3 Wave function and Probabilities

Throughout the lecture course, we focus on a particle of mass $m$ moving in one dimension with potential $V(x)$. In classical mechanics, the particle has definite position and momentum $(x(t),p(t))$, which evolve according to Hamilton’s equations with Hamiltonian $H = \frac{p^2}{2m} +V(x)$.

Introducing the wave function which leads to the probabilistic character of quantum mechanics

Motivated by the the double slit experiment, however, we must give up the idea that a particle as a definite position and momentum $(x(t),p(t))$. Instead, the particle is described by a complex wave function $\psi(x,t)$ that encodes the probability for a measurement of position or momentum to yield values in a given range. In this lecture, we consider measurements of position. In the subsequent lectures, we will consider measurements of momentum and energy.

3.1 The Wave function

The wave function $\psi(x,t)$ is a complex function of position $x$ and time $t$. Putting on our analysis hat, the wave function defines a continuous function \[\begin{aligned} \psi_t & : \mathbb{R} \to \mathbb{C} \\ & : x \mapsto \psi(x,t) \end{aligned}\] at each time $t\in \mathbb{R}$. We will discuss to what extent the wave function should be differentiable later in the course.

The probability density to find a particle at position $x$ at time $t$ is given by $P(x,t)=|\psi(x,t)|^2$.

A basic postulate of quantum mechanics is that the modulus squared of the wave function, \[P(x,t) := |\psi(x,t)|^2 \, ,\] is the ‘probability density’ for a measurement at time $t$ to find the particle at position $x$. There are two equivalent ways to say what this means:

The probability to find the particle between infinitesimally separated points $x$ and $x+d x$ at time $t$ is $P(x,t)d x$.
The probability to find the particle in a finite interval $a<x<b$ is \[\displaystyle \int^b_a P(x,t)dx \, .\]

We find an immediate constraint on the wave function $\psi(x,t)$ from the requirement that the probability to find the particle anywhere must be $1$. In particular, a physical wave function $\psi(x,t)$ should obey \[\int_{-\infty}^{\infty} P(x,t) dx = 1 \label{eq:norm}\] at any time $t$. This constraint typically requires that the wave function should decay fast enough as $x\to \pm \infty$. Some terminology:

If the above integral exists, the wave function is ‘square-normalisable’.
If the above integral is equal to $1$, the wave function is ‘normalised’.

3.2 Expectation Values

The mean of measurements of position, $\langle x\rangle$, for many measurements of particles described by the same wave function, is given by weighing the position by the probability, $\displaystyle \langle x\rangle = \int\, x |\psi(x,t)|^2 {\rm d}x$.

As for any probability distribution, the expectation value of a polynomial function $f(x)$ is given by \[\begin{aligned} \langle f(x) \rangle := & \int_{-\infty}^\infty f(x) P(x,t) \,{\rm d}x \\[1ex] = & \int_{-\infty}^\infty f(x) \big|\psi(x,t)\big|^2\, {\rm d}x \, . \end{aligned}\] There are two important expectation values that will feature in these lectures:

First, the expectation value $\langle x\rangle$ is the mean of position measurements on an ensemble of particles with the same wave function $\psi(x,t)$.
Second, the standard deviation is defined by \[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} \, .\] This is a measure of the spread of the probability distribution around the mean $\langle x\rangle$. In quantum mechanics, it is therefore customary to call $\Delta x$ the ‘uncertainty’ in the position.

3.3 Examples

In the first part of the course, we consider wave functions and expectation values at a fixed point in time $t$ - how the wave function evolves in time will be considered later. With this understood, we denote the wave function by $\psi(x)$.

