Time-independent non-degenerate perturbation theory

By the end of Chapter 6 we owned the complete solution of the hydrogen atom — but it is time to confess something: hydrogen is very nearly the last system that can be solved exactly. The harmonic oscillator, the infinite square well, the hydrogen atom — and the list is essentially over. Take one step beyond it and trouble arrives immediately.

Helium, for instance. It has just one more electron than hydrogen, yet its Hamiltonian becomes

$$\hat H=\underbrace{-\frac{\hbar^2}{2m}\nabla_1^2-\frac{2e^2}{4\pi\varepsilon_0 r_1}}_{\text{electron 1 and nucleus}} \underbrace{-\frac{\hbar^2}{2m}\nabla_2^2-\frac{2e^2}{4\pi\varepsilon_0 r_2}}_{\text{electron 2 and nucleus}} +\underbrace{\frac{e^2}{4\pi\varepsilon_0 r_{12}}}_{\text{electron repulsion}}\tag{7.1.1}$$

The first two blocks are each a hydrogen problem (with nuclear charge $2e$), and those we can solve. All the bad news is in the last term: $r_{12}=|\vec r_1-\vec r_2|$ tangles the two electrons’ coordinates together, and the equation no longer separates. More than a century on, nobody has written down an analytic solution for helium.

If we simply throw the repulsion term away, each electron sits in a $Z=2$ hydrogen-like ground state and the energy comes out to $2\times(-4\times13.6)=-108.8\ \text{eV}$; the experimental value is $-79.0\ \text{eV}$. That is a 30 eV miss — throwing it away is not an option. But notice a ratio: the repulsion term contributes about 34 eV, only a third of the main part’s roughly 109 eV. It cannot be neglected, but it is smaller than the main part.

This is exactly where perturbation theory earns its keep: split the Hamiltonian into “a big piece we can solve” plus “a small tail we cannot”, then let the small tail correct the big piece’s answer one order at a time.

The basic setup

Write the Hamiltonian as

$$\hat H=\hat H_0+\lambda\hat H'\tag{7.1.2}$$

$\hat H_0$ is the part we have already solved: its eigenstates $\ket{n^{(0)}}$ and eigenvalues $E_n^{(0)}$ are all known, and they form a complete orthonormal basis (section 3.5). $\hat H'$ is the “small” term added on top, and $\lambda$ is a bookkeeping parameter introduced by hand: it tags “which order of smallness this term is”, and we set $\lambda=1$ at the end.

“Non-degenerate” means: the level $E_n^{(0)}$ we care about has its state all to itself — no other state shares its energy. The degenerate case needs a different treatment, and that is the subject of the next section.

The strategy is to assume that both the true energy and the true state can be expanded as power series in $\lambda$:

$$E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\cdots,\qquad \ket{n}=\ket{n^{(0)}}+\lambda\ket{n^{(1)}}+\lambda^2\ket{n^{(2)}}+\cdots\tag{7.1.3}$$

Substitute back into the eigenvalue equation and balance order by order — that is the entire machinery of the method.

Balancing order by order: the first- and second-order correctionsbasic~10 minexpand

Step 1: substitute and sort by powers of $\lambda$.

Insert both expansions into $(\hat H_0+\lambda\hat H')\ket{n}=E_n\ket{n}$ and sort the two sides by powers of $\lambda$:

$$\begin{aligned} \lambda^0:&\quad \hat H_0\ket{n^{(0)}}=E_n^{(0)}\ket{n^{(0)}}\\ \lambda^1:&\quad \hat H_0\ket{n^{(1)}}+\hat H'\ket{n^{(0)}} =E_n^{(0)}\ket{n^{(1)}}+E_n^{(1)}\ket{n^{(0)}}\\ \lambda^2:&\quad \hat H_0\ket{n^{(2)}}+\hat H'\ket{n^{(1)}} =E_n^{(0)}\ket{n^{(2)}}+E_n^{(1)}\ket{n^{(1)}}+E_n^{(2)}\ket{n^{(0)}} \end{aligned}$$

Zeroth order is just the original problem — no new information. The information lives in first and second order.

Step 2: hit the first-order equation with $\bra{n^{(0)}}$.

