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Separating the Schrödinger equation, the time-independent equation and stationary states, superpositions with their energy probabilities $|c_n|^2$, and the sloshing of a superposition at $(E_2 - E_1)/h$.
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 separate the Schrödinger equation, explain why stationary states have constant densities, find energy probabilities and averages for a superposition, and find the period with which a superposition oscillates.
You can compute expectation values from a wave function, and you know the Schrödinger equation and the Hamiltonian operator. From differential equations you know separation of variables. This lesson separates the Schrödinger equation, finds the states whose measurable properties never change, and builds every other state from them.
| Term | What it means |
|---|---|
| Separation of variables | Seeking solutions $\Psi(x, t) = \psi(x)\varphi(t)$, which splits the equation into an $x$ part and a $t$ part. |
| Time-independent Schrödinger equation | $\hat{H}\psi = E\psi$: $-\frac{\hbar^2}{2m}\psi'' + V\psi = E\psi$. |
| Stationary state | A solution $\psi_n(x)e^{-iE_nt/\hbar}$, whose probability density and expectation values do not change in time. |
| Energy eigenvalue | $E_n$, an allowed energy: the constant for which the time-independent equation has an acceptable solution. |
| Superposition | A sum $\sum c_n\psi_n$ of stationary states, itself a valid state. |
| Expansion coefficient | $c_n$; $\vert c_n\vert ^2$ is the probability of finding energy $E_n$. |
| Quantum beats | Oscillations in a measurable quantity at frequency $(E_2 - E_1)/h$ from a superposition of two levels. |
When the potential does not depend on time, look for solutions that are a product of a function of $x$ and a function of $t$, $\Psi = \psi(x)\varphi(t)$. Substituting into the Schrödinger equation and dividing by $\psi\varphi$ gives
$$i\hbar\frac{1}{\varphi}\frac{d\varphi}{dt} = \frac{1}{\psi}\left[-\frac{\hbar^2}{2m}\psi'' + V\psi\right].$$
The left side depends only on $t$ and the right only on $x$, so both equal a constant, called $E$. The time part solves at once: $\varphi = e^{-iEt/\hbar}$. The space part is the time-independent Schrödinger equation,
$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi, \qquad \text{or} \qquad \hat{H}\psi = E\psi.$$
Its acceptable solutions exist only for special values $E_n$, the allowed energies, with wave functions $\psi_n$. Each $\Psi_n = \psi_ne^{-iE_nt/\hbar}$ is a stationary state: its density $|\Psi_n|^2 = |\psi_n|^2$ never changes, nor does any expectation value, and its energy is exactly $E_n$ with no spread.
Because the Schrödinger equation is linear, any combination of stationary states is also a solution:
$$\Psi(x, t) = \sum_n c_n\psi_n(x)e^{-iE_nt/\hbar}.$$
The coefficients are fixed by the state at $t = 0$, and they carry the physics: a measurement of energy returns one of the $E_n$, with probability $|c_n|^2$, so $\sum|c_n|^2 = 1$ and $\langle E\rangle = \sum|c_n|^2E_n$. A superposition has an energy spread and a density that moves, because its pieces turn their phases at different rates.
Another way: picture
Picture each stationary state as a clock hand turning at its own speed, $E_n/\hbar$. A single hand turning tells you nothing you can measure — every position of the hand looks the same. Put two hands turning at different speeds side by side and the angle between them changes; that changing angle is what makes the density of a superposition slosh, at the difference of their speeds.
Another way: steps
Checks. The probabilities $|c_n|^2$ must add to one. The average energy must lie between the smallest and largest $E_n$ present, and is constant in time even when the density moves, since energy is conserved. A stationary state's density must not depend on time; if a calculation says it does, a phase has been mishandled. And the oscillation period of a superposition depends only on energy differences, never on the absolute energies.
Take an equal superposition of two real stationary states,
$$\Psi = \frac{1}{\sqrt{2}}\left[\psi_1e^{-iE_1t/\hbar} + \psi_2e^{-iE_2t/\hbar}\right].$$
Its density is
$$|\Psi|^2 = \frac{1}{2}\left[\psi_1^2 + \psi_2^2 + 2\psi_1\psi_2\cos\frac{(E_2 - E_1)t}{\hbar}\right].$$
The first two terms are fixed; the cross term oscillates. In a box, $\psi_1$ is a single hump and $\psi_2$ is positive on the left half and negative on the right, so their product is positive on the left and negative on the right. When the cosine is $+1$ the density piles up on the left; half a period later, on the right. The particle's probability sloshes from wall to wall with period
$$T = \frac{h}{E_2 - E_1},$$
and $\langle x\rangle$ swings back and forth like a classical particle bouncing in the box. For an electron in a $1$ nm box, $E_2 - E_1 = 3 \times 0.376 = 1.13$ eV and $T = 3.7$ fs.
