2.1 The Free Particle

Now that we have done plenty of exposition, we will finally study at what the motion of a quantum particle actually looks like. We will first analyse the simplest case, which is that of a free particle.

2.1.1 Solving the Schrodinger Equation

A free particle is under the influence of no forces, so the potential $V(x)$ is zero. The TISE (equation 1.18) then becomes

	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{\operatorname{\mathrm{d}}\!^{2}u(x)}{% \operatorname{\mathrm{d}}\!x^{2}}=Eu(x)$		(2.1)
	$\displaystyle\implies\frac{\operatorname{\mathrm{d}}\!^{2}u(x)}{\operatorname{% \mathrm{d}}\!x^{2}}=-\frac{2mE}{\hbar^{2}}u(x).$		(2.2)

For a free particle, the total energy $E$ is just the kinetic energy $\frac{p^{2}}{2m}$ , so inputting this gives

\frac{\operatorname{\mathrm{d}}\!^{2}u(x)}{\operatorname{\mathrm{d}}\!x^{2}}=-% \frac{p^{2}}{\hbar^{2}}u(x)=-k^{2}u(x),

(2.3)

where we have defined $k^{2}=\frac{p^{2}}{\hbar^{2}}$ .

This happens to be an equation we know very well, the equation for simple harmonic motion! This implies that the spatial part of the energy eigenfunctions take the form

u(x)=Ae^{ikx}.

(2.4)

These solutions are known as plane wave solutions. Plane waves of the form $e^{-ikx}$ are also solutions, but we will ignore them for now and bring them back later.

As we learned in section 1.2.3, the temporal part of the solution is given by equation 1.23. We multiply this with the spatial part to get the full energy eigenfunctions for the free particle:

\psi(x,t)=Ae^{\frac{i}{\hbar}(px-Et)}=Ae^{i(kx-\omega t)},

(2.5)

where we have introduced the frequency $\omega$ of the temporal oscillation given by

E=\hbar\omega.

(2.6)

This is the well-known Einstein relation. $k$ represents the wavenumber of the oscillation in space, and is related to the particle’s momentum by

p=\hbar k.

(2.7)

This is known as the de Broglie relation.

Having already derived two seminal results of early quantum mechanics, you might think we are not doing too bad, however there are a couple of problems with these energy eigenfunctions.

The first issue is that these plane waves clearly do not represent the most general solution to the Schrodinger equation. In principle, we should be able to choose any valid initial wavefunction $\psi(x,0)$ , which need not even be separable, and then the Schrodinger equation determines its evolution. However, for a plane wave, we can only specify $A$ which is its amplitude (and initial phase if $A$ is complex) at $t=0$ .

The second issue is more serious, and it is that the plane wave solutions are unnormalisable. To see this, we can calculate the integral of probability density over all space:

\int_{-\infty}^{\infty}\lvert\psi(x,t)\rvert^{2}\operatorname{\mathrm{d}}\!x=% \int_{-\infty}^{\infty}\lvert Ae^{i(kx-\omega t)}\rvert^{2}\operatorname{% \mathrm{d}}\!x=\lvert A\rvert^{2}\int_{-\infty}^{\infty}\operatorname{\mathrm{% d}}\!x=\infty.

(2.8)

This happens because plane waves do not decay as $x\to\pm\infty$ . Physically, this happens because plane waves have definite momentum and therefore, by the uncertainty principle, their position is completely undefined. This reflects the fact that if we calculate the probability density for a plane wave, it is constant everywhere in space!

\lvert\psi(x,t)\rvert^{2}=\lvert Ae^{i(kx-\omega t)}\rvert^{2}=\lvert A\rvert^% {2}.

(2.9)

The fact that plane waves have definite momentum is elucidated by the fact that they are eigenstates of the momentum operator. This means that if we measure the momentum of a plane wave, we get back a definite result.

\hat{p}\psi(x,t)=-i\hbar A\frac{\partial}{\partial x}e^{\frac{i}{\hbar}(px-Et)% }=-i\hbar A\frac{ip}{\hbar}e^{\frac{i}{\hbar}(px-Et)}=p\psi(x,t).

(2.10)

To be clear, the fact that momentum eigenstates are unnormalisable means that they are unphysical, a free particle can never be found in a stationary state.

Luckily both of these problems, the lack and generality and the unnormalisability, can be solved using the linearity of the Schrodinger equation, i.e. using the superposition principle.

2.1.2 Wave Packets

For a wavefunction to be normalisable, it must be localised to some extent in space. Using the principle of superposition, we can add together momentum eigenstates to get a normalisable wavefunction. This will have the consequence of the particle no longer having definite momentum, but that means that position will no longer be undefined.

