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Griffiths (Introduction to Quantum Mechanics, 3rd edition, §7.1.2) presents the following derivation for first-order energy perturbations in nondegenerate time-independent perturbation theory.

Suppose we have a time-independent Hamiltonian $H=H^0+\lambda H'$, with known, unperturbed, orthonormal, and possibly degenerate energy eigenstates $H^0\psi_n^0=E_n^0\psi_n^0$. Expanding the time-independent Schrödinger equation for $H$ as a power series in $\lambda$, we have: $$\left(H^0+\lambda H'\right)\sum_{k=0}^\infty\lambda^k\psi_n^k=\left(\sum_{k=0}^\infty\lambda^kE_n^k\right)\left[\sum_{k=0}^\infty\lambda^k\psi_n^k\right]$$ $$\implies H^0\psi_n^0+\lambda\left(H^0\psi_n^1+H'\psi_n^0\right)+\mathcal{O}\left(\lambda^2\right)=E_n^0\psi_n^0+\lambda\left(E_n^0\psi_n^1+E_n^1\psi_n^0\right)+\mathcal{O}\left(\lambda^2\right)$$ To first order in $\lambda$, we then have: $$H^0\psi_n^1+H'\psi_n^0=E_n^0\psi_n^1+E_n^1\psi_n^0$$ $$\implies\langle{\psi_n^0}|{H^0\psi_n^1}\rangle+\langle{\psi_n^0}|{H'\psi_n^0}\rangle=\langle{\psi_n^0}|{E_n^0\psi_n^1}\rangle+\langle{\psi_n^0}|{E_n^1\psi_n^0}\rangle$$ $$\implies\boxed{E_n^1=\langle{\psi_n^0}|H'|{\psi_n^0}\rangle}$$ However, we know that this correction breaks down in degenerate perturbation theory; the actual first-order correction to the energy requires that we use "good" states instead of arbitrary unperturbed states.

At what point has the derivation above failed? What assumptions have we made that we were not allowed to make?

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2 Answers 2

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Consider the equation we are trying to solve, $H\psi_n=E_n\psi_n$. In expanding $\psi_n$ as a power series, we write: $$\psi_n=\sum_{k=0}^\infty\lambda^k\psi_n^k$$ However, in doing so, we have already chosen a basis $\{\psi_n^0\}$. Moreover, we assumed that these states have continuous perturbative corrections. Such a series only exists for "good" states of the perturbative Hamiltonian $H'$. We may not make this series expansion with full generality; hence, the nondegenerate result fails in the degenerate case.

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Perturbation theory is an expansion in inverse powers of the difference $E^{(0)}_i-E^{(0)}_j$. If the two are equal (i.e. if the levels $i$ and $j$ are degenerate), you are dividing by $0$, which is not allowed. Alternatively, the quantity $1/(E^{(0)}_i-E^{(0)}_j)$ is not small hence the perturbative series does not converge.

Alternatively, in the subspace spanned by $\vert i\rangle$ and $\vert j\rangle$, $H_0$ is proportional to the unit matrix: $$ H_0=E_i^{(0)}\left(\begin{array}{cc} 1&0\\ 0&1\end{array}\right) $$ and any similarity transformation will leave this invariant so diagonalizing $H_0+\lambda H_1$ in this subspace is the same as exactly diagonalizing $H_1$, so the diagonalization is exact and does not depend on $H_0$, signalling that a perturbative treatment based on the eigenstates of $H_0$ is pointless.

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