5.1.1 Toy Model#

Worked solutions for the homework problems in the 5.1.1 Toy Model lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Accuracy window. Use the approximation \(E_+^{\text{(2nd)}}(\lambda)=1+\lambda^2/2\) as a model prediction for the upper level. Find the largest real \(\lambda>0\) such that the relative error

is below \(5\%\). Interpret the result as a practical criterion for when second-order perturbation theory is trustworthy.

Solution.

The exact upper level is \(E_+^{\text{exact}} = \sqrt{1+\lambda^2}\) and the second-order truncation is \(E_+^{\text{(2nd)}} = 1 + \lambda^2/2\). Since

the truncation overshoots the exact value for every real \(\lambda \ne 0\) (the first omitted term, \(-\lambda^4/8\), is negative). The relative error is therefore

Set \(r(\lambda) = 0.05\) and write \(u = \lambda^2\):

Square both sides and use the identity \((1+u/2)^2 = (1+u) + u^2/4\):

This is the quadratic \(u^2 - 0.41\,u - 0.41 = 0\), with positive root

so

(Direct check: at \(\lambda = 0.937\), \(\sqrt{1+\lambda^2} = 1.3702\) and \(1+\lambda^2/2 = 1.4387\), giving \(r = 0.0500\).)

Interpretation. Second-order perturbation theory holds the upper level to within \(5\%\) all the way up to \(\lambda \approx 0.94\) — coupling of order unity, even though \(\lambda\) was nominally assumed “small.” The window is this generous because the leading error is \(O(\lambda^4)\): the \(\lambda^1\) and \(\lambda^3\) terms vanish in this model, so the first uncorrected term is \(\lambda^4/8\) and the error stays tiny over a long range of \(\lambda\) before turning on sharply. The practical criterion is not “is \(\lambda\) small?” in the abstract, but “is the relative error below tolerance?” — and the only reliable way to know is to compare the truncation against the exact (or a higher-order) value and watch the error grow as \(\lambda\) increases.

2. Inverse design. Suppose spectroscopy reports an upper-level energy \(E_+=1.18\) for the toy model. Estimate \(\lambda\) using second-order perturbation theory, then compute the exact \(\lambda\) from \(E_+=\sqrt{1+\lambda^2}\). Compare the two inferred couplings and discuss why the inverse problem can amplify approximation error.

Solution.

Second-order estimate. Invert \(E_+^{\text{(2nd)}} = 1 + \lambda^2/2 = 1.18\):

Exact value. Invert \(E_+ = \sqrt{1+\lambda^2} = 1.18\):

The two inferred couplings differ by \(\lambda_{\text{exact}} - \lambda_{\text{2nd}} = 0.026\), a relative discrepancy of about \(4.2\%\).

Why the inverse problem amplifies error. Two effects compound.

First, the informative signal lives in \(E_+ - 1\), not in \(E_+\). The unperturbed energy is \(1\); all the coupling information sits in the deviation \(E_+ - 1 = 0.18\). The forward model error of the second-order truncation at \(\lambda \approx 0.6\) is \(\lambda^4/8 \approx 0.016\). That is only \(\sim 1.4\%\) of \(E_+\), but it is \(\sim 9\%\) of the signal \(E_+ - 1 = 0.18\). Measuring the energy relative to an offset turns a small absolute error into a large fractional error of the quantity that actually carries the answer.

Second, the forward map is locally quadratic, so the inverse map is locally a square root. Near the origin \(E_+ - 1 \approx \lambda^2/2\), hence \(\lambda \approx \sqrt{2(E_+-1)}\) and \(\delta\lambda/\lambda = \tfrac12\,\delta(E_+-1)/(E_+-1)\). Equivalently, the forward slope \(\mathrm{d}E_+/\mathrm{d}\lambda = \lambda/\sqrt{1+\lambda^2}\) is less than one (it vanishes as \(\lambda\to0\)), so a given energy error is divided by a small slope when mapped back to \(\lambda\). The \(9\%\) signal error is roughly halved by the square root to the observed \(\sim 4\%\) in \(\lambda\), but it is still several times larger than the \(1.4\%\) error one would naively quote from \(E_+\). The lesson: a model accurate in the forward direction can still be a poor instrument for inversion, especially when the response is weak (small slope) and the signal sits on a large baseline.

