2.6: Example
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Consider a simple harmonic oscillator in its ground state, to which we apply a perturbation ˆV=λx2. We know the unperturbed wavefunction |n0⟩=[mω0/πℏ]14 exp{−mω0x2/2ℏ}, so we can evaluate the first order shift in energy according to the perturbation theory:
ΔE0=⟨n0|λx2|n0⟩=λ√mω0/πℏ∫x2 exp{−mω0x2/ℏ}dx=λ2ℏmω0
In this case we know the exact shift, since the perturbation is simply an additional harmonic potential, giving a total k=mω20+2λ and an exact ground state energy of 12ℏ√ω20+2λ/m. It is easy to verify that to first order in λ these expressions are identical.
To determine the amount of mixing of states, we need to evaluate matrix elements like ⟨n0|λx2|ni⟩. We won’t evaluate these here, but we will note that for odd i the integral is zero - the symmetric perturbation only mixes in symmetric excited states.