Back to the on-screen lesson ·
Kets, bras, brackets and projectors, and two coupled states whose energies split by $2A$ and whose populations oscillate as $\sin^2(At/\hbar)$.
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 work in Dirac notation, diagonalize a two-state Hamiltonian, and predict how the population of coupled states oscillates in time.
You can represent states as columns and observables as Hermitian matrices, compute probabilities as squared components, and evolve a state by attaching $e^{-iE_nt/\hbar}$ to each energy eigenstate. This lesson adds Dirac's notation, which writes all of that without choosing a basis, and applies it to the simplest system with interesting dynamics: two states and a coupling.
| Term | What it means |
|---|---|
| Ket | $\vert \alpha\rangle$, a state vector, written as a column in a basis. |
| Bra | $\langle\alpha\vert $, the conjugate transpose of the ket, a row of conjugated components. |
| Bracket | $\langle\alpha\vert \beta\rangle$, the inner product, a complex number. |
| Projection operator | $\hat{P}_n = \vert n\rangle\langle n\vert $, which picks out the component along $\vert n\rangle$. |
| Completeness relation | $\sum_n\vert n\rangle\langle n\vert = \hat{1}$ for any orthonormal basis. |
| Coupling | An off-diagonal Hamiltonian element that connects two basis states. |
| Rabi oscillation | The periodic transfer of population between two coupled states. |
Dirac wrote a state as a ket $|\alpha\rangle$ and its conjugate as a bra $\langle\alpha|$. Put together they make a bracket, the inner product $\langle\alpha|\beta\rangle$, with $\langle\beta|\alpha\rangle = \langle\alpha|\beta\rangle^$. Put the other way round they make an operator: $|\alpha\rangle\langle\beta|$ acting on $|\gamma\rangle$ gives $|\alpha\rangle\langle\beta|\gamma\rangle$. The projector* $|n\rangle\langle n|$ keeps the component along $|n\rangle$, and for an orthonormal basis
$$\sum_n|n\rangle\langle n| = \hat{1}.$$
Inserting this identity anywhere expands in the basis: $|\Psi\rangle = \sum|n\rangle\langle n|\Psi\rangle$, and matrix elements are $Q_{mn} = \langle m|\hat{Q}|n\rangle$.
Now take two states $|1\rangle$ and $|2\rangle$ with the same energy $E_0$, coupled by a matrix element $-A$:
$$\hat{H} = \begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix}.$$
The basis states are not stationary. The energy eigenstates are $|\pm\rangle = (|1\rangle \pm |2\rangle)/\sqrt{2}$ with energies $E_0 \mp A$, split by $2A$. A system started in $|1\rangle$ is an equal mixture of them, their relative phase grows as $2At/\hbar$, and the probability of finding it in $|2\rangle$ oscillates:
$$P_2(t) = \sin^2\frac{At}{\hbar}.$$
Another way: picture
Picture two identical pendulums joined by a weak spring. Start one swinging and the other still: the motion gradually passes to the second pendulum, then back, again and again. The normal modes — swinging together and swinging opposite — have slightly different frequencies, and the transfer happens at their difference, the beat frequency. Two coupled quantum states behave the same way: the symmetric and antisymmetric combinations are the normal modes, and the population beats between the two basis states at the rate $2A/\hbar$.
Another way: steps
Checks. $\langle\alpha|\alpha\rangle$ must be positive; $|\langle\alpha|\beta\rangle|^2 \le \langle\alpha|\alpha\rangle\langle\beta|\beta\rangle$. Probabilities at every time must add to one, $P_1 + P_2 = \cos^2 + \sin^2 = 1$. And the oscillation frequency must equal the energy splitting over $h$.
Dirac's notation separates the physics from the choice of basis. The statement $\langle x|\Psi\rangle = \Psi(x)$ says that the wave function is the state's component along the position eigenstate $|x\rangle$; $\langle p|\Psi\rangle$ is the momentum-space wave function; and both describe the same ket. The completeness relation in position, $\int|x\rangle\langle x|\,dx = \hat{1}$, turns every bracket into the familiar integral: $\langle\alpha|\beta\rangle = \int\langle\alpha|x\rangle\langle x|\beta\rangle\,dx = \int\alpha^*\beta\,dx$.
Most manipulations then become the insertion of a well-chosen identity. To find $\langle\alpha|\hat{Q}|\beta\rangle$ in the eigenbasis of $\hat{Q}$, insert $\sum|n\rangle\langle n|$ twice and get $\sum q_n\langle\alpha|n\rangle\langle n|\beta\rangle$. The notation is so efficient that it is used everywhere in quantum physics and quantum computing, where circuits are written as products of operators acting on kets like $|0\rangle$ and $|1\rangle$.
