5.1.3 Degenerate Perturbation Theory#

Worked solutions for the homework problems in the 5.1.3 Degenerate Perturbation Theory lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Why the old formula fails. Start from the non-degenerate first-order state correction formula and explain precisely where divergence appears for a \(d\)-fold degenerate level. Which hidden assumption about labeling eigenstates fails?

Solution.

The non-degenerate first-order state correction reads

\[ \vert n^{(1)}\rangle=\sum_{m\neq n}\vert m\rangle\,\frac{V_{mn}}{E_n-E_m}, \qquad V_{mn}=\langle m\vert\hat V\vert n\rangle. \]

Suppose the level \(E_n\) is \(d\)-fold degenerate, with an orthonormal basis \(\vert n,\alpha\rangle\), \(\alpha=1,\dots,d\), all at the same energy \(E_n\). Pick one of them, \(\vert n,\alpha\rangle\), and apply the formula. The sum “over \(m\neq n\)” must be read as a sum over every basis state other than \(\vert n,\alpha\rangle\) itself — and that includes the \(d-1\) partners \(\vert n,\beta\rangle\), \(\beta\neq\alpha\), in the same manifold. For each of those,

\[ E_n-E_m=E_n-E_n=0, \]

so the coefficient is \(V_{n\beta,n\alpha}/0\). Unless every off-diagonal in-manifold matrix element \(V_{n\beta,n\alpha}\) happens to vanish — which a generic perturbation does not arrange — the correction diverges. The same zero denominator reappears in the second-order energy \(\sum_{m\neq n}\vert V_{mn}\vert^2/(E_n-E_m)\).

The hidden assumption is about labeling. Non-degenerate perturbation theory assumes each unperturbed eigenstate is uniquely picked out by its energy, so that as \(\lambda\to0\) the perturbed state \(\vert n(\lambda)\rangle\) has one and only one limit, namely \(\vert n\rangle\). For a degenerate level this is false: the energy \(E_n\) does not single out a state, only a \(d\)-dimensional subspace, and any orthonormal basis of that subspace is an equally legal choice — the basis is fixed only up to a \(U(d)\) rotation. The perturbation \(\hat V\) generally selects a particular basis as the \(\lambda\to0\) limit of the exact eigenstates, and a generic starting basis is not it. Feeding the wrong basis into a formula that presumes the right one is what produces the \(0/0\). The minimal fix is to first diagonalize \(\hat V\) inside the manifold; in that basis the dangerous numerators \(V_{n\beta,n\alpha}\) (\(\beta\neq\alpha\)) vanish and the indeterminate terms drop out, while the intra-manifold mixing is fixed by the diagonalization itself rather than by the divergent ratio.

2. Block first, levels later. For a degenerate manifold with basis \(\{\vert n,\alpha\rangle\}_{\alpha=1}^d\), define \(W^{(n)}_{\alpha\beta}=\langle n,\alpha\vert\hat V\vert n,\beta\rangle\).

(a) Show that first-order shifts are eigenvalues of \(W^{(n)}\).

(b) Show that eigenvectors of \(W^{(n)}\) define the good zeroth-order basis.

Solution.

Work with the in-manifold energy matrix \(E_{n,\alpha\beta}(\lambda)\), whose eigenvalues are the physical energies that grow out of the manifold under \(\hat H(\lambda)=\hat H_0+\lambda\hat V\).

(a) The first (degenerate) Hellmann-Feynman identity, evaluated at \(\lambda=0\), gives

\[ \partial_\lambda E_{n,\alpha\beta}\big\vert_{\lambda=0} =\langle n,\alpha\vert\hat V\vert n,\beta\rangle=W^{(n)}_{\alpha\beta}. \]

Hence to first order

\[ E_{n,\alpha\beta}(\lambda)=E_n\,\delta_{\alpha\beta}+\lambda\,W^{(n)}_{\alpha\beta}+O(\lambda^2). \]

The first term is \(E_n\) times the \(d\times d\) identity; it commutes with everything and merely shifts all eigenvalues by \(E_n\). The physical energies are the eigenvalues of the full matrix, so they are \(E_n+\lambda w_a+O(\lambda^2)\), where \(w_1,\dots,w_d\) are the eigenvalues of the Hermitian matrix \(W^{(n)}\). The first-order shifts \(E^{(1)}_{n,a}=w_a\) are exactly the eigenvalues of \(W^{(n)}\). Because \(W^{(n)}\) is Hermitian its eigenvalues are real, as energy shifts must be.