3. Basis rotation: energies vs states. Consider

(a) Find a unitary \(\hat U(\varphi)\) that rotates about \(\hat Z\) and satisfies \(\hat U(\varphi)^\dagger\,\hat H(\lambda,\varphi)\,\hat U(\varphi)=\hat H(\lambda,0)\).

(b) Use this unitary equivalence to argue that the eigenvalues \(E_\pm(\lambda,\varphi)\) are independent of \(\varphi\) — the spectrum depends on \(\lambda\) alone.

(c) Show that the eigenstates do depend on \(\varphi\): \(\vert\psi_\pm(\lambda,\varphi)\rangle=\hat U(\varphi)\,\vert\psi_\pm(\lambda,0)\rangle\).

(d) Expand \(E_+(\lambda,\varphi)\) as a Taylor series in \(\varphi\) about \(\varphi=0\). What must every coefficient be? Now expand \(\vert\psi_+(\lambda,\varphi)\rangle\) in \(\varphi\) — is the first-order \(\varphi\)-correction zero? If not, compute it explicitly.

(e) State the general principle illustrated here: when a parameter enters \(\hat H\) only through a unitary similarity transformation, energies are invariant but eigenstates rotate.

Solution.

(a) Take the \(\hat Z\)-rotation

Its conjugation action on the Pauli operators follows from \([\hat Z,\hat X] = 2\mathrm{i}\hat Y\) and \([\hat Z,\hat Y] = -2\mathrm{i}\hat X\). Writing \(f(\varphi) = \mathrm{e}^{\mathrm{i}\varphi\hat Z/2}\hat X\,\mathrm{e}^{-\mathrm{i}\varphi\hat Z/2}\), differentiation gives \(f'(\varphi) = \tfrac{\mathrm{i}}{2}\, \mathrm{e}^{\mathrm{i}\varphi\hat Z/2}[\hat Z,\hat X]\,\mathrm{e}^{-\mathrm{i}\varphi\hat Z/2} = -\,\mathrm{e}^{\mathrm{i}\varphi\hat Z/2}\hat Y\,\mathrm{e}^{-\mathrm{i}\varphi\hat Z/2}\), and likewise for \(\hat Y\); solving the coupled pair with \(f(0)=\hat X\), \(g(0)=\hat Y\) yields

Therefore

So \(\hat U(\varphi) = \mathrm{e}^{-\mathrm{i}\varphi\hat Z/2}\) does the job: geometrically it rotates the in-plane field direction \((\cos\varphi,\sin\varphi)\) back onto the \(x\)-axis.

(b) Rearranging part (a), \(\hat H(\lambda,\varphi) = \hat U(\varphi)\,\hat H(\lambda,0)\,\hat U(\varphi)^\dagger\). This is a unitary similarity transformation, and similar matrices have identical characteristic polynomials, hence identical eigenvalues. Therefore

independent of \(\varphi\) — the spectrum depends on the coupling strength \(\lambda\) alone, not on the in-plane direction \(\varphi\).

(c) Let \(\hat H(\lambda,0)\,\vert\psi_\pm(\lambda,0)\rangle = E_\pm\, \vert\psi_\pm(\lambda,0)\rangle\). Apply \(\hat U(\varphi)\) and insert \(\hat U^\dagger\hat U = \hat 1\):

So \(\hat U(\varphi)\vert\psi_\pm(\lambda,0)\rangle\) is an eigenstate of \(\hat H(\lambda,\varphi)\) with the same eigenvalue, i.e.

(up to an irrelevant overall phase). The eigenstates do depend on \(\varphi\): they are carried around the \(z\)-axis by \(\hat U(\varphi)\).

