5.1.3 Degenerate Perturbation Theory

5.1.3 Degenerate Perturbation Theory#

Prompts

  • Why does non-degenerate perturbation theory fail when multiple unperturbed states share the same energy, and what is the minimal fix?

  • In a degenerate subspace, why must we diagonalize the perturbation first, and how does this define the correct zeroth-order basis?

  • What is the physical difference between first-order splitting inside the degenerate manifold and second-order shifts from virtual coupling to outside states?

  • For a concrete model, how can you diagnose whether degeneracy is fully lifted, partially lifted, or unchanged at first order, and what should be done next in each case?

Lecture Notes#

Overview#

§5.1.2 assumed isolated levels and produced scalar corrections. This subsection handles the complementary case: when unperturbed levels are degenerate, perturbation theory must be organized as a block problem inside each degenerate manifold.

The practical workflow is: identify the degenerate manifold, diagonalize in-manifold matrix elements at first order, then add virtual-coupling corrections from outside manifolds at second order.

Why non-degenerate formulas fail#

For \(\hat{H}(\lambda)=\hat{H}_0+\lambda\hat{V}\), suppose \(\hat{H}_0\) has a \(d_n\)-fold degenerate eigenspace at energy \(E_n\), spanned by \(\vert n,\alpha\rangle\) (\(\alpha=1,\dots,d_n\)):

\[ \hat{H}_0\vert n,\alpha\rangle=E_n\vert n,\alpha\rangle. \]

A quick wrong attempt is to reuse the non-degenerate structure directly:

\[ \vert n,\alpha\rangle^{(1)}\sim\sum_{(m,\beta)\neq(n,\alpha)}\vert m,\beta\rangle\frac{(\cdots)}{E_n-E_m}. \]

Then terms with \(m=n\) but \(\beta\neq\alpha\) give

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

so the denominator diverges.

Core Idea

The proper strategy is divide and conquer:

  • reorganize the Hilbert space into degenerate-manifold blocks,

  • construct the effective Hamiltonian for each block,

  • diagonalize each block,

  • if residual degeneracy remains, continue to the next order/block level.

The central goal of degenerate perturbation theory is to construct the effective Hamiltonian within each degenerate subspace.

Problem Setup#

Degenerate Perturbation Problem

Consider \(\hat H(\lambda)=\hat H_0+\lambda\hat V\), and work in a basis \(\vert n,\alpha\rangle\) where \(n\) labels the manifold and \(\alpha=1,\dots,d_n\) labels orthonormal states inside that manifold:

\[ \hat H_0=\sum_n\sum_\alpha E_n\,\vert n,\alpha\rangle\langle n,\alpha\vert. \]
\[ \hat V=\sum_{m,\alpha}\sum_{n,\beta}\vert m,\alpha\rangle V_{m\alpha,n\beta}\langle n,\beta\vert. \]

Here \(V_{m\alpha,n\beta}=\langle m,\alpha\vert\hat V\vert n,\beta\rangle\), and states inside one degenerate manifold can mix.

The corresponding eigenvalue equation is \(\hat H(\lambda)\vert n,\beta(\lambda)\rangle=\sum_\alpha E_{n,\alpha\beta}(\lambda)\vert n,\alpha(\lambda)\rangle\), and our objective is to construct \(E_{n,\alpha\beta}(\lambda)\) and \(\vert n,\alpha(\lambda)\rangle\) order by order in \(\lambda\).

Hellmann-Feynman Identities (Degenerate Form)#

Differentiate the matrix eigenproblem and project onto \(\langle m,\gamma\vert\) at \(\lambda=0\):

\[ \langle m,\gamma\vert\partial_\lambda\hat H\vert n,\beta\rangle =\partial_\lambda E_{n,\gamma\beta}\,\delta_{mn} +(E_n-E_m)\langle m,\gamma\vert\partial_\lambda n,\beta\rangle. \]

Hellmann-Feynman identities (degenerate)

  • 1st Hellmann-Feynman Identity (energy matrix derivative, within manifold):

\[ \partial_\lambda E_{n,\alpha\beta}=V_{n\alpha,n\beta}. \]

