2.1. Quantum Mechanics of the Free Particle

2.1.1. The wave function

Schrodinger, Heisenberg, Born, and many others, developed the mathematical formalism for treating particles as waves. There are many analogous behaviours between quantum wave mechanics, electromagnetic waves and classical wave mechanics. For this reason it is typically the way most people are introduced to quantum mechanics. We will be no exception and will start our quantum mechanical treatment of particles by defining the wave function

(2.1)\[ \Psi(x,t)=Ae^{i(kx-\omega t)}. \]

Wave function notation

We will use \(\Psi(x,t)\) (The Greek letter psi) for a wave function defined as a function of position and time, \(\psi(x)\) for a wave function defined only by position, and occasionally we will add a subscript e.g. \(\psi_k(x)\) to indicate quantities that are constant values. In the latter example the subscript \(k\) indicates that the wave number has a definite value and should not be treated as a variable. In the former two examples of notation I have included the variables, but I may not always do this. Assume a capital \(\Psi\) or lower case \(\psi\) follows the definition given above.

Where \(k\) is the wave number, \(\omega\) is the angular frequency, \(x\) is the position coordinate, and \(t\) is time. The wave function contains all of the information we know about a particle. This can sound rather grand but in practice we will be solving problems in one dimension (typically \(x\)) where all we need to know is the momentum of a particle and it’s total energy. By total energy I mean the sum of the potential energy (\(V\)) and kinetic energy (\(E_\text{kin}\)). \(\omega\) can then be determined using the Planck-Einstein formula.

The Planck-Einstein formula gives the total energy \(E\)

(2.2)\[ E=hf=\frac{h\omega}{2\pi}=\hbar \omega. \]

Where \(h\) is Planck’s constant, \(\omega\) is the angular frequency, \(\hbar\) is the reduced Planck constant. The deBroglie formula gives the momentum \(p_x\)

(2.3)\[ p_x=\frac{h}{\lambda}=\frac{hk}{2\pi}=\hbar k. \]

where \(\lambda\) is the wavelength and \(k\) is the wavenumber. The kinetic energy \(E_\text{kin}\) can be calculated using

(2.4)\[ E_\text{kin}=\frac{mv^2_x}{2}=\frac{(mv_x)^2}{2m}=\frac{p^2_x}{2m}. \]

Where \(m\) is the mass of the particle, \(v\) is the velocity and \(p\) is the momentum. The subscript \(x\) denotes these are linear in the \(x\) direction, you can make similar equations for the \(y\) and \(z\) directions.

The momentum equation (2.3) is just another form of equation of the de Broglie equation we studied earlier in the course, equation (), and the right most form given for the kinetic energy, equation (2.4), is the one most typically employed in this course. Become as familiar with this form as you are with the \(\frac{1}{2}mv^2\) form. If you compare equation (2.1) to the wave representation we used for de Broglie waves, equation (1.15), you will see they are identical. Similarly, when we are not interested in the time (you are already familiar with this in classical wave mechanics from solving standing wave problems) we will define the wave function as

(2.5)\[ \psi(x)=Ae^{ikx}. \]

At this point it is also worth recalling the Euler’s relationship between exponential an trigonometric functions

(2.6)\[ e^{i\theta}=\cos(\theta)+i\sin(\theta) \]

Which can be employed to write equations (2.1) and (2.5) is a trigonometric form. We will look at this in more detail as the course progresses.

The Born Interpretation

The probability density of a wave function given by equation (2.7) is called the Born interpretation of the wave function. Born’s interpretation relates the square of wave function to the observed behaviour of particles and experiment backs up this interpretation. There are some proofs of this (see Gleason for example) but they are extremely difficult. We will treat this as a postulate and assume no proof is required. However, there is an analogue with electromagnetic waves. Take diffraction as an example, when we observe diffraction patterns on a screen we do not observe the amplitudes from the superposition of the waves but the square of the superposition of the electromagnetic waves.

