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Superpositions of energy states, measurement probabilities $|c_n|^2$, average energies, the beat period $h/\Delta E$, and the harmonic oscillator's evenly spaced levels.
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 predict the outcomes and averages of energy measurements on superpositions, and apply the harmonic oscillator.
From the last two lessons you can normalize wave functions and find the energy levels of a particle in a box. From lesson 12 you know that probabilities are squared amplitudes and that each measurement gives one definite result. This lesson combines them: what happens when a system is in several energy states at once, and what a measurement of its energy gives.
| Term | What it means |
|---|---|
| Superposition | $\psi = \sum c_n\psi_n$, a combination of energy states. |
| Expansion coefficient | $c_n$, the amplitude for state $n$ in a superposition. |
| Expectation value of energy | $\langle E\rangle = \sum\vert c_n\vert ^2E_n$, the average over many measurements. |
| Collapse | The change of the state to the measured energy state after a measurement. |
| Quantum harmonic oscillator | A particle in a parabolic potential, with levels $E_n = (n + \tfrac{1}{2})hf$. |
| Zero-point energy | The oscillator's lowest energy, $\tfrac{1}{2}hf$. |
| Beat period | $h/(E_2 - E_1)$, the time for a two-state superposition's density to repeat. |
Because the Schrödinger equation is linear, any combination of energy states is also a possible state:
$$\psi = c_1\psi_1 + c_2\psi_2 + c_3\psi_3 + \cdots, \qquad \sum|c_n|^2 = 1.$$
Measuring the energy of such a superposition gives one of the allowed values $E_n$, never a value in between, with probability $|c_n|^2$. Afterward the system is in state $\psi_n$, and measuring again gives $E_n$ again. Over many identically prepared systems, the average is
$$\langle E\rangle = \sum|c_n|^2E_n.$$
A superposition of two energies is not static: its probability density oscillates with period $h/(E_2 - E_1)$. The harmonic oscillator, a particle in a parabolic potential, has evenly spaced levels $E_n = (n + \tfrac{1}{2})hf$, and is the model for every vibration in nature, from molecules to crystals.
Another way: picture
Picture a jar of marbles, some red and some blue, in fixed proportions. Each draw gives one color, never a purple marble, and the proportions show up only after many draws. A superposition is like the jar: each energy measurement draws one allowed value, and the squared amplitudes are the proportions. Unlike marbles, the values are not decided until the draw.
Another way: steps
A particle in a box prepared as $0.6\psi_1 + 0.8\psi_2$ is not in level 1 or level 2 but in a combination. Measure its energy and you get $E_1$ with probability $0.36$ or $E_2$ with probability $0.64$. Measure a thousand identically prepared particles and about $360$ give $E_1$; the rest give $E_2$.
After the measurement, the particle is in the state that was found. Measure it again immediately and you get the same energy with certainty. This change, from a superposition to a single state, is often called the collapse of the wave function, and it is the measurement problem of lesson 12 in its sharpest form.
The average of many energy measurements is $\langle E\rangle = \sum|c_n|^2E_n$. For the state above with $E_1 = 1$ eV and $E_2 = 4$ eV, $\langle E\rangle = 0.36 + 2.56 = 2.92$ eV, a value that no single measurement ever gives. The average is a statistical property of the ensemble, not an energy the particle has.
Energy is conserved in a quantum system left alone: the probabilities $|c_n|^2$ do not change with time, so neither does $\langle E\rangle$. What changes are the relative phases of the $c_n$, which move the probability density around without changing the energy distribution.
Each energy state carries a time factor $e^{-iE_nt/\hbar}$. In a single state this factor has no effect on $|\psi|^2$, which is why such states are called stationary. In a superposition of two, the cross term in $|\psi|^2$ contains $\cos((E_2 - E_1)t/\hbar)$, and the probability density sloshes with period $T = h/(E_2 - E_1)$.
