Free Particle

“If anyone aims to have anything like a broad understanding of the sciences that underpin modern technology, as well as obtaining some insight into the modern view of the character of the physical world, then some knowledge and understanding of quantum mechanics is mandatory.” (Cresser (2001))

A material that seems very interesting is the PDF “Quantum Physics Notes” by Professor J. D. Cresser (Cresser (2001)). I will probably be working next with Chapters 3, 4, 6, and 7. But it seems to me that the entire PDF is an excellent introduction; I am just personally too lazy to review (once again) the mathematical formalism of quantum mechanics^[1]. But perhaps one day I will add it as an appendix.

Quantum mechanics provides a framework of physical and mathematical principles through which we seek to understand the physical nature of the world in which we live. These principles are fundamental and are presumed to apply to all physical systems. They embody fundamental physical and philosophical questions that can be abstracted from any particular physical system and studied independently of the specific systems in which they may be manifested. It is this abstract formulation of these principles that constitutes quantum mechanics.

Some historical background¶

Planck’s Black Body Theory (1900): In an attempt to understand the form of the spectrum of the electromagnetic radiation emitted by an object in thermal equilibrium at some temperature (a black body), Planck was forced to propose that the atoms making up the object absorbed and emitted light of (angular) frequency $\omega$ in multiples of a fundamental unit of energy, or quantum of energy, $E=\hbar\omega$, where $\hbar=h/2\pi$ and where $h$ is a new fundamental constant of nature, Planck’s constant, $h=6.622\times10^{-34}Js^{-1}$ though the name Planck’s constant is now often applied to $\hbar$.

Einstein’s Light Quanta (1905): Although Planck believed that the rule for the absorption and emission of light in quanta applied only to black body radiation, and was a property of the atoms, rather than the radiation, Einstein saw it as a property of electromagnetic radiation, whether it was black body radiation or of any other origin.In particular, in his work on the photoelectric effect, he proposed that light of frequency $\omega$ was made up of particles of energy $\hbar\omega$, now known as photons, which could be only absorbed or emitted in their entirety.So light, a form of wave motion, had been given a particle character.Much later, in 1922, the particle nature of light was quite explicitly confirmed in the light scattering experiments of Compton.

De Broglie’s Hypothesis (1924) Inspired by Einstein’s picture of light, a form of wave motion, as also behaving in some circumstances as if it was made up of particles, and inspired also by the success of the Bohr model of the hydrogen atom, de Broglie was lead, by purely aesthetic arguments to make a radical proposal.If light waves of frequency $\omega$ can behave under some circumstances like a collection of particles of energy $E=\hbar\omega$, then by symmetry, a massive particle of energy $E$, an electron say, should behave under some circumstances like a wave of frequency $\omega=E/\hbar$.

But a defining characteristic of a wave is its wavelength. For a photon, the wavelength of the associated wave is $\lambda=c/f$ where $f=\omega/2\pi$ . So what is it for a massive particle? From Einstein’s theory of relativity, which showed that the energy of a photon (moving freely in empty space) is related to its momentum $p$ by $E=pc$, it follows that

So $p=h/\lambda$. This equation then gave the wavelength of the photon in terms of its momentum, but it is also an expression that contains nothing that is specific to a photon.So De Broglie assumed that this relationship applied to all free particles, whether they were photons or electrons or anything else, and so arrived at the pair of equations

which gave the frequency and wavelength of the waves that were to be associated with a free particle of kinetic energy $E$ and momentum $p$. But now, given that particles can exhibit wave like properties, the natural question that arises is: what is doing the ‘waving’ ? Further, as wave motion is usually describable in terms of some kind of wave equation, it is then also natural to ask what the wave equation is for these de Broglie waves.

The latter question turned out to be much easier to answer than the first – these waves satisfy the famous Schr¨ odinger wave equation.But what these waves are is still, largely speaking, an incompletely answered question.Nevertheless, it is possible to say a considerable amount about what these waves tell us about the particle. It is this topic to which we now turn.

