Introduction to Quantum Mechanics

Dr Mark S. D. Read

Table of Contents ...

1 Introduction
1.1 Born’s Statistical Interpretation
1.1.1 Complex Numbers and Conjugates
1.2 Probability of finding a particle
1.3 Normalisation of a wave function
1.3.1 Normalisation of \Psi(x) = 3x^2 where 6 \leqslant x \leqslant 8
1.3.2 Integrals of Common Functions
1.3.2 Normalisation of \Psi(x) = e^{\textup{i}nx} where 0 \leqslant x \leqslant 2\pi
1.3.3 Normalisation of the Particle in a Box Wave Function \Psi(x) = \sin \del[2]{\frac{n\pi x}{L}} where n is a positive integer and 0 \leqslant x \leqslant L

1 Introduction

A wave function \Psi (x,t) is a function that satisfies a wave equation and describes the properties of a wave [1] . For a sound wave, the wave function is associated with the pressure at a time t and position x. For a water wave, \Psi (x,t) is the height of the wave at a timet.

Quantum mechanics treats moving matter as a wave, called a `matter wave’ (by combining the classical wave equation and the de Broglie relation [2]). However, unlike the classical mechanics of wave motion, in quantum mechanics the wave amplitude \Psi (x,t) itself has no physical meaning. It is a complex quantity representing the variation of a matter wave.

[1] Wave functions are commonly denoted by the variable \Psi.
[2] In 1924 Louis de Broglie suggested that a relationship that had been derived to relate momentum and wavelength for light should also apply to particles.

1.1 Born’s Statistical Interpretation

In 1926 Max Born suggested an interpretation of the `wavy nature’ of quantum particles as `waves of probability’. According to Born, the wave equation presented by Schrödinger the previous year was fundamentally a piece of mathematical machinery for calculating the chances of observing a particular outcome in an experiment. Essentially, Born’s rule connects quantum theory to experiment. “The Born rule is the crucial link between the abstract mathematical objects of quantum theory and the world of experience.”

The Born rule is a key postulate of quantum mechanics which gives the probability that measurement of a quantum system produces a given result. In its simplest form, it states that “the probability density of finding a particle at a given point is proportional to the square of the magnitude of the particle’s wave function at that point.”

Born’s postulate that the wave function, (\Psi) describing a particle’s behaviour is related to the probability of finding the particle may be represented mathematically by:

(1)   \begin{equation*} \int_a^b \envert[2]{\Psi(x)} ^2\ \dif x = \cbr[2]{\text{Probability of finding the particle between $a$ and $b$}} \end{equation*}

The implication here is that the wave function itself does not represent the probability, but a probability amplitude, and that the information contained in \Psi only represents the probability that one would measure a certain dynamical quantity, but cannot give pre-determined results in the same manner that deterministic classical mechanics will. Whereas classical mechanics is completely deterministic (if the initial conditions are known, the exact position where the particle will be at a later point in time can be predicted), quantum mechanics only provides statistical information about what the possible measurements will be. This interpretation, although since born out by much experimentation, caused much debate in the history of quantum mechanics.

1.1.1 Complex Numbers and Conjugates

A complex number is a combination of a real and an imaginary number in the form a + b\ \text{i} where a and b are real numbers and i is the “unit imaginary number” \sqrt{-1}

(2)   \begin{equation*} \underbracket{\color{red}{a}}_{\color{red}{\tiny{\text{Real Part}}}} + \underbracket{\color{blue}{b}}_{\color{blue}{\tiny{\text{Imaginary Part}}}} \ \underbracket{\color{black}{\text{i}}}_{\color{black}{\tiny{\sqrt{-1}}}}\end{equation*}

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign. I.e. (if a and b are real, then) the complex conjugate of f = a + b\ \text{i} is f^* = a - b\ \text{i}, complex conjugates are denoted by a superscript asterisk. In polar form, the conjugate of f = r \text{e}^{\text{i}\varphi} is f^* = r \text{e}^{-\text{i}\varphi} which can be shown using Euler’s formula.