For an electron in a $1$ nm box, $E_2 - E_1 = 1.13$ eV and the period is $3.7$ fs. Motion in quantum mechanics always comes from superpositions of different energies; a particle with a definite energy has a frozen probability distribution. The frequency of sloshing equals the frequency of the light emitted when the particle drops between the levels.
Near the bottom of any smooth potential well, the potential is nearly a parabola, $U = \tfrac{1}{2}kx^2$, and classical motion is simple harmonic with frequency $f$. Quantum mechanics gives evenly spaced levels, $E_n = (n + \tfrac{1}{2})hf$, with a zero-point energy $\tfrac{1}{2}hf$ in the ground state.
The even spacing means an oscillator absorbs and emits only photons of frequency $f$, matching the classical vibration frequency. Diatomic molecules vibrate this way: carbon monoxide at $6.42 \times 10^{13}$ Hz, absorbing infrared at $4.67$ μm. Every chemical bond has a vibration frequency, and infrared spectra identify molecules by these frequencies.
Checking an answer. Probabilities must lie between 0 and 1 and add to one. The average energy must lie between the smallest and largest energies involved. A single measurement must give an allowed level. And oscillator levels must be evenly spaced, starting at $\tfrac{1}{2}hf$, not zero.
Expanding any state in energy states is always possible, because the energy states of a system form a complete set, just as any string shape can be written as a sum of its harmonics. The coefficients are found by overlap integrals, $c_n = \int\psi_n^*\psi\,dx$, which the university quantum mechanics course develops.
The rule that $|c_n|^2$ is the probability of $E_n$ is the Born rule extended from positions to energies. It has the same experimental support: in every measurement ever made, the frequencies of results match the squared amplitudes.
The quantum harmonic oscillator describes the vibrations of molecules, the vibrations of atoms in crystals, called phonons, and even the electromagnetic field in a cavity, where each mode is an oscillator and each quantum of energy $hf$ is a photon. The zero-point energy of those field oscillators produces measurable forces between closely spaced metal plates, the Casimir effect.
At low temperature, vibrations freeze into their ground states because $hf$ exceeds the thermal energy $kT$. That is why the heat capacity of solids falls toward zero near absolute zero, a puzzle Einstein solved with quantized oscillators in 1907, one of the first successes of quantum theory outside radiation.
Earth radiates heat as infrared light peaking near $10$ μm. Molecules with vibrations whose quanta match that light absorb it: carbon dioxide's bending vibration absorbs at $15$ μm, a quantum of $83$ meV; water and methane absorb at their own frequencies. The absorbed energy is re-emitted in all directions, some back toward the ground.
Nitrogen and oxygen, the main gases of the atmosphere, have no such effect, because their symmetric vibrations do not change the molecule's charge distribution and so cannot absorb light. The quantum rules for which vibrations absorb explain why a gas making up $0.04$ percent of the air controls so much of the planet's heat balance.
A qubit is a two-level quantum system deliberately held in superpositions: $c_0|0\rangle + c_1|1\rangle$. Measuring it gives 0 with probability $|c_0|^2$ and 1 with probability $|c_1|^2$, exactly the rules of this lesson. The superconducting qubits in quantum computers built by IBM and Google are tiny circuits whose two lowest energy levels, a few GHz apart, serve as 0 and 1.
Those circuits are nonlinear oscillators, deliberately uneven so that only the two lowest levels are used. The computation rotates the amplitudes with microwave pulses timed to fractions of the beat period $h/\Delta E$, a few hundred picoseconds. Everything in the machine rests on the superposition rules above.
Superpositions are not exotic accidents; laboratories make them on purpose. A short pulse of light tuned between two levels, lasting a fraction of the time needed to transfer an atom fully, leaves it partly excited: after a pulse of the right length, an atom that started in the ground state is in an equal superposition of the two levels. Longer pulses shift the balance, and the probability of finding the atom excited rises and falls with pulse length in what physicists call Rabi oscillations.