The Harmonic Wave Function¶

On the basis of de Broglie’s hypothesis, there is associated with a particle of energy $E$ and momentum $p$, a wave of frequency $f$ and wavelength $\lambda$ given by the de Broglie relations:

It more usual to work in terms of the angular frequency $\omega=2\pi f$ and wave number $k=2\pi/\lambda$ so that the de Broglie relations become

With this in mind, we are in a position to make a reasonable guess at a mathematical expression for the wave associated with the particle.The possibilities include (in one dimension)^[2]:

At this stage, we have no idea what the quantity $\Psi\left(x,t\right)$ represents physically.It is given the name the wave function, and in this particular case we will use the term harmonic wave function to describe any trigonometric wave function of the kind listed above.

In order to understand what information may be contained in the wave function, which will lead us to gaining a physical understanding of what it might represent, we will turn things around briefly and look at what we can learn about the properties of a particle if we know what its wave function is.

First, given that the wave has frequency $\omega$ and wave number $k$, then it is straightforward to calculate the phase velocity $v_{p}$ of the wave:

Thus, given the frequency and wave number of a wave function, we can determine the speed of the particle from the phase velocity of its wave function, $v=2v_{p}$.

We could also try to learn from the wave function the position of the particle.However, the wave function above tells us nothing about where the particle is to be found in space.We can make this statement because this wave function is the same everywhere i.e. there is nothing whatsoever to distinguish $\Psi$ at one point in space from any other.

Thus, this particular wave function gives no information on the whereabouts of the particle with which it is associated.So from a harmonic wave function it is possible to learn how fast a particle is moving, but not what the position is of the particle.

Wave Packets¶

From what was said above about a wave function that was constant throughout all space, it would seem that a wave function can only convey information on the position of the particle if the wave function did not have the same amplitude throughout all space.

In fact, since what we mean by a particle is a physical object that is confined to a highly localized region in space, ideally a point, it would be intuitively appealing to be able to devise a wave function that is zero or nearly so everywhere in space except for one localized region.

It is in fact possible to construct, from the harmonic wave functions, a wave function which has this property.To show how this is done, we first consider what happens if we combine together two harmonic waves of very close frequency. The resultis well-known: a ‘beat note’ is produced, i.e. periodically in space the waves add together in phase to produce a local maximum, while midway in between the waves will be totally out of phase and hence will destructively interfere.

Each localized maximum is known as a wave packet, so what is produced is a series of wave packets.Now suppose we add together a large number of harmonic waves with wave numbers $k_{1},k_{2},k_{3},\dots$ all lying in the range:

around a mean value $\overline{k}$, i. e.

where $A\left(k_{n}\right)$ is a function peaked about the mean value $\overline{k}$ with a full width at half maximum $2\Delta k$^[3]. What is found is that in the limit in which the sum becomes an integral:

all the waves interfere constructively to produce only a single beat note (in effect, the ‘beat notes’ or wave packets are infinitely far apart).In other words, the wave function so constructed is found to have essentially zero amplitude everywhere except for a single localized region in space, over a region of width $2\Delta x$, i.e. the wave function $\Psi\left(x,t\right)$ in this case takes the form of a single wave packet.

This wave packet is clearly particle-like in that its region of significant magnitude is confined to a localized region in space.Moreover, this wave packet is constructed out of a group of waves with an average wave number $\overline{k}$, and so these waves could be associated in some sense with a particle of momentum $\overline{p}=h\overline{k}$. If this were true, then the wave packet would be expected to move with a velocity of $\overline{p}/m$.This is in fact found to be the case, as the following calculation shows.

Because a wave packet is made up of individual waves which themselves are moving, though not with the same speed, the wave packet itself will move (and spread as well). The speed with which the wave packet moves is given by its group velocity $v_{g}$:

This is the speed of the maximum of the wave packet i.e. it is the speed of the point on the wave packet where all the waves are in phase.Calculating the group velocity requires determining the relationship between $\omega$ to $k$, known as a dispersion relation.