When we discussed the behaviour of a de Broglie waves ( wave-like particles) passing through a slit we showed that the path of a particle is not predicted. Instead the de Broglie wave can say something about the probability of where a particle will end up after the slit. Hence, the wave function is related to the probability of finding a particle in a particular position. The probability of the particle being between the point \(x\) and \(x+dx\), where \(dx\) is a small change in \(x\), is given by

(2.7)\[ P(x)dx=\psi^*(x)\psi(x)dx=|\psi(x)|^2 dx \]

\(P(x)\) is called the probability density (probability per unit length around the point \(x\)). The wave function is complex and the * indicates that we use the complex conjugate of the wave function. The is Born’s probability interpretation of the wavefunction. For a complex number the conjugate is found by changing the sign of the \(i\) (i.e. in this case the imaginary part of the wave function). Whilst the wave function is always complex the expression for the magnitude used in the probability density function, \(\psi^*(x)\psi(x)\), will always be real.

2.1.2. The postulates of quantum mechanics

The two concepts introduced above are the commonly referred to as the postulates of quantum mechanics. These are the tenants on which everything in quantum mechanics is based:

• The properties of any quantum mechanical system are contained within the wavefunction.

• The wavefunction is interpreted as the probability amplitude with the absolute square interpreted as the probability density.

There are two more postulates we will cover in this course and we will summarise them as they are introduced. The two provided here are sufficient for the time being.

2.1.3. Properties of the wave function

Other properties of the wave function:

• It must be single valued and continuous.

• It’s first derivative must be continuous.

  • This condition can be dropped at infinity potential boundaries.

• It must be normalisable.

The last point is related to the interpretation of the wave function as a probability wave. Hence, the probability of finding the particle represented by the wave function somewhere must be 1. To achieve this we sum over all values of \(x\) by integrating equation (2.7) for \(-\infty\le x\le +\infty\).

2.1.4. Normalisation of a wave function

The equation for normalising a wavefunction is

(2.8)\[ \int_{-\infty}^{+\infty}|\psi(x)|^2 dx=\int_{-\infty}^{+\infty}\psi^*(x)\psi(x)dx=1 \]

Equation (2.8) is the nomalisation condition for our wave function. Notice that this equation is expressed in two ways; \(|\psi(x)|^2\) can be read as the magnitude of \(\psi(x)\) squared, which we calculate be multiplying \(\psi(x)^*\), the complex conjugate of the wavefunction, by \(\psi(x)\), the wavefunction. What does this mean? The integration adds together all of the probabilities for every allowed value of \(x\) giving the total probability of finding a particle at some position. The probability of finding the particle somewhere must be \(1\) hence the right and side of the equation. In practice you will normalise a wavefunction to find the value of the constant \(A\) in equation (2.5) (sometimes called the amplitude or the normalisation constant) that satisfies equation (2.8). We will give several example of this process throughout the course.

2.1.5. Definition of a Free Particle

The easiest way to start this discussion is with a free particle; this is a particle that is not subjected to any external forces. However, recall from mechanics that a force is related to the potential in which the particle is traveling by

\[ F=-\frac{dV}{dx}. \]

Where \(F\) is the force acting on the particle due to the rate at which the potential \(V\) changes with position \(x\). So by free particle we also mean the particle is traveling in a potential that is constant as a function of position. To make this easier still we will assume that the potential \(V\) is zero everywhere so that for the time being the only energy we need to consider is the kinetic energy. We know how such particles behave classically and you will have already encountered many problems in physics that rely on the following two properties.

• They will have a well defined position \(\vec{\mathbf{r}}=(x,y,z)\). Which you will often see them described as being localised to that position.

• They will also have a well defined momentum \(\vec{\mathbf{p}}=(p_x,p_y,p_z)\), which together with the initial position allows you to exactly predict the future location of the particle after some interval of time has passed.

We have already looked at electron diffraction and shown experimentally that neither of these properties is true in quantum physics. In this section of the course we will show that using a quantum wave mechanics description of a free particle means it doesn’t have a well defined position and we can not predict exactly where it will be after some interval of time.

