Uncertainty Principle for Non-Commuting Operators

Let us now derive the uncertainty relation for non-commuting operators \bgroup\color{black}$A$\egroup and \bgroup\color{black}$B$\egroup. First, given a state \bgroup\color{black}$\psi$\egroup, the Mean Square uncertainty in the physical quantity represented is defined as

(\Delta A)^2&=&\langle\psi\vert(A-\langle A\rangle)^2\psi\rang...
...angle B\rangle)^2\psi\rangle=\langle\psi\vert V^2\psi\rangle \\

where we define (just to keep our expressions small)

U&=&A-\langle\psi\vert A\psi\rangle\\
V&=&B-\langle\psi\vert B\psi\rangle .\\

Since \bgroup\color{black}$\langle A\rangle$\egroup and \bgroup\color{black}$\langle B\rangle$\egroup are just constants, notice that

\begin{displaymath}\bgroup\color{black} [U,V]=[A,B] \egroup\end{displaymath}

OK, so much for the definitions.

Now we will dot \bgroup\color{black}$U\psi+i\lambda V\psi$\egroup into itself to get some information about the uncertainties. The dot product must be greater than or equal to zero.

\langle U\psi+i\lambda V\psi\vert U\psi+i\lambda V\psi\rangle\...
...t V\psi\rangle-i\lambda\langle V\psi\vert U\psi\rangle\geq 0 \\

This expression contains the uncertainties, so lets identify them.

\begin{displaymath}\bgroup\color{black} (\Delta A)^2+\lambda^2(\Delta B)^2+i\lambda\langle\psi\vert[U,V]\vert\psi\rangle\geq 0 \egroup\end{displaymath}

Choose a \bgroup\color{black}$\lambda$\egroup to minimize the expression, to get the strongest inequality.

{\partial\over\partial\lambda}=0 \\
2\lambda(\Delta B)^2+i\la...
...{-i\langle\psi\vert[U,V]\vert\psi\rangle\over 2(\Delta B)^2} \\

Plug in that \bgroup\color{black}$\lambda$\egroup.

(\Delta A)^2-{1\over 4}{\langle\psi\vert[U,V]\vert\psi\rangle^...
=\langle\psi\vert{i\over 2}[U,V]\vert\psi\rangle^2 \\

This result is the uncertainty for non-commuting operators.

\bgroup\color{black}$\displaystyle (\Delta A)(\Delta B)\geq{i\over 2}\langle[A,B]\rangle $\egroup
If the commutator is a constant, as in the case of \bgroup\color{black}$[p,x]$\egroup, the expectation values can be removed.

\begin{displaymath}\bgroup\color{black} (\Delta A)(\Delta B)\geq{i\over 2}[A,B] \egroup\end{displaymath}

For momentum and position, this agrees with the uncertainty principle we know.

\bgroup\color{black}$\displaystyle (\Delta p)(\Delta x)\geq{i\over 2}\langle[p,x]\rangle={\hbar\over 2} $\egroup

(Note that we could have simplified the proof by just stating that we choose to dot \bgroup\color{black}$(U+{i\langle[U,V]\rangle\over 2(\Delta B)^2}V)\psi$\egroup into itself and require that its positive. It would not have been clear that this was the strongest condition we could get.)

Jim Branson 2013-04-22