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2.7: Uncertainty Relation

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    08:15, 16 Nov 2014
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How is a momentum space wavefunction related to the corresponding coordinate space wavefunction? To answer this question, let us consider the representative $ \langle x'\vert p_x'\rangle$ of the momentum eigenkets $ \vert p'\rangle$ in the Schrödinger representation for a system with a single degree of freedom. This representative satisfies

 

$\displaystyle p_x'\, \langle x'\vert p_x'\rangle = \langle x'\vert\,p_x\,\vert p_x'\rangle = -{\rm i}\,\hbar\,\frac{d}{dx'}\, \langle x'\vert p_x'\rangle,$ (182)

 

 

where use has been made of Equation (169) (for the case of a system with one degree of freedom). The solution of the above differential equation is

 

$\displaystyle \langle x' \vert p_x'\rangle = c' \exp(\,{\rm i}\, p_x'\, x'/\hbar),$ (183)

 

 

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

 

$\displaystyle \langle p_x'\vert p_x''\rangle = \int_{-\infty}^{+\infty} dx'\,\l... ...ast}\, c'' \int_{-\infty}^{\infty}dx'\,\exp[-{\rm i}\, (p_x'-p_x'')\,x'/\hbar].$ (184)

 

 

The well-known mathematical result

 

$\displaystyle \int_{-\infty}^{+\infty}dx \,\exp(\,{\rm i}\, a\,x) = 2\pi\,\delta (a),$ (185)

 

 

yields

 

$\displaystyle \langle p_x'\vert p_x''\rangle = \vert c'\vert^{\,2} \, h\, \delta(p_x'-p_x'').$ (186)

 

 

This is consistent with Equation (171), provided that $ c' = h^{-1/2}$ . Thus,

 

$\displaystyle \langle x'\vert p_x'\rangle = h^{-1/2}\, \exp(\,{\rm i}\, p_x'\, x'/\hbar).$ (187)

 

 

Consider a general state ket $ \vert A\rangle$ whose coordinate wavefunction is $ \psi(x')$ , and whose momentum wavefunction is $ {\mit\Psi}(p_x')$ . In other words,

 

$\displaystyle \psi(x')$ $\displaystyle = \langle x'\vert A\rangle,$ (188)
$\displaystyle {\mit\Psi}(p_x')$ $\displaystyle = \langle p_x'\vert A\rangle.$ (189)

 

 

It is easily demonstrated that

 

$\displaystyle \psi(x') = \int_{-\infty}^{+\infty} dp_x'\, \langle x'\vert p_x'\... ...{-\infty}^{+\infty} dp_x'\,{\mit\Psi}(p_x') \, \exp(\,{\rm i} \,p_x'\,x'/\hbar)$ (190)

 

 

and

 

$\displaystyle {\mit\Psi}(p_x') = \int_{-\infty}^{+\infty} dx' \,\langle p_x'\ve... ...{1/2}} \int_{-\infty}^{+\infty} dx'\,\psi(x')\, \exp(-{\rm i}\, p_x' x'/\hbar),$ (191)

 

 

where use has been made of Equations (118), (172), (185), and (187). Clearly, the momentum space wavefunction is the Fourier transform of the coordinate space wavefunction.

Consider a state whose coordinate space wavefunction is a wavepacket. In other words, the wavefunction only has non-negligible amplitude in some spatially localized region of extent $ {\mit\Delta} x$ . As is well-known, the Fourier transform of a wavepacket fills up a wavenumber band of approximate extent $ {\mit\Delta} k \sim 1/{\mit\Delta} x$ . Note that in Equation (190) the role of the wavenumber $ k$ is played by the quantity $ p_x'/\hbar$ . It follows that the momentum space wavefunction corresponding to a wavepacket in coordinate space extends over a range of momenta $ {\mit\Delta} p_x \sim \hbar /{\mit\Delta} x$ . Clearly, a measurement of $ x$ is almost certain to give a result lying in a range of width $ {\mit\Delta} x$ . Likewise, measurement of $ p_x$ is almost certain to yield a result lying in a range of width $ {\mit\Delta} p_x$ . The product of these two uncertainties is

 

$\displaystyle {\mit\Delta} x \,{\mit\Delta} p_x \sim \hbar.$ (192)

 

 

This result is called the Heisenberg uncertainty principle.

Actually, it is possible to write the Heisenberg uncertainty principle more exactly by making use of Equation (83) and the commutation relation (138). We obtain

 

$\displaystyle \langle ({\mit\Delta} x)^2\rangle \,\langle ({\mit\Delta} p_x)^2\rangle \geq \frac{\hbar^2}{4}$ (193)

 

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 wavepackets in both coordinate and momentum space.

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