Time-Independent Perturbation Theory

The perturbation Hamiltonian is

H' = \begin{pmatrix} H_{11}' & H_{12}' & \cdots \\ H_{21}' & H_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}

and we find $H_{ij}'$ by

H_{ij}' = \bra{i^{(0)}} H' \ket{j^{(0)}}

Non-degenerate Perturbation Theory

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First Order Correction

The first order energy correction is

E_n^{(1)} = H_{nn}' = \bra{n^{(0)}} H' \ket{n^{(0)}}

And in wave function notation

E_n^{(1)} = H_{nn}' = \int \psi_n^{(0)*}(\textbf{r}) H' \psi_n^{(0)}(\textbf{r}) \, dV

The first order eigenstate correction is

\ket{n^{(1)}} = \sum_{m \neq n} \frac{\bra{m^{(0)}}H'\ket{n^{(0)}}}{\big(E_n^{(0)} - E_m^{(0)}\big)} \ket{m^{(0)}}

With these corrections, we have the new energies and states

E_n = E_n^{(0)} + \bra{n^{(0)}} H' \ket{n^{(0)}}

\ket{n} = \ket{n^{(0)}} + \sum_{m \neq n} \frac{\bra{m^{(0)}}H'\ket{n^{(0)}}}{\big(E_n^{(0)} - E_m^{(0)}\big)} \ket{m^{(0)}}

Second Order Correction

The second order correction to the energy is

E_n^{(2)} = \sum_{m \neq n} \frac{|\bra{n^{(0)}}H'\ket{m^{(0)}}|^2}{\big(E_n^{(0)} - E_m^{(0)} \big)}

There is a second order correction to the eigenstate, but it is complicated, so we will skip it here :).

Degenerate Perturbation Theory

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Diagonalize the Perturbation Hamiltonian in the Degenerate Subspace

Diagonalize the perturbation Hamiltonian in the degenerate subspace.

E_{2a} = E_2^{(0)} + E_{2a}^{(1)}

E_{2b} = E_2^{(0)} + E_{2b}^{(1)}

Perturbation of Atoms

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Fine Structure

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Hyperfine Splitting

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Lamb Shift

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Stark Effect

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Zeeman Effect

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1 of 7

An infinite square well (L=1 nm) has perturbation H'=V₀ for L/2 ≤ x ≤ L, zero elsewhere. V₀ = 0.1 eV. Watch the integral sweep and find E₁⁽¹⁾ in eV.

The right half (yellow) is the perturbation H' = 0.1 eV. Press Sweep to animate the integral E₁⁽¹⁾ = ⟨ψ₁|H'|ψ₁⟩ sweeping across the well. Then enter E₁⁽¹⁾ in eV.

E₁⁽¹⁾ = eV

2 of 7

For a QHO with perturbation H'= cx, the parity visualization shows why ⟨0|cx|0⟩ = 0. What is E₀⁽¹⁾ (the first-order energy correction)?

Step through the parity argument: ψ₀ is even, x is odd, so their product ψ₀·x is odd — its integral vanishes. Therefore E₀⁽¹⁾ = 0.

E₀⁽¹⁾ = eV

3 of 7

For the same linear QHO perturbation H'=cx, the second-order correction is E₀⁽²⁾ = -c²/(2mω²). For c=1.0×10⁻¹⁰ N, m=9.11×10⁻³¹ kg, ω=2×10¹⁴ rad/s, find |E₀⁽²⁾| in eV.

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