2.1.6. Plane wavefunction: free particles with a definite momentum

The function for \(\psi(x)\) given by equation (2.5) is often referred to as a plane wave, you may already have employed this or the equivalent trigonometric form to calculate properties of mechanical or electromagnetic waves. How does this function behave mathematically? It extend out to \(x=\pm\infty\) and has a constant amplitude, hence the area under the probability density function is also infinite and no value of \(A\) satisfies the normalisation condition (2.8). We say \(\psi(x)\) can not be normalised. This is a problem but it is not unexpected, if we use equation (2.5) to represent our particle then it has a momentum determined by the quantity \(k\), i.e. there is no uncertainty in the momentum. Heisenberg tells us that the uncertainty in the position must then go to \(\infty\). We will describe this as a delocalised free particle. In almost all cases we would normalise the wavefunction to find the constant \(A\), in this case we can not so we will plot the function below:

\[ \psi_k(x) = Ae^{ikx} \]

and keep \(A\) as an arbitary constant. The subscript \(k\) is there to remind us to treat this as a single valued constant, i.e. this equation represents a particle with momentum \(\hbar k\) travelling in the positive \(x\) direction.

2.1.7. Visualising the wavefunction and probability density

To visualize the wave function \(\psi_k(x)\) we use Euler relationship, equation (2.6), and graph separately the Real and Imaginary contributions to the wave function. We can see from the figure below that these oscillate as we would expect.

On the other hand, the Probability Density is a constant, since:

\[ \psi^*_k(x) \psi_k(x) = Ae^{-ikx}Ae^{ikx} =A^2 \]

Note: From this point on many of the plots are interactive. If you download this notebook or open in GoogleColab (an online JupyterNotebook provided by google) then you can adjust the parameters for \(k\) and \(x_{max}\) in the plot below. If you don’t do this then the sliders at the top of the figure will do nothing.

Fig. 2.1 (Left) The real and imaginary components of a plane wave wavefunction plotted as a function of position. Note it has a constant amplitude as a function of position. (Right) The probability density function as a function of position. Note it is constant with position.

2.1.8. Interpreting the wavefunction and probability density

Figure 2.1 shows a wavefunction plotted on the left and a probability density for the wavefunction plotted on the right. The fact that this is a complex number can cause some confusion at first and we must learn how to interpret what plots like this show. It is also worth noting at this point that the probability density is a constant real number. How do we interpret such plots? We can think of the wavefunction (left) as showing how the particle is distributed in space and in this case the particle is distributed in such a way that it has no preferred location (we sometimes call this delocalised). We can infer this because the wave extends over all values of \(x\) with the same amplitude but this information is given exactly in the probability density plot (right) which has the same value for all \(x\) positions. Imagine performing an experiment, you are trying to measure the location of the particle so you set up a detector. You make a measurement that shows a particle at say \(x=5\) Angstroms. Now you decide to repeat the measurement, so you set up the experiment again with the same conditions and make a measurement. Will we see the particle at the same location? The answer is maybe but not certainly, this is a bit odd. After all if we were performing an experiment with a ball being thrown several times under the same condition we would expect to find it at the same location. This is where quantum mechanics differs from classical mechanics; we are no longer dealing with certain outcomes given the same starting point but we no have probabilities of an outcome given a given by the probability density (\(|\psi^*_k(x) \psi_k(x) |\)). WE can see by looking at the right had plot the particle has an equal probability of being observed at any value of \(x\) and so in our thought experiment it is more likely to be at some other value of \(x\) that \(x=5\).

In reality we will be looking for the probability of finding our free particle within \(\Delta x\), a small region in space. In our thought experiment this could be the width of the detector and the probability of finding the particle in the region \(\Delta x\) is the area under the probability distribution and when this is constant this is given by

\[ \psi^*_k(x) \psi_k(x) \Delta x = \left|\psi_k(x)\right|^2 \Delta x = \left(A e^{ikx}\right)^* A e^{ikx} \Delta x = A e^{-ikx} A e^{ikx} \Delta x = A^2 \Delta x \]

Mathematically this tells us that there is an equal probability of finding the particle represented by the wavefunction within a \(\Delta x\).