The product of a complex number and its complex conjugate is a real number and is also the complex number analogue to squaring a real function.

    \begin{align*} \Psi &= \num{2 + 3i} \\\therefore \quad \Psi^* &= \num{2 -3i} \quad \text{[complex conjugate]}\\\Psi \times \Psi^* = \Psi \Psi^* &= \left( \num{2 + 3i} \right) \left( \num{2 - 3i} \right) \\&= 4 -6\text{i} +6\text{i} -9\text{i}^2 \\&= 4 + 9 \quad \text{\quad since i = $\sqrt{-1} \ \therefore \ i^2 = -1$ } \\&= 13\end{align*}

1.2 Probability of finding a particle

The interpretation of the wave function in terms of the location of a particle is based on Max Born’s postulate. Born used the wave theory of light as an analogy, in which the square of the amplitude of an electromagnetic wave in a region is interpreted as its intensity and therefore (in quantum terms) as a measure of the probability of finding a photon present in this region.

The Born Interpretation focuses on the square of the wave function (or the square modulus if the wave function is complex):

(3)   \begin{equation*} p(x) = \envert[2]{\Psi(x)} ^2 =\begin{cases}\Psi (x) \times \Psi (x) &\mbox{if wave function is real } \\\Psi (x) \times \Psi^* (x) & \mbox{if wave function is complex }\end{cases}\end{equation*}

For a one-dimensional system, if the wave function of a particle has the value \Psi at some point x, then the probability of finding the particle between x and x + \dif x is proportional to \envert[1]{\Psi} ^2 \dif x. Thus, \envert[1]{\Psi} ^2 is the probability density and to obtain the probability it must be multiplied by the length of the infinitesimal region \dif x. The wave function \Psi itself is called the probability amplitude.

For a particle free to move in three dimensions (e.g. an electron near a nucleus in an atom), the wave function depends on the point r with coordinates (x,y,z) and \Psi(r) is interpreted as follows:

If the wave function of a particle has the value \Psi at some point r, then the probability of finding the particle in an infinitesimal volume \dif \tau = \dif x\ \dif y\ \dif z at that point is proportional to \envert[1]{\Psi} ^2 \dif \tau.

The Born Interpretation negates worrying about the significance of a negative (and, in general, complex) value of \Psi because \envert[1]{\Psi} ^2 is always real and never negative!

There is no direct significance in the negative (or complex) value of a wave function only the square modulus, a positive quantity, is directly physically significant. Indeed, both positive and negative regions of a wave function may correspond to a high probability of finding a particle in a region.

1.3 Normalisation of a wave function

Since we interpret \envert[1]{\Psi (x)} ^2 as defining a probability for finding the particle at some position x. Then as the particle must exist somewhere within the defined region, we are guaranteed to find the particle if we look everywhere! Thus, if we sum up the probabilities over all possible positions of x, they should sum to 1 (or \SI{100}{\percent}).

Normalising a wave function simply means multiplying it by a constant to ensure that the sum of the probabilities for finding that particle equals 1.

Mathematically, this means integrating \envert[1]{\Psi (x)} ^2 over all space should equal 1.

(4)   \begin{equation*} \int_{- \infty}^{+ \infty} \envert[1]{\Psi(x)} ^2\ \dif x = 1\end{equation*}

1.3.1 Normalisation of \Psi(x) = 3x^2 where 6 \leqslant x \leqslant 8

Consider a simple wave function, \Psi(x) = 3x^2 operating between the limits 6 \leqslant x \leqslant 8. Since this wave function is real, the probability density is \envert[1]{\Psi(x)} ^2. Figure 1 shows a plot of the probability density and the integrated area between the limits 6 \leqslant x \leqslant 8. In order for the total probability of finding a particle within this shaded region, the area (or integral) should equal 1.

Figure 1: Plot of \envert[1]{\Psi (x)} ^2 showing the integral between the limits 6 \leqslant x \leqslant 8

Just from looking at the scale of the y axis, this is clearly not the case and the area much greater that 1! So, what is the area of the shaded region, i.e. the integration of the probability density between the limits of 6 \leqslant x \leqslant 8?

    \begin{align*}\text{wave function} \quad \Psi (x) &= 3x^2 \\\text{probability density} \quad \envert[1]{\Psi (x)} ^2 &= \Psi (x) \times \Psi (x) \\&= \del[1]{3x^2} \times \del[1]{3x^2}\intertext{Integrate between the limits $6 \leqslant x \leqslant 8$ to determine the total probability}\int_6^8 \envert[1]{\Psi(x)} ^2\ \dif x &= \int_6^8 \sbr[1]{3x^2} \sbr[1]{3x^2} \dif x \\&= \int_6^8 9x^4 \dif x \\&= \sbr[4]{\frac{9x^5}{5}}_6^8 \\&= \sbr[4]{\frac{9(8)^5}{5} - \frac{9(6)^5}{5}} \\&= \frac{\num{294912} - \num{69984}}{5} \\&= \frac{\num{224928}}{5} = \boxed{\num{44985.6}} \\\end{align*}

