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Standing waves between rigid walls: quantized energies $n^2\pi^2\hbar^2/(2ma^2)$, normalized sine states, nodes, transition wavelengths and probabilities, and the colors of dyes and quantum wells.
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By the end of this lesson you will be able to solve the infinite square well, find its energies and wave functions, compute transition wavelengths and probabilities in the well, and apply the model to dyes, quantum wells and nuclei.
You know the time-independent Schrödinger equation, $\hat{H}\psi = E\psi$, and that its acceptable solutions give stationary states. From differential equations you know that $\psi'' = -k^2\psi$ has solutions $\sin kx$ and $\cos kx$. This lesson solves the equation for the simplest bound system and finds that the boundary conditions alone quantize the energy.
| Term | What it means |
|---|---|
| Infinite square well | $V = 0$ for $0 < x < a$ and $V = \infty$ outside: a particle trapped between perfectly rigid walls. |
| Boundary condition | A requirement on $\psi$ at an edge; here $\psi(0) = \psi(a) = 0$. |
| Wave number | $k = \sqrt{2mE}/\hbar$, with $\psi = \sin kx$ inside the well. |
| Quantum number | The integer $n$ labeling the allowed states, $n = 1, 2, 3, \ldots$ |
| Ground state | The lowest-energy state, $n = 1$, with energy $E_1 = \pi^2\hbar^2/(2ma^2)$. |
| Node | A point where $\psi = 0$; the state $n$ has $n - 1$ inside the well. |
| Orthonormal | Normalized and mutually orthogonal: $\int\psi_m\psi_n\,dx = 0$ for $m \neq n$ and $1$ for $m = n$. |
| Zero-point energy | The ground-state energy $E_1 > 0$: a confined particle can never be at rest. |
Inside the well $V = 0$, so the time-independent equation is
$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \quad\Rightarrow\quad \psi'' = -k^2\psi, \qquad k = \frac{\sqrt{2mE}}{\hbar}.$$
The general solution is $\psi = A\sin kx + B\cos kx$. Outside the well the infinite potential forces $\psi = 0$, and continuity at the walls requires $\psi(0) = 0$ and $\psi(a) = 0$. The first kills the cosine, $B = 0$. The second requires $\sin ka = 0$, so
$$ka = n\pi, \qquad n = 1, 2, 3, \ldots$$
($n = 0$ gives $\psi = 0$ everywhere, no particle at all.) The allowed energies follow from $E = \hbar^2k^2/2m$:
$$E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} = n^2E_1,$$
and normalization fixes $A = \sqrt{2/a}$:
$$\psi_n(x) = \sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a}.$$
Four features carry through to every bound system. The energies are quantized, and the quantization comes from the boundary conditions, not from any assumption. The lowest energy is not zero: $E_1 > 0$, the zero-point energy, because a particle confined to width $a$ has a momentum spread of order $\hbar/a$. The $n$th state has $n - 1$ nodes, points where the particle is never found. And the states are orthonormal, so any state in the well can be expanded as $\sum c_n\psi_n$.
For an electron it is convenient to write $\pi^2\hbar^2/(2m_e) = 0.376$ eV nm², so $E_n = 0.376\,n^2/a^2$ eV with $a$ in nanometers.
Another way: picture
Picture a guitar string pinned at both ends. It can vibrate only in patterns that fit a whole number of half-wavelengths between the pins: the fundamental, the second harmonic with a still point in the middle, and so on. The particle in a box is the same, with the wave function playing the string. Each harmonic has a definite energy, and the energies climb as the square of the harmonic number, because momentum goes as $1/\lambda$ and energy as momentum squared.
Another way: steps
Checks. Energies must scale as $n^2$ and as $1/(ma^2)$: a wider box or a heavier particle gives lower energies. Energy gaps grow with $n$, as $(2n + 1)E_1$. The probability of finding the particle in either half of the box is exactly $\tfrac{1}{2}$ for every $n$, by symmetry. At large $n$ the density $\sin^2(n\pi x/a)$ oscillates so fast that its average over any small region is $1/a$: the uniform distribution a classical ball bouncing at constant speed would give, which is the correspondence principle at work.