Let us consider a superposition of two momentum eigenstates with wavenumbers $k_{1}$ and $k_{2}$ . Note that wavenumber $k$ is basically synonymous with momentum in quantum mechanics, as the two are related by the de Broglie relation 2.7.

\psi(x,t)=Ae^{i(k_{1}x-\omega_{1}t)}+Be^{i(k_{2}x-\omega_{2}t)}.

(2.11)

The temporal frequencies $\omega_{1}$ and $\omega_{2}$ are given by the Einstein relation 2.6, so $\omega_{1}=\frac{E_{1}}{\hbar}=\frac{p_{1}^{2}}{2m\hbar}$ and similarly for $\omega_{2}$ . In the same way that wavenumber is synonymous with momentum, temporal frequency is synonymous with energy. This solution is no longer separable, i.e. it is not of the form $\psi(x,t)=u(t)e^{i\omega t}$ . Now, the probability density is not uniform over all space and this wavefunction is no longer a momentum eigenstate, however this wavefunction is still not normalisable as it is periodic.

In fact, we can add together infinitely many plane waves, and this would still be the case.

\psi(x,t)=\sum_{n}A_{n}e^{i(k_{n}x-\omega_{n}t)}.

(2.12)

Notice that this is a Fourier series, which are always periodic over the domain $(-\infty,\infty)$ .

To fully eliminate the periodicity, we need to make the range of wavenumbers continuous, and so the sum becomes an integral.

\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(k)e^{i(kx-\omega t)}% \operatorname{\mathrm{d}}\!k.

(2.13)

The intuition for why this is the case is that as the spacing between wavenumbers decreases, the spacing between maxima in the Fourier sum increases. So in the limit that the wavenumber spacing goes to zero and becomes continuous, the spacing between maxima goes to infinity. Notice that the amplitudes and initial phases of the plane waves $A_{n}$ have become a function of the wavenumber $A(k)$ . This definition of $\psi(x,t)$ in terms of an integral of plane waves is known as a wave packet. The factor $1/\sqrt{2\pi}$ in front of the integral is placed there because it happens to be the correct normalisation (so the integral of $\lvert\psi(x,t)\rvert^{2}$ over all space is one), which will be explained in a moment.

If we absorb the time evolution into $A(k)$ so it becomes $A(k,t)=A(k)e^{-i\omega t}$ , then this wavefunction becomes

\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}A(k,t)e^{ikx}% \operatorname{\mathrm{d}}\!k,

(2.14)

which shows that the wavefunction $\psi(x,t)$ is the Fourier transform of $A(k,t)$ . Likewise, we can say that the amplitude $A(k,t)$ is the Fourier transform of the wavefunction $\psi(x,t)$ :

A(k,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,t)e^{-ikx}% \operatorname{\mathrm{d}}\!x.

(2.15)

There is a result from mathematics called Parseval’s theorem, which says that if $\psi(x,t)$ is normalised, then its Fourier transform $A(k,t)$ is normalised, and vice versa. This means that if we enforce

\int_{-\infty}^{\infty}\lvert A(k,t)\rvert^{2}\operatorname{\mathrm{d}}\!k=1,

(2.16)

then $\psi(x,t)$ will be a correctly normalised wavefunction.

Note that we can also define this Fourier integral in terms of momentum using the de Broglie relation 2.7:

	$\displaystyle\psi(x,t)$	$\displaystyle=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi(p,t)e^{% \frac{i}{\hbar}px}\operatorname{\mathrm{d}}\!p$		(2.17)
	$\displaystyle\phi(p,t)$	$\displaystyle=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\psi(x,t)e^{-% \frac{i}{\hbar}px}\operatorname{\mathrm{d}}\!x,$		(2.18)

where the normalisation factor has changed to $1/\sqrt{2\pi\hbar}$ .

We can interpret $\lvert A(k,t)\rvert^{2}$ as the wavenumber probability density and $\lvert\phi(p,t)\rvert^{2}$ as the momentum probability density, i.e. $\lvert\phi(p,t)\rvert^{2}\operatorname{\mathrm{d}}\!p$ represents the probability that a measurement of a particle’s momentum will be in the range $(p,p+\operatorname{\mathrm{d}}\!p)$ at time $t$ .

In summary, we have found that by choosing an appropriate range of momenta and amplitudes for their corresponding plane waves (namely a $\phi(p,t)$ or $A(k,t)$ which is normalisable), we can construct a normalised wave packet $\psi(x,t)$ . This all followed from the principle of superposition. Furthermore, note that the Fourier transform relations above hold for any system in quantum mechanics, not just the free particle. However, there is an important caveat that $E=p^{2}/2m$ does not hold in general, only for the free particle.