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For a perturbed Hamiltonian

$$ H = H^{(0)} + H' $$

the perturbation theory approach

$$ \Psi = \Psi^{(0)} + \lambda \Psi^{(1)} + ... \\ E = E^{(0)} + \lambda E^{(1)} + ... $$

leads to the equations

$$ \begin{align} H^{(0)}\Psi^{(0)}& =E^{(0)}\Psi^{(0)} \tag 1 \\ (H^{(0)}-E^{(0)})\Psi^{(1)} & = (E^{(1)}-H')\Psi^{(0)} \tag 2 \end{align} $$

For a perturbed harmonic oscillator

$$ H = \frac{p^2}{2m}+\frac 12 m\omega^2 q^2 + \lambda q^4 $$

we get

$$ \begin{align} H^{(0)} & = \frac{p^2}{2m}+\frac 12 m\omega^2 q^2 \\ E_n^{(0)} & = \hbar\omega(n+\frac 12) \\ \Psi^{(0)} & = |n\rangle \end{align} $$

If we are only interested in the energy shift $\Delta E_n = E_n^{(1)}$ for the first order of $\lambda$, we get from $(2)$:

$$ \tag{3}(H^{(0)}-E_n^{(0)})\Psi^{(1)} = (\Delta E_n-q^4)|n\rangle $$

On the bottom of page 4 of this script, exactly the same is done, but somehow they get

$$ \tag{4}\Delta E_n = \lambda\langle n|q^4|n\rangle + O(\lambda^2) $$

How can I get $(4)$ from $(3)$? Or, if the script does it in another way, how do they get $(4)$?

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    $\begingroup$ Multiply both sides of equation (3) by $\left<n \right|$. $\endgroup$ – Praan Nov 11 '15 at 16:17
  • $\begingroup$ Thanks! Then $H^{(0)}$ can act on the $\langle n|$ on its left because it is self-adjoint, so these terms cancel each other and we have $(4)$. (actually I tried that but I didn't use the self-adjointness of $H$). $\endgroup$ – Bass Nov 11 '15 at 16:32

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