Gaussian Wave function

Consider the wave function \[\psi(x) = C \, e^{-x^2/4\Delta^2} \, ,\] where $\Delta >0$ has units of length and $C$ is a normalisation constant. To determine the normalisation constant $C$, we require that the probability to find the particle anywhere is $1$, \[\begin{aligned} 1 & = \int^\infty_{-\infty} |\psi(x)|^2 dx \\[1ex] & = |C|^2 \int^\infty_{-\infty} e^{-x^2/2\Delta^2} dx \\[1ex] & = |C|^2 \sqrt{2\Delta^2} \int^\infty_{-\infty} e^{-y^2} dy \\[1ex] & = |C|^2 \sqrt{2\pi \Delta^2} \, , \end{aligned}\] where we have used the substitution $y = \sqrt{2\Delta^2}$ and the standard Gaussian integral \[\int^\infty_{-\infty} e^{-y^2} dy = \sqrt{\pi} \, .\] We can therefore choose $C = (2\pi \Delta^2)^{-1/4}$. Note that we could have multiplied the normalisation by a constant phase $e^{i\theta}$. This would not change the probability density or position expectation values, so for convenience we can set it to $1$.

Gaussian probability density with width \Delta. — Gaussian probability density with width $\Delta$.

The normalized probability distribution is \[P(x) = \frac{1}{\sqrt{2\pi \Delta^2}} e^{-x^2/2\Delta^2}\] which is a standard Gaussian probability distribution. Before doing any computations, we can immediately say that:

$\langle x^{2n+1}\rangle = 0$ for $n\in\mathbb{Z}_{\geq 0}$ since the integrand is an odd function of $x$.
Since $\Delta$ is the only length in the problem, dimensional analysis tells us that $\langle x^{2n}\rangle \propto \Delta^{2n}$ for $n\in \mathbb{Z}_{\geq 0}$.

In one of the problems, you are asked to verify that $\langle x^2\rangle = \Delta^2$ and therefore $\Delta x = \Delta$.

Infinite Potential Well

Consider a particle confined to the region $0 < x < L$. You can regard this as a particle in an infinite potential well, \[V(x) = \begin{cases} 0 & 0 < x < L \\ \infty & \mathrm{other} \end{cases} \, .\] The particle would require infinite energy to be found with $x\leq 0$ or $x \geq L$. We therefore require the wave function vanishes in these regions so that the probability to find it there is zero.

A wave function that meets this requirement is \[\psi(x) = \begin{cases} C \sqrt{x(L-x)} & 0 < x < L \\ 0 & \mathrm{otherwise} \end{cases}\, .\] To determine the normalisation $C$, we require that the probability to find the particle anywhere is $1$, \[1 %& = \int^\infty_{-\infty} |\psi(x)|^2 dx \\ =| C|^2 \int^L_0 x(L-x) dx = |C|^2 \frac{L^3}{6} \, ,\] and therefore $C = \sqrt{6/L^3} \, e^{i\theta}$. For convenience, we can choose $e^{i\theta} =1$.

We can now compute some expectation values. Since $L$ is the only length in the problem, dimensional analysis means that $\langle x^n\rangle \propto L^n$ for any $n\in \mathbb{Z}_{\geq 0}$. Computing the first few, we find \[\begin{aligned} \langle x\rangle & = \frac{6}{L^3} \int^L_0 x^2(L-x)\, {\rm d}x = \frac{L}{2}\,, \\[1ex] \langle x^2\rangle & = \frac{6}{L^3} \int^L_0 x^3(L-x)\, {\rm d}x = \frac{3L^2}{10}\,. \, . \end{aligned}\] Since the probability density is symmetric around $L/2$, we should have expected $\langle x\rangle = L/2$. Finally, the uncertainty is \[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \frac{L}{\sqrt{20}} \, .\]

This example shows that we will, at some point, have to be more

The wave function is continuous. It is also differentiable, except at places where $V(x)$ is not finite.

precise about the continuity and differentiability properties of the wave function. For the infinite potential well, we see that $\psi(x)$ is continuous everywhere, and differentiable except at the edges of the well. Jumping ahead a bit, we will find that this holds in general: the wave function $\psi(x)$ will always have to be continuous, and it will have to be differentiable except at positions $x$ where the potential $V(x)$ is not finite. We will discuss and derive these conditions in more detail once we have introduced the Schrödinger equation.

