Skip to main content
\(\require{cancel}\)
Physics LibreTexts

1.14: 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

 

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

 

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

 

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

 

 

which is analogous to

 

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

 

 

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

 

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

 

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

 

 

which is the same as the Schwarz inequality.

Let us substitute

 

$\displaystyle \vert A\rangle$ $\displaystyle = {\mit\Delta} \xi\, \vert~\rangle,$ (74)
$\displaystyle \vert B\rangle$ $\displaystyle = {\mit\Delta} \eta\, \vert~\rangle,$ (75)

 

 

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

 

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

 

 

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

 

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

 

$\displaystyle \left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right]$ $\displaystyle \equiv {\mit\Delta}\xi \,{\mit\Delta} \eta -{\mit\Delta} \eta\, {\mit\Delta} \xi,$ (78)
$\displaystyle \left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}$ $\displaystyle \equiv {\mit\Delta} \xi\, {\mit\Delta} \eta + {\mit\Delta}\eta \, {\mit\Delta}\xi.$ (79)

 

 

The commutator is clearly anti-Hermitian,

 

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

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

 

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

 

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

 

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

Contributors