The same technique works for electrons in quantum dots, nuclear spins in MRI, and the qubits of quantum computers, with pulses of light, radio waves or microwaves matched to each system's level spacing. Ultrafast lasers, with pulses of a few femtoseconds, can even prepare superpositions of electron states in molecules and watch the electrons slosh, the motion computed in this lesson, as it happens. Chemists at national laboratories use such measurements to follow how energy moves through molecules in the first moments after light is absorbed, as in photosynthesis or solar cells.
At NOAA's Global Monitoring Laboratory in Boulder, Colorado, and its observatory on Mauna Loa in Hawaii, scientists measure how much infrared light greenhouse gases absorb. Carbon dioxide's bending vibration has a quantum of $83$ meV, matching infrared at $15$ μm, close to the peak of Earth's heat glow near $10$ μm.
Each absorption lifts a molecule from its vibrational ground state to its first excited state; collisions then pass the energy to the surrounding air or the molecule re-emits it in a random direction. The quantum spacing $hf$ decides which wavelengths each gas blocks, which is why carbon dioxide, methane and water vapor warm the planet while nitrogen and oxygen do not.
IBM's quantum computers, developed at its research center in Yorktown Heights, New York, use superconducting circuits whose two lowest energy levels, about $5$ GHz apart, form a qubit. The circuits are cooled to $0.01$ K so that $kT$ is far below $hf$ and the qubit starts in its ground state.
Microwave pulses rotate each qubit into superpositions with chosen amplitudes, and a readout at the end measures 0 or 1 with probabilities $|c_0|^2$ and $|c_1|^2$. Because each run gives one outcome, a computation is repeated thousands of times and the answer read from the statistics, exactly the rules of this lesson applied to engineering.
It is tempting to think a system in a superposition of $E_1$ and $E_2$ has an energy partway between them. Measure it and you get $E_1$ or $E_2$, exactly, every time. Only the average of many measurements lies between, and no individual system ever shows that average.
A related error is to think the amplitudes themselves are the probabilities, adding $c_1$ and $c_2$ to one. It is their squares that add to one: $0.6$ and $0.8$ are a normalized pair because $0.36 + 0.64 = 1$.
A particle is in $\tfrac{1}{\sqrt{2}}(\psi_1 + \psi_2)$. Find each probability.
$P_1 = P_2 = \tfrac{1}{2}$
Square the amplitudes.
With $E_1 = 2.0$ eV, find $E_2$ for a box.
$E_2 = 4 \times 2.0 = 8.0\ \text{eV}$
$n^2$ scaling.
Find the average energy.
$\langle E\rangle = \tfrac{1}{2}(2.0 + 8.0) = 5.0\ \text{eV}$
Weighted average.
Find the sloshing period.
$T = \dfrac{4.136}{6.0} = 0.69\ \text{fs}$
$h/\Delta E$.
State the result of one measurement.
$2.0\ \text{or}\ 8.0\ \text{eV, never } 5.0$
Only allowed levels.
A state has $c_1 = 0.5$, $c_2 = 0.5$, and $c_3$ unknown, all real. Find $c_3^2$.
$c_3^2 = 1 - 0.25 - 0.25 = 0.50$
Normalization.
Find the probability of $E_3$.
$P_3 = 0.50$
Half the time.
With box energies $E_n = 0.4n^2$ eV, list them.
$0.4, 1.6, 3.6\ \text{eV}$
$n^2$ scaling.
Find the average energy.
$0.25 \times 0.4 + 0.25 \times 1.6 + 0.50 \times 3.6 = 2.3\ \text{eV}$
Weighted average.
Measure and get $1.6$ eV. Find the new state.
$\psi_2$
Collapse to the state found.
Measure again at once.
$1.6\ \text{eV with certainty}$
Repeatable.
Hydrogen chloride vibrates at $8.66 \times 10^{13}$ Hz. Find the quantum $hf$.
$hf = 4.136 \times 10^{-15} \times 8.66 \times 10^{13} = 0.358\ \text{eV}$
Planck's relation.
Find the zero-point energy.
$E_0 = 0.179\ \text{eV}$
$\tfrac{1}{2}hf$.
Find the first two excited levels.