  • 2nd Hellmann-Feynman Identity (state-mixing derivative, across manifolds):

\[ \langle m,\alpha\vert\partial_\lambda n,\beta\rangle =\frac{V_{m\alpha,n\beta}}{E_n-E_m} \text{ for }m\neq n. \]

Derivation: degenerate Hellmann-Feynman identities

Step 1: Differentiate the matrix eigenvalue equation

Start from

\[ \hat H(\lambda)\vert n,\beta(\lambda)\rangle =\sum_\alpha E_{n,\alpha\beta}(\lambda)\,\vert n,\alpha(\lambda)\rangle. \]

Differentiate with respect to \(\lambda\):

\[ \begin{split} \partial_\lambda\hat H\,\vert n,\beta\rangle+\hat H\,\vert\partial_\lambda n,\beta\rangle =\sum_\alpha\Big[ (\partial_\lambda E_{n,\alpha\beta})\vert n,\alpha\rangle +E_{n,\alpha\beta}\vert\partial_\lambda n,\alpha\rangle \Big]. \end{split} \]

Step 2: Project with \(\langle m,\gamma\vert\) and evaluate at \(\lambda=0\)

Left-multiply by \(\langle m,\gamma\vert\):

\[ \begin{split} \langle m,\gamma\vert\partial_\lambda\hat H\vert n,\beta\rangle +\langle m,\gamma\vert\hat H\vert\partial_\lambda n,\beta\rangle =\sum_\alpha\Big[ \partial_\lambda E_{n,\alpha\beta}\,\langle m,\gamma\vert n,\alpha\rangle +E_{n,\alpha\beta}\,\langle m,\gamma\vert\partial_\lambda n,\alpha\rangle \Big]. \end{split} \]

Evaluate at \(\lambda=0\):

\[ V_{m\gamma,n\beta}+E_m\langle m,\gamma\vert\partial_\lambda n,\beta\rangle =\delta_{mn}\,\partial_\lambda E_{n,\gamma\beta}+E_n\langle m,\gamma\vert\partial_\lambda n,\beta\rangle. \]

So

\[ V_{m\gamma,n\beta} =\delta_{mn}\,\partial_\lambda E_{n,\gamma\beta} +(E_n-E_m)\langle m,\gamma\vert\partial_\lambda n,\beta\rangle. \]

Step 3: Split into \(m=n\) and \(m\neq n\)

If \(m=n\),

\[ \partial_\lambda E_{n,\alpha\beta}=V_{n\alpha,n\beta}. \]

If \(m\neq n\),

\[ \langle m,\alpha\vert\partial_\lambda n,\beta\rangle =\frac{V_{m\alpha,n\beta}}{E_n-E_m}. \]

Conjugating both sides — using \(\langle\psi\vert\phi\rangle^{*}=\langle\phi\vert\psi\rangle\) on the left and \(V_{m\alpha,n\beta}^{*}=V_{n\beta,m\alpha}\) (Hermiticity of \(\hat V\)) on the right — and relabeling \(m\leftrightarrow n\), \(\alpha\leftrightarrow\beta\) gives the conjugate identity

\[ \langle\partial_\lambda m,\alpha\vert n,\beta\rangle =\frac{V_{m\alpha,n\beta}}{E_m-E_n}. \]

These are the two degenerate Hellmann-Feynman identities (with the conjugate form) used below.

Energy Corrections#

Use the Taylor expansion of the in-manifold energy matrix:

Energy matrix expansion up to second order

Using degenerate Hellmann-Feynman identities, the energy matrix correction is given by:

\[\begin{split} \begin{split} E_{n,\alpha\beta}(\lambda) &=E_n\delta_{\alpha\beta} +\lambda\,\partial_\lambda E_{n,\alpha\beta} +\frac{\lambda^2}{2}\,\partial_\lambda^2 E_{n,\alpha\beta} +O(\lambda^3)\\ &=E_n\delta_{\alpha\beta} +\lambda V_{n\alpha,n\beta} +\lambda^2\sum_{m\neq n}\sum_\gamma \frac{V_{n\alpha,m\gamma}V_{m\gamma,n\beta}}{E_n-E_m} +O(\lambda^3). \end{split} \end{split}\]

Derivation: first-order energy matrix derivative

Set \(m=n\) in the projected derivative equation:

\[ \langle n,\alpha\vert\partial_\lambda \hat H\vert n,\beta\rangle =\partial_\lambda E_{n,\alpha\beta}. \]

