Back to the on-screen lesson ·
The wave function and Born's rule $dP = |\psi|^2dx$, normalization, probabilities over intervals, expectation values, Schrödinger's equation and exponential tails.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to normalize wave functions and compute probabilities and average positions from them.
From unit 3 you know that matter has waves, that $|\text{amplitude}|^2$ gives a probability, and that confinement costs energy. From calculus you can integrate polynomials and exponentials. This lesson makes the wave precise: a function $\psi(x)$ whose square gives the probability density of finding a particle, the object Schrödinger's equation determines.
| Term | What it means |
|---|---|
| Wave function | $\psi(x)$, the amplitude whose square gives where a particle may be found. |
| Probability density | $\vert \psi(x)\vert ^2$, probability per unit length. |
| Normalization | Scaling $\psi$ so that $\int\vert \psi\vert ^2dx = 1$. |
| Expectation value | $\langle x\rangle = \int x\vert \psi\vert ^2dx$, the average of many position measurements. |
| Schrödinger equation | The equation that determines $\psi$ from the particle's potential energy. |
| Classically forbidden region | A place where the particle's energy is below the potential, reached only by the wave function's tail. |
| Tunneling | Crossing a barrier through the forbidden region. |
A quantum particle is described by a wave function $\psi(x)$. It is not a probability itself; its squared magnitude is the probability density:
$$dP = |\psi(x)|^2\,dx.$$
Since the particle must be somewhere, the total probability is one, which fixes the wave function's overall size, its normalization:
$$\int_{-\infty}^{\infty}|\psi(x)|^2\,dx = 1.$$
The probability of finding the particle between $a$ and $b$ is $\int_a^b|\psi|^2dx$, and the average of many position measurements is the expectation value $\langle x\rangle = \int x|\psi|^2dx$. The wave function must be continuous and finite. Schrödinger's equation, $-\dfrac{\hbar^2}{2m}\psi'' + U\psi = E\psi$, determines which wave functions are allowed for a given potential energy $U(x)$.
Another way: picture
Picture a sheet of sand on a table, piled higher in some places than others. The height of the pile is $|\psi|^2$. Drop a marble blindfolded, with the chance of landing at any spot proportional to the sand there, and many drops will pile up in the same shape. Normalizing the wave function means using exactly one bucket of sand in total.
Another way: steps
When Erwin Schrödinger wrote his equation in 1926, he thought $\psi$ described the electron smeared out like a cloud of charge. Max Born proposed instead that $|\psi|^2$ gives the probability of finding the electron at each place, and experiments bore him out: electrons are always detected whole, one at a time, with a distribution matching $|\psi|^2$.
Born received the 1954 Nobel Prize for this interpretation. It is the same rule met in lesson 12 for photons at a polarizer and particles at two slits: amplitudes are computed, their squares give probabilities, and single outcomes are random.
The Schrödinger equation is linear, so if $\psi$ is a solution, so is $3\psi$ or $-\psi$. Normalization picks the size. For $\psi = Ax$ on $0 \le x \le L$, $\int_0^L A^2x^2dx = A^2L^3/3 = 1$, so $A^2 = 3/L^3$. For $\psi = Ae^{-x/a}$ on $x \ge 0$, $\int_0^{\infty}A^2e^{-2x/a}dx = A^2a/2$, so $A^2 = 2/a$.
Since $|\psi|^2$ is probability per unit length, $\psi$ in one dimension has units of $\text{length}^{-1/2}$. The overall sign or phase of $\psi$ changes nothing observable; only relative phases between parts of a wave function, which cause interference, matter.
Once $\psi$ is normalized, the probability of finding the particle anywhere in an interval is the area under $|\psi|^2$ there. For $\psi = Ax$ on $[0, L]$, the probability in the left half is $\int_0^{L/2}3x^2/L^3\,dx = 1/8$: the particle is seven times more likely to be found in the right half, where the wave function is larger.