(d) Since \(E_+(\lambda,\varphi) = \sqrt{1+\lambda^2}\) is exactly constant in \(\varphi\), its Taylor series about \(\varphi=0\) is just that constant: every coefficient of \(\varphi^k\) with \(k\ge 1\) vanishes. A perturbative expansion of the energy in \(\varphi\) is trivial at all orders.

The state is a different story. Expand the rotation:

The first-order \(\varphi\)-correction is

which is not zero (it vanishes only if \(\vert\psi_+(\lambda,0)\rangle\) is a \(\hat Z\)-eigenstate, i.e. only at \(\lambda=0\)). Explicitly, write the unperturbed-direction eigenstate as \(\vert\psi_+(\lambda,0)\rangle = \cos\tfrac\theta2\,\vert 0\rangle + \sin\tfrac\theta2\,\vert 1\rangle\) with \(\tan\theta = \lambda\) (the field \((\lambda,0,1)\) tilts by \(\theta\) from the \(z\)-axis). Then

So a perturbative series in \(\varphi\) has all-zero energy coefficients but a genuine, computable first-order state correction.

(e) Principle. When a parameter enters the Hamiltonian only through a unitary similarity transformation, \(\hat H(\xi) = \hat W(\xi)\,\hat H_0\,\hat W(\xi)^\dagger\), the spectrum is invariant under \(\xi\) (eigenvalues are similarity invariants) while the eigenstates are simply transported by \(\hat W\): \(\vert\psi_n(\xi)\rangle = \hat W(\xi)\vert\psi_n(0)\rangle\). Such a parameter is geometric, not spectral — it rotates the states without deforming the energies. Perturbation theory in that parameter is “trivial” for energies and entirely about the state rotation.

4. Controlled asymmetry. Add a diagonal perturbation and study

Treat \(\lambda\) and \(\mu\) as small independent parameters and expand \(E_+(\lambda,\mu)\) to second order in both. Which terms are linear in \(\mu\)? Which terms are linear in \(\lambda\)? Use symmetry arguments to justify the pattern.

Solution.

Collecting the diagonal terms, \(\hat H(\lambda,\mu) = (1+\mu)\hat Z + \lambda\hat X\), i.e. in the \(\{\vert 0\rangle,\vert 1\rangle\}\) basis

Write the radicand as \(1 + s\) with \(s = 2\mu + \mu^2 + \lambda^2\) and use \(\sqrt{1+s} = 1 + s/2 - s^2/8 + \cdots\). Keeping all terms of total degree \(\le 2\) in \((\lambda,\mu)\):

The two \(\mu^2/2\) pieces cancel, leaving

Which terms are linear in \(\mu\)? The single term \(+\mu\). Which terms are linear in \(\lambda\)? None — there is no \(\lambda^1\) term at all; the leading \(\lambda\)-dependence is the quadratic \(\lambda^2/2\). (The first \(\lambda\)-\(\mu\) cross term is \(-\lambda^2\mu/2\), which appears only at third order.)

Symmetry justification. The two perturbations play structurally different roles:

• \(\mu\hat Z\) is diagonal in the unperturbed basis — it commutes with \(\hat H_0 = \hat Z\). Its first-order correction to the upper level is \(\langle 0\vert\,\mu\hat Z\,\vert 0\rangle = \mu \ne 0\), so \(\mu\) enters linearly: a diagonal perturbation shifts the bare energy directly.

• \(\lambda\hat X\) is purely off-diagonal: \(\langle 0\vert\hat X\vert 0\rangle = 0\), so its first-order correction vanishes and the leading effect is second order, \(\lambda^2/2\).

