2.1. Quantum Mechanics of the Free Particle

2.1.1. The wave function

Wave function notation

We will use \(\Psi(x,t)\) 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.

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

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 (\(E_\text{kin}\)), however it can be determined using the Planck-Einstein formula.

(2.2)\[\begin{split} \text{Total energy: }& E=hf=\frac{h\omega}{2\pi}=\hbar \omega \\ \text{Momentum: }& p_x=\frac{h}{\lambda}=\frac{hk}{2\pi}=\hbar k \\ \text{Kinetic energy: }& E_\text{kin}=\frac{mv^2_x}{2}=\frac{(mv_x)^2}{2m}=\frac{p^2_x}{2m} \end{split}\]

The momentum equation is just another form of equation (1.10) and the form given for the kinetic energy 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 () to the wave representation we used for de Broglie waves, equation (1.14), 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

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

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

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

Which can be employed to write equations () and () 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 () 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

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

\(P(x)\) 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. 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. Properties of the wave function

Other properties of the wave function:

• It must be 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 () for \(-\infty\le x\le +\infty\).

2.1.3. Normalisation of a wave function

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

Equation () is the nomalisation condition for our wave function.

2.1.4. Wave function of a Free Particle

We start by describing a free particle: this we will define as a particle that is not under the influence of a potential i.e. no forces are acting on the particle. 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 localized to that position.

• They will also have a well defined momentum \(\vec{\mathbf{p}}=(p_x,p_y,p_z)\).

How do free particle behave according to quantum mechanics?

The function for \(\psi(x)\) given by equation () 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 \(x=\pm\infty\), hence \(\psi(x)\) can not be normalized. This is a problem but not unexpected, if we use equation () 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, the constant \(A\) is chosen in such a way (see note below) that the function becomes:

\[ \psi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx} \]

I have also introduced a subscript \(k\) to remind us to treat this as a single valued constant.

Side note on the normalization factor and Dirac’s delta function (For interest not examinable)

The normalization is achived by:

\[\begin{split} 1&=\int^{+\infty}_{-\infty} \psi_{k'}^*(x)\psi_k(x) dx = \int^{+\infty}_{-\infty} c^* e^{-ik'x} ce^{ikx} dx\\ &= |c|^2 \int^{+\infty}_{-\infty} e^{i(k-k')x} dx = |c|^2 2\pi\ \delta(k-k')=1 \end{split}\]

NB: the coefficients of our wavefunction are arbritary complex constants but when they are squared we multiply the complex number say \(c=a+ib\) by it’s complex conjugate which would be \(c^*=a-ib\). The product is always real and given by \(a^2+b^2\) and since the \(c\) can be thought of as a vector in the complex plane then \(|c|=\sqrt{a^2+b^2}\) hence \(c^*c=|c|^2\).

In the last step of the normalization we used the definition of the Dirac’s delta function:

\[ \delta(x) = \frac{1}{2\pi} \int^{+\infty}_{-\infty} e^{ivx} dv \]

which obeys the following properties:

\[ \delta(x-a)=0\ \text{ if } x \neq a \]

\[ \int^\infty_{-\infty} f(x)\ \delta(x-a) \ dx =f(a) \]

The normalization constant \(c\) is chosen as \(c = \frac{1}{\sqrt{2\pi}}\) in such a way that:

\[ \int^{+\infty}_{-\infty} \psi_{k'}^*(x)\psi_k(x) dx = \delta(k-k') \]

\(k-k'\) represents some range of \(k\) values and hence a range of momentum values (NB: \(p=\hbar k\)). We might consider this value \(\Delta p\) hence in this way already see that the quantum mechanics wave approach gives us an uncertainty in the momentum as we would expect from the Heisenberg uncertainty principle. The Dirac delta function acts to make the probability of finding our particle equal to 0 outside this range and 1 within this range. So in this way it is possible to normalize the wavefunction even though all our wavefunction extend to \(x=\pm\infty\).

To visualize the wave function \(\psi_k(x)\) we use Euler relationship, equation (), 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:

\[ \psi^*_k(x) \psi_k(x) = \frac{1}{\sqrt{2 \pi}}e^{-ikx}\frac{1}{\sqrt{2 \pi}}e^{ikx} =\frac{1}{2\pi}\]

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.

