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First-order energy shifts as averages of the perturbation, first-order states as admixtures, checks against exact results, and when the method can be trusted.
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 compute first-order energy shifts for small changes to solvable potentials, use symmetry to find which levels move, and judge when perturbation theory applies.
You can solve the infinite square well and the harmonic oscillator exactly, compute expectation values, and expand states in an orthonormal basis. From calculus you know how to approximate a function with a Taylor series in a small parameter. This lesson brings those together: when a Hamiltonian is a solvable one plus something small, expand the answer in powers of the small part.
| Term | What it means |
|---|---|
| Unperturbed Hamiltonian | $\hat{H}^0$, a Hamiltonian whose eigenstates $\psi_n^0$ and energies $E_n^0$ are known exactly. |
| Perturbation | $\hat{H}'$, a small addition to $\hat{H}^0$. |
| First-order energy correction | $E_n^1 = \langle\psi_n^0\vert \hat{H}'\vert \psi_n^0\rangle$, the perturbation averaged over the unperturbed state. |
| Matrix element | $H'_{mn} = \langle\psi_m^0\vert \hat{H}'\vert \psi_n^0\rangle$, which connects two unperturbed states. |
| Second-order correction | $E_n^2 = \sum_{m \ne n}\vert H'_{mn}\vert ^2/(E_n^0 - E_m^0)$. |
| Admixture | The small amount of another unperturbed state mixed into a perturbed state. |
| Delta-function perturbation | $\alpha\,\delta(x - x_0)$, a very narrow bump whose only property is its area $\alpha$. |
Suppose the Hamiltonian is $\hat{H} = \hat{H}^0 + \lambda\hat{H}'$, where the eigenstates of $\hat{H}^0$ are known and $\lambda$ is a bookkeeping parameter that will be set to one. Expand the unknown energies and states in powers of $\lambda$,
$$E_n = E_n^0 + \lambda E_n^1 + \lambda^2E_n^2 + \cdots, \qquad \psi_n = \psi_n^0 + \lambda\psi_n^1 + \cdots,$$
substitute into $\hat{H}\psi_n = E_n\psi_n$, and match powers of $\lambda$. Taking the inner product of the first-order equation with $\psi_n^0$ gives the central result:
$$E_n^1 = \langle\psi_n^0|\hat{H}'|\psi_n^0\rangle.$$
The first-order energy shift is simply the perturbation averaged over the unperturbed state. A state that spends a lot of time where $\hat{H}'$ is large shifts a lot; one with a node there barely moves. The same equations give the change in the state,
$$\psi_n^1 = \sum_{m \ne n}\frac{H'_{mn}}{E_n^0 - E_m^0}\psi_m^0,$$
which mixes in other states in proportion to how strongly the perturbation connects them and inversely to how far apart they are in energy. The method works when every $|H'_{mn}|$ is much smaller than the corresponding gap $|E_n^0 - E_m^0|$; it fails when levels are degenerate, the subject of the next lesson.
Another way: picture
Picture pressing gently on a vibrating guitar string at one point. If you press at a node of the note being played, the note hardly changes; if you press at an antinode, where the string moves most, it changes the most. First-order perturbation theory says the same for quantum states: the shift is the perturbation weighted by where the state lives. The state itself also bends a little, borrowing the shapes of neighboring notes.
Another way: steps
Checks. A constant perturbation $V_0$ must shift every level by $V_0$. A perturbation that is odd about the center of a symmetric potential gives zero first-order shift for every state. When an exact solution is available — a stiffened spring, a uniform shift — the first-order result must match the first term of its Taylor expansion. And if $E_n^1$ comes out comparable to the level spacing, the approximation is suspect.
Write the first-order equation, $\hat{H}^0\psi_n^1 + \hat{H}'\psi_n^0 = E_n^0\psi_n^1 + E_n^1\psi_n^0$, and take its inner product with $\psi_n^0$. Because $\hat{H}^0$ is Hermitian, $\langle\psi_n^0|\hat{H}^0\psi_n^1\rangle = E_n^0\langle\psi_n^0|\psi_n^1\rangle$, which cancels the matching term on the right. What remains is $\langle\psi_n^0|\hat{H}'|\psi_n^0\rangle = E_n^1$.
Physically, to first order the state has not had time to rearrange itself in response to the perturbation, so the energy changes by the perturbation as seen by the old state. The rearrangement, $\psi_n^1$, affects the energy only at second order. The same logic appears in classical physics: the first-order change in a pendulum's frequency from a small extra force is the force averaged over the unperturbed swing.
