13 Case Study: The Free Particle

13.1 Step 1: Hamiltonian Eigenfunctions

The starting point for understanding is to construct an orthonormal basis of eigenfunctions of the Hamiltonian operator \[\hat H = \frac{\hat p^2}{2m} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \, .\] In this case, it is first convenient to first discuss the momentum operator \(\hat p\). This is because an eigenfunction of the momentum operator \(\hat p\) with eigenvalue \(p\) is automatically an eigenfunction of the Hamiltonian operator \(\hat H\) with eigenvalue \(E = p^2/2m\).

The momentum eigenfunctions are solutions to the differential equation \[-i \hbar \frac{\partial }{\partial x} \phi_p(x) = p \phi_p(x) \, .\] The solutions are \[\phi_p(x) = \frac{1}{\sqrt{2\pi \hbar}} e^{i p x/\hbar}\] for any \(p \in \mathbb{R}\). The momentum operator therefore has a continuous spectrum and correspondingly the normalisation is chosen so that \[\begin{aligned} \langle \phi_p , \phi_{p'}\rangle & = \int^\infty_{-\infty} \overline{\phi_p(x)} \phi_{p'}(x) {\rm d}x \\ & = \frac{1}{2\pi \hbar}\int^{\infty}_{-\infty} e^{-i(p-p')x/\hbar} {\rm d}x \\ & = \delta(p-p') \, . \end{aligned}\] The spectrum of the momentum operator is also non-degenerate: there is a unique eigenfunction \(\phi_p(x)\) for each eigenvalue \(p \in \mathbb{R}\).

The same wave functions are also Hamiltonian eigenfunctions \[\hat H \cdot \phi_p(x) = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \phi_p(x) = E_p \phi_p(x)\] where \[E_p = \frac{p^2}{2m} \, .\] However, the spectrum of the Hamiltonian is degenerate since \(\phi_p(x)\) and \(\phi_{-p}(x)\) have the same energy since \(E_p = E_{-p}\). This is a consequence of “parity" symmetry \(x \mapsto -x\) that sends \[\phi_p(x) \mapsto \phi_p(-x) = \phi_{-p}(x) \, ,\] while leaving the Hamiltonian operator \(\hat H\) invariant. Nevertheless, the wave functions \(\phi_p(x)\) can be taken as an orthonormal basis of Hamiltonian eigenfunctions.

13.2 Step 2: Stationary Solutions

We can immediately promote the Hamiltonian eigenfunctions \(\phi_p(x)\) to stationary solutions of Schrödinger’s equation, \[\begin{aligned} \psi_p(x,t) & = \phi_p(x) e^{- i E_pt / \hbar} \\ & = \frac{1}{\sqrt{2\pi \hbar}} e^{i (p x- E_pt)/ \hbar} \, . \end{aligned}\] These stationary wave functions correspond to plane waves. In particular, \(\psi_p(x,t)\) and \(\psi_{-p}(x,t)\) are plane waves with equal magnitude of momentum in opposite directions and therefore equal energy.

13.3 Step 3: Time Evolution of Wave function

We now return to the problem at hand: how to determine the solution of Schrödinger’s equation \(\psi(x,t)\) given an initial wave function \(\psi(x,0)\).

We expand \(\psi(x,0)\) as a linear combination of the Hamiltonian eigenfunctions. In this case, the spectrum of the Hamiltonian operator is continuous and the expansion becomes an integral \[\begin{aligned} \psi(x,0) & = \int^\infty_{-\infty} {\rm d}p \, c(p) \, \phi_p(x) \\ & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}p \, c(p) \, e^{ipx/\hbar} \, . \end{aligned}\] The coefficients function \(c(p)\) is computed via the inverse relation, \[c(p) = \frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x,0) \, e^{-ipx/\hbar} \, .\] This is of course the Fourier transform between the initial wave function \(\psi(x,0)\) and the initial momentum space wave function, \(c(p) = \widetilde\psi(p,0)\)

To compute the wave function at later times, we promote the Hamiltonian eigenfunctions \(\phi_p(x)\) to stationary wave functions, \[\begin{aligned} \psi(x,t) & =\int^\infty_{-\infty} {\rm d}p \, c(p) \psi_p(x,t) \\ & = \frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty} {\rm d}p \, c(p) \, e^{ipx/\hbar} e^{-ip^2t/2m\hbar} \, . \end{aligned}\] We can express this in terms of the initial wave function \(\psi(x,0)\) by substituting in the inverse Fourier transform for the coefficients \(c(p)\) and interchanging the order of integration, \[\begin{aligned} \psi(x,t) & = \frac{1}{2\pi\hbar}\int^\infty_{-\infty} {\rm d}p \, \left( \int^\infty_{-\infty} {\rm d}x' \, \psi(x',0) \, e^{-ipx'/\hbar} \right) \, e^{ipx/\hbar} e^{-ip^2t/2m\hbar} \\ & = \frac{1}{2\pi\hbar} \int^\infty_{-\infty} {\rm d}x'\psi(x',0) \int^\infty_{-\infty} {\rm d}p \, e^{ip(x-x')/\hbar}e^{-ip^2t/2m\hbar} \, . \end{aligned}\]