The probability of finding the particle between the limits 6 \leqslant x \leqslant 8 for the above wave function is \num{44985.6}. This is unphysical since the probability must equal 1 over this area. Thus, this wave function needs to be normalised so that this integral equals 1. This is achieved by by multiplying the wave function \Psi (x) by a Normalisation Constant N.

(5)   \begin{equation*} \Psi (x) = N\ 3x^2\end{equation*}

Now we have to determine a value for N in order for the probability density to equal 1 over the area between the limits.

    \begin{align*}\text{Normalised wave function} \quad \Psi (x) &= N\ 3x^2 \\\text{Normalised probability density} \quad \envert[1]{\Psi (x)} ^2 &= \Psi (x) \times \Psi (x) \\&= \del[1]{N\ 3x^2} \times \del[1]{N\ 3x^2}\intertext{Integrate between the limits $6 \leqslant x \leqslant 8$ and set the total probability to 1}\int_6^8 \envert[1]{\Psi(x)} ^2\ \dif x &= \int_6^8 \sbr[1]{N\ 3x^2} \sbr[1]{N\ 3x^2} \dif x = 1\\&= \int_6^8 N^2 \sbr[1]{3x^2} \sbr[1]{3x^2} \dif x = 1\\&= N^2 \int_6^8 \sbr[1]{3x^2} ^2 \dif x = 1 \quad \text{As $N$ is a constant, move outside integral}\\&= N^2 \int_6^8 9x^4 \dif x = 1\intertext{Integrate between the limits $6 \leqslant x \leqslant 8$}&= N^2 \sbr[4]{\frac{9x^5}{5}}_6^8 = 1 \\&= N^2 \sbr[4]{\frac{9(8)^5}{5} - \frac{9(6)^5}{5}} = 1 \\&= N^2\ \frac{\num{294912} - \num{69984}}{5} = 1 \\&= N^2 \times \frac{\num{224928}}{5} = 1\\N^2 &= 1 \div \frac{\num{224928}}{5} \\N^2 &= 1 \times \frac{5}{\num{224928}} \\N &= \sqrt{ \frac{5}{\num{224928}} }\end{align*}

Thus, the normalised form of our wave function is:

(6)   \begin{equation*} \Psi (x) = \sqrt{ \frac{5}{\num{224928}} }\ 3x^2\end{equation*}

Figure 2: Plot of Normalised \envert[1]{\Psi (x)} ^2 showing the integral between the limits 6 \leqslant x \leqslant 8 = 1

Visual inspection of Figure 2 indicates that the area of the integral is 1. This is confirmed by multiplying the integral of the square of the un-normalised wave function by the square of the normalisation constant. i.e.

(7)   \begin{equation*} \frac{\num{224928}}{5} \times \frac{5}{\num{224928}} = \boxed{1}\end{equation*}

1.3.2 Integrals of Common Functions

\renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Function & Indefinite Integral \\ $f(x)$ & $\int\ f(x)\ \dif x$ \\ \hline Constant, $k$ & $kx\ + c$ \\ $x$ & $\frac{1}{2}\ x^2\ + c$ \\ $x^2$ & $\frac{1}{3}\ x^3\ + c$ \\ $x^n$ & $\frac{1}{n+1}x^{n+1}\ + c, \quad n \neq -1$ \\ $x^{-1}\ \text{ or }\ \del[2]{\frac{1}{x}}$ & $\ln\ x\ + c$ \\ $\cos\ x$ & $\sin\ x\ + c$ \\ $\sin\ x$ & $-\cos\ x\ + c$ \\ $\cos\ kx$ & $\frac{1}{k}\ \sin\ kx\ + c$ \\ $\sin\ kx$ & $-\frac{1}{k}\ \cos\ kx\ + c$ \\ $\tan\ kx$ & $\frac{1}{k}\ \ln\ \envert[2]{\sec\ kx}\ + c$ \\ $e^x$ & $e^x\ + c$ \\ $e^{-x}$ & $-e^{-x}\ + c$ \\ $e^{kx}$ & $\frac{1}{k}\ e^{kx}\ + c$ \\ \hline $\sin^2\ kx$ & $\frac{x}{2}\ - \frac{\sin\ 2kx}{4k}\ + c$ \\ \hline \end{tabular}