The probability of finding the particle in the state $n$ between $0$ and $b$ is
$$P = \frac{2}{a}\int_0^b\sin^2\frac{n\pi x}{a}\,dx = \frac{b}{a} - \frac{1}{2n\pi}\sin\frac{2n\pi b}{a}.$$
The first term is the classical answer, the fraction of the box; the second is the quantum correction, which shrinks as $n$ grows. For the ground state and the first quarter of the box, $P = \tfrac{1}{4} - \tfrac{1}{2\pi} = 0.091$: the particle avoids the wall, where its wave function must vanish. For $n = 2$ and the first quarter, $\sin(\pi) = 0$ and $P = \tfrac{1}{4}$ exactly, even though the density is far from uniform, because the quarter spans exactly one hump.
These integrals are exact, and they explain a result that surprises many learners: in the state $n = 2$, the particle is never found at the center, yet it is equally likely to be found in either half. It does not "cross" the center, because it is not a ball moving along a path; it is described by a standing wave with a node there.
The box states are orthogonal:
$$\frac{2}{a}\int_0^a\sin\frac{m\pi x}{a}\sin\frac{n\pi x}{a}\,dx = \begin{cases}1 & m = n\\0 & m \neq n\end{cases}$$
which follows from the product-to-sum identity. They are also complete: any reasonable function on $0 < x < a$ that vanishes at the walls is a Fourier sine series $\sum c_n\psi_n$. Together these make the box states a basis in which any state of the particle can be written, with $c_n = \int\psi_n\Psi(x, 0)\,dx$ and energy probabilities $|c_n|^2$. The parabolic well state of the first lessons has $|c_1|^2 = 960/\pi^6 = 0.9986$, $c_2 = 0$ by symmetry, and $|c_3|^2 = 960/(729\pi^6) = 0.0014$: it is almost entirely the ground state. This expansion is exactly Fourier analysis, and the square well is where quantum mechanics and Fourier series meet most directly.
No real well has infinitely high walls, but the model is accurate whenever the walls are much higher than the energies of the states being studied, and it is the first estimate for any confined particle. Electrons in a thin layer of gallium arsenide sandwiched between aluminum gallium arsenide — a quantum well — are confined in one direction by barriers a few tenths of an eV high, and the infinite-well energies, corrected with the electron's effective mass, predict the wavelengths of the lasers built from them. The pi electrons of long molecules with alternating single and double bonds move almost freely along the chain and are held in by its ends, and the infinite well predicts the colors of these dyes to within about ten percent. The nucleons of a nucleus are confined by the strong force to a region a few femtometers across, and the well's $8$ MeV ground-state energy sets the scale of nuclear physics. The finite well, later in this unit, shows what changes when the walls are low enough to matter.
Real potentials never have infinitely high walls, so the infinite well is always an approximation. It works well when the walls are much higher than the energies of the states you care about: an electron in a semiconductor quantum dot sits in a well a few tenths of an electron volt deep, and its lowest one or two levels are close to the box formula, while higher levels, which approach the rim, are pushed down because the wave function leaks into the walls. The finite well, two lessons ahead, makes that correction exact.
The box also ignores the interactions between particles. For the pi electrons of a dye molecule the free-electron model gets the color roughly right because the electrons fill levels two at a time, as the Pauli principle requires, and the lowest empty and highest filled levels set the absorption. Treat such answers as estimates good to perhaps twenty percent, and use them to see trends: longer molecules, like larger dots, absorb at longer wavelengths.
Cyanine dyes, used in photographic film, in the DVD-R discs of the 2000s and as fluorescent labels in biology, owe their colors to particle-in-a-box physics. Their molecules have a chain of alternating single and double bonds along which $N$ pi electrons move almost freely, confined by the ends of the chain.
Filling the box levels two electrons at a time puts the highest electrons in level $N/2$. The lowest-energy absorption lifts one to level $N/2 + 1$, a gap of $[(N/2 + 1)^2 - (N/2)^2]h^2/(8mL^2) = (N + 1)h^2/(8mL^2)$. The absorbed wavelength is
$$\lambda = \frac{8mcL^2}{h(N + 1)} = \frac{3297L^2}{N + 1}\ \text{nm}.$$
A chain $1.0$ nm long with $6$ pi electrons absorbs near $471$ nm, blue light, and looks orange-red; lengthening the chain to $1.4$ nm with $10$ electrons shifts the absorption to about $590$ nm and the color to blue. Adding one double bond adds two electrons and about $0.14$ nm of length, moving the absorption about $100$ nm to the red — exactly the trend chemists observe, and the rule they use to design dyes for particular colors.