Thus when the particle has a well-defined momentum (\(\hbar k\)), the probability of finding it anywhere in space is the same.

We have already discussed that the wave function used above, whether in the form \(\psi(x)=Ae^{ikx}\) or \(\psi(x)=A\cos(kx)+B\sin(kx)\) is a complex function. In the first case is it easier to see as the argument of the exponential contains an \(i\), in the second case the amplitudes of each component \(A\) or \(B\) can be complex or the Euler identity can be employed to rewrite the function in its exponential form. Below is a two dimensional plot of the wavefunction showing the real and the imaginary parts of the plane wave wave-function \(\psi(x)=Ae^{ikx}\), the projections of the real and the imaginary parts are also provided. Don’t forget we can have a time component also, \(\Psi(x,t)=Ae^{i(kx-\omega t)}\), press play and observe how the \(i\omega t\) time component affects the wavefunction.

Once Loop Reflect

Fig. 2.2 An animation of the wavefunction showing both the real and imaginary components, press play and observe the behaviour of the wavefunction with time.

Question: What would happen if you increased the energy \(E\) of the particle?

2.1.9. Wavepackets: Localisation of a free particle

In reality when we perform an experiment to observe a particle we have some knowledge of where it will be, that is we can say that the particle has some preferred location. We will refer to these particles as being localised so that the probability density function should be larger for some values of \(x\) that others. To see how this might be achieved we will take a lesson from sound waves and beats; let’s remind ourselves what happens when two waves with slightly different wavelength are added together. The cell below will plot the superposition of the two waves represented by the function

\[ \psi(x)=\sin((k\pm\Delta k) x). \]

Try changing plot parameters for \(k\) and \(\Delta k\). \(x_{max}\) can be used to extend the range of the x-axis. Note: you must download or open the notebook in GoogleColab in order to interact with the plot.

Fig. 2.3 (Right) Beats caused by the superposition of two waves, this is a well known acoustic phenomena. (Left) The two waves that superpose to create the beats.

The pattern formed should be familiar from your study of beats with sound waves. As the two waves, see the left figure, move in and out of phase with each other the effect is to create maxima and minima in the amplitude of the combined wave, right figure.

In quantum mechanics we can interpret this as a particle that has a momentum of either \(p=\hbar(\bar{k}-\Delta k)\) or \(p=\hbar(\bar{k}+\Delta k)\). This means that if we take repeated measurements of the momentum then 50% of the results (50% because the amplitudes of the two component wavefunctions are the same) would be \(p=\hbar(\bar{k}-\Delta k)\) and the other 50% of the results would be \(p=\hbar(\bar{k}+\Delta k)\). The amplitude of the combined wavefunctions falls periodically to zero and maximises; we can interpret this as there being a higher probability of observing the particle at a position where the wavefunction has the largest amplitude.

Now we will introduced more wavefunctions in the superposition, in figure 2.4 several waves of the same amplitude are superposed. If you use the interactive plot you can add or remove the number of component wavefunction.

Fig. 2.4 (Top) Individual waves contributing to the superposition (Bottom). Note, if we interpret a wave function as being related to the probability of observing a particle then the most likely location for the plot above is \(x = 0\).

We are beginning to localise the position of our particle so that it is close to \(x = 0\) Angstroms. Notice that each wave has the same amplitude, from this we can infer that all momenta between the upper and lower limits are equally likely. Also, we are inferring the probability by looking at the wave. In the next section we will address these two limitations by:

• considering a mathematical method for using a continuous function for the uncertainty in the momentum.

• change weighting for each contribution so that momenta that are far from the average contribute less than those that are close to the average value.

• calculate probability density function for the wave function.