Many first-order integrals vanish by symmetry, saving work and revealing physics. In a potential symmetric about the origin, each unperturbed state is even or odd, so $|\psi_n^0|^2$ is even. An odd perturbation, such as a uniform electric field $eEx$, then has zero average in every state: its first-order shift vanishes, and the leading effect is second order. This is why the energies of a symmetric molecule or of hydrogen's ground state shift quadratically, not linearly, in a weak electric field.
Nodes work similarly. A delta bump at the center of a box shifts the odd-$n$ states, which peak there, by $2\alpha/a$, and leaves the even-$n$ states, which have a node there, untouched to first order. A physicist who can see where a state lives can often predict which levels move before calculating anything.
A stiffened spring, $k \to (1 + \varepsilon)k$, is still a harmonic oscillator, with frequency $\omega\sqrt{1 + \varepsilon}$, so its energies are exactly $(n + \tfrac{1}{2})\hbar\omega\sqrt{1 + \varepsilon}$. Expanding the square root, $\sqrt{1 + \varepsilon} = 1 + \tfrac{1}{2}\varepsilon - \tfrac{1}{8}\varepsilon^2 + \cdots$, the first-order change is $\tfrac{\varepsilon}{2}(n + \tfrac{1}{2})\hbar\omega$, exactly what perturbation theory gives from $\tfrac{1}{2}\varepsilon k\langle x^2\rangle$.
The second-order term, $-\tfrac{\varepsilon^2}{8}(n + \tfrac{1}{2})\hbar\omega$, is negative, and the next lesson's second-order formula reproduces it. For $\varepsilon = 0.05$, the first-order result is off by only $0.03$ percent; for $\varepsilon = 0.5$, by about $2$ percent. This is the pattern to expect: perturbation theory is excellent when the perturbation is small and degrades smoothly as it grows.
Most of the precise numbers of atomic physics come from perturbation theory. Hydrogen's fine structure — the relativistic correction to the kinetic energy, spin–orbit coupling and the Darwin term — is computed as first-order shifts of the Bohr levels, each of order $\alpha^2$ times the binding energy, where $\alpha = 1/137$. The Zeeman and Stark effects are perturbations by external fields. The finite size of the proton shifts hydrogen's $s$ states by a first-order amount proportional to the proton's radius squared, which is how that radius is measured spectroscopically.
Beyond atoms, perturbation theory computes the band structures of nearly free electrons in metals, the frequency shifts of molecular vibrations from anharmonicity, and, in quantum field theory, the scattering of particles as a series in the coupling constant — the Feynman diagrams of particle physics are a systematic way of writing out its higher orders.
The expansion assumes that the perturbation changes each state only slightly. Two things can break that. If two unperturbed levels are degenerate or nearly so, the denominators $E_n^0 - E_m^0$ vanish or become tiny, and even a small perturbation can mix the states completely. The fix, degenerate perturbation theory, diagonalizes the perturbation within the nearly degenerate set first.
And if the perturbation is not small — a well so deep that its bump rivals the level spacing, or an electric field strong enough to ionize an atom — the series may converge slowly or not at all. Then other methods take over: the variational principle, which bounds the ground energy from above for any strength, and the WKB approximation, which handles potentials that vary slowly. The next three lessons develop those tools.
The lasers in fiber-optic links, Blu-ray players and barcode scanners are built from quantum wells: layers of one semiconductor a few nanometers thick sandwiched between layers of another with a larger band gap. The well's lowest level sets the color of the light. Engineers tune it not only by changing the well's width but by inserting thin barriers or dips of other compositions at chosen positions.
First-order perturbation theory tells them where to put these layers. A thin barrier at the center pushes up the ground state, which peaks there, while leaving the second level, which has a node there, almost unchanged: a barrier $0.28$ nm thick and $1$ eV high in a $10$ nm well raises the ground state by about $2 \times 0.28/10 = 0.056$ eV. Placing the same barrier off center moves both levels. Designers of quantum cascade lasers, which produce mid-infrared light for gas sensing and are now made in the United States for environmental monitoring, use exactly these estimates to lay out dozens of layers before refining them numerically.
