The 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

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

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

$$\langle(\Delta \xi)^2\rangle = \langle \xi^2\rangle - \langle \xi\rangle^2.$$ (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

$$\vert A\rangle = \Delta \xi \vert \rangle,$$ (74)

$$\vert B\rangle = \Delta \eta \vert \rangle,$$ (75)

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

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

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

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

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

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

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

The commutator is clearly anti-Hermitian,

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

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

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

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

$$\vert\langle\Delta \xi \Delta \eta\rangle\vert^2= \frac{1}... ...t\langle\left\{ \Delta \xi, \Delta \eta\right\}\rangle\vert^2,$$ (82)

where use has been made of $\langle \Delta\xi\rangle = 0$, etc. The final term in the above expression is positive definite, so we can write

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

where use has been made of Eq. (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 $\eta$.