Without coupling, $|1\rangle$ and $|2\rangle$ have the same energy and any combination is stationary. The coupling picks out two special combinations. In the symmetric one, the two amplitudes reinforce through the coupling, lowering the energy by $A$; in the antisymmetric one, they oppose, raising it by $A$. This is the origin of chemical bonding in its simplest form: an electron shared between two protons in the hydrogen molecular ion has a symmetric, bonding state below the separate-atom energy and an antisymmetric, antibonding state above it.
The sign of the coupling determines which combination is lower but not the splitting, which is $2|A|$. The dynamics follow from the splitting alone: the system oscillates between the basis states at angular frequency $2A/\hbar$ in probability, completing a transfer in time $\pi\hbar/2A$. The stronger the coupling, the faster the exchange.
If the two basis states have different energies, $E_0 + \varepsilon$ and $E_0 - \varepsilon$, the Hamiltonian becomes $\begin{pmatrix} E_0 + \varepsilon & -A \\ -A & E_0 - \varepsilon \end{pmatrix}$, with eigenvalues $E_0 \pm \sqrt{\varepsilon^2 + A^2}$. Starting in $|1\rangle$, the probability of $|2\rangle$ reaches at most $A^2/(A^2 + \varepsilon^2)$. A large mismatch, $\varepsilon \gg A$, nearly freezes the system in its initial state: the eigenstates are then almost the basis states themselves.
This is the principle of resonance. Driving an atom with light near its transition frequency is, in a rotating frame, a two-state problem whose detuning is the frequency mismatch and whose coupling is set by the light's intensity. On resonance the population swings fully between ground and excited states — Rabi oscillations, named for Isidor Rabi, who used them in 1938 to measure nuclear magnetic moments. Off resonance, the swings shrink. Every magnetic resonance scanner and every qubit gate relies on this formula.
Ammonia, NH₃, is a pyramid with the nitrogen above or below the plane of the three hydrogens. The two positions, $|1\rangle$ and $|2\rangle$, have equal energy, and the nitrogen can tunnel through the plane between them, providing a coupling. The energy eigenstates are the symmetric and antisymmetric combinations, split by $9.87 \times 10^{-5}$ eV, which corresponds to a frequency of $23.87$ GHz.
In 1954 Charles Townes and his students used this pair of levels to build the first maser. An electric field gradient separated molecules in the upper state from those in the lower, and the upper-state molecules, sent into a cavity tuned to $23.87$ GHz, amplified the microwave field by stimulated emission. The laser followed six years later. The same splitting makes ammonia a thermometer for interstellar clouds, where radio astronomers measure the populations of its levels.
Many systems have far more than two states, yet the two-state model describes them well whenever two levels are close together in energy and far from all the others. An atom driven by a laser tuned to one transition behaves as a two-level atom, because the other transitions are badly out of resonance. A spin in a magnetic field has exactly two states. The lowest two levels of a superconducting circuit, whose third level is detuned by a few hundred megahertz, serve as a qubit. Even the benzene molecule's two Kekulé structures, with the double bonds in alternating positions, act as two coupled states whose symmetric combination is the molecule's real, extra-stable ground state.
In each case the same questions arise and the same answers follow. What are the diagonal energies and the coupling? The eigenstates and their splitting come from a two-by-two matrix. How does a prepared state evolve? It oscillates at the splitting divided by Planck's constant, fully if the diagonal energies match and partly if they do not. Learning this one model thoroughly pays off across atomic physics, chemistry, magnetic resonance and quantum technology.
Neutrinos come in three flavors — electron, muon and tau — but the states with definite mass are different superpositions of them. A muon neutrino produced in a particle accelerator is therefore a superposition of mass states, each of whose phases advance at slightly different rates. Treated as a two-state system, the probability that it arrives as a different flavor after traveling a distance $L$ is $\sin^2 2\theta\,\sin^2(1.267\,\Delta m^2L/E)$, with $\Delta m^2$ in eV², $L$ in km and $E$ in GeV.
Experiments are designed around this formula. Japan's T2K sends $0.6$ GeV neutrinos $295$ km from Tokai to the Super-Kamiokande detector, a ratio chosen to put the phase near $\pi/2$ for $\Delta m^2 = 2.5 \times 10^{-3}$ eV², where nearly all the muon neutrinos have oscillated. The observation of oscillations, by Super-Kamiokande and by the Sudbury Neutrino Observatory, proved that neutrinos have mass and earned Takaaki Kajita and Arthur McDonald the 2015 Nobel Prize. The U.S. Deep Underground Neutrino Experiment will send a beam $1300$ km from Fermilab to South Dakota to measure the remaining parameters.