A point-like proton would give hydrogen's $s$ states a potential of exactly $-e^2/(4\pi\varepsilon_0r)$ all the way in. A real proton has a charge radius of about $0.84$ fm, and inside it the potential is weaker. Treating the difference as a perturbation, the first-order shift is the difference averaged over the electron's density, which is essentially constant, $|\psi(0)|^2$, over the tiny proton. The result is proportional to $|\psi(0)|^2r_p^2$: only $s$ states, which have density at the origin, are affected.
In muonic hydrogen, where a muon 207 times heavier than an electron orbits much closer to the proton, $|\psi(0)|^2$ is $207^3$ times larger and the shift is correspondingly enormous. Measuring it with lasers in 2010, a team at the Paul Scherrer Institute found a radius four percent smaller than earlier values, the "proton radius puzzle." Later measurements in ordinary hydrogen, including at York University in Toronto, have largely confirmed the smaller value. All of it rests on a first-order energy shift.
It is tempting to think that adding a bump of a given size raises every level by about that much. First-order perturbation theory says otherwise: each level moves by the perturbation averaged over its own probability density. A bump at a node leaves that state untouched; a bump where the state is concentrated moves it the most. Only a perturbation that is constant everywhere shifts all levels by the same amount.
A second misconception is that the perturbed state is the same as the unperturbed one. The energy to first order uses the old state, but the state itself changes, mixing in neighbors weighted by $H'_{mn}/(E_n^0 - E_m^0)$. Those admixtures give the second-order energy and are responsible for effects such as an atom's induced dipole moment in an electric field.
Write the perturbation.
$\hat{H}' = \alpha\,\delta\left(x - \dfrac{a}{2}\right)$
A thin, tall bump at the center of a box of width $a$.
Write the first-order shift as an integral.
$E_n^1 = \displaystyle\int_0^a|\psi_n^0|^2\alpha\,\delta\left(x - \dfrac{a}{2}\right)dx$
The perturbation averaged over the state.
Use the delta function's sifting property.
$E_n^1 = \alpha\left|\psi_n^0\left(\dfrac{a}{2}\right)\right|^2 = \dfrac{2\alpha}{a}\sin^2\dfrac{n\pi}{2}$
The integral picks out the density at the center.
Evaluate for odd and even levels.
$E_n^1 = \dfrac{2\alpha}{a} \ (n \text{ odd}), \qquad 0 \ (n \text{ even})$
Even states have a node at the center.
Evaluate for $\alpha = 20$ meV nm and $a = 2$ nm.
$E_1^1 = E_3^1 = 20\ \text{meV}, \qquad E_2^1 = E_4^1 = 0$
The bump moves only half of the levels.
Write the perturbation for $k \to (1 + \varepsilon)k$.
$\hat{H}' = \tfrac{1}{2}\varepsilon k\hat{x}^2$
The extra spring energy.
Find $\langle x^2\rangle$ in the ground state.
$\langle x^2\rangle_0 = \dfrac{\hbar}{2m\omega}$
The Gaussian's variance.
Compute the first-order shift.
$E_0^1 = \tfrac{1}{2}\varepsilon m\omega^2 \cdot \dfrac{\hbar}{2m\omega} = \dfrac{\varepsilon}{4}\hbar\omega$
With $k = m\omega^2$.
Write the exact energy.
$E_0 = \tfrac{1}{2}\hbar\omega\sqrt{1 + \varepsilon}$
Still an oscillator, with a new frequency.
Expand the exact result.
$E_0 = \tfrac{1}{2}\hbar\omega\left(1 + \tfrac{\varepsilon}{2} - \tfrac{\varepsilon^2}{8} + \cdots\right)$
Taylor series of the square root.
Compare the first-order terms.
$\tfrac{1}{2}\hbar\omega \cdot \tfrac{\varepsilon}{2} = \tfrac{\varepsilon}{4}\hbar\omega$
Perturbation theory reproduces the exact expansion.
Write a uniform field across a box centered on the origin.
$\hat{H}' = eE\hat{x}, \qquad -\dfrac{a}{2} < x < \dfrac{a}{2}$
A tilt of the floor.
Note the symmetry of the states.
$|\psi_n^0(-x)|^2 = |\psi_n^0(x)|^2$
Every state of a symmetric box has a symmetric density.
Write the first-order shift.
$E_n^1 = eE\displaystyle\int_{-a/2}^{a/2}x|\psi_n^0|^2\,dx$
The field times the average position.
Evaluate the integral by symmetry.