(b) The “good” zeroth-order basis is the basis of unperturbed states that the exact perturbed eigenstates \(\vert n,a(\lambda)\rangle\) actually reduce to as \(\lambda\to0\). Diagonalize \(W^{(n)}\),

\[ W^{(n)}\vert n,a\rangle=w_a\,\vert n,a\rangle, \]

and adopt \(\{\vert n,a\rangle\}\) as the manifold basis. In this basis the effective block is \(E_n\hat I_d+\lambda\,\mathrm{diag}(w_1,\dots,w_d)+O(\lambda^2)\) — already diagonal at order \(\lambda\). Its eigenvectors are the fixed vectors \(\vert n,a\rangle\), independent of \(\lambda\) to leading order, so each perturbed eigenstate emerges continuously from one definite \(\vert n,a\rangle\). Any other choice is a \(U(d)\) rotation \(\vert n,\alpha\rangle=\sum_a U_{a\alpha}\vert n,a\rangle\) of this one; in a rotated basis the order-\(\lambda\) block is not diagonal, and to diagonalize it one needs an order-\(1\) (not small) rotation even though the perturbation is infinitesimal. A basis that requires a discontinuous jump to track the eigenstates cannot be a valid zeroth-order limit. Only the eigenbasis of \(W^{(n)}\) connects smoothly, so it is the good basis — provided the eigenvalues \(w_a\) are distinct (the degenerate case is Problem 5).

(c) In the good basis \(W^{(n)}\) is diagonal, \(V_{na,nb}=w_a\,\delta_{ab}\), so every off-diagonal in-manifold matrix element vanishes. These off-diagonal elements are precisely the numerators that sat over the zero denominator \(E_n-E_n\) in Problem 1. With the numerator forced to zero, the indeterminate \(0/0\) becomes a plain absent term: the higher-order sums need only run over states \(m\) in other manifolds, where \(E_n-E_m\neq0\) and every contribution is finite. Diagonalizing \(W^{(n)}\) before touching any higher-order correction is what guarantees this — it resolves all intra-manifold mixing in advance, leaving second-order perturbation theory only the well-behaved cross-manifold transitions. Applying a higher-order formula in a generic basis would leave the divergence in place, so the block diagonalization must come first.

3. Effective Hamiltonian and dark state. Consider a three-level system with \(\hat H_0=\Delta\,\vert 3\rangle\langle 3\vert\) (\(\Delta>0\)), so the ground manifold \(\{\vert 1\rangle,\vert 2\rangle\}\) is doubly degenerate at \(E=0\). Add

\[ \hat V=\mu\,(\vert 1\rangle\langle 2\vert+\vert 2\rangle\langle 1\vert)+\lambda\,(\vert 3\rangle\langle 1\vert+\vert 3\rangle\langle 2\vert+\mathrm{h.c.}), \]

with real \(\mu,\lambda\) and \(\vert\mu\vert,\vert\lambda\vert\ll\Delta\).

(a) Compute \(\hat P_d\hat V\hat P_d\) in \(\{\vert 1\rangle,\vert 2\rangle\}\), where \(\hat P_d\) projects onto the degenerate subspace. Read off the first-order splitting, and identify the good zeroth-order basis \(\vert\pm\rangle=(\vert 1\rangle\pm\vert 2\rangle)/\sqrt 2\).

(b) The lecture’s worked Example on the three-level bright/dark mechanism constructs the second-order effective Hamiltonian for the case \(\mu=0\) by summing virtual transitions through \(\vert 3\rangle\). Quote that result, specialized to equal cross-manifold couplings, and write the second-order block \((\hat H^{\text{eff}})^{(2)}\) as a \(2\times 2\) matrix in \(\{\vert 1\rangle,\vert 2\rangle\}\) — do not rebuild it from the virtual-transition sum.

(c) Restore \(\mu\ne 0\). Show that the first-order matrix from (a) and the second-order matrix from (b) commute, so a single basis diagonalizes both. Write the two ground-manifold energies through order \(\mu\) and \(\lambda^2\).

(d) Show that the dark state \(\vert-\rangle=(\vert 1\rangle-\vert 2\rangle)/\sqrt 2\) is an exact eigenstate of \(\hat H_0+\hat V\) for any \(\mu,\lambda\), and find its exact eigenvalue. Contrast this with the Example’s \(\mu=0\) dark state, and explain in one sentence what protects \(\vert-\rangle\) from the coupling to \(\vert 3\rangle\).

Solution.