Every quantum computer drives its qubits with pulses that couple $|0\rangle$ and $|1\rangle$. In the qubit's rotating frame the pulse acts as a coupling $A$, exactly the two-state problem of this lesson, and the population oscillates as $\sin^2(At/\hbar)$. A pulse that lasts until $At/\hbar = \pi/2$ — a pi pulse, named for the full rotation angle $2At/\hbar = \pi$ — flips $|0\rangle$ to $|1\rangle$: the quantum NOT gate. Half as long gives an equal superposition, the starting point of most algorithms.
Engineers calibrate each qubit by measuring these Rabi oscillations: they vary the pulse length, record the probability of $|1\rangle$, and fit a sine squared. Superconducting qubits reach a pi pulse in about $20$ to $40$ ns; trapped ions, driven by lasers, take a few microseconds. Detuning matters too: a pulse a little off the qubit's frequency transfers less than the full population, and gate errors of a tenth of a percent require frequencies tuned to within kilohertz.
It is natural to assume that a system prepared in $|1\rangle$ stays there unless something disturbs it. That is true only if $|1\rangle$ is an eigenstate of the Hamiltonian. When $\hat{H}$ has an off-diagonal element connecting $|1\rangle$ to $|2\rangle$, the stationary states are combinations of the two, and $|1\rangle$ is a superposition of them whose relative phase keeps changing: the system oscillates. Nothing external is needed; the coupling is part of the system.
A second error is to forget to conjugate when forming a bra. The bra of $a|1\rangle + b|2\rangle$ is $a^\langle 1| + b^\langle 2|$. Without the conjugate, $\langle\alpha|\alpha\rangle$ can come out negative or complex, which no squared length can be.
Let $|\alpha\rangle = \tfrac{1}{\sqrt{2}}(|1\rangle + i|2\rangle)$. Write its bra.
$\langle\alpha| = \tfrac{1}{\sqrt{2}}(\langle 1| - i\langle 2|)$
Conjugate each coefficient.
Check its normalization.
$\langle\alpha|\alpha\rangle = \tfrac{1}{2}(1 + (-i)(i)) = \tfrac{1}{2}(1 + 1) = 1$
Cross terms vanish by orthonormality.
Find its overlap with $|2\rangle$.
$\langle 2|\alpha\rangle = \tfrac{i}{\sqrt{2}}$
The component along the second basis state.
Build the projector onto $|\alpha\rangle$ as a matrix.
$|\alpha\rangle\langle\alpha| = \tfrac{1}{2}\begin{pmatrix} 1 \\ i \end{pmatrix}\begin{pmatrix} 1 & -i \end{pmatrix} = \tfrac{1}{2}\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$
Column times row.
Check that projecting twice changes nothing.
$\hat{P}^2 = |\alpha\rangle\langle\alpha|\alpha\rangle\langle\alpha| = |\alpha\rangle\langle\alpha| = \hat{P}$
Because $\langle\alpha|\alpha\rangle = 1$.
Write the Hamiltonian for two degenerate coupled states.
$\hat{H} = E_0(|1\rangle\langle 1| + |2\rangle\langle 2|) - A(|1\rangle\langle 2| + |2\rangle\langle 1|)$
The same matrix in Dirac notation.
Act on the symmetric combination.
$\hat{H}(|1\rangle + |2\rangle) = E_0(|1\rangle + |2\rangle) - A(|2\rangle + |1\rangle)$
The coupling swaps the two states.
Read off its energy.
$E_+ = E_0 - A$
The symmetric state is lowered.
Act on the antisymmetric combination.
$\hat{H}(|1\rangle - |2\rangle) = E_0(|1\rangle - |2\rangle) - A(|2\rangle - |1\rangle) = (E_0 + A)(|1\rangle - |2\rangle)$
The swap flips its sign.
Find the splitting.
$E_- - E_+ = 2A$
Twice the coupling.
Evaluate for ammonia.
$2A = 9.87 \times 10^{-5}\ \text{eV}, \qquad f = \dfrac{2A}{h} = \dfrac{9.87 \times 10^{-5}}{4.1357 \times 10^{-15}} = 23.9\ \text{GHz}$
The frequency of the ammonia maser.
Start in $|1\rangle$ and expand in the eigenstates.
$|1\rangle = \tfrac{1}{\sqrt{2}}(|+\rangle + |-\rangle)$
Insert the identity $|+\rangle\langle +| + |-\rangle\langle -|$.
Evolve each eigenstate.
$|\Psi(t)\rangle = \tfrac{1}{\sqrt{2}}\left(e^{-i(E_0 - A)t/\hbar}|+\rangle + e^{-i(E_0 + A)t/\hbar}|-\rangle\right)$
Stationary states only acquire phases.
Factor out the common phase.
$|\Psi(t)\rangle = \tfrac{1}{\sqrt{2}}e^{-iE_0t/\hbar}\left(e^{iAt/\hbar}|+\rangle + e^{-iAt/\hbar}|-\rangle\right)$
An overall phase has no effect on probabilities.
