Chapter 5: Momentum in Quantum Mechanics

Video: Momentum in Quantum Mechanics I

In the previous section we determined how free particles behave according to quantum mechanics. We saw that the stationary states of a free particle are unnormalisable, but by carefully constructing superpositions of stationary states it is possible to find normalised wavefunctions. In this section we will take a closer look at the significance of these superpositions, which will provide a way to define momentum in quantum mechanics.

Much like a particle doesn’t have a well-defined position, but is rather in a superposition of locations, in quantum mechanics particles do not have well-defined momenta, but rather have a superposition of momenta. We will see below that we can define probability amplitudes for a particle to have a momentum $p$ , and that this amplitude is specified by a momentum wavefunction.

5.1Plane waves and De Broglie¶

In the previous section, we found that the separable solutions to the Schrödinger equation for a free particle were given by (4.13), i.e.

\Psi_k(x,t) = \frac{1}{\sqrt{2\pi}} e^{-i\hbar k^2 t/2M} e^{ikx} ,

(5.1)

where $E = \hbar^2 k^2 / 2M$ was the constant which appeared in the TISE. Our starting point will be to take a closer look at these solutions, to better understand what they represent.

Similarly to how it was instructive to look at the angular frequency of the temporal part of a stationary state (see Section 3.1.1) it will also be instructive to look at the wavelength of the spatial part of these wavefunctions. In particular, for the function $e^{ikx}$ we immediately see that the wavelength is $\lambda = 2\pi/k$ since

e^{ik(x + \lambda)} = e^{i(kx + 2\pi)} = e^{ikx}

(5.2)

which means that $k$ is in fact the wave number of the wave (which is the reason why we chose to call it $k$ in the first place). Such functions are in fact complex plane waves. The real and imaginary parts of such functions oscillate sinusoidally with wavelength $\lambda = 2\pi/k$ .

We can now invoke de Broglie’s hypothesis (1.8), but in reverse. Whereas de Broglie’s hypothesis is that if a particle has a definite momentum $p$ , then it has definite wavelength $\lambda = h/p$ and therefore definite wavenumber $k = 2\pi/\lambda = p/\hbar$ , we can go in reverse, and say that if a wavefunction has a definite wavenumber $k$ , then it has a definite momentum $p = \hbar k$ .

We thus tentatively view the stationary wavefunctions $\Psi_k(x,t)$ in (4.13) as corresponding to particles with momentum $p = \hbar k$ , since they are states with a definite wavelength $\lambda = 2\pi/k$ .

A first indication that this is a good idea is to recall the relation between $E$ and $k$ from (4.5). We see that, in conjunction with the de Broglie relation $p = \hbar k$ , we obtain

\vph E = \frac{\hbar^2 k^2}{2M} = \frac{p^2}{2M},

(5.3)

which shows that the identification of $\hbar k$ with the momentum fits together nicely with the identification of $E$ in the TISE with the energy. In particular, if $E$ is the energy, and $p$ is the momentum, this equation is precisely the equation defining the kinetic energy of a particle.

5.2Superpositions of momenta¶

In the previous section we saw that the wavefunctions $\Psi_k(x,t)$ for a free particle are unnormalisable, and hence are inadmissible states for a particle. Now that these wavefunctions are seen to correspond to states of definite momentum, we arrive at the absolutely fundamental conclusion that it is impossible for a particle to have a definite momentum in quantum mechanics.

In order to get around this problem, we realised that a superposition of the states $\Psi_k(x,t)$ could in fact form a normalised wavefunction. We now see that such superpositions correspond to superpositions of momenta. In particular, when we introduced the superposition principle, we stressed that if the different wavefunctions being superposed have different physical properties, then it is natural to say that in the superposed wavefunction this property is in superposition. This is exactly what we are doing here, but now for the momentum of a particle.

5.3The momentum wavefunction¶

Video: Momentum in Quantum Mechanics II

We will now show how to go one step further, and describe the superposition of momentum quantitatively by introducing the idea of a probability density for momentum and an associated probability amplitude, in direct analogy to the way we defined these quantities for the position of the particle. In particular, previously we said that a particle with wavefunction $\Psi(x,t_0)$ is in a superposition of locations, with the probability amplitude at the position $x$ at time $t_0$ given by $\Psi(x,t_0)$ , and the probability density given by $P(x,t_0) = |\Psi(x,t_0)|^2$ .