Stationary states with different energies are orthogonal: $\int\psi_m^\psi_n\,dx = 0$ for $m \neq n$. This follows from the Hamiltonian being a Hermitian operator, the subject of the formalism unit, and it is what makes the expansion easy to use. Multiply $\Psi(x, 0) = \sum c_n\psi_n$ by $\psi_m^$ and integrate: every term but one vanishes, leaving
$$c_m = \int\psi_m^*(x)\Psi(x, 0)\,dx.$$
So the probability of finding energy $E_m$ is the squared overlap of the initial state with the $m$th stationary state. The parabolic well state of the earlier lessons, for instance, overlaps the box's ground state so strongly that $|c_1|^2 = 960/\pi^6 = 0.9986$: an energy measurement on it finds the ground-state energy more than ninety-nine times in a hundred. That high overlap is why its average energy came within two percent of the ground-state energy.
The word stationary describes what can be measured, not what is happening. The wave function of a stationary state turns in phase at angular frequency $E/\hbar$ — for an electron in a hydrogen atom's ground state, about $2 \times 10^{16}$ radians a second. The electron has kinetic energy and a spread of momenta; it simply has a density that does not change.
That is why an atom in its ground state does not radiate, answering the puzzle that sank the classical planetary model: a stationary state has no oscillating charge density to produce radiation. Only a superposition of two levels has a density that oscillates, at $(E_2 - E_1)/h$, and that oscillating charge radiates light of exactly that frequency — Bohr's frequency condition, now explained.
Stationary states are the eigenstates of the Hamiltonian, and energy is singled out because the Hamiltonian is also what drives time evolution: the Schrödinger equation is $i\hbar\,\partial\Psi/\partial t = \hat{H}\Psi$. A state with a definite energy is therefore one that time can only multiply by a phase. A state with a definite position or a definite momentum, in general, is not an eigenstate of $\hat{H}$ and begins to change at once: a particle localized at a point spreads, and a particle with a definite momentum in a potential speeds up or slows down.
This gives the standard strategy for any time-dependent problem with a time-independent potential. Expand the initial state in energy eigenstates, attach to each its phase $e^{-iE_nt/\hbar}$, and add them up. All the dynamics lives in the relative phases, and all the difficulty lies in finding the eigenstates — the job of the next unit, for the potentials that can be solved exactly. Once they are known, the evolution of any initial state is a matter of bookkeeping.
Ahmed Zewail won the 1999 Nobel Prize in Chemistry for watching atoms move during chemical reactions, using laser pulses a few femtoseconds long. A first pulse puts a molecule into a superposition of vibrational levels; a second, delayed pulse probes where the resulting wave packet is. Scanning the delay traces the packet as it sloshes back and forth — exactly the superposition dynamics of this lesson, in a real molecule.
In sodium iodide, Zewail's group saw the packet oscillate between a covalent and an ionic arrangement with a period of about $1.25$ ps, leaking a little toward dissociation on each swing. That period corresponds to a vibrational level spacing of $h/T = 4.136 \times 10^{-15}/1.25 \times 10^{-12} = 3.3$ meV, matching the spectroscopy of the molecule.
Semiconductor physicists do the same with electrons in quantum wells, where two levels $20$ meV apart give quantum beats with a period of $h/0.020\ \text{eV} = 207$ fs. Such measurements test the superposition principle directly: a mixture of molecules, some in one level and some in the other, would show no beats at all.
Bohr's 1913 model of hydrogen asserted, without explanation, that an atom in a stationary state does not radiate and that it emits light of frequency $f = (E_2 - E_1)/h$ when it jumps between levels. The Schrödinger equation explains both. A stationary state's charge density is constant, so it does not radiate. A superposition of two levels has a charge density oscillating at exactly $(E_2 - E_1)/h$, and an oscillating charge radiates at its own frequency.
For hydrogen's red Balmer line, the levels $n = 3$ and $n = 2$ differ by $1.89$ eV, so the light has frequency $1.89/4.136 \times 10^{-15} = 4.57 \times 10^{14}$ Hz and wavelength $656$ nm. Each atom's spectrum is a set of such differences between its stationary energies, which is why every element has its own fingerprint of sharp lines — the basis of the spectroscopy that identifies elements in distant stars, in the exhaust of a car engine and in a forensic laboratory's samples.