A probability density can be larger than one per unit length, as long as the total area is one: a particle confined to $0.1$ nm has $|\psi|^2$ around $10$ nm⁻¹. What must never exceed one is a probability, an integral over an interval.
Measuring position many times on identically prepared particles gives a spread of results whose average is $\langle x\rangle = \int x|\psi|^2dx$. For the uniform wave function on $[0, L]$ that is $L/2$; for $\psi = Ax$, it is $3L/4$. The spread around the average, $\Delta x$, is found from $\langle x^2\rangle - \langle x\rangle^2$.
Expectation values are what connects quantum mechanics to classical physics. For a large object, the spread is tiny and the expectation values of position and momentum follow Newton's laws, a result known as Ehrenfest's theorem. The quantum description reduces to the classical one in its averages.
For a particle of energy $E$ in a potential $U(x)$, the time-independent Schrödinger equation is $-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2} + U\psi = E\psi$. Where $E > U$, the particle can be found classically, and $\psi$ oscillates like a wave with wavelength $h/\sqrt{2m(E - U)}$, de Broglie's relation.
Where $E < U$, classically forbidden, $\psi$ does not stop at zero but decays exponentially as $e^{-\kappa x}$ with $\kappa = \sqrt{2m(U - E)}/\hbar$. That tail lets particles tunnel through thin barriers, the basis of the scanning tunneling microscope and of radioactive alpha decay. The next lesson solves the equation for a particle trapped in a box.
Checking an answer. Total probability must be one. Probabilities over intervals must lie between 0 and 1. Averages must lie within the region where $\psi$ is nonzero, and nearer where $|\psi|^2$ is larger. Units of $|\psi|^2$ must be inverse length.
Integrating $|\psi|^2$ over an interval adds the probabilities of the particle being in each small piece, since those events are mutually exclusive. That is why areas under $|\psi|^2$ are probabilities, and why the total area must be one.
Requiring $\psi$ to be continuous and finite keeps probabilities sensible and, through the Schrödinger equation, keeps the kinetic energy finite. These requirements, applied at boundaries and at infinity, are what produce quantized energies, as the next lesson shows.
The wave function is not a physical wave like sound or water. It can be negative or complex, it is not directly measurable, and for two particles it is a function of both their positions at once, living in a space of six dimensions, not three. It is best thought of as a tool for computing probabilities.
What happens to it during a measurement, when a particle spread over a region is suddenly found at one point, is the measurement problem of lesson 12, and lesson 20 returns to it. The rules for using $\psi$, though, are clear and have never failed a test, from single electrons to the behavior of whole superconductors, and the chips in every phone and computer made today.
Chemists draw orbitals as clouds, which are pictures of $|\psi|^2$ for electrons in atoms and molecules. The shape of a bond, the angle between atoms in water, and the reason carbon forms four bonds all come from the wave functions of electrons shared between nuclei.
Computational chemists at national laboratories such as Pacific Northwest National Laboratory in Richland, Washington, solve approximate Schrödinger equations for molecules with thousands of electrons to design catalysts, batteries and drugs. Their programs compute wave functions and then, from $|\psi|^2$, the electron densities that decide how molecules react.
A graph of $\psi(x)$ and a graph of $|\psi(x)|^2$ tell different stories, and it pays to sketch both. Where $\psi$ crosses zero, a node, the particle is never found, however large $\psi$ is on either side. Where $\psi$ is large and negative, the particle is just as likely to be found as where $\psi$ is large and positive, since squaring removes the sign. The most probable position is where $|\psi|^2$ peaks, and it need not coincide with the average position, which weighs the whole distribution.
The curvature of $\psi$ also carries information. By the Schrödinger equation, $\psi''$ is proportional to $(U - E)\psi$. Where the particle's energy exceeds the potential, $\psi$ curves back toward the axis and oscillates, more tightly where the kinetic energy is larger, since that means a shorter de Broglie wavelength. Where the energy is below the potential, $\psi$ curves away from the axis, and an acceptable wave function must then decay toward zero rather than blow up. Physicists sketch wave functions by these rules long before solving any equation, and the sketches are usually right about the number of nodes and where the particle is most likely to be.
