The uncertainty relation

How is a momentum space wave-function related to the corresponding coordinate space wave-function? To answer this question, let us consider the representative $\langle x'\vert p'\rangle$ of the momentum eigenkets $\vert p'\rangle$ in Schrödinger's representation for a system with a single degree of freedom. This representative satisfies

$$p' \langle x'\vert p'\rangle = \langle x'\vert p\vert p'\rangle = -{\rm i} \hbar\frac{d}{dx'} \langle x'\vert p'\rangle,$$ (181)

where use has been made of Eq. (168) (for the case of a system with one degree of freedom). The solution of the above differential equation is

$$\langle x' \vert p'\rangle = c' \exp({\rm i} p' x'/\hbar),$$ (182)

where $c' = c'(p')$. It is easily demonstrated that

$$\langle p'\vert p''\rangle = \int_{-\infty}^{+\infty} \langl... ...nt_{-\infty}^{\infty}\exp[-{\rm i} (p'-p'') x'/\hbar] dx'.$$ (183)

The well-known mathematical result

$$\int_{-\infty}^{+\infty} \exp({\rm i} a x) dx = 2\pi \delta (a),$$ (184)

yields

$$\langle p'\vert p''\rangle = \vert c'\vert^2 h \delta(p'-p'').$$ (185)

This is consistent with Eq. (170), provided that $c' = h^{-1/2}$. Thus,

$$\langle x'\vert p'\rangle = h^{-1/2} \exp({\rm i} p' x'/\hbar).$$ (186)

Consider a general state ket $\vert A\rangle$ whose coordinate wave-function is $\psi(x')$, and whose momentum wave-function is $\Psi(p')$. In other words,

$$\psi(x') = \langle x'\vert A\rangle,$$ (187)

$$\Psi(p') = \langle p'\vert A\rangle.$$ (188)

It is easily demonstrated that

$$\psi(x') = \int_{-\infty}^{+\infty} dp' \langle x'\vert p'\rangle \langle p'\vert A\rangle$$

$$= \frac{1}{h^{1/2}} \int_{-\infty}^{+\infty} \Psi(p') \exp({\rm i} p' x'/\hbar) dp'$$ (189)

and

$$\Psi(p') = \int_{-\infty}^{+\infty} dx' \langle p'\vert x'\rangle \langle x'\vert A\rangle$$

$$= \frac{1}{h^{1/2}} \int_{-\infty}^{+\infty} \psi(x') \exp(-{\rm i} p' x'/\hbar) dx',$$ (190)

where use has been made of Eqs. (117), (171), (184), and (186). Clearly, the momentum space wave-function is the Fourier transform of the coordinate space wave-function.

Consider a state whose coordinate space wave-function is a wave-packet. In other words, the wave-function only has non-negligible amplitude in some spatially localized region of extent $\Delta x$. As is well-know, the Fourier transform of a wave-packet fills up a wave-number band of approximate extent $\delta k \sim 1/\Delta x$. Note that in Eq. (189) the role of the wave-number $k$ is played by the quantity $p'/\hbar$. It follows that the momentum space wave-function corresponding to a wave-packet in coordinate space extends over a range of momenta $\Delta p \sim \hbar /\Delta x$. Clearly, a measurement of $x$ is almost certain to give a result lying in a range of width $\Delta x$. Likewise, measurement of $p$ is almost certain to yield a result lying in a range of width $\Delta p$. The product of these two uncertainties is

$$\Delta x \Delta p \sim \hbar.$$ (191)

This result is called Heisenberg's uncertainty principle.

Actually, it is possible to write Heisenberg's uncertainty principle more exactly by making use of Eq. (83) and the commutation relation (137). We obtain

$$\langle (\Delta x)^2\rangle \langle (\Delta p)^2\rangle \geq \frac{\hbar^2}{4}$$ (192)

for any general state. It is easily demonstrated that the minimum uncertainty states, for which the equality sign holds in the above relation, correspond to Gaussian wave-packets in both coordinate and momentum space.