Take the inner product of the $\lambda^1$ line with $\bra{n^{(0)}}$. The first term on the left is $\braket{n^{(0)}}{\hat H_0 n^{(1)}}$; since $\hat H_0$ is Hermitian we can let it act to the left, giving $E_n^{(0)}\braket{n^{(0)}}{n^{(1)}}$ — which cancels the first term on the right exactly. What survives is

$$E_n^{(1)}=\bra{n^{(0)}}\hat H'\ket{n^{(0)}}\equiv H'_{nn}\tag{7.1.4}$$

The first-order energy shift is just the average of the perturbation in the unperturbed state. This is one of the most heavily used formulas in all of quantum mechanics: no new equation to solve — take the old wavefunction, compute one expectation value, done.

Step 3: hit the first-order equation with $\bra{m^{(0)}}$ ($m\neq n$).

Repeat the same move projecting in a different direction. The cancellation no longer happens, and we get

$$\braket{m^{(0)}}{n^{(1)}}=\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}, \qquad H'_{mn}\equiv\bra{m^{(0)}}\hat H'\ket{n^{(0)}}\tag{7.1.5}$$

This gives every component of the first-order correction in the old basis ($\braket{n^{(0)}}{n^{(1)}}$ may be set to zero, which amounts to choosing a normalisation convention):

$$\ket{n^{(1)}}=\sum_{m\neq n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\ket{m^{(0)}}\tag{7.1.6}$$

Read it like this: the perturbation “mixes” other states in, and how much of each depends on two factors — the matrix element $H'_{mn}$ (can the perturbation connect the two states at all?) and the energy difference (the farther away, the less gets mixed in).

Step 4: the second-order energy.

Take the inner product of the $\lambda^2$ line with $\bra{n^{(0)}}$. After the same cancellation what remains is $E_n^{(2)}=\bra{n^{(0)}}\hat H'\ket{n^{(1)}}$. Insert the $\ket{n^{(1)}}$ from above:

$$E_n^{(2)}=\sum_{m\neq n}\frac{|H'_{mn}|^2}{E_n^{(0)}-E_m^{(0)}}\tag{7.1.7}$$

The numerator is a modulus squared, never negative; the sign is decided entirely by the denominator.

A worked example: a ramp inside the well

Let us practise on the system we know best: an infinite square well of width $L$ (section 2.7) holding a charged particle. Now apply a weak uniform electric field along the well — put the well inside a capacitor, say. The potential picks up a linear ramp:

$$\hat H'=V_0\,\frac{x}{L},\qquad 0<x<L\tag{7.1.10}$$

$V_0$ is the drop of the ramp across the well, assumed far smaller than the level spacing.

First- and second-order energy shifts for the rampadvanced~8 minexpand

First order. Average with $\psi_n=\sqrt{2/L}\sin(n\pi x/L)$:

$$E_n^{(1)}=\frac{V_0}{L}\langle x\rangle_n=\frac{V_0}{L}\cdot\frac{L}{2}=\frac{V_0}{2}\tag{7.1.11}$$

($|\psi_n|^2$ is symmetric about the centre of the well, so $\langle x\rangle=L/2$ for every $n$.) Every level rises by the same $V_0/2$ — equivalent to shifting the zero of potential energy up by half the drop, with the level spacings completely untouched. At first order the correction here is a mere “translation”; the real physics sits at second order.

Second order. We need the off-diagonal matrix elements. The $x$ matrix element of the infinite well is a standard integral (integrate by parts twice, or look it up):

$$\bra{m}x\ket{n}=-\frac{8L}{\pi^2}\,\frac{mn}{(m^2-n^2)^2} \quad(m+n\ \text{odd}),\qquad \bra{m}x\ket{n}=0\quad(m+n\ \text{even},\ m\neq n)\tag{7.1.12}$$

Half the matrix elements vanish outright — parity doing us a favour: $x-L/2$ is odd about the centre of the well, so it can only connect states of opposite symmetry. For the ground state, the nearest non-zero term is $m=2$:

$$\bra{2}x\ket{1}=-\frac{8L}{\pi^2}\cdot\frac{2}{9}=-\frac{16L}{9\pi^2}\tag{7.1.13}$$

Insert into the second-order formula (with $E_m^{(0)}=m^2E_1^{(0)}$):

$$E_1^{(2)}=\frac{V_0^2}{L^2}\sum_{m=2,4,6\dots} \frac{|\bra{m}x\ket{1}|^2}{E_1^{(0)}(1-m^2)}\tag{7.1.14}$$

The $m=2$ term gives $\dfrac{V_0^2}{E_1^{(0)}}\cdot\dfrac{(16/9\pi^2)^2}{1-4}\approx-0.01082\,\dfrac{V_0^2}{E_1^{(0)}}$. Adding up all of $m=4,6,\dots$ (a numerical sum):

$$E_1^{(2)}\approx-0.01083\,\frac{V_0^2}{E_1^{(0)}}\tag{7.1.15}$$

The first term alone accounts for 99.9%. The $1-m^2$ in the denominator together with the $(m^2-1)^{-4}$ decay in the numerator makes the series converge ferociously fast — “distant states mix in less”, now in actual numbers.