The scanning tunneling microscope, invented by Gerd Binnig and Heinrich Rohrer at IBM in 1981, brings a sharp metal tip within a nanometer of a surface. Electrons cannot classically cross the gap, but their wave functions leak across it, falling off as $e^{-\kappa d}$ with $\kappa$ about $10$ nm⁻¹. A small voltage drives a tunneling current proportional to $|\psi|^2$, which falls by about $7$ times for each $0.1$ nm of extra gap.
That extreme sensitivity lets a feedback loop hold the current constant while the tip scans, tracing single atoms. At IBM's Almaden Research Center in San Jose, researchers used the tip to push xenon atoms into the letters I-B-M in 1989, and later to build a stop-motion movie from moved atoms.
The flash memory in phones and USB drives stores each bit as charge on a tiny floating electrode surrounded by insulating oxide a few nanometers thick. To write or erase a bit, a strong voltage lets electrons tunnel through the oxide, riding the tail of their wave function across a barrier they could not climb.
Because tunneling probability falls exponentially with barrier thickness, a few nanometers of oxide can hold charge for ten years when no voltage is applied, yet pass it in microseconds when the barrier is tilted by a field. Memory engineers at companies like Micron in Boise, Idaho, design the oxide thickness from exactly these exponentials.
It is easy to read $\psi$ itself as the probability of finding the particle. It is not: $\psi$ can be negative or complex, and only $|\psi|^2$ is a probability density. Doubling $\psi$ at a point makes the particle four times as likely to be found there, not twice.
A related error is to picture $\psi$ as a physical wave filling space, like a sound wave. It is a mathematical amplitude for computing probabilities. A single electron is never found spread out; it is found whole at one place, with odds set by $|\psi|^2$.
A particle is equally likely anywhere in $0 \le x \le 4$ nm. Find $|\psi|^2$.
$|\psi|^2 = \dfrac{1}{4}\ \text{nm}^{-1}$
Area one over width $4$.
Find the probability between $1$ and $2$ nm.
$P = \dfrac{1}{4} \times 1 = 0.25$
Density times width.
Find the average position.
$\langle x\rangle = 2\ \text{nm}$
The middle.
Find $\langle x^2\rangle$.
$\displaystyle\int_0^4 \dfrac{x^2}{4}dx = \dfrac{16}{3} = 5.33\ \text{nm}^2$
Weighted average of $x^2$.
Find the spread.
$\Delta x = \sqrt{5.33 - 4} = 1.15\ \text{nm}$
$L/\sqrt{12}$.
Normalize $\psi = Ax$ on $0 \le x \le 2$ nm.
$A^2\dfrac{8}{3} = 1 \Rightarrow A^2 = 0.375\ \text{nm}^{-3}$
$\int_0^2 x^2dx = 8/3$.
Write the density.
$|\psi|^2 = 0.375x^2$
In nm⁻¹.
Find the probability in $0$ to $1$ nm.
$P = \dfrac{1^3}{8} = 0.125$
$b^3/L^3$.
Find the probability in $1$ to $2$ nm.
$P = 1 - 0.125 = 0.875$
The rest.
Find the average position.
$\langle x\rangle = \dfrac{3}{4} \times 2 = 1.5\ \text{nm}$
$3L/4$.
Find the most likely position.
$x = 2\ \text{nm}$
Where $|\psi|^2$ is largest.
Normalize $\psi = Ae^{-x/a}$ for $x \ge 0$, with $a = 0.5$ nm.
$A^2\dfrac{a}{2} = 1 \Rightarrow A^2 = 4\ \text{nm}^{-1}$
$\int_0^{\infty}e^{-2x/a}dx = a/2$.
Write the density.
$|\psi|^2 = 4e^{-4x}$
In nm⁻¹, $x$ in nm.