$E_1 = 0.537\ \text{eV}, \quad E_2 = 0.896\ \text{eV}$
Evenly spaced.
Find the absorbed wavelength.
$\lambda = \dfrac{c}{f} = \dfrac{3.00 \times 10^8}{8.66 \times 10^{13}} = 3.46\ \mu\text{m}$
Infrared.
Compare $hf$ with room-temperature $kT$.
$0.358\ \text{eV} \gg 0.025\ \text{eV}$
Almost all molecules in the ground state.
Estimate the fraction in level 1 at $300$ K.
$e^{-0.358/0.0259} = 10^{-6}$
Boltzmann factor.
Explain why the zero-point energy matters.
$\text{the bond is never at rest}$
It shifts bond strengths measurably.
Find the probabilities.
$P_1 = 0.64, \quad P_2 = 0.36$
Squared amplitudes.
Weight the energies.
$\langle E\rangle = 0.64 \times 3.0 + 0.36 \times 12$
Average.
Evaluate the average.
A particle in a box is in the state $\psi = 0.8\psi_1 + 0.6\psi_2$. Its energy is measured. What is the probability the result is $E_1$?
Complete the worked solution: the vibration of hydrogen chloride is a quantum harmonic oscillator of frequency $8.66 \times 10^{13}$ Hz. With $h = 4.136 \times 10^{-15}$ eV·s and $c = 3.00 \times 10^8$ m/s, find its zero-point energy and first excited energy in eV, and the wavelength of the photon it absorbs in the $0 \to 1$ jump, in μm.
Find the zero-point energy.
$E_0 = \tfrac{1}{2}hf =$ z
Never zero.
Find the first excited energy.
$E_1 = \tfrac{3}{2}hf =$ e
One quantum higher.
Find the absorbed wavelength.
$\lambda = \dfrac{c}{f} =$ l
In μm: infrared.
Note the even spacing.
$E_{n+1} - E_n = hf \text{ for every } n$
The harmonic oscillator's signature.
Match each rule for energy measurements to its statement.
| $|c_n|^2$ | $\sum|c_n|^2 = 1$ | $\sum|c_n|^2E_n$ | in the state found | |
|---|---|---|---|---|
| probability of E_n | ||||
| normalization | ||||
| average energy | ||||
| after a measurement |
A particle in a box, with $E_1 = 2$ eV and $E_2 = 8$ eV, is in the state $0.6\psi_1 + 0.8\psi_2$. Fill in the probability of measuring $E_1$, the probability of measuring $E_2$, and the average energy of many measurements in eV.
| value | |
|---|---|
| probability of E₁ | |
| probability of E₂ | |
| average energy (eV) |
A particle in a box has $E_1 = 5$ eV and $E_2 = 20$ eV. It is prepared in a superposition of just these two levels, with probability $p$ of being found in level 2. Write its average measured energy, in eV, as a formula in $p$.
Answer:
An electron in a box $0.5$ nm wide is placed in an equal superposition of its two lowest states, which differ in energy by $4.512$ eV. Its probability density sloshes from side to side. With $h = 4.136$ eV·fs, what is the period of the sloshing, in fs?
Answer: fs
Climate scientists at NOAA in Boulder, Colorado, measure the infrared absorbed by greenhouse gases. water vapor's bending vibration absorbs at $6.27$ μm. With $hc = 1240$ meV·μm, how large is the vibrational quantum, in meV?
Answer: meV
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle in a box has $E_1 = 8$ eV and $E_2 = 32$ eV. It is prepared in a superposition of just these two levels, with probability $p$ of being found in level 2. Write its average measured energy, in eV, as a formula in $p$.
Answer:
You can reason about superpositions. Explain to someone why an energy measurement never gives the average energy of a superposition.
21. Your turn: a particle is in $0.8\psi_1 + 0.6\psi_2$ with $E_1 = 3.0$ eV and $E_2 = 12$ eV. Find the average energy., step 3
$\langle E\rangle = 6.24\ \text{eV}$
Never itself measured.