The red laser in a DVD player or a laser pointer, and the infrared lasers that carry signals through fiber-optic networks, are quantum well lasers. A layer of gallium arsenide a few nanometers thick is sandwiched between layers of aluminum gallium arsenide, which has a larger band gap and so confines electrons and holes to the thin layer, like walls of a box.
In gallium arsenide the electron's effective mass is $0.067m_e$, so its ground-state confinement energy in a well $8$ nm wide is about $0.376/(0.067 \times 64) = 0.088$ eV. Added to the band gap of $1.42$ eV, and with a smaller contribution from the heavier holes, it raises the emitted photon energy and shifts the wavelength from $873$ nm toward about $830$ nm. Engineers choose the well width to set the laser's color within a few nanometers.
The quantization also makes the laser efficient: confining electrons to discrete levels concentrates them at the energy where they emit, rather than spreading them across a band. Quantum well lasers, developed at Bell Labs and elsewhere in the 1970s, now power the internet's optical links.
The classical picture of a particle in a box is a ball bouncing back and forth, equally likely to be anywhere. The quantum stationary states are nothing like it. In the ground state the particle is most likely found in the middle and almost never near the walls; in the state $n = 2$ it is never found at the center at all, though it is found on both sides with equal probability. There is no path along which it travels, and so no puzzle about how it "gets across" a node.
A second misconception is that the lowest energy could be zero — the particle simply sitting still. The boundary conditions forbid $n = 0$, since that would make $\psi$ zero everywhere, and the smallest allowed energy is $E_1 = \pi^2\hbar^2/(2ma^2)$. A confined particle can never be at rest; this zero-point energy is the uncertainty principle made exact.
An electron is confined to a well $1.0$ nm wide. Write the ground-state energy.
$E_1 = \dfrac{\pi^2\hbar^2}{2ma^2}$
The lowest standing wave has one half-wavelength across the box.
Put in the electron's constant.
$\dfrac{\pi^2\hbar^2}{2m_e} = \dfrac{\pi^2(\hbar c)^2}{2m_ec^2} = \dfrac{\pi^2 \times (197.3\ \text{eV nm})^2}{2 \times 511\,000\ \text{eV}} = 0.376\ \text{eV nm}^2$
Multiplying top and bottom by $c^2$ gives convenient units.
Divide by the width squared.
$E_1 = \dfrac{0.376}{(1.0)^2} = 0.376\ \text{eV}$
A fraction of an electronvolt.
Find the next two levels.
$E_2 = 4 \times 0.376 = 1.504\ \text{eV}, \qquad E_3 = 9 \times 0.376 = 3.384\ \text{eV}$
$E_n = n^2E_1$.
Compare with thermal energy.
$kT = 0.026\ \text{eV} \ll E_2 - E_1 = 1.13\ \text{eV}$
At room temperature the electron sits almost always in the ground state.
The electron in the $1.0$ nm box drops from $n = 2$ to $n = 1$. Find the energy released.
$\Delta E = E_2 - E_1 = (4 - 1)E_1 = 3 \times 0.376 = 1.128\ \text{eV}$
The photon carries the difference of the levels.
Write the photon's wavelength.
$\lambda = \dfrac{hc}{\Delta E}$
Photon energy and wavelength are related by $E = hc/\lambda$.
Substitute with $hc = 1239.8$ eV nm.
$\lambda = \dfrac{1239.8}{1.128}$
A convenient form of $hc$.
Evaluate the result.
$\lambda = 1099\ \text{nm}$
Near infrared, just beyond red.
Find the width that would give green light, $520$ nm.
$\Delta E = \dfrac{1239.8}{520} = 2.384\ \text{eV} = \dfrac{3 \times 0.376}{a^2} \quad\Rightarrow\quad a = \sqrt{\dfrac{1.128}{2.384}} = 0.688\ \text{nm}$
Narrower boxes give bluer light.
Say why it matters.
$\lambda \propto a^2$
The color depends on the square of the size, which is how quantum dots are tuned.
Write the probability of finding the ground-state particle between $0$ and $a/4$.
$P = \dfrac{2}{a}\int_0^{a/4}\sin^2\dfrac{\pi x}{a}\,dx$
Born's rule with the normalized ground state.
Use the half-angle identity.
$\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$
It turns the square into something easy to integrate.
Substitute the identity.
