Uncertainty Relation

We have seen that if $\xi$ and $\eta$ are two noncommuting observables then a determination of the value of $\xi$ leaves the value of $\eta$ uncertain, and vice versa. It is possible to quantify this uncertainty. For a general observable $\xi$ , we can define a Hermitian operator

$${\mit\Delta} \xi = \xi - \langle \xi \rangle,$$ (68)

where the expectation value is taken over the particular physical state under consideration. It is obvious that the expectation value of ${\mit\Delta} \xi$ is zero. The expectation value of $({\mit\Delta} \xi)^2 \equiv {\mit\Delta} \xi\,{\mit\Delta} \xi$ is termed the variance of $\xi$ , and is, in general, non-zero. In fact, it is easily demonstrated that

$$\langle({\mit\Delta} \xi)^2\rangle = \langle \xi^{\,2}\rangle - \langle \xi\rangle^2.$$ (69)

The variance of $\xi$ is a measure of the uncertainty in the value of $\xi$ for the particular state in question (i.e., it is a measure of the width of the distribution of likely values of $\xi$ about the expectation value). If the variance is zero then there is no uncertainty, and a measurement of $\xi$ is bound to give the expectation value, $\langle\xi\rangle$ .

Consider the Schwarz inequality

$$\langle A\vert A\rangle \langle B\vert B\rangle \geq \vert\langle A\vert B\rangle\vert^{\,2},$$ (70)

which is analogous to

$$\vert{\bf a}\vert^{\,2} \,\vert\,{\bf b}\vert^{\,2} \geq \vert{\bf a}\cdot {\bf b}\vert^{\,2}$$ (71)

in Euclidian space. This inequality can be proved by noting that

$$(\langle A\vert + c^\ast\, \langle B\vert)\, (\vert A\rangle + c\, \vert B\rangle) \geq 0,$$ (72)

where $c$ is any complex number. If $c$ takes the special value $-\langle B\vert A\rangle/\langle B\vert B\rangle$ then the above inequality reduces to

$$\langle A\vert A\rangle \langle B\vert B\rangle - \vert\langle A\vert B\rangle\vert^{\,2} \geq 0,$$ (73)

which is the same as the Schwarz inequality.

Let us substitute

$$\vert A\rangle = {\mit\Delta} \xi\, \vert~\rangle,$$ (74)

$$\vert B\rangle = {\mit\Delta} \eta\, \vert~\rangle,$$ (75)

into the Schwarz inequality, where the blank ket $\vert~\rangle$ stands for any general ket. We find

$$\langle ({\mit\Delta} \xi)^2\rangle \langle ({\mit\Delta} \eta)^2\rangle \geq \vert\langle {\mit\Delta} \xi \,{\mit\Delta} \eta\rangle \vert^{\,2},$$ (76)

where use has been made of the fact that ${\mit\Delta} \xi$ and ${\mit\Delta} \eta$ are Hermitian operators. Note that

$${\mit\Delta} \xi \,{\mit\Delta} \eta= \frac{1}{2} \left[ {\mit\De... ...} \eta\right] +\frac{1}{2} \left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\},$$ (77)

where the commutator, $\left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right]$ , and the anti-commutator, $\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}$ , are defined

$$\left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right] \equiv {\mit\Delta}\xi \,{\mit\Delta} \eta -{\mit\Delta} \eta\, {\mit\Delta} \xi,$$ (78)

$$\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\} \equiv {\mit\Delta} \xi\, {\mit\Delta} \eta + {\mit\Delta}\eta \, {\mit\Delta}\xi.$$ (79)

The commutator is clearly anti-Hermitian,

$$(\left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right])^{\dag } = ({\... ...a}\xi\, {\mit\Delta}\eta = - \left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right],$$ (80)

whereas the anti-commutator is obviously Hermitian. Now, it is easily demonstrated that the expectation value of an Hermitian operator is a real number, whereas the expectation value of an anti-Hermitian operator is an imaginary number. It follows that the right-hand side of

$$\langle{\mit\Delta} \xi \,{\mit\Delta} \eta\rangle= \frac{1}{2} \... ...frac{1}{2}\, \langle\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}\rangle,$$ (81)

consists of the sum of an imaginary and a real number. Taking the modulus squared of both sides gives

$$\vert\langle{\mit\Delta} \xi \,{\mit\Delta} \eta\rangle\vert^{\,2... ...rt\langle\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}\rangle\vert^{\,2},$$ (82)

where use has been made of $\langle {\mit\Delta}\xi\rangle = 0$ , etc. The final term on the right-hand side of the above expression is positive definite, so we can write

$$\langle ({\mit\Delta} \xi)^{\,2}\rangle\, \langle ({\mit\Delta} \... ...{\,2}\geq \frac{1}{4} \, \vert\langle\left[ \xi, \eta\right]\rangle\vert^{\,2},$$ (83)

where use has been made of Equation (76). The above expression is termed the uncertainty relation. According to this relation, an exact knowledge of the value of $\xi$ implies no knowledge whatsoever of the value of $\eta$ , and vice versa. The one exception to this rule is when $\xi$ and $\eta$ commute, in which case exact knowledge of $\xi$ does not necessarily imply no knowledge of $\eta$ .