle \langle T \rangle$ and $\langle \langle T \rangle^2 \rangle = \langle T \rangle^2 \langle \psi,\psi \rangle = \langle T \rangle^2 \cdot 1= \langle T \rangle\langle T \rangle$.

Finally we get $$\langle A \rangle = \langle T^2 \rangle - 2 \langle T\rangle \langle T \rangle + \langle T \rangle\langle T \rangle = \langle T^2 \rangle - \langle T\rangle \langle T \rangle = \langle T^2 \rangle - \langle T\rangle ^2.$$

This number is real if $T$ is hermetian.