It is tempting to read stationary state as a particle standing still. A stationary state has a definite energy, including kinetic energy, and a wave function whose phase turns continuously; only its measurable quantities are constant. An electron in the ground state of hydrogen moves, in the sense that its momentum spread is large, but nothing about its probability distribution changes.
A second trap is to think a superposition $c_1\psi_1 + c_2\psi_2$ means the particle is "really" in state 1 with probability $|c_1|^2$ and in state 2 otherwise, and we merely do not know which. If that were so, the density would be $|c_1|^2\psi_1^2 + |c_2|^2\psi_2^2$, with no cross term, and nothing would slosh. The measured quantum beats — oscillations at $(E_2 - E_1)/h$ in the light a superposition emits — show the cross term is real. Until an energy measurement is made, the particle is in the superposition itself.
Write a stationary state.
$\Psi_n(x, t) = \psi_n(x)e^{-iE_nt/\hbar}$
A solution of the time-independent equation times its phase factor.
Write its complex conjugate.
$\Psi_n^ = \psi_n^(x)e^{+iE_nt/\hbar}$
Replace $i$ by $-i$; $E_n$ and $t$ are real.
Multiply to get the density.
$|\Psi_n|^2 = \psi_n^*\psi_ne^{+iE_nt/\hbar}e^{-iE_nt/\hbar}$
The density is $\Psi^*\Psi$.
Combine the exponentials.
$e^{+iE_nt/\hbar}e^{-iE_nt/\hbar} = e^{0} = 1$
Equal and opposite exponents cancel.
Draw the conclusion.
$|\Psi_n(x, t)|^2 = |\psi_n(x)|^2$
No time dependence: the state is stationary.
A particle in a box is in $\Psi = \sqrt{0.6}\,\psi_1 + \sqrt{0.4}\,\psi_2$ at $t = 0$. Find the energy probabilities.
$P(E_1) = |c_1|^2 = 0.6, \qquad P(E_2) = |c_2|^2 = 0.4$
Squared magnitudes of the coefficients.
Check the normalization.
$0.6 + 0.4 = 1$
The state is normalized because the $\psi_n$ are orthonormal.
Write the energies.
$E_2 = 4E_1$
Box energies grow as $n^2$.
Find the average energy.
$\langle E\rangle = 0.6E_1 + 0.4 \times 4E_1 = 2.2E_1$
Each energy weighted by its probability.
Find the mean square energy.
$\langle E^2\rangle = 0.6E_1^2 + 0.4 \times 16E_1^2 = 7.0E_1^2$
Square each energy before weighting.
Find the energy spread.
$\sigma_E^2 = 7.0E_1^2 - (2.2E_1)^2 = 2.16E_1^2 \quad\Rightarrow\quad \sigma_E = 1.47E_1$
A superposition has an energy spread; a stationary state has none.
An electron in a $1.0$ nm box is in $\Psi = (\psi_1e^{-iE_1t/\hbar} + \psi_2e^{-iE_2t/\hbar})/\sqrt{2}$. Write its density.
$|\Psi|^2 = \tfrac{1}{2}\left[\psi_1^2 + \psi_2^2 + 2\psi_1\psi_2\cos\omega t\right]$
The cross terms combine into a cosine of the phase difference.
Identify the angular frequency.
$\omega = \dfrac{E_2 - E_1}{\hbar}$
The rate at which the two phases drift apart.
Find the ground-state energy.
$E_1 = \dfrac{\pi^2\hbar^2}{2ma^2} = \dfrac{0.3760\ \text{eV nm}^2}{(1.0\ \text{nm})^2} = 0.376\ \text{eV}$
The box result, derived in the next unit.
Find the energy difference.
$E_2 - E_1 = 4E_1 - E_1 = 3 \times 0.376 = 1.128\ \text{eV}$
$E_2 = 4E_1$.
Find the period.
$T = \dfrac{h}{E_2 - E_1} = \dfrac{4.136 \times 10^{-15}\ \text{eV s}}{1.128\ \text{eV}} = 3.67 \times 10^{-15}\ \text{s}$
Planck's constant in eV s.
Describe the motion.
$t = 0: \ \text{piled left}; \qquad t = T/2: \ \text{piled right}$
The product $\psi_1\psi_2$ is positive on the left half of the box and negative on the right.