With \(\partial_\lambda\hat H=\hat V\), this gives

\[ \partial_\lambda E_{n,\alpha\beta}=V_{n\alpha,n\beta}. \]

Derivation: second-order energy matrix derivative

Differentiate \(\partial_\lambda E_{n,\alpha\beta}=\langle n,\alpha\vert\hat V\vert n,\beta\rangle\) once more:

\[ \partial_\lambda^2 E_{n,\alpha\beta} =\langle\partial_\lambda n,\alpha\vert\hat V\vert n,\beta\rangle +\langle n,\alpha\vert\hat V\vert\partial_\lambda n,\beta\rangle. \]

Insert a complete basis explicitly:

\[ \partial_\lambda^2 E_{n,\alpha\beta} =\sum_{m,\gamma}\Big(\langle\partial_\lambda n,\alpha\vert m,\gamma\rangle V_{m\gamma,n\beta} +V_{n\alpha,m\gamma}\langle m,\gamma\vert\partial_\lambda n,\beta\rangle\Big). \]

Now split into \(m\neq n\) and \(m=n\) pieces. For \(m\neq n\), use the second Hellmann-Feynman identity:

\[ \langle m,\gamma\vert\partial_\lambda n,\beta\rangle =\frac{V_{m\gamma,n\beta}}{E_n-E_m}, \]

and its conjugate relation, giving

\[ 2\sum_{m\neq n}\sum_\gamma \frac{V_{n\alpha,m\gamma}V_{m\gamma,n\beta}}{E_n-E_m}. \]

For \(m=n\), terms involve in-manifold overlaps \(A_{\gamma\beta}=\langle n,\gamma\vert\partial_\lambda n,\beta\rangle\). These are gauge/basis-choice dependent within the degenerate manifold and can be set to zero by parallel-transport gauge after fixing the first-order in-manifold basis.

Therefore

\[ \partial_\lambda^2 E_{n,\alpha\beta} =2\sum_{m\neq n}\sum_\gamma \frac{V_{n\alpha,m\gamma}V_{m\gamma,n\beta}}{E_n-E_m}. \]

State Corrections#

The corrected basis vectors are also expanded in \(\lambda\).

State expansion and first-order mixing

Using the second Hellmann-Feynman identity, the state correction is given by:

\[\begin{split} \begin{split} \vert n,\alpha(\lambda)\rangle &=\vert n,\alpha\rangle +\lambda\vert\partial_\lambda n,\alpha\rangle +O(\lambda^2)\\ &=\vert n,\alpha\rangle +\lambda\sum_{m\neq n}\sum_\beta \vert m,\beta\rangle\frac{V_{m\beta,n\alpha}}{E_n-E_m} +O(\lambda^2). \end{split} \end{split}\]

Derivation: first-order state derivative

Expand in the unperturbed basis and separate the two parts:

\[ \vert\partial_\lambda n,\alpha\rangle =\sum_{\beta}\vert n,\beta\rangle\langle n,\beta\vert\partial_\lambda n,\alpha\rangle +\sum_{m\neq n,\beta}\vert m,\beta\rangle\langle m,\beta\vert\partial_\lambda n,\alpha\rangle. \]

For \(m\neq n\), apply

\[ \langle m,\beta\vert\partial_\lambda n,\alpha\rangle =\frac{V_{m\beta,n\alpha}}{E_n-E_m}. \]

For \(m=n\), define in-manifold overlaps \(A_{\beta\alpha}=\langle n,\beta\vert\partial_\lambda n,\alpha\rangle\). From orthonormality, \(A\) is anti-Hermitian, and by basis/gauge choice inside the degenerate manifold (after first-order diagonalization) one may set \(A=0\).