Find the probability beyond $0.5$ nm.
$P = e^{-2 \times 0.5/0.5} = e^{-2} = 0.135$
$e^{-2x_0/a}$.
Find the probability within $0.25$ nm.
$P = 1 - e^{-1} = 0.632$
Complement.
Find the average position.
$\langle x\rangle = \displaystyle\int_0^{\infty}4xe^{-4x}dx = 0.25\ \text{nm}$
$a/2$.
Find the median, where $P = \tfrac{1}{2}$.
$e^{-4x} = 0.5 \Rightarrow x = 0.173\ \text{nm}$
Below the average: a long tail.
Relate it to tunneling.
$\text{such tails reach into forbidden regions}$
Where $E < U$.
Check the normalization.
$0.5 \times 2 = 1$
It is normalized.
Multiply the density by the width.
$P = 0.5 \times 0.6$
Width $0.6$ nm.
Evaluate the probability.
At point P a particle's wave function is $4$ times as large as at point Q. How does the probability of finding the particle in a tiny interval at P compare with an equal interval at Q?
Complete the worked solution: a particle is equally likely anywhere between $x = 0$ and $x = 5$ nm, and never outside. Find $|\psi|^2$ inside in nm⁻¹, the probability of finding it between $1$ nm and $4$ nm, and its average position in nm.
Normalize the constant density.
$|\psi|^2 = \dfrac{1}{L} =$ d
Its integral over the width is one.
Multiply by the interval's width.
$P = |\psi|^2 \times 3 =$ p
Constant density times width.
Find the average position.
$\langle x\rangle = \displaystyle\int_0^L \dfrac{x}{L}dx = \dfrac{L}{2} =$ m
The middle, by symmetry.
Check the probability's range.
$0 \le P \le 1$
Any interval inside gives a fraction.
Match each wave-function idea to its statement.
| $|\psi|^2$ | $\int|\psi|^2dx = 1$ | $\int x|\psi|^2dx$ | continuous and finite | |
|---|---|---|---|---|
| probability density | ||||
| normalization | ||||
| average position | ||||
| conditions on ψ |
A particle confined between $x = 0$ and $x = L = 2$ nm has wave function $\psi = Ax$ there and zero elsewhere. Fill in $A^2$ in nm⁻³, the probability of finding it in the left half, and its average position in nm.
| value | |
|---|---|
| A² (nm⁻³) | |
| probability in left half | |
| average position (nm) |
A particle confined between $x = 0$ and $x = 5$ nm has the normalized wave function $\psi = Ax$ there, with $A^2 = \tfrac{3}{125}$ nm⁻³. Write the probability of finding it between $0$ and $b$ nm, as a formula in $b$.
Answer:
A particle's normalized wave function for $x \ge 0$ is $\psi = \sqrt{2/a}\,e^{-x/a}$ with $a = 2$ nm, and zero for $x < 0$. What is the probability of finding it beyond $x_0 = 0.5$ nm?
Answer: probability
A scanning tunneling microscope, invented at IBM and used at its Almaden Research Center in San Jose, senses surfaces through the tail of the electrons' wave function in the gap. The tail falls as $e^{-\kappa d}$ with $\kappa = 10$ nm⁻¹. If the tip is raised by $0.05$ nm, by what factor does the tunneling current change?
Answer: factor
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle confined between $x = 0$ and $x = 3$ nm has the normalized wave function $\psi = Ax$ there, with $A^2 = \tfrac{3}{27}$ nm⁻³. Write the probability of finding it between $0$ and $b$ nm, as a formula in $b$.
Answer:
You can work with wave functions. Explain to someone why doubling $\psi$ at a point makes the particle four times as likely to be found there.
21. Your turn: a particle has $|\psi|^2 = 0.5$ nm⁻¹ between $0$ and $2$ nm and zero elsewhere. Find the probability between $0.5$ and $1.1$ nm., step 3
$P = 0.30$
Thirty percent.