VII Momentum probabilities and the uncertainty principle

In analogy to the case of position $x$, we can introduce a momentum wave function $\tilde{\psi }(p)$, which delivers the probability to find the value $p$ in a momentum measurement as $P(p)={|\tilde{\psi }(p)|}^2$. This momentum wave function is given by

$$\tilde{\psi }(p)={(2\pi \mathrm{\hslash })}^{-1/2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp (-ipx/\mathrm{\hslash })\psi (x)𝑑x.$$

(98)

Mathematically, this is known as the Fourier transformation of $\psi (x)$ and simply decomposes $\psi (x)$ into plane waves $\exp (ipx/\mathrm{\hslash })$. Note that such plane waves are eigenfunction of the momentum operator, $\widehat{p}\exp (ipx/\mathrm{\hslash })=p\exp (ipx/\mathrm{\hslash })$. The Fourier transform in Eq. (98) involves the complex conjugate $\exp (-ipx/\mathrm{\hslash })$ of this function. The factor ${(2\pi \mathrm{\hslash })}^{-1/2}$ ensures that the resulting probability density of momentum $P(p)={|\tilde{\psi }(p)|}^2$ is properly normalised. In other words, we can write Eq. (98) also as

$$\tilde{\psi }(p)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\psi }_p^{*}(x)\psi (x)𝑑x,$$

(99)

where ${\psi }_p(x)$ is the momentum eigenfunction given in (33).

Observe that according to Eq. (98), $\psi (x)$ and $\tilde{\psi }(p)$ are not independent of each other. Hence, $P(x)$ and $P(p)$ are also not independent of each other, which has direct physical consequences such as Heisenberg’s uncertainty relation, to which we come shortly.

The averaged result of many experiments measuring some observable $\widehat{A}$ is called the expectation value of $\widehat{A}$, denoted by $⟨A⟩$. According to the general rules for probabilities, the expectation value is obtained from the probability density $P(A)$ by weighting it with the result $A$:

$$⟨A⟩=\int AP(A)𝑑A.$$

(100)

We can also calculated expectation values of functions of $A$, such as $A^2$:

$$⟨A^2⟩=\int A^{2}P(A)𝑑A.$$

(101)

Hence, the expectation value of position is

$$⟨x⟩=\int x{|\psi (x)|}^2𝑑x,$$

(102)

and the expectation value of position-squared is

$$⟨x^2⟩=\int x^2{|\psi (x)|}^2𝑑x.$$

(103)

The expectation values of momentum are given by $⟨p⟩=\int pP(p)𝑑p=\int p{|\tilde{\psi }(p)|}^2𝑑p$, $⟨p^2⟩=\int p^{2}P(p)𝑑p=\int p^2{|\tilde{\psi }(p)|}^2𝑑p$. Now, using standard theorems about the Fourier transformation (see PHYS213), one can express these expectation values directly in terms of the wave function $\psi (x)$:

	$⟨p⟩$	$=$	${\displaystyle \int {\psi }^{*}(x)[\widehat{p}\psi (x)]𝑑x}$		(104)
		$=$	$-i\mathrm{\hslash }{\displaystyle \int {\psi }^{*}(x){\psi }^{\prime }(x)𝑑x},$		(105)

	$⟨p^2⟩$	$=$	${\displaystyle \int {\psi }^{*}(x)[{\widehat{p}}^2\psi (x)]𝑑x}$		(106)
		$=$	$-{\mathrm{\hslash }}^2{\displaystyle \int {\psi }^{*}(x){\psi }^{\prime \prime }(x)𝑑x}.$		(107)

Hence, in order to calculate these expectation values one does not need to carry out a Fourier transformation; instead, one only needs to evaluate an ordinary integral.

A particularly useful quantity related to the expectation values is the standard deviation

$$\mathrm{\Delta }A=\sqrt{⟨A^2⟩-{⟨A⟩}^2}.$$

(108)

In quantum mechanics, the standard deviation is known as the uncertainty of the observable $A$.

In simple situations, $P(A)$ is peaked around some value. The expectation value $⟨A⟩$ then gives us an indication of the position of this peak, while $\mathrm{\Delta }A$ gives us an indication of the width of this peak.

Using again standard theorems about the Fourier transformation (see PHYS213), a strongly peaked position wave function $\psi (x)$ corresponds to a very broad momentum wave function $\tilde{\psi }(p)$, and vice versa. Mathematically, this can be expressed as a relation for the uncertainties of momentum and position:

$$\mathrm{\Delta }x\mathrm{\Delta }p\ge \mathrm{\hslash }/2.$$

(109)

This is the celebrated Heisenberg uncertainty relation. As a consequence, the better the position of a particle is determined, the less precisely determined is its momentum, and vice versa.

This relation also explains the zero-point energy of a particle in a potential, which we already have encountered for the particle in the square well. Classically, the minimal total energy $E=V_{\min }$ is obtained when the particle rests at the minimum of the potential. A quantum-mechanical particle never can be at rest at a potential minimum, since then both its position (at the minimum) and its momentum (which vanishes at rest) would be fixed.

