University Quantum Mechanics II · Time-Independent Perturbation Theory · 6.1
Expanding the eigenvalue problem in powers of λ
Few Hamiltonians left in this course can be solved exactly, but most sit close to one that can. This lesson builds the engine every later approximation runs on: a dial λ from a solved Hamiltonian to the real one, one equation per power of λ, and a projection that solves each equation in turn.
Build the model
Connect the measurement to the mechanism.
Perturbation theory trades one eigenvalue problem you cannot solve for an infinite ladder of problems you can. Write H = H₀ + λH′ with H₀ solved exactly, assume the level you follow moves smoothly as the dial λ turns from 0 to 1, and expand both its energy and its state in powers of λ. The eigenvalue equation holds for every λ, so each power must balance on its own, and every rung of the ladder has the same shape: H₀ − Eₙ⁽⁰⁾ acting on the newest state correction equals a right side built from corrections already found, plus one unknown number, the newest energy.
The operator on the left is singular, because it annihilates |n⁽⁰⁾⟩, and that is the whole trick. Projecting onto ⟨n⁽⁰⁾| wipes out the left side and leaves an equation for the energy alone; projecting onto every other ⟨m⁽⁰⁾| gives the state's components, each divided by an energy gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. The one component the ladder never fixes, along |n⁽⁰⁾⟩, is the eigenvector's free length and phase, and it is spent by keeping every correction orthogonal to |n⁽⁰⁾⟩.
The method stops where its assumptions do: when another level shares Eₙ⁽⁰⁾ and a gap is zero, when the level is not isolated but lies inside a continuum, and when λ = 1 lies outside the series' radius of convergence, which the Hamiltonian sets, not the size of λ.
- Simple definition
- Rayleigh–Schrödinger perturbation theory writes H = H₀ + λH′, expands an eigenvalue and its eigenstate in powers of λ, and solves the eigenvalue equation one power of λ at a time, starting from the known eigenstates of H₀.
- Example
- Levels 0 and Δ = 1.00 eV coupled by V = 0.100 eV: the lower level's series is 0 − 10.0 meV λ² + 0.100 meV λ⁴ − 0.002 meV λ⁶ + …, and at λ = 1 its three terms give −9.902 meV, the exact 2 × 2 root.
A definition, not a result. The power of λ on any term counts the factors of H′ in it, so terms can be sorted by order before a single one is evaluated.
H₀ and H′ in J; λ dimensionless, set to 1 at the end; H₀|m⁽⁰⁾⟩ = Eₘ⁽⁰⁾|m⁽⁰⁾⟩ known exactly for every m
Derived: the eigenvalue equation holds for every λ, so each power balances alone. The right side holds only lower orders plus one new number, Eₙ⁽ᵏ⁾.
Coefficient of λᵏ once Eₙ = Σ λᵏEₙ⁽ᵏ⁾ and |n⟩ = Σ λᵏ|n⁽ᵏ⁾⟩ are inserted; every term is J times a dimensionless ket
Derived: ⟨n⁽⁰⁾|(H₀ − Eₙ⁽⁰⁾) = 0 because H₀ is Hermitian, so the right side has no component along |n⁽⁰⁾⟩. The kth energy needs the state only to order k − 1.
In J; holds when ⟨n⁽⁰⁾|n⁽ʲ⁾⟩ = 0 for all j ≥ 1; k = 1 gives Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩
Derived. It needs Eₘ⁽⁰⁾ ≠ Eₙ⁽⁰⁾ for every m ≠ n. At k = 1 it is ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), and that ratio, not λ, is what must be small.
m ≠ n; numerator and denominator are both in J, so every component is dimensionless
Derived: the order-λ cross terms vanish, so |n⟩ is normalised through first order. Energies never depend on the choice; second-order probabilities need a division by ⟨n|n⟩.
Dimensionless; follows from the choice ⟨n⁽⁰⁾|n⁽ᵏ⁾⟩ = 0 for every k ≥ 1, that is ⟨n⁽⁰⁾|n⟩ = 1
The exact root is quoted from the 2 × 2 problem and its series derived by Taylor expansion. The ladder must return every coefficient; odd orders vanish because V → −V is a relabelling.
