Perturbation theory: approximate methods that split a problem into a solvable part and a small correction.
\(H=H_0+H_P\), \(H_P=\lambda W\), \(\lambda\ll1\); \(H_0\left|\phi_n\right\rangle=E_n^0\left|\phi_n\right\rangle\) with non-degenerate \(E_n^0\).
Expand in powers of \(\lambda\): \[E_n=E_n^0+\lambda E_n^1+\lambda^2E_n^2+\cdots,\qquad \left|\psi_n\right\rangle=\left|\phi_n\right\rangle+\lambda\left|\psi_n^1\right\rangle+\lambda^2\left|\psi_n^2\right\rangle+\cdots\]
Insert in \(H\left|\psi_n\right\rangle=E_n\left|\psi_n\right\rangle\) and equate powers of \(\lambda\):
\(\lambda^0\): \(H_0\left|\phi_n\right\rangle=E_n^0\left|\phi_n\right\rangle\)
\(\lambda^1\): \(H_0\left|\psi_n^1\right\rangle+W\left|\phi_n\right\rangle=E_n^0\left|\psi_n^1\right\rangle+E_n^1\left|\phi_n\right\rangle\)
\(\lambda^2\): \(H_0\left|\psi_n^2\right\rangle+W\left|\psi_n^1\right\rangle=E_n^0\left|\psi_n^2\right\rangle+E_n^1\left|\psi_n^1\right\rangle+E_n^2\left|\phi_n\right\rangle\)
Normalization convention \(\left\langle \phi_n\middle|\psi_n\right\rangle =1\Rightarrow\left\langle \phi_n\middle|\psi_n^k\right\rangle =0\) for \(k\ge1\).
Results
First-order energy: \(E_n^1=\left\langle \phi_n\middle|W\middle|\phi_n\right\rangle\) — the expectation value of the perturbation.
First-order state: \(\left|\psi_n^1\right\rangle=\displaystyle\sum_{m\ne n}\frac{\left\langle \phi_m\middle|W\middle|\phi_n\right\rangle }{E_n^0-E_m^0}\left|\phi_m\right\rangle\).
Second-order energy: \(E_n^2=\left\langle \phi_n\middle|W\middle|\psi_n^1\right\rangle =\displaystyle\sum_{m\ne n}\frac{|\left\langle \phi_m\middle|W\middle|\phi_n\right\rangle |^2}{E_n^0-E_m^0}\).
For the ground state every denominator is negative \(\Rightarrow E_0^2\le0\) always.
Example 12.1 Two-level system
\(H=\begin{pmatrix}E_1&0\\0&E_2\end{pmatrix}+\lambda\begin{pmatrix}0&E_0\\E_0&0\end{pmatrix}\), \(E_1<E_2\). Find (i) \(E_n^0,\left|\phi_n\right\rangle\); (ii) \(E_n^1\); (iii) \(\left|\psi_n^1\right\rangle\); (iv) \(E_n^2\).
Solution
(i) \(E_1^0=E_1\), \(\left|\phi_1\right\rangle=\binom10\); \(E_2^0=E_2\), \(\left|\phi_2\right\rangle=\binom01\).
(ii) \(W\) has zero diagonal \(\Rightarrow E_1^1=E_2^1=0\).
(iii) \(\left|\psi_1^1\right\rangle=\dfrac{E_0}{E_1-E_2}\binom01\), \(\left|\psi_2^1\right\rangle=\dfrac{E_0}{E_2-E_1}\binom10\).
(iv) \(E_1^2=\dfrac{E_0^2}{E_1-E_2}<0\), \(E_2^2=\dfrac{E_0^2}{E_2-E_1}>0\) — the levels repel.
Example 12.2 Perturbations in an infinite well
Box \([0,a]\), \(\phi_n=\sqrt{\frac2a}\sin\frac{n\pi x}{a}\). Find \(E_n^1\) for: (i) \(\lambda V_0\) throughout; (ii) \(\frac{\lambda V_0}{2}\) on \([0,\frac a2]\), \(\lambda V_0\) on \([\frac a2,a]\); (iii) \(\frac{\lambda V_0x}{a}\); (iv) \(\lambda V_0a\,\delta(x-\frac a2)\); (v) \(\lambda V_0a\,\delta(x-\frac a4)\).
Solution
(i) \(E_n^1=\lambda V_0\) (a constant shift; wavefunction unchanged).