Our goal is to determine analogous quantities for momentum, namely a momentum wavefunction $\tilde{\Psi}(p,t_0)$ such that the probability amplitude for the particle to have momentum $p$ at time $t_0$ is $\tilde{\Psi}(p,t_0)$ , and the probability density for momentum is $|\tilde{\Psi}(p,t_0)|^2$ .

To guide us, let us look once again at the expression (4.30) from the previous section, for a free particle at $t = 0$ , namely

\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\infint c(k) e^{ikx} dk,

(5.4)

which we showed was a normalised wavefunction as long as

\infint |c(k)|^2dk= 1.

(5.5)

If we substitute the de Broglie relation $p = \hbar k$ into this condition, taking care that $dp = \hbar dk$ , we obtain

\infint \frac{1}{\hbar}|c(p/\hbar)|^2 dp= 1.

(5.6)

This suggests that we should identify the integrand $\frac{1}{\hbar}|c(p/\hbar)|^2$ with $P(p,0)$ , the probability density for momentum at $t = 0$ . Going one step further, it also suggests that we identify $\frac{1}{\sqrt{\hbar}}c(p/\hbar)$ with $\tilde{\Psi}(p,0)$ , the momentum wavefunction at $t = 0$ , since the modulus square of this function is precisely $P(p,0)$ .

Having the momentum wavefunction and probability density defined in terms of $c(p/\hbar)$ is however not fully satisfactory, and hides the final message, since ultimately we would like to have them expressed in terms of the wavefunction $\Psi(x,0)$ itself, which, after all, is a complete description of the state of the particle at $t = 0$ . We can achieve this by using the relation (4.33) from the previous section. In particular, we find

\tilde{\Psi}(p,0) = \frac{1}{\sqrt{2\pi \hbar}}\infint \Psi(x,0) e^{-ipx/\hbar}dx.

(5.7)

This form also suggests how to extend the definition from $t = 0$ to all times $t$ : We simply replace $\Psi(x,0)$ on the right-hand side, with $\Psi(x,t)$ , arriving at

\tilde{\Psi}(p,t) = \frac{1}{\sqrt{2\pi \hbar}}\infint \Psi(x,t) e^{-ipx/\hbar}dx.

(5.8)

This is the momentum wavefunction of a particle. We see that at any given time $t_0$ it is completely specified by the wavefunction $\Psi(x,t_0)$ at the same time. That is, the momentum wavefunction is not independent of the wavefunction, but is a function of it. This was to be expected, since the wavefunction is a complete specification of the state of the particle at any given time. What we see is that a (rather complicated) function of the wavefunction specifies the probability amplitude and the corresponding probability density for the particle to have momentum $p$ .

We have thus achieved a very important goal — we now understand, in full generality, how momentum is specified in quantum mechanics.

5.3.1The momentum wavefunction as a different representation of the state¶

Interestingly, the momentum wavefunction is also a complete specification of the state of a particle. To see this, consider, just as we did in Section 4.4, multiplying both sides of (5.8) by $e^{ipx'/\hbar}$ and integrating over $p$ , to obtain

\begin{align*} \infint \tilde{\Psi}(p,t) e^{ipx'/\hbar}dp &= \frac{1}{\sqrt{2\pi \hbar}}\infint dp \infint dx \Psi(x,t) e^{-ipx/\hbar}e^{ipx'/\hbar},\\ &= \frac{1}{\sqrt{2\pi \hbar}}\infint \Psi(x,t) 2\pi \delta\left(\tfrac{x'-x}{\hbar}\right)dx , \\ &= \sqrt{2\pi \hbar}\Psi(x',t), \end{align*}

(5.9)

where we used the definition of the Dirac Delta function (4.17) to arrive at the second line, and the change of scale property (4.19), i.e. that $\delta\left(\tfrac{x'-x}{\hbar}\right) = \hbar \delta(x'-x)$ , to arrive at the third line. This shows that, after replacing everywhere $x'$ with $x$ , that

\vph\Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \infint \tilde{\Psi}(p,t) e^{ipx/\hbar}dp.

(5.10)

Thus, if the momentum wavefunction is known at a time $t_0$ , then it is possible to recover the spatial wavefunction $\Psi(x,t_0)$ at that time. Both wavefunctions are complete specifications of the state of a particle. We say that they constitute two different representations of the state.

