D.19 The generalized uncertainty relationship

This note derives the generalized uncertainty relationship.

For brevity, define $A'=A-\langle A\rangle$ and $B'=B-\langle B\rangle$, then the general expression for standard deviation says

$$\sigma_A^2 \sigma_B^2 = \langle A'^2\rangle \langle B'^2\r... ...e\Psi \vert A'^2\Psi\rangle \langle\Psi \vert B'^2\Psi\rangle$$

Hermitian operators can be taken to the other side of inner products, so

$$\sigma_A^2 \sigma_B^2 = \langle A'\Psi \vert A'\Psi\rangle \langle B'\Psi \vert B'\Psi\rangle$$

Now the Cauchy-Schwartz inequality says that for any $f$ and $g$,

$$\vert\langle f\vert g \rangle\vert \le \sqrt{\langle f\vert f\rangle}\sqrt{\langle g\vert g\rangle}$$

(See the notations for more on this theorem.) Using the Cauchy-Schwartz inequality in reversed order, you get

$$\sigma_A^2 \sigma_B^2 \ge \vert\langle A'\Psi \vert B'\Psi\rangle\vert^2 = \vert\langle A' B'\rangle\vert^2$$

Now by the definition of the inner product, the complex conjugate of $\langle A'\Psi \vert B'\Psi\rangle$ is $\langle B'\Psi \vert A'\Psi\rangle$, so the complex conjugate of $\langle A' B'\rangle$ is $\langle B' A'\rangle$, and averaging a complex number with minus its complex conjugate reduces its size, since the real part averages away, so

$$\sigma_A^2 \sigma_B^2 \ge \left\vert\frac{\langle A' B'\rangle-\langle B' A'\rangle}2\right\vert^2$$

The quantity in the top is the expectation value of the commutator $[A',B']$. Writing it out shows that $[A',B']=[A,B]$.