This dispersion relation is obtained from

Substituting in the de Broglie relations ($f=\frac{E}{h},\lambda=\frac{h}{p}$) gives

Where we also use: $\omega=2\pi f$ and wave number $k=2\pi/\lambda$. From which follows the dispersion relation

The group velocity of the wave packet is then

Substituting $\overline{p}=\hbar\overline{k}$, this becomes $v_{g}=\overline{p}/m$ .i.e.the packet is indeed moving with the velocity of a particle of momentum $\overline{p}$, as suspected.

This is a result of some significance, i.e. we have constructed a wave function of the form of a wave packet which is particle-like in nature.But unfortunately this is done at a cost. We had to combine together harmonic wave functions $\cos\left(kx-\omega t\right)$ with a range of $k$ values $2\Delta k$ to produce a wave packet which has a spread in space of size $2\Delta k$.The two ranges of $k$ and $x$ are not unrelated – their connection is embodied in an important result known as the Heisenberg Uncertainty Principle.

The Heisenberg Uncertainty Principle¶

The wave packet constructed in the previous section obviously has properties that are reminiscent of a particle, but it is not entirely particle-like – the wave function is non-zero over a region in space of size $2\Delta x$.In the absence of any better way of relating the wave function to the position of the atom, it is intuitively appealing to suppose that where $\Psi\left(x,t\right)$ has its greatest amplitude is where the particle is most likely to be found, i.e the particle is to be found somewhere in a region of size $2\Delta x$.

More than that, however, we have seen that to construct this wavepacket, harmonic waves having $k$ values in the range $\left(k-\Delta k,k+\Delta k\right)$ were adding together.These ranges $\Delta x$ and $\Delta k$ are related by the bandwidth theorem, which applies when adding together harmonic waves, which tell us that

Using $p=\hbar k$, we have $\Delta p=\hbar\Delta k$ so that

[A more rigorous derivation, based on a more precise definition of $\Delta x$ and $\Delta k$ leads to

A closer look at this result is warranted.A wave packet that has a significant amplitude within a range $2\Delta x$ was constructed from harmonic wave functions which represent a range of momenta $\overline{p}-\Delta p$ to $\overline{p}+\Delta p$.We can say then say that the particle is likely to be found somewhere in the region $2\Delta x$, and given that wave functions representing a range of possible momenta were used to form this wave packet, we could also say that the momentum of the particle will have a value in the range $\overline{p}-\Delta p$ to $\overline{p}+\Delta p$.

The quantities $\Delta x$ and $\Delta p$ are known as uncertainties for reasons that will become increasingly apparent, and the relation above is known as the Heisenberg uncertainty relation for position and momentum.

The Statistical Interpretation¶

From this point on, we will continue with Professor Paul Skrzypczyk’s work (Skrzypczyk (2026)).

According to quantum mechanics, in general, particles do not have well-defined properties such as position, momentum or energy. Quantum mechanics says that if at time $t=t_{0}$ a particle has a wavefunction $\Psi\left(x,t_{0}\right)$ then the probability to find the particle between $x=x_{a}$ and $x=x_{b}$ is

That is, the modulus square of the wavefunction specifies, in the form of probabilities, where the particle is at time $t_{0}$. We can interpret $\left|\Psi\left(x,t_{0}\right)^{2}\right|$ as the probability density $P\left(x,t_{0}\right)$ to find the particle at position $x$ at time $t=t_{0}$. That is,

is the probability to find the particle between $x$ and $x+dx$ at time $t_{0}$, for $dx$ infinitesimally small. We can think of this as being the probability to find the particle in the vicinity of $x$ at time $t_{0}$.

It is important to note that probabilities enter quantum mechanics in a fundamental way: Even if the most careful and ideal experiment were to be carried out in order to measure the position of a particle, then it would still be found in different places in different runs of the experiment. This means that in quantum mechanics, identical experiments lead to non-identical outcomes.

The probabilities that arise in quantum mechanics have nothing to do with imperfections or experimental limitations. This is contrary to classical mechanics, where in principle all experiments can be made perfectly repeatable, such that identical experiments lead to identical outcomes if sufficient care is taken (this is true even for chaotic systems, which are very sensitive to changes in conditions).

The probabilistic nature of quantum mechanics is arguably the biggest departure from classical mechanics, from which many of its counter-intuitive predictions arise.