What do we learn from this graphical representation?: The first plot shows that that the wave function is completely delocalized in the x coordinate. The value of the probability density (\(|\psi^*_k(x) \psi_k(x) |\)) is the same for all values of \(x\).

An important question to ask is: what is the probability to find a free particle within a \(\Delta x\) region in space?

This is crucial for measurements, since we would never measure ALL space at the same time, but instead, our instrument will measure a section of \(x\) space

\[ \psi^*_k(x) \psi_k(x) \Delta x = \left|\psi_k(x)\right|^2 \Delta x = \left(\frac{1}{\sqrt{2\pi}} e^{ikx}\right)^*\frac{1}{\sqrt{2\pi}} e^{ikx} \Delta x = \frac{1}{2\pi} \Delta x \]

The probability of finding it within a \(\Delta x\), at any place in space, is the same.

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

The wave function as a complex function: 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 wave function 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.

Once Loop Reflect

By Localization we mean that certain values of \(x\) have a high probability of the particle being present. This is of course a more realistic situation that we looked at above, typically we will have some knowledge of the location of the particle. We can localise a the particle by considering not just a single value of momentum (\(\hbar k\)), but a range of values. We will consider only two cases

• The momentum is a constant within some range of values.

• The momentum follows a Gaussian distribution within some range of values.

Consider the superposition of waves with a range of momenta given by \(p=p_0 \pm \Delta p\) \(\hspace{0.25cm}\) or \(\hspace{0.25cm}\) \(\hbar k = \hbar(k_0 \pm \Delta k)\)

First 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.

The pattern formed should be familiar from your study of beats. Notice that the amplitude falls periodically to zero.

Now consider the situation where the momentum is a constant have of \(\hbar k\) between \(k-\Delta k\) and \(k+\Delta k\). For the superposition we can integrate for the waves:

\[\begin{split} \psi_{\Delta k}(x) &=N\int^{k_o+\Delta k}_{k_o-\Delta k}\ e^{ikx}dk \\ &= \frac{2N\sin(\Delta kx)}{x} e^{ik_ox} \end{split}\]

where \(N\) is a normalization constant and is equal to

\[ N=\frac{1}{2\sqrt{\pi\Delta k}} \]

Don’t forget that the wavefunction is a complex number so to understand this new wavefunction, we plot its Real and Imaginary components separately.

Try running the cell below with \(k_0\) between 1 and 10 and \(\Delta k < k_0\).

Assuming \(k_o > \Delta k\), the \(\cos(k_o x)\) and \(\sin(k_o x)\) components oscillate with a larger frequency (\(k_o\)) and they are modulated by a sinusoidal component with smaller frequency (\(\Delta k\)).

If \(\Delta k\) is very small, the particle is still delocalized. As \(\Delta k\) increases, the particle becomes more localized in a specific region around \(x = 0\). This 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.

The superposition of waves can be accomplished by giving different contributions to different values of momenta.

For example, we can construct the superposition with a Gaussian weighting function. In this case, we modulate the contribution of each k-wave with a value given by a Gaussian distribution (maximum contribution from \(k_o\), with smaller contributions from waves with other \(k\) values).

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

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

After integration, we obtain:

\[ \psi_{\Delta k}(x) = \sqrt{2\pi}N\Delta k e^{-\frac{1}{2}x^2\Delta k^2} e^{ik_ox} \]

where \(N\) is a normalization constant and is equal to

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

We can compare the effect of the Gaussian distribution with the equally-weigthed \(k\) values:

Here are a couple questions to consider graphically:

• Q1: What do you expect to see for a particle with a momentum (\(k_o\)) with contributions from waves with very different momenta? (large value of (\(\Delta k\))

• Q2: What do you expect to see for a a particel with a momentum (\(k_o\)) with contributions from waves with similar momenta? (small value of (\(\Delta k\))

What is the probability density of finding the particle as a function of position? 

What can we conclude: From the graphical representation we can learn that for \(\Delta k\) very small compared to \(k_o\) (try plotting for \(k_o = 4\) and \(\Delta k = 0.04\)), the particle is still delocalized. As \(\Delta k\) increases the particle becomes more localized in a specific region around \(x=0\) (i.e \(k_o = 4\) and \(\Delta k = 0.4\)). When \(\Delta k\) is large in comparison to \(k_o\) (i.e. \(k_o = 4\) \( \Delta k = 16\)), we oberve the probability density concentrated around \(x=0\).

This 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.