D.19 The generalized uncertainty relationship

This note derives the generalized uncertainty relationship.

For brevity, define $A'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $A-\big\langle A\big\rangle $ and $B'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $B-\langle{B}\rangle$, then the general expression for standard deviation says

\begin{displaymath}
\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
\end{displaymath}

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

\begin{displaymath}
\sigma_A^2 \sigma_B^2
= \langle A'\Psi \vert A'\Psi\rangle \langle B'\Psi \vert B'\Psi\rangle
\end{displaymath}

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

\begin{displaymath}
\vert\langle f\vert g \rangle\vert \mathrel{\raisebox{-.7p...
... \sqrt{\langle f\vert f\rangle}\sqrt{\langle g\vert g\rangle}
\end{displaymath}

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

\begin{displaymath}
\sigma_A^2 \sigma_B^2 \mathrel{\raisebox{-1pt}{$\geqslant$...
...\vert B'\Psi\rangle\vert^2 = \vert\langle A' B'\rangle\vert^2
\end{displaymath}

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

\begin{displaymath}
\sigma_A^2 \sigma_B^2 \mathrel{\raisebox{-1pt}{$\geqslant$...
...frac{\langle A' B'\rangle-\langle B' A'\rangle}2\right\vert^2
\end{displaymath}

The quantity in the top is the expectation value of the commutator $[A',B']$. Writing it out shows that $[A',B']$ $\vphantom0\raisebox{1.5pt}{$=$}$ $[A,B]$.