Uncertainty Principle for Non-Commuting Operators

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

$$\begin{eqnarray*} (\Delta A)^2&=&\langle\psi\vert(A-\langle A\rangle)^2\psi\rang... ...angle B\rangle)^2\psi\rangle=\langle\psi\vert V^2\psi\rangle \\ \end{eqnarray*}$$

where we define (just to keep our expressions small)

$$\begin{eqnarray*} U&=&A-\langle\psi\vert A\psi\rangle\\ V&=&B-\langle\psi\vert B\psi\rangle .\\ \end{eqnarray*}$$

Since $\langle A\rangle$ and $\langle B\rangle$ are just constants, notice that

$$\bgroup\color{black} [U,V]=[A,B] \egroup$$

OK, so much for the definitions.

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

$$\begin{eqnarray*} \langle U\psi+i\lambda V\psi\vert U\psi+i\lambda V\psi\rangle\... ...t V\psi\rangle-i\lambda\langle V\psi\vert U\psi\rangle\geq 0 \\ \end{eqnarray*}$$

This expression contains the uncertainties, so lets identify them.

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

Choose a $\lambda$ to minimize the expression, to get the strongest inequality.

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

Plug in that $\lambda$.

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

This result is the uncertainty for non-commuting operators.

$$(\Delta A)(\Delta B)\geq{i\over 2}\langle[A,B]\rangle$$

If the commutator is a constant, as in the case of $[p,x]$, the expectation values can be removed.

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

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

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

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