A useful procedure for finding the normalisation constant (N) (and thus writing the normalised wave function, \Psi_{\text{Normailsed}} ) is as follows:

  1. Determine \Psi^* (depending on whether \Psi is real or imaginary)
    If \Psi is real, i.e. does not contain i in the expression, then \Psi^* = \Psi
    If \Psi is imaginary, i.e. does contain i in the expression, then the complex conjugate, \Psi^* is the number
    with an equal real part and an imaginary part equal in magnitude but opposite in sign,
    e.g. a + b\ \text{i} \Rightarrow a - b\ \text{i},\ e^{ik\phi} \Rightarrow e^{-ik\phi}
  2. Introduce N and set the integral of \Psi^* \times \Psi equal to 1
  3. Integrate and solve for N
  4. Write the final Normalised wave function

Both real and imaginary wave function examples are given below:

1.3.2 Normalisation of \Psi(x) = e^{\textup{i}nx} where 0 \leqslant x \leqslant 2\pi

Determine the value of N such that the following wave function is normalised from 0 to 2\pi

  1. Determine \Psi^* (depending on whether \Psi is real or imaginary)
  2. Introduce N and set the integral of \Psi^* \times \Psi equal to 1
  3. Integrate and solve for N
  4. Write the final Normalised wave function

    \begin{align*}\intertext{Step 1: Determine $\Psi^*$ (depending on whether $\Psi$ is real or imaginary) }\intertext{Since $\Psi (x)$ is imaginary (contains i which = $\sqrt{-1}$)}\Psi (x) &= e^{\textup{i}nx} \\\Psi^* (x) &= e^{-\textup{i}nx}\quad \text{$\Psi$ is imaginary,\ $\therefore$ change sign of i} \\\intertext{Step 2: Introduce N and set the integral of $\Psi^* \times \Psi$ equal to 1 :}\Psi (x) &= N\ e^{\textup{i}nx} \\\Psi^* (x) &= N\ e^{-\textup{i}nx} \\\envert[1]{\Psi (x)} ^2 &= \Psi^* \times \Psi = N\ e^{-\textup{i}nx} \times N\ e^{\textup{i}nx} \\\int_0^{2\pi} \envert[1]{\Psi (x)} ^2 &= \int_0^{2\pi} N^2\ e^{-\textup{i}nx\ +\ \textup{i}nx} \dif x = 1 \quad \text{multiply two exponentials by adding their powers}\\&= N^2\ \int_0^{2\pi} e^0 \dif x = 1 \quad \text{adding a positive term to the same negative term equals zero}\\&= N^2\ \int_0^{2\pi} 1 \dif x = 1 \quad \text{anything raised to power 0 equals 1!}\\\intertext{Step 3: Integrate between the limits and solve for N}\int_0^{2\pi} \envert[1]{\Psi (x)} ^2 &= N^2\ \sbr[1]{x}_0^{2\pi} = 1 \quad \text{integral of 1 equals $x$} \\&= N^2\ \sbr[1]{2\pi\ -\ 0} = 1 \\N^2 \times 2\pi &= 1 \\N^2 &= 1 \div 2\pi \\N &= \boxed{\sqrt{\frac{1}{2\pi}}} \\\intertext{Step 4: Write the final Normalised wave function}\Psi_{\text{Normailsed}} (x) &= \sqrt{\frac{1}{2\pi}}\ e^{\textup{i}nx}\end{align*}

1.3.3 Normalisation of the Particle in a Box Wave Function \Psi(x) = \sin \del[2]{\frac{n\pi x}{L}} where n is a positive integer and 0 \leqslant x \leqslant L

Determine the value of N such that the following wave function is normalised from 0 to 1