With the above statistical interpretation in place, it is useful to introduce some terminology. We will say that $\Psi\left(x,t_{0}\right)$ is the probability amplitude for the particle to be at position x at time $t_{0}$. When there is no loss of confusion, we will also simply use amplitude in place of probability amplitude. Probability amplitudes are thus complex numbers, whose modulus squares are probabilities.

In order for the statistical interpretation of the wavefunction to be possible, $\Psi\left(x,t_{0}\right)$ is required to satisfy a normalisation condition which ensures that the total probability for the particle to be found somewhere is one. Mathematically, because $\left|\Psi\left(x,t_{0}\right)^{2}\right|$ is the probability density $P\left(x,t_{0}\right)$, this condition is given by

Normalisation is not a very restrictive property^[4], since we can normalise functions to turn them into valid wavefunctions. In particular, if we have some unnormalised function $\Psi\left(x,t_{0}\right)$ that satisfies

where $N<\infty$, that is, if this integral is finite, then

will be a normalised wavefunction. This condition is known as being square-integrable. It simply says that the integral of the modulus squared of the function should be a finite number, i.e. that the total area under the curve is not infinite.

If $N$ would have been infinite then it would not have been possible to divide by $\sqrt{N}$ in order to obtain a normalised wavefunction, as dividing by infinity would have left us with a wavefunction that vanishes everywhere, which doesn’t make sense physically. We can also note that the normalization is independent of a phase factor $\theta$ in $\Psi'=\frac{e^{i\theta}}{\sqrt{N}}\Psi$. This shows that there is freedom in choosing how to normalise a wavefunction, since no matter how we normalise it we end up with the same probability density. A standard choice is to pick $\theta=0$, such that $e^{i\theta}=1$. That is, the standard choice is to normalise by a real and positive number (even though a negative, or complex number would also be an equally valid choice).

The requirement that the wavefunction be normalised imposes constraints on the behaviour of particles at $x=\pm\infty$. Physically, the normalisation says that the probability to find the particle somewhere must be one. If there is a non-zero probability to find the particle at infinity, then such a wavefunction necessarily cannot be normalised, as the area under the curve will be infinite.

This statement can be made more precise, as we can show that the wavefunction must decay to zero sufficiently fast in order to be normalised. In particular, $\Psi\left(x,t_{0}\right)$ must decay faster than $1/\sqrt{\left|x\right|}$ in order that the probability to find the particle at large $x$ does not diverge.

Let’s try to demonstrate this, starting with $\Psi=x^{-1/2}$ for $x\geq x_{0}$:

The total area under the curve of $\left|\Psi\left(x,t_{0}\right)\right|^{2}$ in the region $x\geq x_{0}$ is infinite. This shows that this function is not square integrable, and we cannot form a valid wavefunction from it. Let’s assume now that for $x\geq x_{0}$ , the wavefunction has the form $\Psi=x^{-\left(1/2+a/2\right)}=$ where $a\neq0$:

If $a>0$, the total area under the curve of $\left|\Psi\left(x,t_{0}\right)\right|$ in the region $x\geq x_{0}$is finite.This shows that this function is potentially square integrable (it is only potentially so since we don’t know how it behaves in other regions).

As we discussed briefly above, in classical mechanics we have a lot of freedom in specifying the state of the particle at some time $t_{0}$. In quantum mechanics, we have even more freedom. In general, within reason, any normalised wavefunction $\Psi\left(x,t_{0}\right)$ is a valid state for a particle. This is rather remarkable, as the wavefunction specifies a complex number at every point in space. There is thus significantly more freedom than just specifying the initial position $x_{0}$ and initial momentum $p_{0}$ in classical mechanics.

The Free Particle¶

We had already obtained:

Using$k^{2}=2mE/\hbar^{2}$ and $\omega=\frac{E}{\hbar}$, with some manipulation and by renaming the constants, we can write:

This is a separable solution to the free-particle Schrödinger equation for all values of $A$ and $B$, and all values of $k>0$. It might appear that we are thus done, as we have managed to solve the Schrödinger equation, and thus have found how a free particle evolves in time. There are however a number of obstacles that we still need to overcome.