The absence of every odd power of \(\lambda\) (not just the linear one) follows from an exact symmetry. Conjugating by \(\hat Z\) flips the sign of \(\hat X\) while leaving \(\hat Z\) alone (\(\hat Z\hat X\hat Z = -\hat X\)):

This unitary equivalence forces \(E_\pm(\lambda,\mu) = E_\pm(-\lambda,\mu)\) — \(E_+\) is an even function of \(\lambda\), so all odd powers of \(\lambda\) are forbidden. No analogous symmetry sends \(\mu\to-\mu\), so nothing forbids the linear-in-\(\mu\) term, and indeed it is present. This is the general pattern: a perturbation that commutes with \(\hat H_0\) contributes at first order, while a perturbation that anticommutes with (or is purely off-diagonal in) \(\hat H_0\) is parity-protected and contributes only at even orders.

5. Avoided crossing width. For

show that the minimum gap over all \(\delta\) is \(2|v|\). Then solve the inverse question: if an experiment measures a minimum gap of \(0.12\,\mathrm{eV}\), what is \(|v|\)? Explain why this is a direct operational meaning of level repulsion.

Solution.

The matrix \(\hat H = \delta\,\hat Z + v\,\hat X\) has eigenvalues

so the level gap is

As a function of the diagonal detuning \(\delta\), \(\Delta(\delta)\) is minimized where the radicand is smallest. Since \(\delta^2 \ge 0\) with equality only at \(\delta = 0\),

The bare diagonal energies \(\pm\delta\) cross at \(\delta = 0\), but the true eigenvalues never meet: their closest approach is exactly \(2|v|\).

Inverse question. A measured minimum gap of \(0.12\,\mathrm{eV}\) gives

Operational meaning. \(v\) is the off-diagonal coupling between the two bare states. Without it (\(v=0\)) the levels would simply cross at \(\delta=0\); with it they “repel” and stay split. Crucially, the size of that splitting at the crossing point is not some complicated function of the coupling — it is precisely \(2|v|\). So sweeping \(\delta\) through the avoided crossing and reading off the closest approach is a direct measurement of the coupling strength: \(|v| = \Delta_{\min}/2\). This is the concrete, operational content of “level repulsion” — the minimum gap is (twice) the coupling, and a nonzero \(v\) guarantees the two levels can never be exactly degenerate.

6. Misconception test. One might claim: “If the unperturbed gap is nonzero, non-degenerate perturbation theory should remain valid even when \(\lambda\) is large.” Test this claim quantitatively by comparing exact and second-order energies at \(\lambda=0.2,0.8,1.5\). Identify where the claim fails and explain the failure using error growth as \(\lambda\) increases.

Solution.

For the toy model the unperturbed gap is fixed at \(E_+^{(0)} - E_-^{(0)} = 2\) for all \(\lambda\). Compare the exact upper level \(E_+ = \sqrt{1+\lambda^2}\) with the second-order truncation \(E_+^{\text{(2nd)}} = 1+\lambda^2/2\):

At \(\lambda = 0.2\) the truncation is essentially exact (\(0.02\%\)). By \(\lambda = 0.8\) it is marginal (\(\sim 3\%\)). At \(\lambda = 1.5\) it is wrong by nearly \(18\%\) — and adding more terms does not rescue it.

Where the claim fails and why. The claim fails for any \(\lambda\) of order one or larger, despite the unperturbed gap being nonzero and constant (equal to \(2\)) throughout. A nonzero gap is therefore necessary but not sufficient. The reason is that the perturbative series is the Taylor expansion of \(\sqrt{1+\lambda^2}\), and that function has branch points at \(\lambda = \pm\mathrm{i}\). The radius of convergence is fixed by the nearest singularity:

For \(|\lambda| < 1\) the series converges and successive truncations approach the exact answer; for \(|\lambda| > 1\) the series diverges — the terms eventually grow, and no order of truncation is reliable. That is exactly why \(\lambda = 1.5\) (outside the disk of convergence) is hopeless while \(\lambda = 0.2\) (well inside) is excellent. Even inside the disk the truncation error grows steeply, the leading relative error scaling as \(\sim\lambda^4/8\).