Under that gauge,

\[ \vert\partial_\lambda n,\alpha\rangle =\sum_{m\neq n,\beta}\vert m,\beta\rangle\frac{V_{m\beta,n\alpha}}{E_n-E_m}. \]

Note: Operator Form

The same result can be written with projectors \(\hat{P}_n=\sum_\alpha \vert n,\alpha\rangle\langle n,\alpha\vert\) and \(\hat{Q}_n=\hat I-\hat P_n\):

\[ \hat H^{\text{eff}}_n =E_n\hat I_d +\lambda\hat P_n\hat V\hat P_n +\lambda^2\hat P_n\hat V\hat Q_n \frac{1}{E_n-\hat Q_n\hat H_0\hat Q_n} \hat Q_n\hat V\hat P_n +O(\lambda^3). \]

This is exactly the same content as the component formulas above:

  • \(\hat P_n\hat V\hat P_n \leftrightarrow V_{n\alpha,n\beta}\) (first-order block),

  • second-order projector term \(\leftrightarrow \sum_{m\neq n,\gamma}V_{n\alpha,m\gamma}V_{m\gamma,n\beta}/(E_n-E_m)\).

Use component form for calculations; use operator form as a compact summary.

Example: Three-level bright/dark mechanism

Problem. Consider

\[ \hat H_0=\Delta\vert 3\rangle\langle3\vert, \]
\[ \hat V=\lambda_1\vert 1\rangle\langle3\vert+\lambda_2\vert 2\rangle\langle3\vert+\mathrm{h.c.}, \]

with \(\vert\lambda_{1,2}\vert\ll\Delta\). The degenerate manifold is \(\{\vert 1\rangle,\vert 2\rangle\}\) at energy \(0\).

Step 1. First-order block check. Inside the manifold,

\[ (H^{\text{eff}})_{ij}^{(1)}=\langle i\vert\hat V\vert j\rangle=0,\qquad i,j\in\{1,2\}, \]

so first-order splitting is absent.

Step 2. Build second-order block element-by-element. Because the only outside state is \(\vert 3\rangle\) with energy \(\Delta\),

\[\begin{split} \begin{split} (H^{\text{eff}})_{ij}^{(2)} &=\sum_{m\notin\{1,2\}}\frac{\langle i\vert\hat V\vert m\rangle\langle m\vert\hat V\vert j\rangle}{0-E_m}\\ &=-\frac{1}{\Delta}\langle i\vert\hat V\vert 3\rangle\langle3\vert\hat V\vert j\rangle. \end{split} \end{split}\]

Now evaluate four entries explicitly:

\[ (H^{\text{eff}})_{11}^{(2)}=-\frac{\vert\lambda_1\vert^2}{\Delta}, \]
\[ (H^{\text{eff}})_{22}^{(2)}=-\frac{\vert\lambda_2\vert^2}{\Delta}, \]
\[ (H^{\text{eff}})_{12}^{(2)}=-\frac{\lambda_1\lambda_2^*}{\Delta}, \]
\[ (H^{\text{eff}})_{21}^{(2)}=-\frac{\lambda_2\lambda_1^*}{\Delta}. \]

Hence

\[\begin{split} \hat H^{\text{eff}}=-\frac{1}{\Delta} \begin{pmatrix} \vert\lambda_1\vert^2 & \lambda_1\lambda_2^*\\ \lambda_2\lambda_1^* & \vert\lambda_2\vert^2 \end{pmatrix} =-\frac{1}{\Delta}\vert\lambda\rangle\langle\lambda\vert, \end{split}\]

with \(\vert\lambda\rangle=\lambda_1\vert 1\rangle+\lambda_2\vert 2\rangle\).

Step 3. Diagonalize and identify bright/dark states. This is a rank-1 projector form, so eigenvalues are immediate:

\[ E_B=-\frac{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}{\Delta}, \]
\[ E_D=0. \]

A normalized bright state is

\[ \vert B\rangle= \frac{\lambda_1\vert 1\rangle+\lambda_2\vert 2\rangle}{\sqrt{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}}, \]

and a normalized dark state orthogonal to \(\vert B\rangle\) is

\[ \vert D\rangle= \frac{\lambda_2^*\vert 1\rangle-\lambda_1^*\vert 2\rangle}{\sqrt{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}}. \]

Step 4. Exact zero-mode check (not only perturbative).