Plug in numbers: an electron in a well of $L=1.0\ \text{nm}$ ($E_1^{(0)}=0.376\ \text{eV}$, computed back in section 2.7), with a ramp of $V_0=0.1\ \text{eV}$. First order lifts everything by $0.05\ \text{eV}$; second order presses down by $0.01083\times0.01/0.376\approx2.9\times10^{-4}\ \text{eV}$. Second order is two orders of magnitude below first — the series looks perfectly healthy.

When it works, and when it crashes

The condition for trusting the perturbation series can be read straight off the expansion. The mixing coefficient $H'_{mn}/(E_n^{(0)}-E_m^{(0)})$ must be small:

$$\left|\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\right|\ll1 \qquad\text{for all } m\neq n\tag{7.1.16}$$

“Small” is not the perturbation’s own call — it is measured against the level spacing. The very same perturbation is tiny when acting on a system with sparse levels, and a disaster on one with dense levels.

Back to helium: the repulsion term is about $1/3$ of the main part — not exactly small — and first-order perturbation theory gives $E\approx-108.8+34.0=-74.8\ \text{eV}$, about 4 eV away from the experimental $-79.0\ \text{eV}$ — the right direction, but mediocre accuracy. To do better, either grind out higher orders or switch to a method that does not depend on a “small parameter”. The latter is the variational method of section 7.3.

∎Key formulas

• First-order energy shift

  $$E_n^{(1)}=\bra{n^{(0)}}\hat H'\ket{n^{(0)}}$$

  The average of the perturbation in the old state — no new equation to solve

• First-order wavefunction

  $$\ket{n^{(1)}}=\sum_{m\neq n}\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\ket{m^{(0)}}$$

  Mixing = matrix element ÷ energy difference

• Second-order energy shift

  $$E_n^{(2)}=\sum_{m\neq n}\frac{|H'_{mn}|^2}{E_n^{(0)}-E_m^{(0)}}$$

  Level repulsion; the ground-state correction is always ≤ 0

• Validity criterion

  $$\left|\frac{H'_{mn}}{E_n^{(0)}-E_m^{(0)}}\right|\ll1$$

  "Small" is relative to the level spacing; fails at degeneracy

?Self-check4 questions

1. 1.

   The first-order energy correction E⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩ uses which wavefunction?

   • The old, unperturbed wavefunction — so no new equation needs solving
   • The new, corrected wavefunction
   • Half old and half new
   • The eigenfunctions of the perturbation operator itself

   Submit
2. 2.

   Why is the second-order energy correction to the ground state never positive?

   • Because every denominator E₁⁽⁰⁾−Eₘ⁽⁰⁾ in the sum is negative while the numerator |H′ₘ₁|² is non-negative
   • Because perturbations are always attractive
   • Because the ground-state wavefunction has no nodes
   • The claim is false — the second-order correction can have either sign

   Submit
3. 3.

   The correct criterion for whether perturbation theory applies is:

   • The ratio of the matrix element H′ₘₙ to the level spacing Eₙ⁽⁰⁾−Eₘ⁽⁰⁾ is much less than 1
   • H′ is numerically small in absolute terms (say, below 1 eV)
   • H′ commutes with H₀
   • The system has an analytic solution

   Submit
4. 4.

   An electron sits in an infinite well of L = 1.0 nm (E₁⁽⁰⁾ = 0.376 eV), with a linear ramp H′ = V₀x/L of drop V₀ = 0.10 eV added. Find the first-order correction to the ground-state energy, in eV.

   eV1% relative tolerance

   Submit

What comes next

There is a landmine buried in the derivation: both the first-order wavefunction and the second-order energy have $E_n^{(0)}-E_m^{(0)}$ in the denominator. The 2s and the three 2p states of hydrogen at $n=2$ have exactly the same energy — the denominator is strictly zero, and every formula above blows up on the spot. And this is precisely the most common situation: the more symmetric the system, the more degeneracy it has. The next section repairs perturbation theory so that it works inside a degenerate subspace.