(a) The projector onto the degenerate manifold is \(\hat P_d=\vert1\rangle\langle1\vert+\vert2\rangle\langle2\vert\). Split \(\hat V\) into its two pieces. The \(\lambda\) piece, \(\lambda(\vert3\rangle\langle1\vert+\vert3\rangle\langle2\vert+\mathrm{h.c.})\), always has one leg on \(\vert3\rangle\); sandwiched between two \(\hat P_d\) it gives zero because \(\hat P_d\vert3\rangle=0\). Only the \(\mu\) piece survives:

\[\begin{split} \hat P_d\hat V\hat P_d=\mu\,(\vert1\rangle\langle2\vert+\vert2\rangle\langle1\vert) \simeq\mu\begin{pmatrix}0&1\\1&0\end{pmatrix}=\mu\hat X, \end{split}\]

in the ordered basis \((\vert1\rangle,\vert2\rangle)\). Its eigenvalues are \(\pm\mu\), with eigenvectors \(\vert\pm\rangle=(\vert1\rangle\pm\vert2\rangle)/\sqrt2\). The first-order energies in the manifold are \(E^{(1)}_\pm=\pm\mu\): the degeneracy is lifted already at first order, with splitting \(2\mu\). Because the two eigenvalues \(\pm\mu\) are distinct, the diagonalizing basis is forced — the good zeroth-order basis is \(\vert\pm\rangle=(\vert1\rangle\pm\vert2\rangle)/\sqrt2\).

(b) With \(\mu=0\) the first-order block vanishes, and the surviving perturbation \(\lambda(\vert3\rangle\langle1\vert+\vert3\rangle\langle2\vert+\mathrm{h.c.})\) is exactly the cross-manifold coupling of the lecture’s worked Example (Three-Level Bright/Dark Mechanism), with the Example’s two couplings \(\lambda_1,\lambda_2\) both set equal to \(\lambda\). That Example sums the virtual transitions through \(\vert3\rangle\) once and for all and returns the rank-one projector form

\[ (\hat H^{\text{eff}})^{(2)}=-\frac{1}{\Delta}\vert\lambda\rangle\langle\lambda\vert, \qquad \vert\lambda\rangle=\lambda_1\vert1\rangle+\lambda_2\vert2\rangle. \]

Specializing to equal couplings \(\lambda_1=\lambda_2=\lambda\) gives \(\vert\lambda\rangle=\lambda(\vert1\rangle+\vert2\rangle)\), so

\[\begin{split} (\hat H^{\text{eff}})^{(2)} =-\frac{\lambda^2}{\Delta}(\vert1\rangle+\vert2\rangle)(\langle1\vert+\langle2\vert) =-\frac{\lambda^2}{\Delta}\begin{pmatrix}1&1\\1&1\end{pmatrix} \end{split}\]

in the ordered basis \((\vert1\rangle,\vert2\rangle)\) — the Example’s result read off at \(\lambda_1=\lambda_2\), with no need to re-sum the virtual transitions.

(c) Restore \(\mu\neq0\). From (a) the first-order block is \(\mu\hat X\); from (b) the second-order block is \(-\tfrac{\lambda^2}{\Delta}M\) with \(M=\begin{pmatrix}1&1\\1&1\end{pmatrix}=\hat I+\hat X\). The two matrices commute:

\[ \Big[\mu\hat X,\;-\tfrac{\lambda^2}{\Delta}(\hat I+\hat X)\Big] =-\frac{\mu\lambda^2}{\Delta}\big([\hat X,\hat I]+[\hat X,\hat X]\big)=0. \]

Commuting Hermitian matrices share an eigenbasis. Here it is \(\{\vert+\rangle,\vert-\rangle\}\), the eigenbasis of \(\hat X\), which also diagonalizes \(M\) (\(M\vert+\rangle=2\vert+\rangle\), \(M\vert-\rangle=0\)). So the single basis \(\{\vert+\rangle,\vert-\rangle\}\) diagonalizes both blocks, and the first-order (\(\mu\)) and second-order (\(\lambda^2\)) shifts simply add on each vector. The two ground-manifold energies, through order \(\mu\) and \(\lambda^2\), are

\[ E_+=+\mu-\frac{2\lambda^2}{\Delta}, \qquad E_-=-\mu. \]

The bright state \(\vert+\rangle\) collects both the \(+\mu\) first-order shift and the \(-2\lambda^2/\Delta\) second-order shift; the dark state \(\vert-\rangle\) collects only the \(-\mu\) first-order shift.

(d) The dark state \(\vert-\rangle=(\vert1\rangle-\vert2\rangle)/\sqrt2\) is an exact eigenstate of \(\hat H_0+\hat V\) for any \(\mu,\lambda\) — no truncation in perturbation order. Check the three pieces directly.