For \(\hat H_0\): since \(\hat H_0=\Delta\vert 3\rangle\langle 3\vert\) and \(\vert D\rangle\in\mathrm{span}\{\vert 1\rangle,\vert 2\rangle\}\) has no \(\vert 3\rangle\) component (so \(\langle 3\vert D\rangle=0\)),

\[ \hat H_0\vert D\rangle=\Delta\vert 3\rangle\langle 3\vert D\rangle=0. \]

For \(\hat V\), rewrite the perturbation in factored form,

\[ \hat V=(\lambda_1\vert 1\rangle+\lambda_2\vert 2\rangle)\langle 3\vert+\vert 3\rangle(\lambda_1^*\langle 1\vert+\lambda_2^*\langle 2\vert), \]

and act on \(\vert D\rangle=(\lambda_2^*\vert 1\rangle-\lambda_1^*\vert 2\rangle)/\sqrt{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}\):

\[ \hat V\vert D\rangle=(\lambda_1\vert 1\rangle+\lambda_2\vert 2\rangle)\langle 3\vert D\rangle+\frac{\vert 3\rangle(\lambda_1^*\langle 1\vert+\lambda_2^*\langle 2\vert)(\lambda_2^*\vert 1\rangle-\lambda_1^*\vert 2\rangle)}{\sqrt{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}}. \]

The first term vanishes because \(\langle 3\vert D\rangle=0\). Expanding the second term with \(\langle 1\vert 1\rangle=\langle 2\vert 2\rangle=1\) and \(\langle 1\vert 2\rangle=\langle 2\vert 1\rangle=0\),

\[ \hat V\vert D\rangle=\frac{\vert 3\rangle(\lambda_1^*\lambda_2^*-\lambda_2^*\lambda_1^*)}{\sqrt{\vert\lambda_1\vert^2+\vert\lambda_2\vert^2}}=0. \]

Therefore \((\hat H_0+\hat V)\vert D\rangle=0\) exactly, regardless of \(\lambda_{1,2}\).

Interpretation. First-order splitting is absent, but second-order virtual coupling generates an effective interaction with one bright mode and one protected dark mode.

Example: Spin-1 splitting from a degenerate manifold

Problem. Consider a spin-1 system in basis \(\{\vert+1\rangle,\vert 0\rangle,\vert-1\rangle\}\) with

\[ \hat H(\lambda)=\hat H_0+\lambda\hat V, \]
\[ \hat H_0=(\hat S^z)^2, \]
\[ \hat V=\hat S^x+\hat S^z. \]

In units where \(\hbar=1\), the operator matrices are

\[\begin{split} \hat S^x= \frac{1}{\sqrt2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}, \end{split}\]
\[\begin{split} \hat S^z= \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}, \end{split}\]

so

\[\begin{split} \hat H_0= \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&1 \end{pmatrix}, \end{split}\]
\[\begin{split} \hat V= \begin{pmatrix} 1&\frac{1}{\sqrt2}&0\\ \frac{1}{\sqrt2}&0&\frac{1}{\sqrt2}\\ 0&\frac{1}{\sqrt2}&-1 \end{pmatrix}. \end{split}\]

Unperturbed manifolds: \(E_1=1\) with \(\{\vert+1\rangle,\vert-1\rangle\}\) (degenerate), and \(E_0=0\) with \(\vert 0\rangle\).

Step 1 (degenerate PT): construct effective blocks.

For the \(E_1\) manifold (\(\alpha,\beta\in\{+1,-1\}\)), the only state outside the manifold is \(\vert 0\rangle\), with \(E_1-E_0=1\). The relevant matrix elements of \(\hat V\), read off from the matrix above, are \(V_{+1,+1}=1\), \(V_{-1,-1}=-1\), \(V_{+1,-1}=V_{-1,+1}=0\), and \(V_{\pm 1,0}=V_{0,\pm 1}=1/\sqrt 2\). Substituting into the Energy matrix expansion up to second order formula entry by entry:

\[ (\hat H^{\text{eff}}_1)_{+1,+1}=E_1+\lambda V_{+1,+1}+\lambda^2\,\frac{V_{+1,0}V_{0,+1}}{E_1-E_0}=1+\lambda+\frac{\lambda^2}{2}, \]
\[ (\hat H^{\text{eff}}_1)_{-1,-1}=E_1+\lambda V_{-1,-1}+\lambda^2\,\frac{V_{-1,0}V_{0,-1}}{E_1-E_0}=1-\lambda+\frac{\lambda^2}{2}, \]
\[ (\hat H^{\text{eff}}_1)_{+1,-1}=\lambda V_{+1,-1}+\lambda^2\,\frac{V_{+1,0}V_{0,-1}}{E_1-E_0}=\frac{\lambda^2}{2}, \]
\[ (\hat H^{\text{eff}}_1)_{-1,+1}=\lambda V_{-1,+1}+\lambda^2\,\frac{V_{-1,0}V_{0,+1}}{E_1-E_0}=\frac{\lambda^2}{2}. \]