To construct out free particle wavefunction we add together contributions from waves representing different momenta. Mathematically, this can be achieved using integration. Hence, out wavefunction can be found by

\[ \psi_{\Delta k}={\int_\text{k low}^\text{k high}A(k)e^{i(kx-\omega t)}dk} \]

where \(A(k)\) is the amplitude of the wavefunction as a function of \(k\).

To address the second point in the list we will assume that the average momentum of the free particle is \(p=\hbar \bar{k}\) and is the most probable; as we move away from this value the probability of observing a particular value of momentum will fall away with a Gaussian (normal) distribution. We know the normal distribution works in many experimental situations so we will assume it is appropriate in this case. Mathematically, we can achieve this by integrating using the following conditions,

• the limits of the integral are \(k=\pm\infty\), that is we can observe any value of momentum it just become increasingly less likely the further away from the mean value we go.

• the amplitude will have a Gaussian function that reduces the amplitude of the wavefunction as we move further from the mean value of the momentum.

\[ A = A^\prime e^{-\frac{1}{2}(\frac{k-\bar{k}}{\Delta{k}})^2} \]

Applying these changes we integrate to obtain our new wavefunction

\[ \psi_{\Delta k}(x) =\int^{+\infty}_{-\infty}{A^\prime e^{-\frac{1}{2}(\frac{k-\bar{k}}{\Delta{k}})^2} \ e^{ikx}}dk, \]

where \(\Delta k\) is related to the width of the Gaussian distribution.

After integration, we obtain:

\[ \psi_{\Delta k}(x) = \sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2/\Delta k^2} e^{i\bar{k}x} \]

where \(A^\prime\) is a normalization constant and is equal to

\[ A^\prime=\frac{1}{\sqrt{2\Delta k\sqrt{\pi^3}}} \]

Try using equation (2.7) to calculate the probability density function. Don’t be put off by the maths, it’s only algebra.

The probability density function of a gaussian wave packet

Using equation (2.7) we can write the probability density function as

\[\psi_{\Delta k}^*\psi_{\Delta k} = \sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2\Delta k^2} e^{-i\bar{k}x} \times \sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2\Delta k^2} e^{i\bar{k}x} \]

which simplifies to

\[\begin{split} \psi_{\Delta k}^*\psi_{\Delta k} &= \sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2\Delta k^2} \times \sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2\Delta k^2}\\ &=(\sqrt{2\pi}A^\prime \Delta k e^{-\frac{1}{2}x^2\Delta k^2})^2 \\ &=2\pi(A^\prime \Delta k)^2 e^{-x^2\Delta k^2} \end{split}\]

Note that the terms before the exponential are constants and will determine the amplitude of the function, the exponential has some constants and a \(-x^2\) term. This makes it a gaussian shape centred on \(x=0\) with a width of \(\sigma=1/2\Delta k\).

Fig. 2.5 A Gaussian wave packet formed by a superposition of waves with average wavenumber \(\bar{k}\) and extending to \(\Delta k\) either side. Each wave has an amplitude determined by a Gaussian function such that show furthest from \(\bar{k}\) contribute the least. (left) the real and imaginary components of the wavefunction, (right) the probability density function.

Try running the interactive notebook (or study the next figure) and consider the following:

• For a fixed value of \(\bar{k}\) had does changing \(\Delta k\) effect the probability of observing the particle

  • Close to the origin?

  • Far from the origin?

• Does the momentum \(\bar{k}\) effect the probability distribution?

These are sometimes called wave packets and we can now compare the effect of the Gaussian distribution with different values of \(\Delta k\):

Fig. 2.6 The probability density functions for the superposition of wavefunctions with amplitude determined by a Gaussian function. For each plot the value of \(\bar{k}\) is the same but the uncertainty \(\Delta k\) increases from left to right. Notice that the average position is the same but as the uncertainty in the momentum increases the uncertainty in the position decreases.

Looking at Fig. 2.6 leads us to conclude that as the uncertainty in the momentum increases (larger \(\Delta k\)), the uncertainty in the position decreases, and the particle becomes more localized. This is the exactly the behavior predicted by the Heisenberg’s uncertainty principle.