For \(\hat H_0\): \(\hat H_0\vert-\rangle=\Delta\vert3\rangle\langle3\vert-\rangle=0\), since \(\vert-\rangle\) has no \(\vert3\rangle\) component.

For the \(\lambda\) part, group its terms by which bra they carry:

\[ \lambda(\vert3\rangle\langle1\vert+\vert3\rangle\langle2\vert+\mathrm{h.c.}) =\lambda\,\vert3\rangle(\langle1\vert+\langle2\vert) +\lambda\,(\vert1\rangle+\vert2\rangle)\langle3\vert. \]

Acting on \(\vert-\rangle\): the \((\vert1\rangle+\vert2\rangle)\langle3\vert\) term gives \(0\) because \(\langle3\vert-\rangle=0\); the \(\vert3\rangle(\langle1\vert+\langle2\vert)\) term gives

\[ \lambda\,\vert3\rangle(\langle1\vert+\langle2\vert)\frac{\vert1\rangle-\vert2\rangle}{\sqrt2} =\lambda\,\vert3\rangle\,\frac{1-1}{\sqrt2}=0. \]

So the \(\lambda\) part annihilates \(\vert-\rangle\).

For the \(\mu\) part: \(\mu\hat X\vert-\rangle=-\mu\vert-\rangle\), since \(\hat X\vert-\rangle=-\vert-\rangle\).

Adding the three pieces,

\[ (\hat H_0+\hat V)\vert-\rangle=-\mu\,\vert-\rangle, \]

so \(\vert-\rangle\) is an exact eigenstate with exact eigenvalue \(E_-=-\mu\).

Contrast with the Example, where \(\mu=0\) and the dark state is an exact zero-energy eigenstate; here \(\mu\neq0\) slides the dark level down to \(-\mu\), but it stays exact — the perturbative \(E_-=-\mu\) of (c) receives no further corrections. In one sentence: the \(\lambda\) coupling reaches \(\vert3\rangle\) only through the symmetric combination \(\vert1\rangle+\vert2\rangle\), so the antisymmetric \(\vert-\rangle\) lies in the kernel of the \(\lambda\) part of \(\hat V\) — the two paths \(3\to1\) and \(3\to2\) interfere destructively — and never mixes with \(\vert3\rangle\) at any order.

4. Hydrogen Stark splitting. In hydrogen (ignoring spin), use basis \(\{\vert 2,0,0\rangle,\vert 2,1,0\rangle,\vert 2,1,1\rangle,\vert 2,1,-1\rangle\}\) and \(\hat V=e\mathcal E_0\hat{z}\).

(a) Use selection rules \(\Delta\ell=\pm1\), \(\Delta m=0\) to write the effective matrix structure.

(b) Explain why two states split linearly while two remain unsplit at first order.

Solution.

(a) The perturbation \(\hat V=e\mathcal E_0\hat{z}\) is built from \(\hat{z}\), a parity-odd, rank-one spherical tensor with zero \(m\)-component, so \(\langle n\ell m\vert\hat{z}\vert n\ell'm'\rangle\) is nonzero only when \(\ell'=\ell\pm1\) and \(m'=m\). Scan the \(n=2\) basis \((\vert200\rangle,\vert210\rangle,\vert211\rangle,\vert21,-1\rangle)\):

\(\vert200\rangle\) (\(\ell=0,m=0\)) couples to \(\vert210\rangle\) (\(\ell=1,m=0\)): \(\Delta\ell=+1\), \(\Delta m=0\) — allowed.
\(\vert211\rangle\) (\(m=1\)) and \(\vert21,-1\rangle\) (\(m=-1\)) would each need a same-\(m\) partner with \(\ell\) shifted by one; the \(n=2\) shell has no \(\ell=0\) or \(\ell=2\) state at \(m=\pm1\), so neither couples to anything.
Diagonal elements vanish: \(\langle2\ell m\vert\hat{z}\vert2\ell m\rangle=0\) because a state of definite \(\ell\) has definite parity while \(\hat{z}\) is parity-odd.

So in the basis order above

\[\begin{split} W=e\mathcal E_0\begin{pmatrix}0&z_0&0&0\\ z_0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}, \qquad z_0=\langle200\vert\hat{z}\vert210\rangle=-3a_0, \end{split}\]

with \(a_0\) the Bohr radius. The matrix is a single \(2\times2\) off-diagonal block in \(\{\vert200\rangle,\vert210\rangle\}\) plus a \(2\times2\) zero block in \(\{\vert211\rangle,\vert21,-1\rangle\}\).