$\displaystyle\int_{-a/2}^{a/2}x|\psi_n^0|^2\,dx = 0$
An odd integrand over a symmetric interval.
Conclude the first-order result.
$E_n^1 = 0$
The tilt raises one side as much as it lowers the other.
Identify the leading effect.
$E_n^2 = \displaystyle\sum_{m \ne n}\dfrac{|eE\,x_{mn}|^2}{E_n^0 - E_m^0}$
The shift is quadratic in the field.
Find its sign for the ground state.
$E_1^0 - E_m^0 < 0 \quad\Rightarrow\quad E_1^2 < 0$
The ground state is always lowered at second order.
Interpret the lowering.
$\psi_1 \approx \psi_1^0 + c\,\psi_2^0$
The state shifts toward the low side of the tilt, gaining energy from the field.
Write the first-order shift.
$E_3^1 = \displaystyle\int|\psi_3|^2V_0\,dx$
The average of a constant.
Use the normalization.
$E_3^1 = V_0\displaystyle\int|\psi_3|^2\,dx = V_0$
The total probability is one.
Evaluate the shift.
The floor of an infinite square well is raised by $V_0 = 27$ meV over its left half only. To first order, how much does the energy of level $n = 4$ rise?
Complete the worked solution: an oscillator with $\hbar\omega = 44$ meV has its spring stiffened by $\varepsilon = 0.07$. Find its unperturbed ground energy, the first-order shift, and the corrected energy.
Take half a quantum for the ground state.
$E_0^0 = \tfrac{1}{2}\hbar\omega =$ g meV
The zero-point energy.
Multiply by half the stiffening.
$E_0^1 = \dfrac{\varepsilon}{2}E_0^0 =$ s meV
From $\tfrac{1}{2}\varepsilon k\langle x^2\rangle$.
Add the shift to the unperturbed energy.
$E_0 \approx E_0^0 + E_0^1 =$ t meV
Correct to first order in $\varepsilon$.
For $\hat{H} = \hat{H}^0 + \hat{H}'$ with nondegenerate unperturbed levels, match each result to its formula.
| $\langle\psi_n^0|H'|\psi_n^0\rangle$ | $\sum_{m \ne n}\frac{H'_{mn}}{E_n^0 - E_m^0}\psi_m^0$ | $\sum_{m \ne n}\frac{|H'_{mn}|^2}{E_n^0 - E_m^0}$ | $|H'_{mn}| \ll |E_n^0 - E_m^0|$ | |
|---|---|---|---|---|
| the first-order energy | ||||
| the first-order state | ||||
| the second-order energy | ||||
| the condition for validity |
A narrow bump $\hat{H}' = \alpha\,\delta(x - a/2)$ with $\alpha = 28$ meV nm sits at the center of an infinite well of width $a = 2$ nm. Fill in the first-order shift of each level, in meV.
| shift (meV) | |
|---|---|
| $n = 1$ | |
| $n = 2$ | |
| $n = 3$ | |
| $n = 4$ |
An infinite well has ground energy $E_1 = 14$ meV. Its floor is raised by $V_0 = 14$ meV over the right half. What is the energy of level $n = 4$, to first order, in meV?
Answer: meV
An oscillator with $\hbar\omega = 46$ meV has its spring constant increased by $3$ percent, $k \to (1 + \varepsilon)k$ with $\varepsilon = 0.03$. Using first-order perturbation theory, by how much does level $n = 2$ shift, in meV?
Answer: meV
Engineers tune the color of a semiconductor laser by growing a thin barrier, $0.56$ nm thick and $0.5$ eV high, at the center of a quantum well $14$ nm wide. Treating the barrier as a delta function and the well as infinitely deep, by how much does the ground level rise, in meV?
Answer: meV
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A narrow bump $\hat{H}' = \alpha\,\delta(x - a/2)$ with $\alpha = 2$ meV nm sits at the center of an infinite well of width $a = 2$ nm. Fill in the first-order shift of each level, in meV.
| shift (meV) | |
|---|---|
| $n = 1$ | |
| $n = 2$ | |
| $n = 3$ | |
| $n = 4$ |
You can estimate how a small change to a potential shifts its energy levels. Explain to someone why a bump at the center of a box moves only half the levels.
17. Your turn: a box's whole floor is raised by $V_0 = 12$ meV. What is the first-order shift of level $n = 3$?, step 3
$E_3^1 = 12\ \text{meV}$
Exact, not just first order: a constant shifts everything.