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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 an Hermitian operator

$\displaystyle {\mit\Delta} \xi = \xi - \langle \xi \rangle,$ (1.71)

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

$\displaystyle \langle({\mit\Delta} \xi)^2\rangle = \langle \xi^{\,2}\rangle - \langle \xi\rangle^2.$ (1.72)

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,

$\displaystyle \langle A\vert A\rangle \langle B\vert B\rangle \geq \vert\langle A\vert B\rangle\vert^{\,2},$ (1.73)

which is analogous to

$\displaystyle \vert{\bf a}\vert^{\,2} \,\vert{\bf b}\vert^{\,2} \geq \vert{\bf a}\cdot {\bf b}\vert^{\,2}$ (1.74)

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

$\displaystyle (\langle A\vert + c^\ast\, \langle B\vert)\, (\vert A\rangle + c\, \vert B\rangle) \geq 0,$ (1.75)

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

$\displaystyle \langle A\vert A\rangle \langle B\vert B\rangle - \vert\langle A\vert B\rangle\vert^{\,2} \geq 0,$ (1.76)

which is equivalent to the Schwarz inequality.

Let us substitute

$\displaystyle \vert A\rangle$ $\displaystyle = {\mit\Delta} \xi\, \vert~\rangle,$ (1.77)
$\displaystyle \vert B\rangle$ $\displaystyle = {\mit\Delta} \eta\, \vert~\rangle,$ (1.78)

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

$\displaystyle \langle ({\mit\Delta} \xi)^2\rangle \langle ({\mit\Delta} \eta)^2\rangle \geq \vert\langle {\mit\Delta} \xi \,{\mit\Delta} \eta\rangle \vert^{\,2},$ (1.79)

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

$\displaystyle {\mit\Delta} \xi \,{\mit\Delta} \eta= \frac{1}{2} \left[ {\mit\De...
...} \eta\right] +\frac{1}{2} \left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\},$ (1.80)

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

$\displaystyle \left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right]$ $\displaystyle \equiv {\mit\Delta}\xi \,{\mit\Delta} \eta -{\mit\Delta} \eta\, {\mit\Delta} \xi,$ (1.81)
$\displaystyle \left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}$ $\displaystyle \equiv {\mit\Delta} \xi\, {\mit\Delta} \eta + {\mit\Delta}\eta \, {\mit\Delta}\xi,$ (1.82)

respectively. The commutator is clearly anti-Hermitian,

$\displaystyle (\left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right])^{\dag } = ({\...
...a}\xi\, {\mit\Delta}\eta = - \left[ {\mit\Delta} \xi, {\mit\Delta} \eta\right],$ (1.83)

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. (See Exercise 11.) It follows that the right-hand side of

$\displaystyle \langle{\mit\Delta} \xi \,{\mit\Delta} \eta\rangle= \frac{1}{2} \...
...frac{1}{2}\, \langle\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}\rangle,$ (1.84)

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

$\displaystyle \vert\langle{\mit\Delta} \xi \,{\mit\Delta} \eta\rangle\vert^{\,2...
...rt\langle\left\{ {\mit\Delta} \xi, {\mit\Delta} \eta\right\}\rangle\vert^{\,2},$ (1.85)

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

$\displaystyle \langle ({\mit\Delta} \xi)^{\,2}\rangle\, \langle ({\mit\Delta} \...
...{\,2}\geq \frac{1}{4} \, \vert\langle\left[ \xi, \eta\right]\rangle\vert^{\,2},$ (1.86)

where use has been made of Equation (1.79). The previous expression is termed the uncertainty relation. According to this relation, an exact knowledge of the value of $ \xi$ (i.e., $ \langle ({\mit\Delta} \xi)^{\,2}\rangle\rightarrow 0$ ) implies no knowledge whatsoever of the value of $ \eta$ (i.e., $ \langle ({\mit\Delta} \eta)^{\,2}\rangle\rightarrow \infty$ ), 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$ .


next up previous
Next: Continuous Spectra Up: Fundamental Concepts Previous: Compatible Observables
Richard Fitzpatrick 2016-01-22