  1. Determine \Psi^* (depending on whether \Psi is real or imaginary)
  2. Introduce N and set the integral of \Psi^* \times \Psi equal to 1
  3. Integrate and solve for N
  4. Write the final Normalised wave function

    \begin{align*}\intertext{Step 1: Determine $\Psi^*$ (depending on whether $\Psi$ is real or imaginary) }\intertext{Since $\Psi (x)$ is real, complex conjugate is the same as $\Psi (x)$}\Psi (x) &= \sin \del[2]{\frac{n\pi x}{L}} \\\Psi^* (x) &= \sin \del[2]{\frac{n\pi x}{L}} \\\intertext{Step 2: Introduce N and set the integral of $\Psi^* \times \Psi$ equal to 1 :}\Psi (x) &= N\ \sin \del[2]{\frac{n\pi x}{L}} \\\Psi^* (x) &= N\ \sin \del[2]{\frac{n\pi x}{L}} \\\envert[1]{\Psi (x)} ^2 &= \Psi^* \times \Psi = N\ \sin \del[2]{\frac{n\pi x}{L}} \times N\ \sin \del[2]{\frac{n\pi x}{L}}\\\int_0^{L} \envert[1]{\Psi (x)} ^2 &= \int_0^{L} N^2\ \sin^2 \del[2]{\frac{n\pi x}{L}} \dif x = 1 \quad \text{N.B.\ $\sin^2 (x) = [\sin(x)]^2$}\end{align*}

    \begin{align*}\intertext{Step 3: Integrate between the limits and solve for $N$}\text{Given that}\quad \int\ \sin^2\ kx &= \frac{x}{2}\ - \frac{\sin\ 2kx}{4k}\ + c \quad\text{Here, $k = \frac{n\pi}{L}$}\\\int_0^{L} \envert[1]{\Psi (x)} ^2 &= N^2\ \sbr[4]{\frac{x}{2}\ - \frac{\sin\ \del[1]{\frac{2n\pi x}{L}}}{4\del[1]{\frac{n\pi}{L}}}}_0^{L} = 1 \\&= N^2\ \sbr[4]{\del[4]{\frac{L}{2}\ - \frac{\sin\ \del[1]{\frac{2n\pi L}{L}}}{4\del[1]{\frac{n\pi}{L}}}} - \del[4]{\frac{0}{2}\ - \frac{\sin\ \del[1]{\frac{2n\pi 0}{L}}}{4\del[1]{\frac{n\pi}{L}}}}} = 1 \\&= N^2\ \sbr[4]{\del[4]{\frac{L}{2}\ - \frac{\sin\ \del[1]{\frac{2n\pi \cancel{L}}{\cancel{L}}}}{4\del[1]{\frac{n\pi}{L}}}} - \del[4]{0 - 0}} = 1 \\&= N^2\ \sbr[4]{\del[4]{\frac{L}{2}\ - \frac{\sin\ \del[1]{2n\pi}}{4\del[1]{\frac{n\pi}{L}}}} - \del[4]{0}} = 1 \\&= N^2\ \sbr[4]{\del[4]{\frac{L}{2}\ - \frac{0}{4\del[1]{\frac{n\pi}{L}}}} - \del[4]{0}} = 1 \quad\text{Since the sine of any even no. of $\pi$ is zero!}\\&= N^2\ \sbr[4]{\del[4]{\frac{L}{2}\ - 0} - \del[4]{0}} = 1 \\&= N^2\ \sbr[4]{\frac{L}{2}} = 1 \\N^2 \times \frac{L}{2} &= 1 \\N^2 &= 1 \div \frac{L}{2} \\N &= \boxed{\sqrt{\frac{2}{L}}} \\\intertext{Step 4: Write the final Normalised wave function}\Psi_{\text{Normalised}} (x) &= \sqrt{\frac{2}{L}}\ \sin \del[3]{\frac{n\pi x}{L}}\end{align*}

Note that the Normalisation Constant is dependent only on the length of the box (L). The Normalised wave function provides a series of functions for n = 1,2,3,4,\cdots. The first five Normalised wave functions are plotted in Figure 3 over the length of the 1D box where L has boundaries at 0 and 1.

Figure 3: Plot of Normalised Wave Functions \envert[1]{\Psi (x)} For a Particle in a 1D Box, n=1-5 L=1

Figure 4 plots the n = 2 state for a particle in a box of length L = 1. Integrating the square of the un-normalised wave function over the dimensions of the box, (purple) the probabilities sum to 0.5 which is unphysical. Whereas, using the square of the normalised wave function probabilities total 1, which is physical and correct.

Figure 4: Plot of Probability Density \envert[1]{\Psi (x)} ^2 For a Particle in a 1D Box, n=2 L=1 Un-Normalised in Purple, Normalised in Orange