(b) The \(2\times2\) block \(\begin{pmatrix}0&w\\w&0\end{pmatrix}\) with \(w=-3ea_0\mathcal E_0\) has eigenvalues \(\pm\vert w\vert=\pm3ea_0\mathcal E_0\) and eigenvectors \((\vert200\rangle\pm\vert210\rangle)/\sqrt2\). These two states acquire energy shifts \(\pm3ea_0\mathcal E_0\), linear in the field — the linear Stark effect. The states \(\vert211\rangle\) and \(\vert21,-1\rangle\) lie in the zero block; \(W\) gives them no first-order shift, so they stay degenerate at first order. Their response is quadratic, \(O(\mathcal E_0^2)\), arising only from second-order virtual coupling to states in other \(n\) shells.

(c) In one sentence: \(\hat{z}\) is odd under parity and conserves \(m\), so within the \(n=2\) shell it can connect only the opposite-parity, equal-\(m\) pair \(\vert200\rangle\leftrightarrow\vert210\rangle\) — the accidental \(\ell\)-degeneracy of hydrogen places states of both parities at the same energy, which is what permits a first-order (linear) effect — whereas \(\vert211\rangle\) and \(\vert21,-1\rangle\) have no opposite-parity partner at the same \(m\) in the shell and so cannot be split until second order.

5. Residual degeneracy. For \(\hat H^{\text{eff}}=\begin{pmatrix}a&b\\b^*&c\end{pmatrix}\):

(a) find eigenvalues,

(b) give the condition for no first-order splitting,

Solution.

(a) For \(\hat H^{\text{eff}}=\begin{pmatrix}a&b\\b^*&c\end{pmatrix}\) (Hermitian, so \(a,c\) real), the characteristic equation is \((a-E)(c-E)-\vert b\vert^2=0\), i.e. \(E^2-(a+c)E+(ac-\vert b\vert^2)=0\). Hence

\[ E_\pm=\frac{a+c}{2}\pm\sqrt{\left(\frac{a-c}{2}\right)^2+\vert b\vert^2}. \]

(b) The first-order splitting is \(E_+-E_-=2\sqrt{\left(\tfrac{a-c}{2}\right)^2+\vert b\vert^2}\). It vanishes only when the square root is zero, and since both terms under it are non-negative, that requires both

\[ a=c\qquad\text{and}\qquad b=0, \]

i.e. \(\hat H^{\text{eff}}=a\,\hat I\) is a multiple of the identity. Then the manifold stays degenerate: first-order perturbation theory produces no splitting at all.

(c) When \((\hat H^{\text{eff}})^{(1)}\propto\hat I\), the first-order matrix is itself degenerate and carries no information — every basis diagonalizes it, so it can pick out neither a good basis nor a splitting. One must build the second-order effective Hamiltonian, the matrix of virtual excursions to states outside the manifold,

\[ (H^{\text{eff}})_{\alpha\beta}^{(2)} =\sum_{m\notin d}\frac{\langle\alpha\vert\hat V\vert m\rangle\langle m\vert\hat V\vert\beta\rangle}{E_d-E_m}, \]

and diagonalize that; its eigenvalues are the second-order shifts and its eigenvectors the good basis. If the second-order matrix lifts the degeneracy, that resolves it. If it too is \(\propto\hat I\), the exact degeneracy is not an accident of low order but is enforced by a symmetry: one should then identify the conserved quantity or group representation responsible — a multidimensional irreducible representation of a (possibly non-abelian) symmetry group, or a Kramers doublet protected by time-reversal — since in that case no finite order of perturbation theory will split it.

6. When to switch methods. For each Hamiltonian below, decide whether non-degenerate or degenerate perturbation theory is appropriate at first order, and justify in one sentence:

(a) Hydrogen \(n=2\) manifold in a small uniform electric field \(\mathcal E\hat{z}\).

(b) The \(n=1\) ground state of hydrogen in the same field.

(d) Two nearly-degenerate bands with gap \(\ll V\).

Solution.

(a) Degenerate perturbation theory. The hydrogen \(n=2\) level is four-fold degenerate (ignoring spin), and \(\hat V=e\mathcal E z\) has a nonzero matrix element inside the manifold, \(\langle200\vert z\vert210\rangle\neq0\), so the in-manifold block must be diagonalized first — this is the linear Stark effect.

(b) Non-degenerate perturbation theory. The \(n=1\) ground state is a single non-degenerate level (ignoring spin), so the ordinary expansion applies; parity forces \(\langle100\vert z\vert100\rangle=0\), so the shift is purely second-order (the quadratic Stark effect).

(c) Non-degenerate perturbation theory. With the gap much larger than the coupling, \(V/\text{gap}\ll1\), so the coupling-over-gap expansion converges and the cross-gap mixing is a small, controlled correction.