H₀ = diag(0, Δ) and H′ = Vσₓ, with Δ and V in J; the series converges only while |λV| stays below Δ/2
A dial from the solved problem to the real one
Split the Hamiltonian as H = H₀ + λH′, where H₀ is a problem Quantum Mechanics I solved completely, with orthonormal eigenstates |m⁽⁰⁾⟩ and energies Eₘ⁽⁰⁾, and H′ is everything left over. λ is a dimensionless dial: λ = 0 is the solved problem and λ = 1 the real one. It belongs to no physical system, and its only job is to count factors of H′, so a term of order λᵏ contains k matrix elements of H′. Often a physical parameter can play its part: for hydrogen in a field F the Stark term eFz brings one power of F per order, and for fine structure in Unit 8 each order brings one more power of α² ≈ 5.3 × 10⁻⁵. The one real assumption is that the level being followed, Eₙ(λ) with its state |n(λ)⟩, is smooth enough in λ to be written as a power series: Eₙ = Eₙ⁽⁰⁾ + λEₙ⁽¹⁾ + λ²Eₙ⁽²⁾ + … and |n⟩ = |n⁽⁰⁾⟩ + λ|n⁽¹⁾⟩ + λ²|n⁽²⁾⟩ + …. Dimensions say what each coefficient is: every Eₙ⁽ᵏ⁾ is an energy in J, and every |n⁽ᵏ⁾⟩ is a ket with dimensionless components in the |m⁽⁰⁾⟩ basis.
Collect the powers of λ, one equation each
Substitute both series into (H₀ + λH′)|n⟩ = Eₙ|n⟩. The equation holds for every λ between 0 and 1, and two power series that agree on an interval agree term by term, so each power of λ must balance by itself. At order λᵏ the left side gives H₀|n⁽ᵏ⁾⟩ + H′|n⁽ᵏ⁻¹⁾⟩; the right side, a product of two series, gives Σⱼ Eₙ⁽ʲ⁾|n⁽ᵏ⁻ʲ⁾⟩ with j running from 0 to k. Move the j = 0 term across and the first three rungs read: order 0, (H₀ − Eₙ⁽⁰⁾)|n⁽⁰⁾⟩ = 0, the solved problem; order 1, (H₀ − Eₙ⁽⁰⁾)|n⁽¹⁾⟩ = (Eₙ⁽¹⁾ − H′)|n⁽⁰⁾⟩; order 2, (H₀ − Eₙ⁽⁰⁾)|n⁽²⁾⟩ = (Eₙ⁽¹⁾ − H′)|n⁽¹⁾⟩ + Eₙ⁽²⁾|n⁽⁰⁾⟩. Every rung has the same operator on the left. On the right sit corrections already found plus exactly one new unknown number, Eₙ⁽ᵏ⁾. The ladder is triangular: it is climbed one rung at a time, and nothing found at order k ever feeds back into a lower order.
The operator that cannot be inverted fixes the energy
H₀ − Eₙ⁽⁰⁾ has no inverse, because it annihilates |n⁽⁰⁾⟩. That looks like an obstacle and is in fact the key. Since H₀ is Hermitian, ⟨n⁽⁰⁾|(H₀ − Eₙ⁽⁰⁾) = 0 too, so projecting any rung onto ⟨n⁽⁰⁾| removes its left side whatever the unknown |n⁽ᵏ⁾⟩ may be. The right side must then have no component along |n⁽⁰⁾⟩, and that one condition is an equation for the one new number: 0 = −⟨n⁽⁰⁾|H′|n⁽ᵏ⁻¹⁾⟩ + Σⱼ₌₁ᵏ Eₙ⁽ʲ⁾⟨n⁽⁰⁾|n⁽ᵏ⁻ʲ⁾⟩. With the choice made below, ⟨n⁽⁰⁾|n⁽ʲ⁾⟩ = 0 for every j ≥ 1, only the j = k term survives and Eₙ⁽ᵏ⁾ = ⟨n⁽⁰⁾|H′|n⁽ᵏ⁻¹⁾⟩. At k = 1 this is ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩, the subject of the next lesson; at k = 2 it is ⟨n⁽⁰⁾|H′|n⁽¹⁾⟩. The kth energy needs the state only to order k − 1, so energies always run at least one order ahead of states. The units check: H′ is in J and the kets are dimensionless.