First, we can specify an arbitrary initial state of the particle, i.e. an arbitrary wavefunction $\Psi\left(x,t_{0}\right)$ at $t=t_{0}$. For stationary states of the form above, at $t=t_{0}$, we only have two complex numbers $A$ and $B$ to specify, and hence this definitely does not amount to an arbitrary wavefunction. We therefore seem to have much less generality than we need.

More worryingly, as we will see next, the solutions found don’t appear to be permissible at all, due to their normalisation. We will see below that both of these problems can be overcome, by using the superposition principle.

Wavefunctions must be normalised in order that the probability to find the particle somewhere is one. Let us therefore attempt to normalise the wavefunction:

unless $A=B=0$. Indeed, when $A\neq0$ or $B\neq0$, the first two terms are positive and constant, and hence contribute an infinite amount to the integral. Moreover, the wavefunction does not decay for large $x$, as we already showed was necessary for normalised wavefunctions. On the other hand, $A=B=0$ leads to the solution $\Psi\left(x,t\right)=0$, which also cannot be normalised, and is inadmissible.

At first sight, this appears like a major problem with the separation of variables method. It is true that our assumption that the wavefunction is separable, for a free particle, is in fact inconsistent^[5], and free particles cannot be in stationary states. However, it does not mean that the method of separation of variables is a bad one, but only that we have to use it in a smarter way. We must further make use of the superposition principle^[6] in order to turn it into a powerful technique for solving the Schrödinger equation.

Consider two distinct separable solutions to the Schrödinger equation, $\Psi_{1}\left(x,t\right)$ and $\Psi_{2}\left(x,t\right)$, each with a corresponding constant $E_{1}>0$ and $E_{2}>0$, such that $E_{1}\neq E_{2}$. That is, consider

where $k_{i}=\frac{\sqrt{2mE_{i}}}{\hbar}$ and $A_{i},B_{i}$ are arbitrary complex constants. Both of these wavefunctions solve the Schrödinger equation separately. Now, because of the superposition principle we know that

is also a solution to the Schrödinger equation, for arbitrary complex numbers $a_{i}$, where $a_{0}=0$, and $k_{-j}=-k_{j}$.

This new superposed wavefunction has an interesting form. First, it is no longer separable. That is, $\Psi\left(x,t\right)$ is not of the form $\Psi\left(x,t\right)=X\left(x\right)T\left(t\right)$, for some functions $T\left(t\right)$ and $X\left(x\right)$. Second, the normalisation of $\Psi\left(x,t\right)$ is not directly related to the normalisation of $\Psi_{i}\left(x,t\right)$. We will not calculate the normalisation of $\Psi\left(x,t\right)$ just yet, but will show next that we can in fact form normalised wavefunctions by carefully superposing separable solutions.

Before doing so, it will turn out to be important to first ask ourselves what form the most general superposition of separable solutions can take. In particular, whereas we superposed two solutions in the above, more generally we can in fact superpose as many solutions as we wish.

In the above example it is instructive to realise that although we superposed two solutions, in fact superpose infinitely many separable solutions, one for every value of $k$ — positive or negative. Such a wavefunction can be written

where $c\left(k\right)$ is a complex function which collects together all of the constants that multiply each function $\Psi_{k}\left(x,t\right)$. We have written this as an integral as $k$ can vary continuously. I also have introduced the constant $1/\sqrt{2\pi}$ as it turns out to be the ‘correct’ normalisation. It is much easier to introduce it at this stage, even if this appears somewhat arbitrary and unjustified, rather than to wait and fix the normalisation later.

We will now show that by choosing the function $c\left(k\right)$ appropriately — i.e. by carefully superposing separable solutions — we can obtain a normalised solution to the Schrödinger equation. This is in spite of the fact that the separable solutions themselves cannot be normalised.