(d) Degenerate (quasi-degenerate) perturbation theory. With gap \(\ll V\) the two bands form a near-degenerate manifold; the coupling-over-gap ratio is large, so one must diagonalize \(V\) exactly within the \(2\times2\) near-degenerate block (treating the small gap as an in-block term) and only then perturb in the coupling to remote bands.

★ 7. Flux ring with cosine perturbation. Recall the flux ring of §4.2.3: a particle on a ring of radius \(R\) threaded by flux \(\Phi\) has angular-momentum eigenstates \(\vert n\rangle\) with wavefunctions \(\psi_n(\theta) = \mathrm{e}^{\mathrm{i}n\theta}/\sqrt{2\pi}\), \(n \in \mathbb{Z}\), and energies

\[ E_n(\Phi) = E_0\,(n - \varphi)^{2}, \qquad E_0 = \frac{\hbar^{2}}{2mR^{2}}, \qquad \varphi = \Phi/\Phi_0. \]

Add a static azimuthal potential

\[ \hat V = V_0\cos\theta, \qquad V_0 \ll E_0. \]

At \(\varphi = 1/2\) the two lowest branches \(\vert 0\rangle\) and \(\vert 1\rangle\) cross at \(E_0(1/2) = E_1(1/2) = E_0/4\) — a two-fold degeneracy that the perturbation may lift.

Scope. Treat only the degenerate doublet \(\{\vert 0\rangle,\vert 1\rangle\}\) at first order in \(V_0\). The next-nearest unperturbed levels at this flux are \(\vert -1\rangle\) and \(\vert 2\rangle\) with \(E_{-1}(1/2) = E_2(1/2) = 9E_0/4\) — separated from the doublet by a gap of \(2E_0 \gg V_0\). They couple to the doublet through \(\hat V\) but only via virtual excursions that contribute at order \(V_0^{2}/E_0\), which is smaller than the first-order shift \(V_0\) by the small ratio \(V_0/E_0 \ll 1\); ignore them.

(a) Compute the matrix elements \(\langle n\vert\cos\theta\vert m\rangle\) in the angular-momentum basis. Show that \(\hat V\) couples only nearest neighbours \(n \leftrightarrow n \pm 1\), and identify the value of the coupling.

(b) At \(\varphi = 1/2\), build the \(2 \times 2\) effective Hamiltonian \(\hat H^{\text{eff}}\) for the doublet at first order in \(\hat V\) and write it in the form \(\hat H^{\text{eff}} = (\text{const})\,\hat I + (\text{coupling})\,\hat\sigma^{x}\).

(c) Diagonalise \(\hat H^{\text{eff}}\) to obtain the first-order energy shifts and the gap \(\Delta E\) that opens at the crossing. Identify the good zeroth-order basis — the symmetric and antisymmetric combinations of \(\vert 0\rangle\) and \(\vert 1\rangle\) — and write the corresponding probability densities \(\vert\psi_{\pm}(\theta)\vert^{2}\).

(d) Slightly off the crossing, set \(\varphi = 1/2 + \delta\) with \(\vert\delta\vert\) small. Build the \(2 \times 2\) matrix at general \(\delta\) (within the same doublet truncation) and find the perturbed energies \(E_{\pm}(\delta)\). Describe in words the shape of \(E_{\pm}(\varphi)\) near \(\varphi = 1/2\) — the avoided crossing.

(e) Move away from the crossing. For \(\vert\varphi\vert < 1/2\) the ground state is the non-degenerate angular-momentum state \(\vert 0\rangle\) with energy \(E_0(\varphi) = E_0\varphi^{2}\). Apply the second-order non-degenerate perturbation formula

\[ E_n(\varphi) = E_n^{(0)}(\varphi) + \langle n\vert\hat V\vert n\rangle + \sum_{m\neq n}\frac{\vert\langle m\vert\hat V\vert n\rangle\vert^{2}}{E_n^{(0)}(\varphi) - E_m^{(0)}(\varphi)} + \mathcal{O}(V_0^{3}) \]

to compute the perturbed ground-state energy \(E_0(\varphi)\) through order \(V_0^{2}\). Show that the expression diverges at \(\varphi = \pm 1/2\), recovering the breakdown signalled by the doublet treatment in (b)–(c).

Solution.