The other components fix the state, one gap at a time
Project the same rung onto ⟨m⁽⁰⁾| with m ≠ n. Now the left side survives as (Eₘ⁽⁰⁾ − Eₙ⁽⁰⁾)⟨m⁽⁰⁾|n⁽ᵏ⁾⟩, and dividing gives each new component from quantities already known. At first order ⟨m⁽⁰⁾|n⁽¹⁾⟩ = ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾); at order k the numerator also carries −Σⱼ Eₙ⁽ʲ⁾⟨m⁽⁰⁾|n⁽ᵏ⁻ʲ⁾⟩ from the lower rungs. Two things follow. The division needs Eₘ⁽⁰⁾ ≠ Eₙ⁽⁰⁾ for every m ≠ n: one degenerate partner with a non-zero matrix element breaks the formula, and later lessons in this unit rebuild the method for that case. And the real expansion parameter appears: every order brings one more ratio of a matrix element of H′ to an energy gap. For hydrogen's ground state in a laboratory field of 1.0 × 10⁶ V m⁻¹, eFa₀ = 5.3 × 10⁻⁵ eV sets the matrix elements and 10.2 eV separates n = 1 from n = 2, a ratio of about 5 × 10⁻⁶. The same field acting inside n = 2, where the gap is zero, is not small at all.
Spend the leftover freedom on orthogonality
One component of each rung is never fixed: ⟨n⁽⁰⁾|n⁽ᵏ⁾⟩. Adding c|n⁽⁰⁾⟩ to |n⁽ᵏ⁾⟩ changes the left side by c(H₀ − Eₙ⁽⁰⁾)|n⁽⁰⁾⟩ = 0, so the equations cannot see it. This is not physics but the freedom every eigenvector has: |n(λ)⟩ multiplied by any N(λ) = 1 + λc₁ + λ²c₂ + … is still an eigenvector with the same eigenvalue. At first order the real part of c₁ stretches the vector and the imaginary part is a phase, since 1 + iλ Im c₁ ≈ e(iλ Im c₁). The standard choice spends all of it at once: ⟨n⁽⁰⁾|n⁽ᵏ⁾⟩ = 0 for every k ≥ 1, equivalently ⟨n⁽⁰⁾|n⟩ = 1. The order-λ terms of ⟨n|n⟩ are ⟨n⁽⁰⁾|n⁽¹⁾⟩ and its complex conjugate, both now zero, so ⟨n|n⟩ = 1 + λ²⟨n⁽¹⁾|n⁽¹⁾⟩ + …: normalised through first order and no further. Energies never notice, because an eigenvalue does not depend on the length of its eigenvector. Probabilities at second order do, so divide by ⟨n|n⟩ before quoting one. In the figure the first-order state ends on the tangent line ⟨1⁽⁰⁾|n⟩ = 1, and its overshoot past the unit circle is that λ² term.
Test the machine on a problem already solved
Take H₀ = diag(0, Δ) and H′ = V(|1⁽⁰⁾⟩⟨2⁽⁰⁾| + |2⁽⁰⁾⟩⟨1⁽⁰⁾|), with Δ and V in J. The exact lower root, quoted from the 2 × 2 problem of Quantum Mechanics I, is Δ/2 − √(Δ²/4 + λ²V²), with Taylor series −λ²V²/Δ + λ⁴V⁴/Δ³ − 2λ⁶V⁶/Δ⁵ + …. The ladder must reproduce every coefficient. Order 1: E⁽¹⁾ = ⟨1⁽⁰⁾|H′|1⁽⁰⁾⟩ = 0, and ⟨2⁽⁰⁾|1⁽¹⁾⟩ = V/(0 − Δ) = −V/Δ. Order 2: E⁽²⁾ = ⟨1⁽⁰⁾|H′|1⁽¹⁾⟩ = −V²/Δ, as required, in J²/J = J. Every odd order vanishes, and a symmetry says why: replacing |2⁽⁰⁾⟩ by −|2⁽⁰⁾⟩ relabels the basis and flips the sign of V, so the energy is even in λV. As V → 0 the unperturbed level returns. The exact root also shows where the machine stops: its branch points sit at λV = ±iΔ/2, so the series converges only while |λV| stays below Δ/2. At V = 0.70Δ the level at λ = 1 is a perfectly finite −0.360Δ, yet the partial sums through λ⁴, λ⁶ and λ⁸ miss it by 0.110Δ, 0.125Δ and 0.163Δ. The end of this unit takes that up.
Change one variable at a time
Make the relationship visible.
At V/Δ = 0.40 and λ = 1 the dashed arrow overshoots the circle by the λ² term, and the order-2 curve lies below the exact one; step K to 4 to close the gap. Then set V/Δ to 0.70, past the radius 0.50, and step K from 4 to 8: the λ = 1 end drifts away.