3.4 Phases

In is important to note that if we multiply the wave function $\psi(x)$ by a position-dependent phase, the probability density is unchanged \[\begin{aligned} \psi(x) & \to e^{i\theta(x)} \psi(x)\,, \qquad P(x) & \to P(x) \, . \end{aligned}\] This means that measurements of position cannot detect the difference between the wave functions $\psi(x)$ and $e^{i\theta(x)}\psi(x)$. In the next lecture, however, we will show that measurements of momentum can detect the difference between these wave functions, unless $\theta(x)$ is constant. On the other hand, a constant phase $e^{i\theta}$ cannot be detected by any measurement and the wave functions $\psi(x)$ and $e^{i\theta}\psi(x)$ describe the same physical state.

3.5 Collapse of the Wave function

We now mention an important subtlety with the interpretation of the probability density $P(x,t)$ in quantum mechanics compared to other areas of mathematical sciences.

Suppose a particle has wave function $\psi(x,t)$ for $t<t_0$. Then at $t = t_0$ the position of the particle is measured and the particle is found at $x = x_0$. What is the wave function after the measurement?

It turns out that another measurement immediately after the first will find the particle at $x = x_0$ with probability $1$.

When a particle's position is measured, its wave function ‘collapses' to a wave function which is sharply peaked at the position where we found the particle.

This is known as wave function collapse. It means that the act of measurement modifies the wave function to something that is tightly localised around $x=x_0$. We will describe this phenomenon more precisely in a future lecture.

This is why we must carefully specify the meaning of expectation values such as $\langle x\rangle$. It is not obtained by averaging over repeated measurements of the same wave function. Instead, it is the average of measurements made on an ensemble of particles with the same wave function $\psi(x,t)$.

