5.1.1 Toy Model#

Prompts

Why do we need perturbation theory? What strategy does it use to approximate unsolvable quantum systems?
For the qubit \(H(\lambda) = \hat{\sigma}^z + \lambda\hat{\sigma}^x\), find the exact eigenvalues and Taylor-expand them. What do the expansion coefficients represent?
What is level repulsion? Why do the two eigenvalues of the qubit model never cross for real \(\lambda\)?
When does the perturbative series converge, and what sets the convergence radius? What happens beyond it?

Lecture Notes#

Overview#

Most quantum systems cannot be solved exactly. Perturbation theory provides a systematic approximation: start from a solvable Hamiltonian \(H_0\), add a small perturbation \(\lambda V\), and expand eigenvalues and eigenstates as power series in \(\lambda\). The qubit model \(H(\lambda) = \hat{\sigma}^z + \lambda\hat{\sigma}^x\) is the ideal testing ground — a \(2\times 2\) system with an exact solution that we can Taylor-expand and compare order-by-order.

The Qubit Hamiltonian#

(77)#\[\begin{split}H(\lambda) = \hat{\sigma}^z + \lambda\hat{\sigma}^x = \begin{pmatrix} 1 & \lambda \\ \lambda & -1 \end{pmatrix}\end{split}\]

Here \(H_0 = \hat{\sigma}^z\) is the unperturbed Hamiltonian with eigenstates \(|0\rangle\) (energy \(+1\)) and \(|1\rangle\) (energy \(-1\)), and \(V = \hat{\sigma}^x\) induces spin flips between them.

Exact Solution#

Diagonalizing:

\[\det(H - E\hat{I}) = (1-E)(-1-E) - \lambda^2 = 0 \quad\Rightarrow\quad E^2 = 1 + \lambda^2\]

Exact Eigenvalues

(78)#\[E_\pm(\lambda) = \pm\sqrt{1 + \lambda^2}\]

The energy gap \(\Delta E = 2\sqrt{1+\lambda^2}\) never closes for real \(\lambda\) — the two levels repel rather than cross.

Taylor Expansion: Recovering the Perturbative Series#

Expanding \(\sqrt{1+\lambda^2}\) for small \(\lambda\):

(79)#\[E_\pm(\lambda) = \pm\left(1 + \frac{\lambda^2}{2} - \frac{\lambda^4}{8} + O(\lambda^6)\right)\]

Reading off the perturbative coefficients for the upper state:

\[E_+^{(0)} = 1, \quad E_+^{(1)} = 0, \quad E_+^{(2)} = \frac{1}{2}, \quad E_+^{(3)} = 0, \quad E_+^{(4)} = -\frac{1}{8}\]

All odd-order corrections vanish because \(H(\lambda)\) has a \(\lambda \to -\lambda\) symmetry (equivalently, \(\hat{\sigma}^x\) has zero diagonal elements in the \(|0\rangle, |1\rangle\) basis, so \(E_n^{(1)} = \langle n|\hat{\sigma}^x|n\rangle = 0\)).

The exact eigenstates expanded to low order:

(80)#\[|\psi_+(\lambda)\rangle = |0\rangle + \frac{\lambda}{2}|1\rangle - \frac{\lambda^2}{8}|0\rangle + O(\lambda^3)\]

The admixture of \(|1\rangle\) grows linearly with \(\lambda\), controlled by the ratio coupling/gap.

Level Repulsion#

Level Repulsion

Two energy levels coupled by a perturbation \(V\) repel each other rather than crossing. The minimum gap at any \(\lambda\) is proportional to the matrix element \(|\langle 1|V|0\rangle|\). In the qubit model, \(\langle 1|\hat{\sigma}^x|0\rangle = 1\), so the levels never touch.

This is a general phenomenon: for a \(2\times 2\) block with off-diagonal coupling \(v\), the eigenvalues are \(E_\pm = \bar{E} \pm \sqrt{\delta^2 + v^2}\), where \(\delta\) is the half-gap and \(\bar{E}\) the midpoint. The minimum splitting \(2|v|\) occurs at \(\delta = 0\).