EXACT E/Δ-0.1403
SERIES TO ORDER K, E/Δ-0.1600
EXACT ⟨2⁽⁰⁾|n⟩ AT ⟨1⁽⁰⁾|n⟩ = 1-0.351
FIRST-ORDER NORM SQUARED1.1600
Live interpretationEXACT E/Δ: −0.1403. SERIES TO ORDER K, E/Δ: −0.1600. EXACT ⟨2⁽⁰⁾|n⟩ AT ⟨1⁽⁰⁾|n⟩ = 1: −0.351. FIRST-ORDER NORM SQUARED: 1.1600
Catch the common trap
Explain before calculating.
Collecting the λ² terms for a non-degenerate level gives (H₀ − Eₙ⁽⁰⁾)|n⁽²⁾⟩ = (Eₙ⁽¹⁾ − H′)|n⁽¹⁾⟩ + Eₙ⁽²⁾|n⁽⁰⁾⟩, with |n⁽¹⁾⟩ already known and orthogonal to |n⁽⁰⁾⟩. How does this equation give Eₙ⁽²⁾ while |n⁽²⁾⟩ is still unknown?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo unperturbed levels sit at E₁⁽⁰⁾ = 0 and E₂⁽⁰⁾ = 0.80 eV, and H′ has ⟨1⁽⁰⁾|H′|2⁽⁰⁾⟩ = 0.12 eV with no diagonal elements. Following the lower level, find |1⁽¹⁾⟩ from the order-λ rung, the norm of the first-order state at λ = 1, and the probability of finding |1⁽⁰⁾⟩, and compare with the exact value.
- Order-λ rung: (H₀ − 0)|1⁽¹⁾⟩ = (E₁⁽¹⁾ − H′)|1⁽⁰⁾⟩. Projected onto ⟨1⁽⁰⁾| it gives E₁⁽¹⁾ = ⟨1⁽⁰⁾|H′|1⁽⁰⁾⟩ = 0, since H′ has no diagonal elements.
- Projected onto ⟨2⁽⁰⁾|: (0.80 eV)⟨2⁽⁰⁾|1⁽¹⁾⟩ = −0.12 eV, so ⟨2⁽⁰⁾|1⁽¹⁾⟩ = −0.15, a dimensionless number as it must be. The component along |1⁽⁰⁾⟩ is zero by choice, so |1⁽¹⁾⟩ = −0.15|2⁽⁰⁾⟩.
- At λ = 1 the state is |1⁽⁰⁾⟩ − 0.15|2⁽⁰⁾⟩, with ⟨1|1⟩ = 1 + 0.15² = 1.0225. There is no order-λ term in the norm, as the orthogonality choice promised; the whole excess is the λ² term.
- Normalise before reading a probability: P = |⟨1⁽⁰⁾|1⟩|²/⟨1|1⟩ = 1/1.0225 = 0.978, so 2.2% of the probability has moved into |2⁽⁰⁾⟩.
- Exact check: the mixing angle obeys tan 2θ = 2V/Δ = 0.30, so θ = 8.35° and cos²θ = 0.979. The two differ by about 2(V/Δ)⁴ = 0.001, the contribution of the third-order admixture a first-order state leaves out.
Answer|1⁽¹⁾⟩ = −0.15|2⁽⁰⁾⟩; the first-order state has ⟨1|1⟩ = 1.0225 at λ = 1, and P(|1⁽⁰⁾⟩) = 0.978 against the exact 0.979.
MediumUnperturbed levels E₁⁽⁰⁾ = 0 and E₂⁽⁰⁾ = 2.00 eV are perturbed by an H′ with ⟨1⁽⁰⁾|H′|1⁽⁰⁾⟩ = 0.30 eV, ⟨1⁽⁰⁾|H′|2⁽⁰⁾⟩ = 0.40 eV and ⟨2⁽⁰⁾|H′|2⁽⁰⁾⟩ = 0. Climb the ladder for the lower level to E₁⁽³⁾, and compare the partial sums at λ = 1 with the exact eigenvalue.
- Order 1: E⁽¹⁾ = ⟨1⁽⁰⁾|H′|1⁽⁰⁾⟩ = 0.30 eV. The ⟨2⁽⁰⁾| projection gives ⟨2⁽⁰⁾|1⁽¹⁾⟩ = 0.40 eV/(0 − 2.00 eV) = −0.20.
- Order 2: E⁽²⁾ = ⟨1⁽⁰⁾|H′|1⁽¹⁾⟩ = (0.40 eV)(−0.20) = −0.080 eV.