Minimal-uncertainty wave packets are states that satisfy Eq. (109) as an equality

$$\mathrm{\Delta }x\mathrm{\Delta }p=\frac{\mathrm{\hslash }}{2}\mathit{\hspace{1em}\hspace{1em}}\text{(for minimal uncertainty).}$$

(110)

Their general form is

$$\psi (x)={(2\pi {\sigma }^2)}^{-1/4}\exp \left[-\frac{{(x-x_0)}^2}{4{\sigma }^2}+\frac{i}{\mathrm{\hslash }}{p}_0(x-x_0)\right],$$

(111)

where $x_0$, $p_0$, and $\sigma$ are real parameters.

The probability density $P(x)={|\psi (x)|}^2$ associated with a minimal-uncertainty wave packet takes the form of a normal distribution

$$P(x)={|\psi (x)|}^2={(2\pi {\sigma }^2)}^{-1/2}\exp [-{(x-x_0)}^2/2{\sigma }^2].$$

(112)

The distribution is symmetric around $x_0$, hence $⟨\widehat{x}⟩=x_0$. The uncertainty is given by $\mathrm{\Delta }x=\sigma$, and is independent of $x_0$.

The momentum wave function $\tilde{\psi }(p)$ is obtained via the Fourier transformation (98). The result is

$$\tilde{\psi }(p)={({\mathrm{\hslash }}^2\pi /2{\sigma }^2)}^{-1/4}\exp \left[-\frac{{\sigma }^2{(p-p_0)}^2}{{\mathrm{\hslash }}^2}-\frac{i}{\mathrm{\hslash }}p{x}_0\right],$$

(113)

which gives

$$P(p)={({\mathrm{\hslash }}^2\pi /2{\sigma }^2)}^{-1/2}\exp \left[-2{\sigma }^2{(p-p_0)}^2/{\mathrm{\hslash }}^2\right].$$

(114)

This is again a normal distribution, shifted away from the origin by $⟨\widehat{p}⟩=p_0$. The widths can be read off by comparing to Eq. (112): ${\sigma }^2$ is replaced by ${\mathrm{\hslash }}^2/4{\sigma }^2$. Hence $\mathrm{\Delta }p=\frac{\mathrm{\hslash }}{2\sigma }$, and the product of uncertainties $\mathrm{\Delta }x\mathrm{\Delta }p$ indeed satisfies Eq. (110).

Minimal-uncertainty wave packets have many applications, especially since they are the best approximation to the ”classical” situation in which both the position and the momentum of a particle are specified at the same time. A prominent example of a minimal-uncertainty wave packet is the ground state of a harmonic oscillator, which is discussed on a worksheet.

Let us now consider the ground state wave function of a particle in a square well with two hard walls at $x=\pm L/2$. This wave function is given by $\psi =\sqrt{\frac{2}{L}}\cos (\pi x/L)$ in between the walls, and vanishes inside the walls.

We find the expectation values $⟨x⟩=0$, $⟨p⟩=0$, $⟨x^2⟩=L^2(1/12-1/2{\pi }^2)$, and $⟨p^2⟩={\mathrm{\hslash }}^2{\pi }^2/L^2$. Hence $\mathrm{\Delta }x\approx 0.181L$ and $\mathrm{\Delta }p=\mathrm{\hslash }\pi /L$, so that $\mathrm{\Delta }x\mathrm{\Delta }p\approx 0.57\mathrm{\hslash }$, which clearly fulfils the uncertainty relation.

The maximal position uncertainty $\mathrm{\Delta }{x}_{\max }=L/2$ is obtained for a particle that sits with equal probability at $x=\pm L/2$. Hence, the uncertainty in momentum in any state fulfils $\mathrm{\Delta }p\ge \mathrm{\hslash }/L$.

This can be used to give a lower bound for the ground state (or zero-point) energy: $E_1=⟨p^2⟩/2m\ge {(\mathrm{\Delta }p)}^2/2m\ge \frac{{\mathrm{\hslash }}^2}{2m{L}^2}$. The exact result $E_1=\frac{{\mathrm{\hslash }}^2{\pi }^2}{2m{L}^2}$ clearly fulfills this condition.

In the Schrödinger equation, time is not an observable but a parameter. Hence, no operator is associated with time. However, it is still useful to talk about the duration of events in quantum mechanics. A good example is an atomic excitation, which decays over a certain time $\mathrm{\Delta }t$. If one looks at the outgoing radiation, the frequency is not sharp but spread over a range $\mathrm{\Delta }\omega \sim 1/\mathrm{\Delta }t$ (you will verify this on a worksheet). In quantum mechanics, frequency is related to energy via $E=\mathrm{\hslash }\omega$, and one ends up with the energy-time uncertainty relation

$$\mathrm{\Delta }t\mathrm{\Delta }E\ge \mathrm{\hslash }/2.$$

(115)

The specific value $\mathrm{\hslash }/2$ of the lower bound can be motivated from the similar forms of the energy and momentum operators $\widehat{E}=i\mathrm{\hslash }\frac{\partial }{\partial t}$ and $\widehat{p}=-i\mathrm{\hslash }\frac{\partial }{\partial x}$, which differ (besides the sign) only by the variable which they differentiate ($t$ or $x$).

Even though the energy-time uncertainty relation stands on a different footing than the position-momentum uncertainty relation, it is as useful as the latter. In many cases a quantum system is excited into a state of a finite lifetime. Fluorescent atoms are one example, radioactive nuclei another. In solids, charge carriers occupy the available states only with a finite lifetime because they are disturbed by disorder and interactions. In these systems, often a detailed calculation of the quantum dynamics is complicated, but the energy-time uncertainty relation captures many important effects.