(a) Matrix element of \(\cos\theta\) in the plane-wave basis:

\[ \langle n\vert\cos\theta\vert m\rangle = \frac{1}{2\pi}\int_0^{2\pi}\cos\theta\,\mathrm{e}^{\mathrm{i}(m-n)\theta}\,\mathrm{d}\theta. \]

Using \(\cos\theta = \tfrac{1}{2}(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta})\),

\[ \langle n\vert\cos\theta\vert m\rangle = \frac{1}{4\pi}\int_0^{2\pi}\!\!\bigl(\mathrm{e}^{\mathrm{i}(m-n+1)\theta} + \mathrm{e}^{\mathrm{i}(m-n-1)\theta}\bigr)\mathrm{d}\theta = \tfrac{1}{2}\,(\delta_{n,m+1} + \delta_{n,m-1}). \]

So \(\langle n\vert\cos\theta\vert m\rangle = 1/2\) when \(\vert n - m\vert = 1\), and zero otherwise: \(\hat V = V_0\cos\theta\) couples each angular-momentum eigenstate only to its nearest neighbours \(\vert n\pm 1\rangle\), with matrix element \(V_0/2\).

(b) Within the doublet \(\{\vert 0\rangle,\vert 1\rangle\}\) at \(\varphi = 1/2\), the unperturbed Hamiltonian is just \((E_0/4)\hat I\). The perturbation block is

\[\begin{split} W = V_0\begin{pmatrix} \langle 0\vert\cos\theta\vert 0\rangle & \langle 0\vert\cos\theta\vert 1\rangle \\ \langle 1\vert\cos\theta\vert 0\rangle & \langle 1\vert\cos\theta\vert 1\rangle\end{pmatrix} = \frac{V_0}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} = \frac{V_0}{2}\hat\sigma^{x}, \end{split}\]

using \(\langle 0\vert\cos\theta\vert 1\rangle = 1/2\) from (a) and zero diagonal entries. The first-order effective Hamiltonian on the doublet is therefore

\[ \hat H^{\text{eff}} = \frac{E_0}{4}\hat I + \frac{V_0}{2}\hat\sigma^{x}. \]

(c) The \(\hat\sigma^{x}\) block has eigenvalues \(\pm 1\) with eigenstates \(\vert\pm\rangle = (\vert 0\rangle \pm \vert 1\rangle)/\sqrt 2\). So

\[ E_\pm = \frac{E_0}{4} \pm \frac{V_0}{2}, \qquad \Delta E = E_+ - E_- = V_0, \]

the gap that the perturbation opens at the crossing. The good zeroth-order basis is the pair of symmetric / antisymmetric combinations,

\[ \vert +\rangle = \frac{\vert 0\rangle + \vert 1\rangle}{\sqrt 2},\qquad \vert -\rangle = \frac{\vert 0\rangle - \vert 1\rangle}{\sqrt 2}, \]

which diagonalise \(\hat V\) on the doublet and emerge continuously from the perturbed eigenstates as \(V_0 \to 0\). Their real-space wavefunctions are \(\psi_\pm(\theta) = (1 \pm \mathrm{e}^{\mathrm{i}\theta})/\sqrt{4\pi}\), with probability densities

\[ \vert\psi_+(\theta)\vert^{2} = \frac{1 + \cos\theta}{2\pi}\quad(\text{peaked at }\theta = 0), \qquad \vert\psi_-(\theta)\vert^{2} = \frac{1 - \cos\theta}{2\pi}\quad(\text{peaked at }\theta = \pi). \]

The two eigenstates are standing waves localised at opposite sides of the ring: \(\vert +\rangle\) at \(\theta = 0\) where \(\cos\theta\) is maximal (so \(\hat V\) is most repulsive, raising its energy), and \(\vert -\rangle\) at \(\theta = \pi\) where \(\cos\theta\) is minimal (lowering its energy). The first-order shifts \(\pm V_0/2\) are exactly the matrix-element-weighted overlap of these standing waves with the potential.

(d) Set \(\varphi = 1/2 + \delta\) and expand the diagonal energies:

\[ E_0(\varphi) = E_0\varphi^{2} = E_0\bigl(\tfrac{1}{4} + \delta + \delta^{2}\bigr), \qquad E_1(\varphi) = E_0(1-\varphi)^{2} = E_0\bigl(\tfrac{1}{4} - \delta + \delta^{2}\bigr). \]

The \(2 \times 2\) matrix in the doublet basis (still truncating to these two levels) is

\[\begin{split} \hat H_{2\times 2}(\delta) = \begin{pmatrix} E_0(1/4 + \delta + \delta^{2}) & V_0/2 \\ V_0/2 & E_0(1/4 - \delta + \delta^{2})\end{pmatrix} = \bigl[\tfrac{E_0}{4} + E_0\delta^{2}\bigr]\hat I + E_0\delta\,\hat\sigma^{z} + \frac{V_0}{2}\hat\sigma^{x}. \end{split}\]