Collecting these four entries gives, to \(O(\lambda^2)\),

\[\begin{split} \hat H^{\text{eff}}_{1}= \begin{pmatrix} 1+\lambda+\frac{\lambda^2}{2} & \frac{\lambda^2}{2}\\ \frac{\lambda^2}{2} & 1-\lambda+\frac{\lambda^2}{2} \end{pmatrix} \end{split}\]

(on the first-round corrected basis). For the \(E_0\) manifold,

\[ \hat H^{\text{eff}}_{0}=(-\lambda^2). \]

At this stage, the original degeneracy is already lifted at \(O(\lambda)\).

Step 2 (second round: non-degenerate PT inside the lifted block).

In \(\hat H^{\text{eff}}_{1}\), the off-diagonal element is \(\lambda^2/2\), while the diagonal splitting is

\[ \Delta E'=(1+\lambda+\lambda^2/2)-(1-\lambda+\lambda^2/2)=2\lambda. \]

So the induced mixing scale is

\[ \frac{\lambda^2/2}{\Delta E'}=\frac{\lambda}{4}=O(\lambda). \]

Hence the induced second-round energy correction is

\[ \frac{(\lambda^2/2)^2}{\Delta E'}\sim O(\lambda^3), \]

which is negligible through \(O(\lambda^2)\). The induced second-round state correction, however, is \(O(\lambda)\) — the same order as the first-round cross-manifold mixing — so it must be carried out.

Therefore, up to \(O(\lambda^2)\),

\[ E_{+1}=1+\lambda+\frac{\lambda^2}{2}+O(\lambda^3), \]
\[ E_{-1}=1-\lambda+\frac{\lambda^2}{2}+O(\lambda^3), \]
\[ E_0=-\lambda^2+O(\lambda^3). \]

A representative first-order state structure follows by substituting the \(\hat V\) matrix elements above into the State expansion and first-order mixing formula. For \(\vert\pm1\rangle\) (in the \(E_1\) manifold), the only state outside the manifold is \(\vert 0\rangle\) with \(E_1-E_0=1\):

\[ \vert\pm1\rangle' =\vert\pm1\rangle+\lambda\,\vert 0\rangle\,\frac{V_{0,\pm1}}{E_1-E_0}+O(\lambda^2) =\vert\pm1\rangle+\frac{\lambda}{\sqrt2}\vert 0\rangle+O(\lambda^2). \]

For \(\vert 0\rangle\) (the \(E_0\) singlet), the sum over \(m\neq 0\) runs over \(\beta\in\{+1,-1\}\) with \(E_0-E_1=-1\):

\[ \vert 0\rangle' =\vert 0\rangle+\lambda\,\vert+1\rangle\,\frac{V_{+1,0}}{E_0-E_1}+\lambda\,\vert-1\rangle\,\frac{V_{-1,0}}{E_0-E_1}+O(\lambda^2) =\vert 0\rangle-\frac{\lambda}{\sqrt2}(\vert+1\rangle+\vert-1\rangle)+O(\lambda^2). \]

The second round mixes the two corrected basis vectors \(\vert+1\rangle'\) and \(\vert-1\rangle'\) via standard non-degenerate PT inside the lifted block. Reading the matrix elements of \(\hat H_1^{\text{eff}}\) off Step 1, \(V'_{\mp1,\pm1}=\lambda^2/2\) and \(E'_{\pm1}-E'_{\mp1}=\pm 2\lambda\), so

\[ \vert\pm1\rangle'' =\vert\pm1\rangle'+\vert\mp1\rangle'\,\frac{V'_{\mp1,\pm1}}{E'_{\pm1}-E'_{\mp1}}+O(\lambda^2) =\vert\pm1\rangle'\pm\frac{\lambda}{4}\vert\mp1\rangle'+O(\lambda^2). \]