Convergence#

Convergence Criterion

The perturbative series converges when the perturbation is small compared to the energy gap:

\[\lambda \|V\| \ll \Delta E\]

For the qubit, the convergence radius is \(R = 1\), set by the branch points at \(\lambda = \pm\mathrm{i}\) in the complex plane where \(E^2 = 1 + \lambda^2 = 0\).

Beyond \(|\lambda| = R\), the series diverges even though the exact solution remains valid. Perturbation theory breaks down precisely when the coupling becomes comparable to the energy gap — the perturbation can no longer be treated as “small.”

Discussion

The first-order corrections \(E_\pm^{(1)} = 0\) vanish because \(\hat{\sigma}^x\) has zero diagonal elements. This is not a coincidence but a consequence of parity symmetry. More generally, when does symmetry guarantee that certain perturbative corrections vanish? Can you think of physical examples where recognizing this saves enormous computational effort?

Summary#

Perturbation theory expands eigenvalues as \(E_n(\lambda) = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots\).
The qubit model provides an exact benchmark: \(E_\pm = \pm\sqrt{1+\lambda^2}\) reproduces all perturbative coefficients when Taylor-expanded.
Level repulsion: coupled levels repel rather than cross; the gap is proportional to \(|\langle 1|V|0\rangle|\).
Convergence: the series converges for \(\lambda\|V\| \ll \Delta E\); breakdown signals that the perturbation is no longer “small.”

Homework#

1. For the qubit \(H(\lambda) = \hat{\sigma}^z + \lambda\hat{\sigma}^x\), compute the exact eigenvalues \(E_\pm(\lambda) = \pm\sqrt{1+\lambda^2}\) by diagonalizing the \(2\times 2\) matrix. Taylor-expand to \(O(\lambda^4)\) and identify the perturbative coefficients \(E_+^{(0)}\) through \(E_+^{(4)}\).

2. Verify that \(E_+^{(1)} = \langle 0|\hat{\sigma}^x|0\rangle = 0\). Explain why this vanishing is guaranteed by the \(\lambda \to -\lambda\) symmetry of \(H(\lambda)\).

3. Compute \(E_+^{(2)}\) using the formula \(E_n^{(2)} = \sum_{m\neq n} |\langle m|V|n\rangle|^2/(E_n^{(0)} - E_m^{(0)})\). Verify it matches the coefficient from the Taylor expansion of \(\sqrt{1+\lambda^2}\).

4. Write the first-order state correction \(|\psi_+^{(1)}\rangle = \frac{\langle 1|\hat{\sigma}^x|0\rangle}{E_+^{(0)} - E_-^{(0)}}|1\rangle\). Compare with the exact eigenstate expanded to \(O(\lambda)\).

5. The convergence radius is \(R = 1\), set by branch points at \(\lambda = \pm\mathrm{i}\). For \(\lambda = 0.5\) and \(\lambda = 0.8\), compare the second-order approximation \(E_+ \approx 1 + \lambda^2/2\) to the exact value. At what \(\lambda\) does the error exceed 10%?

6. For a general \(2\times 2\) Hamiltonian \(H = \begin{pmatrix} E_0 + \delta & v \\ v & E_0 - \delta \end{pmatrix}\) (real \(v\), \(\delta\)), show that the eigenvalues are \(E_\pm = E_0 \pm \sqrt{\delta^2 + v^2}\). Explain why the levels never cross for \(v \neq 0\) (level repulsion).

7. A modified qubit has gap \(\Delta E = \epsilon\) (very small) and coupling \(\langle 1|V|0\rangle = v\). Show that \(E_+^{(2)} \sim v^2/\epsilon\) diverges as \(\epsilon \to 0\). What does this signal about the validity of non-degenerate perturbation theory near degeneracy?