Measurement uncertainty is bigger than the expectation squared

i.e. ${⟨X^2⟩}_{\psi }\ge {⟨X⟩}_{\psi }^2$ , where ${⟨X⟩}_{\psi }$ is the Expectation of a Measurement.

Proof

We will expand the positive scalar ${⟨({⟨X⟩}_{\psi }-X{)}^2⟩}_{\psi }$ , from which the original statement follows.

\begin{array}{rll} 0 &\geq \braket{ ( \braket{X}_\psi - X )^2 }_\psi & \text{Expanding the various terms}\\ &= \braket{ \braket{X}_\psi^2 - 2 X \braket{X}_\psi + X^2 }_\psi & \text{Since expectations are linear} \\ &= \braket{ \braket{X}_\psi^2 }_\psi - 2 \braket { X \braket{X}_\psi }_\psi + \braket{ X^2 }_\psi$ & \text{As $\braket{X}_\psi$ is a scalar} \\ &= \braket{X}_\psi^2 - 2 \braket { X \braket{X}_\psi }_\psi + \braket{ X^2 }_\psi & \text{As $\braket{X}_\psi$ is a scalar} \\ &= \braket{X}_\psi^2 - 2 \braket{X}_\psi^2 + \braket{ X^2 }_\psi\\ &= \braket{X}_\psi^2 - \braket{X}_\psi^2 \end{array}