$P = \dfrac{1}{a}\int_0^{a/4}\left(1 - \cos\dfrac{2\pi x}{a}\right)dx$
The factor $2/a$ times $\tfrac{1}{2}$.
Integrate each term.
$P = \dfrac{1}{a}\left[x - \dfrac{a}{2\pi}\sin\dfrac{2\pi x}{a}\right]_0^{a/4}$
The antiderivative of $\cos(2\pi x/a)$ is $(a/2\pi)\sin(2\pi x/a)$.
Evaluate at the limits.
$P = \dfrac{1}{a}\left[\dfrac{a}{4} - \dfrac{a}{2\pi}\sin\dfrac{\pi}{2}\right] = \dfrac{1}{4} - \dfrac{1}{2\pi}$
$\sin(\pi/2) = 1$, and both terms vanish at $0$.
Evaluate the number numerically.
$P = 0.25 - 0.159 = 0.091$
Less than one in ten.
Compare with the classical answer.
$P_{\text{classical}} = \dfrac{1}{4}$
The ground state keeps the particle away from the walls, where $\psi$ must vanish.
Compare with the parabolic trial state.
$P_{\text{parabola}} = \dfrac{53}{512} = 0.104$
Close to the true ground state's $0.091$, as the overlap $|c_1|^2 = 0.9986$ suggests.
Write the normalization integral.
$A^2\int_0^a\sin^2\dfrac{\pi x}{a}\,dx = 1$
Total probability one.
Use the average of $\sin^2$ over whole humps.
$\int_0^a\sin^2\dfrac{\pi x}{a}\,dx = \dfrac{a}{2}$
Over a whole number of half-periods, $\sin^2$ averages to one half.
Solve for the constant.
An electron's ground-state energy in an infinite well is $9$ eV. The well is made twice as wide. What is the new ground-state energy?
Complete the worked solution: an electron in a well $5$ nm wide, with $E_1 = 0.01504$ eV, drops from $n = 3$ to $n = 2$. Find the two energies and the photon energy.
Multiply the ground-state energy by nine.
$E_3 = 3^2 \times 0.01504 =$ u eV
The third level is nine times the first.
Multiply the ground-state energy by four.
$E_2 = 2^2 \times 0.01504 =$ w eV
The second level is four times the first.
Subtract the final energy from the initial one.
$\Delta E = E_3 - E_2 =$ d eV
The photon carries the difference: $5E_1$.
For a particle in an infinite square well of width $a$, match each quantity for the state $n = 6$ to its expression.
| $36E_1$ | $\sqrt{2/a}\,\sin(6\pi x/a)$ | $5$ | $13E_1$ | |
|---|---|---|---|---|
| the energy | ||||
| the wave function | ||||
| the number of interior nodes | ||||
| the energy gap to the next level up |
An electron in an infinite well $0.5$ nm wide has ground-state energy $E_1 = 1.504$ eV. Fill in the energies of the next three levels, in eV.
| energy | |
|---|---|
| $E_2$ (eV) | |
| $E_3$ (eV) | |
| $E_4$ (eV) |
An electron in an infinite well $0.5$ nm wide drops from level $2$ to level $1$, emitting a photon. What is the photon's wavelength, in nm? Use $E_1 = 0.376/a^2$ eV and $hc = 1239.8$ eV nm.
Answer: nm
An electron is trapped in an infinite square well of width $a = 0.5$ nm. What is the energy of its state $n = 5$, in eV? Use $\pi^2\hbar^2/(2m_e) = 0.376$ eV nm².
Answer: eV
A cyanine dye molecule has a chain of alternating bonds $0.8$ nm long along which $4$ pi electrons move freely. Treating them as electrons in a box, filled two to a level, what wavelength of light does the dye absorb most strongly, in nm?
Answer: nm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An electron in an infinite well $2$ nm wide has ground-state energy $E_1 = 0.094$ eV. Fill in the energies of the next three levels, in eV.
| energy | |
|---|---|
| $E_2$ (eV) | |
| $E_3$ (eV) | |
| $E_4$ (eV) |
You can find the energies and states of a particle in a box and the light it emits. Explain to someone how a particle in the state $n = 2$ can be found on both sides of the box but never at its center.
16. Your turn: find the normalization constant of $\psi = A\sin(\pi x/a)$ on $0 < x < a$., step 3
$A^2\dfrac{a}{2} = 1 \quad\Rightarrow\quad A = \sqrt{\dfrac{2}{a}}$
The same constant for every $n$.