Substituting the first-round results \(\vert\pm1\rangle'=\vert\pm1\rangle+(\lambda/\sqrt2)\vert 0\rangle+O(\lambda^2)\) and dropping terms beyond \(O(\lambda)\),

\[ \vert+1\rangle'' =\vert+1\rangle+\frac{\lambda}{\sqrt2}\vert 0\rangle+\frac{\lambda}{4}\vert-1\rangle+O(\lambda^2), \]
\[ \vert-1\rangle'' =\vert-1\rangle+\frac{\lambda}{\sqrt2}\vert 0\rangle-\frac{\lambda}{4}\vert+1\rangle+O(\lambda^2). \]

The \(\vert 0\rangle\) singlet has no second-round mixing partner, so \(\vert 0\rangle'\) above is the final state through \(O(\lambda)\).

Interpretation. This example shows why degenerate perturbation theory is naturally hierarchical: first build and diagonalize effective blocks, then iterate with non-degenerate perturbation inside the lifted blocks. The two-round structure is required even when the second-round energy shift is negligible, because the second-round state mixing is the same order as the first-round mixing.

Discussion: residual degeneracy at second order

If diagonalizing \(V_{n\alpha,n\beta}\) still leaves repeated eigenvalues, first order is not enough. Then inspect second-order block terms (and symmetry constraints) to decide whether residual degeneracy is lifted.

Summary#

  • Why non-degenerate formulas fail: Inside a degenerate manifold the denominator \(E_n-E_m=0\) blows up, so perturbation theory must be reorganized as a block algorithm — partition the Hilbert space by manifold, build the effective Hamiltonian inside each block, then diagonalize.

  • Degenerate Hellmann-Feynman identities drive every formula: the in-manifold identity \(\partial_\lambda E_{n,\alpha\beta}=V_{n\alpha,n\beta}\) controls splitting, and the cross-manifold identity \(\langle m,\alpha\vert\partial_\lambda n,\beta\rangle=V_{m\alpha,n\beta}/(E_n-E_m)\) for \(m\neq n\) controls mixing.

  • Good zeroth-order basis = eigenvectors of \(V_{n\alpha,n\beta}\): diagonalizing the perturbation inside each manifold first yields the first-order energy splittings and selects the unique basis to which perturbation theory applies.

  • Second-order shifts come from virtual transitions to other manifolds and back, weighted by \(1/(E_n-E_m)\); the same physics is written in component form or in projector form.

  • First-order state correction is purely cross-manifold: the second Hellmann-Feynman identity gives the inter-manifold mixing, while intra-manifold mixing is fixed by the good-basis choice, not by a formula.

  • Hierarchical iteration: if the first-order matrix \(V_{n\alpha,n\beta}\) still has repeated eigenvalues, residual degeneracy must be lifted at the next order via second-order block terms (or by symmetry constraints).

See Also

Homework#

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?

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.

(c) Explain why this removes the divergence problem before applying higher-order corrections.

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.

(b) Set \(\mu=0\). Build the second-order effective Hamiltonian \(\hat H^{(2)}_{\mathrm{eff}}\) in \(\{\vert 1\rangle,\vert 2\rangle\}\) from virtual transitions through \(\vert 3\rangle\). Diagonalize and identify the bright state with shift \(-2\lambda^2/\Delta\) and the dark state with zero shift.

(c) Still with \(\mu=0\), show that the dark state is an exact zero-energy eigenstate of \(\hat H_0+\hat V\) (not only at second order), and explain in one sentence why.

(d) Restore \(\mu\ne 0\). Show that for the symmetric coupling here (equal \(\vert 3\rangle\langle 1\vert\) and \(\vert 3\rangle\langle 2\vert\) matrix elements), \(\hat P_d\hat V\hat P_d=\mu\hat X\) and the second-order rank-1 matrix \(\propto\begin{pmatrix}1&1\\1&1\end{pmatrix}\) commute, so the two splittings share an eigenbasis and combine independently. Write the two ground-manifold energies to order \(\mu+\lambda^2\).

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.

(c) State the symmetry reason in one sentence.

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,

(c) explain what physical information must then be checked at second order (or via symmetry).

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.

(c) Two bands with gap \(\gg V\) and \(V\) mixing across the gap.

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