The misconception conflates two distinct requirements. A nonzero gap ensures non-degenerate perturbation theory is well-defined — no energy denominator \(E_n^{(0)} - E_m^{(0)}\) vanishes, so the order-by-order coefficients exist. But for the truncated series to be accurate, \(\lambda\) must additionally be small compared to the scale set by the gap (here \(|\lambda| < R = 1\)). “Defined” and “accurate” are not the same statement; the claim assumes the first implies the second.

7. Near-degeneracy diagnosis. Consider the Hamiltonian (with \(\epsilon>0\))

(a) Compute the exact eigenvalues and expand for small \(\lambda\) at fixed \(\epsilon\).

(b) Determine the condition on \(\lambda, \epsilon\) under which the second-order term is a controlled correction.

(c) Use your condition to explain why the limits \(\epsilon\to 0\) and \(\lambda\to 0\) do not commute in practice for perturbation theory.

Solution.

(a) The matrix \(\hat H = \epsilon\hat Z + \lambda\hat X\) has exact eigenvalues

The unperturbed Hamiltonian (\(\lambda=0\)) is \(\epsilon\hat Z\) with levels \(\pm\epsilon\) and gap \(2\epsilon\). Expanding the upper level for small \(\lambda\) at fixed \(\epsilon\):

The second-order term is \(E_+^{(2)} = \lambda^2/(2\epsilon)\). This agrees with the non-degenerate perturbation-theory formula, \(E_+^{(2)} = \vert\langle -\vert \hat V\vert +\rangle\vert^2 / (E_+^{(0)} - E_-^{(0)})\) with off-diagonal matrix element \(\lambda\) and energy denominator equal to the unperturbed gap \(2\epsilon\).

(b) The second-order term is a controlled (small) correction when it is much smaller than the zeroth-order structure it corrects — here the half-gap \(\epsilon\):

Equivalently, the dimensionless expansion parameter is \(\lambda/\epsilon\) (it appears as \(\sqrt{1 + (\lambda/\epsilon)^2}\)); the series in \(\lambda\) converges for \(|\lambda|/\epsilon < 1\), i.e. \(|\lambda| < \epsilon\), and the correction is a small controlled one when \(|\lambda| \ll \epsilon\) — the perturbation matrix element must be much smaller than the half-gap \(\epsilon\). (Consistently, \(E_+\) has branch points at \(\lambda = \pm\mathrm{i}\,\epsilon\), so the radius of convergence is exactly \(R = \epsilon\).)

(c) The validity condition involves only the ratio \(\lambda/\epsilon\), not either parameter alone — and that is precisely why the two limits do not commute:

• \(\lambda\to 0\) first, at fixed \(\epsilon>0\): the ratio \(\lambda/\epsilon \to 0\), perturbation theory is valid, \(E_+ \to \epsilon\), and the eigenstates approach the unperturbed \(\vert 0\rangle, \vert 1\rangle\). Then taking \(\epsilon\to 0\) smoothly closes the gap with the system always in the perturbative regime.

• \(\epsilon\to 0\) first, at fixed \(\lambda\): the ratio \(\lambda/\epsilon \to \infty\), the energy denominator in \(E_+^{(2)} = \lambda^2/\epsilon\) blows up, and perturbation theory breaks down. At \(\epsilon = 0\) exactly, \(E_\pm = \pm|\lambda|\) and the eigenstates are the fully rotated states \(\vert\pm\rangle\) — nothing like \(\vert 0\rangle, \vert 1\rangle\). Taking \(\lambda\to 0\) afterward cannot undo a rotation that already happened.

Hence

the double limit \((\lambda,\epsilon)\to(0,0)\) depends on the path, i.e. on the value of \(\lambda/\epsilon\) along the way. This is the near-degenerate regime. When \(\lambda \gtrsim \epsilon\), the two levels are too close for \(\lambda\) to count as a small correction, non-degenerate perturbation theory fails, and one must diagonalize the \(2\times 2\) block exactly (degenerate perturbation theory). The non-commuting limits are the sharp mathematical signature of the lesson in Problem 6: “the gap is nonzero” is not enough — what governs perturbation theory is the size of the perturbation relative to the gap.