We can also use the insight that normalisation is conserved in time, in order to simplify our calculation. In particular, we need only calculate the normalisation of $\Psi\left(x,0\right)$, which has a simpler form than $\Psi\left(x,t\right)$. If $\Psi\left(x,0\right)$ is normalised, we know that $\Psi\left(x,t\right)$ will also be normalised, so we should carry out the simplest calculation we can. We see that

where we introduced a new variable $k'$ for the second integration (which is just a dummy variable that we are integrating over). This expression looks formidable, but it can in fact be substantially simplified.

The Dirac Delta function $\delta\left(x\right)$ is defined such that:

The delta function is an infinitely thin, infinitely tall ‘spike’. Two key properties of the delta function are that it is an even function, $\delta\left(-x\right)=\delta\left(x\right)$ and that under a change of scale $\delta\left(ax\right)=\delta\left(x\right)/\left|a\right|$. One representation of the Dirac Delta function is $\delta\left(x\right)=\int^{+\infty}_{-\infty}\frac{e^{iax}}{2\pi}da$.

We therefore see that:

Therefore the wavefunction $\Psi\left(x,0\right)$ will be normalised if $\int^{+\infty}_{-\infty}\left|c\left(k\right)\right|^{2}dk=1$. To summarise, a wavefunction of the form 30 is a normalised solution to the Schrödinger equation whenever $\int^{+\infty}_{-\infty}\left|c\left(k\right)\right|^{2}dk=1$ is satisfied.

Example: Beginning

We are going now to study a solution to the Schrödinger equation known as a Gaussian wavepacket. This is a very special solution, whereby the particle is seen to be roughly localised in a packet, which evolves in time in a rather simple way, maintaining the same general form for all time. Consider a superposition specified by

which is a Gaussian function, centred at $k=k_{0}$, with width $\sqrt{2}/a$ (distance between inflection points):

Where we substitute $x=k-k_{0}$ ($\frac{dx}{dk}=1$) and $\alpha=2a^{2}$. This is a Gaussian integral with a well-known result (solution). We then define an integral $G=\int^{+\infty}_{-\infty}e^{-ax^{2}}dx$. We can then calculate:

Since $x\rightarrow y$ is the variable of integration in each of the integrals—that is, merely a label—we can substitute $x\rightarrow y$ in the second one, and we are then left with:

We can now perform a change of coordinate system, obviously also changing the area element. Using the polar coordinate system (definition):

To cover the entire space, our limits are $0\leq r<\infty$ and $0\leq\theta<2\pi$. Substituting:

Replacing $u=-ar^{2}$, then $\frac{du}{dr}=-a2r\rightarrow-\frac{du}{a}=2rdr$, and the limits of integration are now $u=-a0^{2}=0$ e $u=-ar\infty^{2}=-\infty$

Thus, if $G^{2}=\pi/a$, then $G=\sqrt{\pi/a}$ e:

Returning to our integral, then:

Specifying $c\left(k\right)$ can be seen to be equivalent to specifying an initial condition $\Psi\left(x,0\right)$. At $t=0$, the wavefunction is found to be

Replacing $u=k-k_{0}$:

And given that:

So:

Replacing again $v=u-\frac{ix}{2a^{2}}$:

This is also a Gaussian function, now centred at the origin, $x=0$, and multiplied by a phase factor $\exp\left(ik_{0}x\right)$.

The probability density to find the particle at position $x$ is:

Which is a normal distribution with mean^[7]:

In this case, since the integrand is an odd function, that is, if $I\left(x\right)=xe^{-bx^{2}}$, then $G\left(-x\right)=-G\left(x\right)$, the integral is simply zero, and we have $\left\langle x\right\rangle =0$. Now let’s calculate: $\left\langle x^{2}\right\rangle$.

We can find this using the standard Gaussian integral $\int^{+\infty}_{-\infty}e^{-bx^{2}}dx=\sqrt{\frac{\pi}{a}}$. Differentiating both sides with respect to the parameter $b$:

Then:

The standard deviation is then:

The particle is therefore localised close to the origin, and the smaller $a$ is, the smaller the standard deviation is, and hence the particle is more tightly localised. Based on the initial conditions $\Psi\left(x,0\right)$, we are then in a position to determine how the wavefunction $\Psi\left(x,t\right)$ evolves in time from this initial condition. But for our discussion, this is sufficient for now.

Example: end