Project onto the first basis state.
$\langle 1|\Psi(t)\rangle = \tfrac{1}{2}e^{-iE_0t/\hbar}\left(e^{iAt/\hbar} + e^{-iAt/\hbar}\right) = e^{-iE_0t/\hbar}\cos\dfrac{At}{\hbar}$
$\langle 1|\pm\rangle = 1/\sqrt{2}$.
Project onto the second basis state.
$\langle 2|\Psi(t)\rangle = ie^{-iE_0t/\hbar}\sin\dfrac{At}{\hbar}$
$\langle 2|\pm\rangle = \pm 1/\sqrt{2}$.
Square both amplitudes.
$P_1 = \cos^2\dfrac{At}{\hbar}, \qquad P_2 = \sin^2\dfrac{At}{\hbar}$
They add to one at every moment.
Find the time for complete transfer.
$\dfrac{At}{\hbar} = \dfrac{\pi}{2} \quad\Rightarrow\quad \tau = \dfrac{\pi\hbar}{2A}$
Half the period of the beat.
Evaluate for ammonia.
$\tau = \dfrac{\pi \times 6.582 \times 10^{-16}\ \text{eV s}}{9.87 \times 10^{-5}\ \text{eV}} = 2.1 \times 10^{-11}\ \text{s}$
The nitrogen flips through the plane in about $21$ ps.
Write the transfer time.
$\tau = \dfrac{\pi\hbar}{2A}$
When $At/\hbar = \pi/2$.
Substitute the values.
$\tau = \dfrac{\pi \times 0.6582}{2 \times 2}$
Microelectronvolts cancel, leaving nanoseconds.
Evaluate the time.
A ket is $|\alpha\rangle = 1|1\rangle + 7i|2\rangle$. What is the corresponding bra $\langle\alpha|$?
Complete the worked solution: two states of energy $E_0 = 40$ meV are coupled by $-A$ with $A = 5$ meV. Find the two energy eigenvalues and the splitting.
Evaluate the symmetric state's energy.
$E_+ = E_0 - A =$ l meV
The coupling lowers the symmetric combination.
Evaluate the antisymmetric state's energy.
$E_- = E_0 + A =$ h meV
It raises the antisymmetric one by the same amount.
Subtract to find the splitting.
$E_- - E_+ =$ s meV
Twice the coupling.
Match each Dirac expression to the kind of object it is.
| a state vector (column) | a conjugated row vector | a complex number | the identity operator | |
|---|---|---|---|---|
| $|\alpha\rangle$ | ||||
| $\langle\alpha|$ | ||||
| $\langle\alpha|\beta\rangle$ | ||||
| $\sum_n|n\rangle\langle n|$ |
Two degenerate states are coupled with strength $A = 7$ μeV, and the system starts in $|1\rangle$. With $\tau = \pi\hbar/2A$, the time for complete transfer, fill in the probability of $|2\rangle$ at each time.
| $P_2$ | |
|---|---|
| $t = \tau/3$ | |
| $t = \tau/2$ | |
| $t = 2\tau/3$ | |
| $t = \tau$ |
Let $|\alpha\rangle = 2|1\rangle + i|2\rangle$ and $|\beta\rangle = 4|1\rangle + 2|2\rangle$. What is $|\langle\alpha|\beta\rangle|^2$?
Answer:
Two degenerate states $|1\rangle$ and $|2\rangle$ of energy $E_0$ are coupled, so that $\hat{H} = \begin{pmatrix} E_0 & -A \\ -A & E_0 \end{pmatrix}$ with $A = 1$ μeV. The system starts in $|1\rangle$. What is the probability of finding it in $|2\rangle$ after $1.2$ ns? Use $\hbar = 0.6582$ μeV ns.
Answer:
The $T2K$ experiment sends muon neutrinos of about $0.6$ GeV to a detector $295$ km away. Treating the flavors as a two-state system with maximal mixing and $\Delta m^2 = 2.5 \times 10^{-3}$ eV², what fraction have changed flavor on arrival?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two degenerate states are coupled with strength $A = 3$ μeV, and the system starts in $|1\rangle$. With $\tau = \pi\hbar/2A$, the time for complete transfer, fill in the probability of $|2\rangle$ at each time.
| $P_2$ | |
|---|---|
| $t = \tau/3$ | |
| $t = \tau/2$ | |
| $t = 2\tau/3$ | |
| $t = \tau$ |
You can solve any two-state system. Explain to someone why an ammonia molecule prepared with its nitrogen above the plane does not stay that way.
17. Your turn: a coupled pair has $A = 2$ μeV. How long does complete transfer take? Use $\hbar = 0.6582$ μeV ns., step 3
$\tau = 0.517\ \text{ns}$
About half a nanosecond.