- State at order 2: ⟨2⁽⁰⁾|1⁽²⁾⟩ = [⟨2⁽⁰⁾|H′|1⁽¹⁾⟩ − E⁽¹⁾⟨2⁽⁰⁾|1⁽¹⁾⟩]/(0 − 2.00 eV). Here ⟨2⁽⁰⁾|H′|1⁽¹⁾⟩ = (0)(−0.20) = 0, so the bracket is −(0.30 eV)(−0.20) = +0.060 eV and ⟨2⁽⁰⁾|1⁽²⁾⟩ = −0.030.
- Order 3: E⁽³⁾ = ⟨1⁽⁰⁾|H′|1⁽²⁾⟩ = (0.40 eV)(−0.030) = −0.012 eV. Count factors of H′ as a check: one diagonal and two off-diagonal elements over two gaps, −(0.30)(0.40)²/(2.00)² eV = −0.012 eV, with dimensions J³/J² = J.
- Exact root of the 2 × 2 matrix: 1.15 eV − √((0.85 eV)² + (0.40 eV)²) = 1.15 − 0.9394 = 0.2106 eV. The partial sums 0.300, 0.220 and 0.208 eV miss it by +0.089, +0.0094 and −0.0026 eV; the fourth-order term, +0.0014 eV, roughly halves the last error.
AnswerE⁽¹⁾ = 0.300 eV, E⁽²⁾ = −0.080 eV and E⁽³⁾ = −0.012 eV, summing to 0.208 eV against the exact 0.2106 eV.
HardThe hydrogen ground-state levels |F = 0, m = 0⟩ and |F = 1, m = 0⟩ are split by Δ = hν₀, with ν₀ = 1420.405751768 MHz measured. A field B along z adds H′ = (gₑμB Sz − gₚμN Iz)B/ħ, with gₑ = 2.00232 and gₚ = 5.58569. Treating B as the expansion parameter, find the shift of the transition between the two states in B = 50.0 μT, the size of the next non-zero order, and the field at which the series in B stops converging.
- Matrix elements first. With |1, 0⟩ = (|↑⇓⟩ + |↓⇑⟩)/√2 and |0, 0⟩ = (|↑⇓⟩ − |↓⇑⟩)/√2 from Unit 5, electron arrow first, Sz|1, 0⟩ = (ħ/2)|0, 0⟩ and Iz|1, 0⟩ = −(ħ/2)|0, 0⟩. Both diagonal elements of H′ vanish and the off-diagonal one is V = (gₑμB + gₚμN)B/2: exactly the two-level machine, with B in the role of λ.
- Numbers: gₑμB = 1.85695 × 10⁻²³ J T⁻¹ and gₚμN = 2.8212 × 10⁻²⁶ J T⁻¹, so at 50.0 μT, V = 4.6494 × 10⁻²⁸ J, or V/h = 701.7 kHz. Δ = 9.4117 × 10⁻²⁵ J = 5.874 μeV, and the leading formula (4/3)gₚ(mₑ/mₚ)α⁴mₑc², quoted from Unit 9, gives 5.877 μeV. V/Δ = 4.940 × 10⁻⁴.
- Order B: zero for both states. Order B²: Eₙ⁽²⁾ = ⟨n⁽⁰⁾|H′|n⁽¹⁾⟩ gives −V²/Δ for the lower |0, 0⟩ and +V²/Δ for the upper |1, 0⟩, a repulsion of (V/h)²/ν₀ = 346.6 Hz each. The transition frequency rises by 693.3 Hz, a fractional shift of 4.88 × 10⁻⁷.
- Order B⁴: +V⁴/Δ³ for the lower level and −V⁴/Δ³ for the upper, of size (V/h)⁴/ν₀³ = 346.6 Hz × (V/Δ)² = 8.5 × 10⁻⁵ Hz, four million times smaller. Odd orders vanish because the energy is even in V.
- Convergence: the exact root has branch points at V = ±iΔ/2, so the series in B converges only while B stays below hν₀/(gₑμB + gₚμN) = 9.4117 × 10⁻²⁵ J/(1.85977 × 10⁻²³ J T⁻¹) = 50.6 mT, a field with units J/(J T⁻¹) = T. Above it the exact Breit–Rabi root is still finite, but no truncation of the series approaches it.
AnswerThe transition rises by 693 Hz, quadratic in B (fractional 4.88 × 10⁻⁷); the B⁴ term is only 8.5 × 10⁻⁵ Hz per level; the series in B converges only below 50.6 mT.