The integral over momentum can now be computed using the standard Gaussian integral formula \[\int^\infty_{-\infty} \, {\rm d}y \, e^{-\alpha y^2+ \beta y} = \sqrt{\frac{\pi}{\alpha}} \, e^{\beta^2/4\alpha} \, .\] With the substitution \[\begin{aligned} \alpha = \frac{it}{2m\hbar} \qquad \beta = i\frac{(x-x')}{\hbar} \, , \end{aligned}\] we find \[\int^\infty_{-\infty} {\rm d}p \, e^{ip(x-x')/\hbar}e^{-ip^2t/2m\hbar} =\left(\frac{2\pi \hbar m}{i t}\right)^{1/2}e^{im(x-x')^2 / 2\hbar t} \, .\]

• Important Subtlety. The Gaussian integral above converges for \(\mathrm{Re}(\alpha)>0\) whereas our \(\alpha\) is purely imaginary. A more careful analysis would add a small imaginary time \(t \to t - i \epsilon\) and consider the limit \(\epsilon \to 0\), or deform the contour of integration slightly into the complex \(p\)-plane. This will not be necessary in this course - the above formulae will be correct for all examples we come across. An example is the Gaussian wave function, which we now turn to.

13.4 Example: Time-Evolution of Gaussian

Consider an initial Gaussian wave function \[\psi(x,0) = C e^{ip_0x/\hbar}e^{-x^2/4\Delta^2}\] with normalisation \(C = (2\pi\Delta^2)^{-1/4}\). The initial probability distribution is \[\begin{aligned} P(x,0) & =|\psi(x,0)|^2 \\ & = \frac{1}{\sqrt{2\pi\Delta^2}} e^{-x^2/2\Delta^2} \end{aligned}\] The initial wave function has the characteristic properties \[\begin{aligned} \langle x \rangle & = 0 \qquad && \Delta x = \Delta \\ \langle p\rangle & = p_0 \qquad && \Delta p = \frac{\hbar}{2\Delta} \, . \end{aligned}\] In particular, Heisenberg’s uncertainty principle is saturated, \(\Delta x \Delta p = \hbar/2\).

The wave function at later times is \[\begin{aligned} \psi(x,t) & = \left(\frac{m}{2\pi\hbar i t}\right)^{1/2} \int^\infty_{-\infty} {\rm d}x' \, \exp\left( \frac{im(x-x')^2}{2\hbar t}\right) \, \psi(x',0) \\ & = C \left(\frac{m}{2\pi\hbar i t}\right)^{1/2} \int^\infty_{-\infty} {\rm d}x' \, \exp\left( \frac{im(x-x')^2}{2\hbar t}\right) \, e^{ip_0x'/\hbar}e^{-x'^2/4\Delta^2} \, . \end{aligned}\] This can be evaluated by performing another Gaussian integral with parameters \[\begin{aligned} \alpha & =\frac{m}{2i\hbar t}\left( 1 + \frac{i\hbar t}{2m\Delta^2} \right)\\ \beta & = -\frac{im}{\hbar t}\left( x - \frac{p_0t}{m} \right) \, . \end{aligned}\] Notice that \(\mathrm{Re}(\alpha)>0\) so the Gaussian integral converges and there is no subtlety in computing it. After some simplification, the final result for the wave function at later times is \[\psi(x,t) = \left[ 2\pi\Delta^2\left(1+\frac{i\hbar t}{2m\Delta^2}\right) \right]^{1/4} \exp\left[ ip_0\left(x-\frac{p_0t}{2m}\right)\right] \exp\left[ \frac{-(x-p_0t/m)^2}{4\Delta^2(1+i\hbar t / 2m\Delta^2)} \right] \, .\] The probability distribution is particularly simple, \[P(x,t) = \frac{1}{\sqrt{2\pi\Delta(t)^2}} \exp\left[ -\frac{x(t)^2}{4\Delta(t)^2} \right]\] where \[x(t) = x - \frac{p_0t}{m} \qquad \Delta(t) = \Delta\sqrt{1+\frac{\hbar^2t^2}{4m^2\Delta^4}} \, .\] This is a time-dependent Gaussian characterised by \[\begin{aligned} \langle x \rangle & = \frac{p_0}{m} t \qquad && \Delta x = \Delta(t) \\ \langle p\rangle & = p_0 \qquad && \Delta p = \frac{\hbar}{2\Delta(t)} \end{aligned}\]

• \(\langle x\rangle\), \(\langle p\rangle\) obey the classical equations of motion: the centre of the wavepacket \(\langle x\rangle\) is moving with constant velocity \(p_0 /m\). In a later lecture, we will prove this result in generality.

• The position uncertainty \(\Delta(t)\) is increasing in time. Intuitively, the initial uncertainty in momentum (and therefore velocity) leads to an increasing uncertainty in position over time.