3.6 Problems

PROBLEMS CLASS 1
Practice with wave functions The wave function of a particle at time $t=0$ is \[\psi(x) = \begin{cases} \; C \, \dfrac{x}{a} & \quad 0 \leq x \leq a \\[10pt] \; C \, \dfrac{b-x}{b-a} & \quad a < x \leq b \\[10pt] \; 0 & \quad \mathrm{otherwise} \end{cases}\] for constants $0<a<b$.
1. What principle fixes the normalisation $C$?
2. Find $C$ in terms of $a$ and $b$.
3. Sketch the probability density $P(x) = |\psi(x)|^2$.
4. Where is the particle most likely to be found?
5. Compute the position expectation value $\langle x\rangle$. Does it coincide with your previous answer?
6. What is the probability of finding the particle in the region $x < a$?
Solution ▶
1. The probability to find the particle anywhere should be 1.
2. The total probability is \[1 = \int^\infty_{-\infty}dx | \psi(x)|^2 = |C|^2 \int^a_0 dx \left(\frac{x}{a}\right)^2 + |C|^2 \int^b_a dx \left( \frac{b-x}{b-a} \right)^2 = \frac{b}{3} | C|^2 \, .\] Therefore $C = \sqrt{\frac{3}{b}}$ up to a phase.
3. Plot of probability density:
4. The most likely value (mode) is $x = a$.
5. The expectation value of $x$ (mean) is \[\langle x \rangle = \int^b_0 dx \, x |\psi(x)|^2 = \frac{3}{b} \int^a_0 dx \, x \left(\frac{x}{a}\right)^2 + \frac{3}{b} \int^b_a dx\, x \left( \frac{b-x}{b-a} \right)^2 = \frac{1}{4}(2a+b) \, .\] The probability distribution is skewed for generic constants $0<a<b$ and therefore the mode and mean to not coincide. However, for $a = \frac{b}{2}$ the probability distribution is symmetric and both answers give the same result, $\frac{b}{2}$.
6. The probability to find the particle in the region $x<a$ is \[P(x<a) = \int^a_{0} |\psi(x)|^2 = \frac{3}{b} \int^a_0 \left(\frac{x}{a}\right)^2 =\frac{a}{b} \, .\]
More practice with wave functions:
Consider the wave function \[\psi(x,t) = C e^{-\lambda|x|} e^{- i \omega t}\] where $C$, $\lambda$, $\omega \in \mathbb{R}_{>0}$.
1. Find the normalization $C$.
2. Find the expectation values $\langle x\rangle$ and $\langle x^2\rangle$ and the uncertainty $\Delta x$.
3. Sketch the probability density $P(x)$ and mark the points $x_\pm := \langle x\rangle \pm \Delta x$.
4. What is the probability to find the particle in the region $x_- < x < x_+$?
5. Can the wave function be physical if $\omega$ has an imaginary part?
Solution ▶
1. The probability to find the particle anywhere must be $1$. \[1 = \int^\infty_\infty |\psi(x,t)|^2 dx = |C|^2 \int^\infty_{-\infty} e^{-2\lambda|x|} dx = 2 |C|^2 \int^\infty_0 e^{-2\lambda x} dx = \frac{ |C|^2}{\lambda}\, .\] Therefore $C = \sqrt{\lambda}$ up to a phase.
2. First, $\langle x\rangle = 0$ since the integrand is an odd function of $x$. Second, \[\begin{aligned} \langle x^2\rangle & = \lambda \int^\infty_{-\infty} x^2 e^{-2\lambda |x|} dx \\ \nonumber & = 2 \lambda \int^\infty_{0} x^2 e^{-2\lambda x} dx \\ & = \frac{\lambda}{2}\frac{\partial^2}{\partial \lambda^2} \int^\infty_0e^{-2\lambda x} dx\\ & = \frac{1}{2\lambda^2}\, . \end{aligned}\] Finally, $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} =1 /\sqrt{2}\lambda$.
3. The probability density is $P(x,t) = \lambda e^{-2\lambda |x|}$.
4. We have $x_\pm = \pm 1 /\sqrt{2}\lambda$. The probability to find the particle in the region $x_- < x < x_+$ is therefore \[\int^{1 /\sqrt{2}\lambda}_{-1 /\sqrt{2}\lambda} \lambda e^{-2\lambda |x|} dx = 2 \lambda \int^{1 /\sqrt{2}\lambda}_{0} e^{-2\lambda x} dx = 1-e^{-\sqrt{2}}\, .\]
5. If $\omega$ had an imaginary part, the probability density would become \[\begin{aligned} \nonumber P(x,t) & = |C|^2 e^{-2\lambda|x|} e^{-i(\omega-\omega^*) t} \\ & = |C|^2 e^{-2\lambda|x|} e^{2 \Im(\omega) t} \, . \end{aligned}\] Attempting to normalize the wave function we would find \[1 = \frac{|C|^2}{\lambda} e^{2 \Im(\omega) t} \, ,\] which cannot be satisfied for all time $t$ unless $\Im(\omega) = 0$. Therefore the wave function is not physical if $\Im(\omega) \neq 0$.