Find the frequency of the light such a superposition emits.
$f = \dfrac{1}{T} = 2.73 \times 10^{14}\ \text{Hz}, \qquad \lambda = \dfrac{c}{f} = 1.10\ \mu\text{m}$
The sloshing charge radiates at exactly the frequency $(E_2 - E_1)/h$: Bohr's condition.
Check that the average energy is constant.
$\langle E\rangle = \tfrac{1}{2}(E_1 + E_2) = 2.5E_1 = 0.94\ \text{eV at all times}$
The density moves, but energy is conserved.
Differentiate the function twice.
$\psi'' = -\left(\dfrac{\pi}{a}\right)^2\sin\dfrac{\pi x}{a} = -\dfrac{\pi^2}{a^2}\psi$
Each derivative of the sine brings out a factor $\pi/a$.
Substitute into the equation with $V = 0$.
$-\dfrac{\hbar^2}{2m}\psi'' = \dfrac{\hbar^2\pi^2}{2ma^2}\psi$
The equation has the form $\hat{H}\psi = E\psi$.
Read off the energy.
A particle is in the stationary state $\Psi = \psi_{1}(x)e^{-iE_{1}t/\hbar}$. As time passes, what about it changes?
Complete the worked solution: a particle is found with energy $12$ meV with probability $0.5$ and $48$ meV otherwise. Find its average energy, its mean squared energy and the variance of energy.
Weight each energy by its probability and add.
$\langle E\rangle = 0.5 \times 12 + 0.5 \times 48 =$ m meV
The average energy.
Weight each squared energy by its probability and add.
$\langle E^2\rangle = 0.5 \times 144 + 0.5 \times 2304 =$ s meV²
The mean of the squares.
Subtract the square of the mean.
$\sigma_E^2 = \langle E^2\rangle - \langle E\rangle^2 =$ v meV²
A superposition of two energies has a nonzero energy spread; a stationary state has none.
For a time-independent potential, with states labeled $1$ to $5$ and beyond, match each piece of the solution to its equation.
| $e^{-iEt/\hbar}$ | $\hat{H}\psi = E\psi$ | $\sum c_n\psi_ne^{-iE_nt/\hbar}$ | $|c_n|^2$ | |
|---|---|---|---|---|
| the time factor of a separable solution | ||||
| the time-independent Schrödinger equation | ||||
| the general solution | ||||
| the probability of measuring energy $E_n$ |
A particle in a box, with $E_1 = 48$ meV, is in the state $\sqrt{0.4}\,\psi_1 + \sqrt{0.4}\,\psi_2 + \sqrt{0.2}\,\psi_3$. Fill in the probability of each energy and the average energy in meV.
| value | |
|---|---|
| probability of $E_1$ | |
| probability of $E_2$ | |
| probability of $E_3$ | |
| average energy (meV) |
An electron in a box of length $2$ nm is in an equal superposition of its two lowest states. With what period, in fs, does its probability density slosh back and forth? Use $E_1 = 0.3760/a^2$ eV and $h = 4.1357 \times 10^{-15}$ eV s.
Answer: fs
An electron in a box has ground-state energy $E_1 = 54$ meV. It is in a superposition of the three lowest states, with $|c_1|^2 = 0.2$, $|c_2|^2 = 0.5$ and $|c_3|^2 = 0.3$. What is its average energy, in meV?
Answer: meV
An ultrafast laser pulse puts electrons in a semiconductor quantum well into a superposition of two levels $50$ meV apart, and the emitted light flickers as the wave packet sloshes. What is the period of the flicker, in fs? Use $h = 4.1357 \times 10^{-15}$ eV s.
Answer: fs
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle in a box, with $E_1 = 59$ meV, is in the state $\sqrt{0.4}\,\psi_1 + \sqrt{0.4}\,\psi_2 + \sqrt{0.2}\,\psi_3$. Fill in the probability of each energy and the average energy in meV.
| value | |
|---|---|
| probability of $E_1$ | |
| probability of $E_2$ | |
| probability of $E_3$ | |
| average energy (meV) |
You can evolve a state in time and predict its energy measurements. Explain to someone why an atom in its ground state does not radiate, while one in a superposition of two levels does.
16. Your turn: show that $\psi = \sin(\pi x/a)$ solves the time-independent equation inside a box where $V = 0$, and find its energy., step 3
$E = \dfrac{\pi^2\hbar^2}{2ma^2}$
The ground-state energy of a box of length $a$.