• Heisenberg’s uncertainty principle is saturated \(\Delta x \Delta p = \frac{\hbar}{2}\) for all \(t\): the wave function remains Gaussian.

• For a macroscopic object of mass \(1g = 10^{-3}kg\) with initial position determined up to uncertainty \(\Delta \sim 10^{-15}m\) equal to the width of a proton, it would take \(300,000\) years for the uncertainty in position to grow to \(\Delta\sim 1mm = 10^{-3}m\). It is therefore reasonable to treat macroscopic objects classically over long periods of time.

13.5 Problems

1. Time evolution of a free particle:

   Consider a free particle with initial wavefunction \(\psi(x,0)\).

   1. Assuming all relevant integrals converge, show that \[\psi(x,t) = \left( \frac{m}{2\pi i \hbar t} \right)^{1/2} \int_{-\infty}^\infty {dx'} \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0)\] is a solution of Schrödinger’s equation for a free particle, \[i \hbar \frac{\partial}{\partial t}\psi(x,t) = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)\, .\]

   2. Consider the initial wavefunction \[\psi(x,0) = \begin{cases} C & \quad \mathrm{if} -a < x < a \\ 0 & \quad \mathrm{else} \end{cases}\] where \(C\) is a normalization constant you should determine. Write down an integral for the wavefunction \(\psi(x,t)\) at later times.

   Solution ▶

   1. We compute both sides of Schrödinger’s equation by differentiating under the integral and compare.
      
      First computing derivatives with respect to \(x\): \[\begin{aligned} \partial_x\psi(x,t) & = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty dx' \, \partial_x \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0) \\ \nonumber & = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty dx' \, \frac{im}{\hbar t}(x-x') \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0) \end{aligned}\] \[\begin{aligned} \partial^2_x\psi(x,t) & = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty dx' \, \left[ \frac{im}{\hbar t} + \left( \frac{im}{\hbar t}\right)^2(x-x')^2 \right] \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0) \\ \nonumber \end{aligned}\] and hence \[-\frac{\hbar^2}{2m} \partial^2_x\psi(x,t) = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty dx' \, \left[ -\frac{i\hbar}{2t} +\frac{m}{2 t^2}(x-x')^2 \right] \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0)\] Second computing the derivative with respect to \(t\): \[\begin{aligned} \partial_t \psi(x,t) & = \partial_t\sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty {dx'} \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0) \\ \nonumber & \quad + \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty {dx'} \,\partial_t\exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \psi(x',0) \\ & = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty {dx'} \, \left[ -\frac{1}{2t} - \frac{im}{2\hbar t^2}(x-x')^2 \right] \exp\left( \frac{im}{2\hbar t}(x-x')^2 \right) \psi(x',0) \, , \end{aligned}\] and hence \[i\hbar \partial_t \psi(x,t) = \sqrt{ \frac{m}{2\pi i \hbar t} } \int_{-\infty}^\infty {dx'} \, \left[ -\frac{i\hbar}{2t} + \frac{m}{2 t^2}(x-x')^2 \right] \exp\left( \frac{im}{2\hbar t}(x-x')^2 \right) \psi(x',0)\] We therefore see explicitly that \[i \hbar \partial_t \psi(x,t) = -\frac{\hbar^2}{2m}\partial_x\psi(x,t) \, .\]

   2. The normalization constant is \(C = 1/\sqrt{2a}\).
      
      The wavefunction at \(t>0\) is \[\psi(x,t) = \left( \frac{m}{4\pi i \hbar a t} \right)^{1/2} \int_{-a}^a {dx'} \exp\left( \frac{im}{2\hbar t}(x-x')^2\right) \, .\] You are not expected to know how to compute this integral.

2. Numerical wave packets:

   In the notebook linked in the margin, you will find some basic Python code to numerically solve the Schrödinger equation. It is used to compute the time-evolution of a Gaussian wave packet, based on the discretisation scheme \[\begin{gathered} \psi(x,t + {\rm d}t) = \psi(x, t-{\rm d}t) \\[1ex] + i {\rm d}t \Big[ \psi(x-{\rm d}x, t) + \psi(x+{\rm d}x, t) - 2 \big(1 + V(x)\big) \psi(x,t) \Big]\,. \end{gathered}\] For more information on this, see section 3.3 of Schroeder’s book.

   Sample notebook to numerically evolve the Schrödinger equation.

   Use this code to verify that width of the Gaussian wave function spreads as \[\Delta(t) = \Delta\sqrt{1+\frac{\hbar^2t^2}{4m^2\Delta^4}} \, ,\] as we derived analytically.

3. Degenerate or non-degenerate?:

   In an earlier chapter we have argued that the spectrum of the Hamiltonian operator on the real line is non-degenerate. However, the present chapter starts by arguing that there are two wave functions for every energy eigenvalue. Which assumption in that proof of non-degeneracy does not hold in the present chapter?