For small \(\vert\delta\vert\) the \(E_0\delta^{2}\) shift is negligible. The eigenvalues are

\[ E_\pm(\delta) = \frac{E_0}{4} \pm \sqrt{(E_0\delta)^{2} + (V_0/2)^{2}}. \]

This is the standard avoided-crossing profile:

At \(\delta = 0\): the gap is exactly \(\Delta E = V_0\), with eigenstates \(\vert\pm\rangle\) from (c).
For \(\vert E_0\delta\vert \gg V_0/2\): \(E_\pm \approx E_0/4 \pm E_0\vert\delta\vert\) — the unperturbed linear branches \(E_0(\varphi) = E_0(1/4 + \delta)\) and \(E_1(\varphi) = E_0(1/4 - \delta)\), with eigenstates approaching \(\vert 0\rangle\) and \(\vert 1\rangle\).

In the \((\varphi, E)\) plane, what would have been a sharp X-shaped crossing of the two linear unperturbed branches at \(\varphi = 1/2\) has been rounded into a hyperbolic anti-crossing with minimum gap \(V_0\) at the centre. The perturbation has lifted the degeneracy and replaced level crossing by level repulsion — the same mechanism that opens band gaps in periodic potentials at the boundary of every Brillouin zone.

(e) For \(\vert\varphi\vert < 1/2\) the diagonal element vanishes by the selection rule of (a), \(\langle 0\vert\hat V\vert 0\rangle = 0\), so \(E_0^{(1)}(\varphi) = 0\). The second-order sum is restricted to \(m = \pm 1\) by the same rule, with \(\vert\langle\pm 1\vert\hat V\vert 0\rangle\vert^{2} = V_0^{2}/4\). The unperturbed gaps are

\[ E_0^{(0)}(\varphi) - E_{+1}^{(0)}(\varphi) = E_0\bigl[\varphi^{2} - (1 - \varphi)^{2}\bigr] = E_0(2\varphi - 1), \]

\[ E_0^{(0)}(\varphi) - E_{-1}^{(0)}(\varphi) = E_0\bigl[\varphi^{2} - (-1 - \varphi)^{2}\bigr] = -E_0(2\varphi + 1). \]

Hence

\[\begin{split} \begin{split} E_0(\varphi) &= E_0\varphi^{2} + \frac{V_0^{2}}{4}\left[\frac{1}{E_0(2\varphi - 1)} + \frac{1}{-E_0(2\varphi + 1)}\right] + \mathcal{O}(V_0^{3})\\ &= E_0\varphi^{2} + \frac{V_0^{2}}{4 E_0}\cdot\frac{(2\varphi + 1) - (2\varphi - 1)}{(2\varphi - 1)(2\varphi + 1)} + \mathcal{O}(V_0^{3})\\ &= E_0\varphi^{2} - \frac{V_0^{2}}{2 E_0\,(1 - 4\varphi^{2})} + \mathcal{O}(V_0^{3}). \end{split} \end{split}\]

The second-order shift is finite throughout the open interval \(\vert\varphi\vert < 1/2\) and is negative — the bottom branch always bends down under a perturbation that couples it only to higher levels (each term in the sum carries a negative denominator).

Breakdown at \(\varphi = \pm 1/2\). The factor \(1 - 4\varphi^{2}\) vanishes there, and the non-degenerate formula diverges. This divergence is not a calculational mistake but a signal that the basic premise — non-degenerate level \(\vert 0\rangle\) with all other levels well separated — fails: at \(\varphi = +1/2\) the level \(\vert 0\rangle\) becomes degenerate with \(\vert 1\rangle\), and at \(\varphi = -1/2\) with \(\vert -1\rangle\). The first-order matrix element \(V_0/2\) between \(\vert 0\rangle\) and its degenerate partner can no longer be treated as a small denominator-suppressed correction; one must instead diagonalise it exactly within the degenerate doublet, exactly as in (b)–(c).

Order-of-magnitude contrast. Away from the crossing, the perturbative shift is of order \(V_0^{2}/E_0\) — quadratic in the perturbation strength and suppressed by the (open) gap. At the crossing the splitting is of order \(V_0\) — linear in the perturbation strength. This promotion from \(\mathcal{O}(V_0^{2}/E_0)\) to \(\mathcal{O}(V_0)\) is the qualitative fingerprint of degeneracy lifting and explains why states near a level crossing respond so much more strongly to small perturbations than states far from any crossing — a phenomenon that underlies, for example, the giant susceptibility of materials near quantum critical points and the high responsivity of qubits operated at sweet-spot flux degeneracies.