Gaussian wave function:
The quantum mechanical wave function of a particle at time $t=0$ is \[\psi(x) = C e^{-(x-x_0)^2 / 4 \Delta^2}\]
1. Find the normalization $C$.
2. Sketch the probability density $P(x) = |\psi(x)|^2$.
3. Find the expectation values $\langle x \rangle$ and $\langle x^2\rangle$ and the uncertainty $\Delta x$.
4. Which parameter controls how well the particle is localized in position space?
You may use the Gaussian integrals \[\int^{\infty}_{-\infty} e^{-y^2} dy = \sqrt{\pi} %\qquad \int^{\infty}_{-\infty} y e^{-y^2} dy = 0 \qquad \int^{\infty}_{-\infty} y^2e^{-y^2} dy = \frac{\sqrt{\pi}}{2}\, .\]
Solution ▶
1. Probability to find the particle anywhere should be $1$. We find, \[\begin{aligned} 1 & = |C|^2 \int^\infty_{-\infty} dx \, e^{-(x-x_0)^2 / 2 \Delta^2} \\ \nonumber & = \sqrt{2\Delta^2} |C|^2 \int^\infty_{-\infty} dy \, e^{-y^2} \\ & = \sqrt{2\pi \Delta^2} \, |C|^2 \, , \end{aligned}\] where we defined $y = (x-x_0) / \sqrt{2\Delta^2}$. Therefore \[C = 1/(2\pi\Delta^2)^{1/4}\] up to a phase.
2. Sketch of the probability density.
3. First, \[\begin{aligned} \langle x\rangle & = \frac{1}{\sqrt{2\pi \Delta^2}} \int^\infty_{-\infty} dx \, x \, e^{-(x-x_0)^2 / 2 \Delta^2} \\ \nonumber & = \frac{1}{\sqrt{\pi}} \int^\infty_{-\infty} dy \, ( x_0 + \sqrt{2\Delta^2} \, y ) \, e^{-y^2} \\ & = \frac{x_0}{\sqrt{\pi}} \, \int^\infty_{-\infty} dy \, e^{-y^2} \\ %+ \frac{1}{\sqrt{\pi}} \int^\infty_{-\infty} dy \, y e^{-y^2} \\ & = x_0 \, . \end{aligned}\] Second, \[\begin{aligned} \langle x^2\rangle & = \frac{1}{\sqrt{2\pi \Delta^2}} \int^\infty_{-\infty} dx \, x^2 \, e^{-(x-x_0)^2 / 2 \Delta^2} \\ \nonumber & = \frac{1}{\sqrt{\pi}} \int^\infty_{-\infty} dy \, ( x_0 + \sqrt{2\Delta^2} \, y )^2 \, e^{-y^2} \\ & = \frac{x_0^2}{\sqrt{\pi}} \int^\infty_{-\infty} dy \, e^{-y^2} % + \frac{\sqrt{2\Delta^2} }{\sqrt{\pi}}x_0 \int^\infty_{-\infty} dy \, y \, e^{-y^2} + \frac{2\Delta^2}{\sqrt{\pi}} \int^\infty_{-\infty} dy \, y^2 \, e^{-y^2} \\ & = x_0^2 + \Delta^2 \, . \end{aligned}\] Finally, \[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \Delta \, .\]
4. $\Delta x = \Delta$ is a measure of how well localised the particle is in space.
TUTORIAL 1
Another wave packet:
Consider the following wave function \[\psi(x) = \sqrt{ \frac{1}{4\pi\hbar\delta p} } \int^{p_0+\delta p}_{p_0-\delta p} e^{i p x / \hbar} dp \, .\]
1. Compute and sketch the probability density $P(x)$.
2. Verify that the wave function is normalized.
3. What happens if you try to compute $\Delta x$?
4. Instead compute the distance $\delta x$ between the zeroes of $P(x)$ closest to the origin and show that \[\delta x \, \delta p = 2\pi \hbar \, .\] You may assume the following integral, \[\int^\infty_{-\infty} \frac{\sin^2(y)}{y^2} = \pi \, .\]
Solution ▶
1. We first compute the integral \[\begin{aligned} \psi(x) & = \sqrt{ \frac{1}{4\pi\hbar\delta p} } \int^{p_0+\delta p}_{p_0-\delta p} e^{i p x / \hbar} dp \\ \nonumber & = \sqrt{ \frac{\hbar}{\pi \delta p}} \frac{e^{i p_0 x / h} \sin \left( x \delta p / \hbar\right)}{x} \, . \end{aligned}\] The probability density is therefore \[P(x) = \frac{\hbar}{\pi \delta p} \frac{\sin^2 \left( x \delta p / \hbar\right)}{x^2} \, .\]
2. Changing variables to $y = x \delta p/\hbar$, we have \[\begin{aligned} \nonumber \int^\infty_{-\infty} P(x) dx & = \frac{\hbar}{\pi \delta p} \int_{-\infty}^\infty \frac{\sin^2 \left( x \delta p / \hbar\right)}{x^2} dx \\ & = \frac{1}{\pi}\int^\infty_{-\infty} \frac{\sin^2(y)}{y^2} = 1 \, . \end{aligned}\]
3. The integrals required to compute $\langle x^n\rangle$ for $n>0$ do not converge. Therefore we cannot compute the uncertainty $\Delta x$.
4. The zeroes closest to the origin are at $\pm \pi \hbar / \delta p$. We therefore take $\delta x = 2\pi \hbar / \delta p$.
5. Hence $\delta x \delta p = 2\pi \hbar$.