In many problems we are interested primarily in the position of the particle, hence the wavefunction $\Psi(x,t)$ is a more useful representation. If however we care more about the momentum of the particle, which happens for instance often in scattering experiments in particle physics, or often in condensed matter physics, then it is more useful to work with the momentum wavefunction $\tilde{\Psi}(p,t)$ . You will see this later in your physics education.

5.4Gaussian wavepacket¶

Video: Momentum in Quantum Mechanics: Gaussian Wavepacket

We end this section by returning to our example of the Gaussian wavepacket of a free particle from Section 4.3, to look at it’s momentum.

Example 5.1 (Gaussian wavepacket of a free particle)

Recall that in Example 4.1 we considered the following superposition

c(k) = \left(\frac{2a^2}{\pi}\right)^{1/4}e^{-a^2(k-k_0)^2}.

(5.11)

which was a Gaussian function, centred at $k = k_0$ . We saw that it lead to the wavefunction

\Psi(x,0) = \left(\frac{1}{a\sqrt{2\pi}}\right)^{1/2} e^{-x^2/4a^2}e^{ik_0 x}.

(5.12)

Using the identity $\tilde{\Psi}(p,0) = \frac{1}{\sqrt{\hbar}}c(p/\hbar)$ , we see that the corresponding momentum wavefunction is

\tilde{\Psi}(p,0) = \left(\frac{2a^2}{\pi \hbar^2}\right)^{1/4} e^{-a^2(p-p_0)^2/\hbar^2},

(5.13)

where we have defined $p_0 = \hbar k_0$ . In Exercise 5.1 below, we show that at later times the momentum wavefunction is

\tilde{\Psi}(p,t) = \left(\frac{2a^2}{\pi \hbar^2}\right)^{1/4} e^{-a^2(p-p_0)^2/\hbar^2} e^{-ip^2t/2M\hbar},

(5.14)

i.e. the wavefunction at later times is multiplied by the phase factor $e^{-ip^2t/2M\hbar}$ . We say that it accumulates a phase in time. In a figure

The probability density for momentum at time $t$ , $P(p,t) = |\tilde{\Psi}(p,t)|^2$ , is therefore

P(p,t) = \left(\frac{2a^2}{\pi \hbar^2}\right)^{1/2} e^{-2a^2(p-p_0)^2/\hbar^2}.

(5.15)

Notably this is independent of time — the probability density of the momentum of the particle does not change in time for a free particle. This is the quantum mechanical manifestation of conservation of momentum that we expect to hold for a free particle from classical physics.

Moreover, we see that, just like the probability density in space $P(x,t)$ , the probability density for momentum is also a normal distribution, with mean and standard deviation

\begin{align*} \langle p(t) \rangle &= p_0, \\ \Delta p(t) &=~\frac{\hbar}{2a}. \end{align*}

(5.16)

This now explains why the wavepacket spreads in time. The particle has a superposition of momenta at all times. The evolution of the particle is the superposition of the different evolutions, one for each momentum. If the particle has larger momentum, then it travels further in a given time, and if it has small momentum, it travels less. It is the superposition of the initial wavepacket travelling all of the different distances that leads to a wider Gaussian wavepacket at later times.

Exercise 5.1 (Momentum wavefunction of a Gaussian wavepacket at time

t

)

In this exercise, we will derive the momentum wavefunction $\tilde{\Psi}(p,t)$ of a free particle at time $t$ , and apply it to the Gaussian wavepacket from Example 4.1.

By making the substitution $k = p/\hbar$ , show that expression for the wavefunction $\Psi(x,t)$ at time $t$ for a free particle, as given in (4.14), can be rewritten as
$\vph \Psi(x,t) = \frac{1}{\hbar\sqrt{2\pi}}\infint c(p/\hbar) e^{-i p^2 t/2M\hbar} e^{ipx/\hbar} dp.$
(5.17)
By appealing to the relationship between $\Psi(x,t)$ and $\tilde{\Psi}(p,t)$ given in (5.10), show that the above implies that the momentum wavefunction $\tilde\Psi(p,t)$ for a free particle is
$\tilde\Psi(p,t) = \frac{1}{\sqrt{\hbar}}c(p/\hbar)e^{-ip^2t/2M\hbar}$
(5.18)
Substitute $c(k)$ for a Gaussian wavepacket, given in (5.11), into this expression, to reproduce (5.14).