(ii) Each half carries probability \(\frac12\): \(E_n^1=\frac12\cdot\frac{\lambda V_0}{2}+\frac12\lambda V_0=\frac{3\lambda V_0}{4}\).
(iii) \(E_n^1=\frac{\lambda V_0}{a}\left\langle x\right\rangle=\frac{\lambda V_0}{2}\).
(iv) \(E_n^1=\lambda V_0a\cdot\frac2a\sin^2\frac{n\pi}{2}=2\lambda V_0\) (\(n\) odd), \(0\) (\(n\) even).
(v) \(E_n^1=2\lambda V_0\sin^2\frac{n\pi}{4}\): \(\lambda V_0\) (\(n\) odd), \(2\lambda V_0\) (\(n=2,6,10,\ldots\)), \(0\) (\(n=4,8,\ldots\)).
Example 12.3 Oscillator with a linear perturbation \(\lambda x\)
Find \(E_n^1\), \(\left|\psi_n^1\right\rangle\) and \(E_n^2\).
Solution
\(x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)\) \(\Rightarrow E_n^1=\lambda\left\langle n\middle|x\middle|n\right\rangle =0\).
\(\left|\psi_n^1\right\rangle=\lambda\sqrt{\dfrac{\hbar}{2m\omega}}\,\dfrac{1}{\hbar\omega}\Big[\sqrt n\left|n-1\right\rangle-\sqrt{n+1}\left|n+1\right\rangle\Big]\).
\(E_n^2=\lambda^2\dfrac{\hbar}{2m\omega}\Big[\dfrac{n}{\hbar\omega}-\dfrac{n+1}{\hbar\omega}\Big]=-\dfrac{\lambda^2}{2m\omega^2}\) — exact (complete the square; cf. Volume I).
Example 12.4 Oscillator with \(\lambda x^2\)
Find \(E_n^1\) and \(\left|\psi_n^1\right\rangle\).
Solution
\(x^2=\frac{\hbar}{2m\omega}\big(a^2+a^{\dagger2}+2N+1\big)\) \(\Rightarrow E_n^1=\lambda\dfrac{\hbar}{2m\omega}(2n+1)\).
\(\left|\psi_n^1\right\rangle=\dfrac{\lambda}{4m\omega^2}\Big[\sqrt{n(n-1)}\left|n-2\right\rangle-\sqrt{(n+1)(n+2)}\left|n+2\right\rangle\Big]\).
Example 12.5 More first-order shifts
Find the first-order ground-state shift for: (a) \(bx^2\) in the delta well \(V=-\delta(x)\), \(\psi=\sqrt\alpha e^{-\alpha|x|}\); (b) \(\varepsilon\cos\frac{2r}{a_0}\) on hydrogen; (c) \(g\cos kx\) on an oscillator; (d) \(\lambda x\) for \(x>0\) only on an oscillator; (e) two particles in a box \([0,a]\) with \(\lambda\delta(x_1-x_2)\); (f) \(bx^4\) on an oscillator (state \(n\)).
Solution
(a) \(E^1=2b\alpha\displaystyle\int_0^\infty x^2e^{-2\alpha x}dx=\frac{b}{2\alpha^2}\).
(b) \(E^1=\dfrac{4\varepsilon}{a_0^3}\mathrm{Re}\displaystyle\int_0^\infty r^2e^{-2r(1-i)/a_0}dr=\frac{\varepsilon}{2}\,\mathrm{Re}\frac{1}{(1-i)^3}=-\frac{\varepsilon}{4}\).
(c) \(E^1=g\,\mathrm{Re}\left\langle 0\middle|e^{ikx}\middle|0\right\rangle =g\exp\Big(-\dfrac{k^2\hbar}{4m\omega}\Big)\).
(d) \(E^1=\lambda\sqrt{\dfrac{m\omega}{\pi\hbar}}\displaystyle\int_0^\infty xe^{-m\omega x^2/\hbar}dx=\frac{\lambda}{2}\sqrt{\frac{\hbar}{\pi m\omega}}\).
(e) Both in \(\phi_1\): \(E^1=\lambda\dfrac{4}{a^2}\displaystyle\int_0^a\sin^4\frac{\pi x}{a}dx=\frac{3\lambda}{2a}\).
(f) \(\left\langle n\middle|x^4\middle|n\right\rangle =\Big(\dfrac{\hbar}{2m\omega}\Big)^2(6n^2+6n+3)\) \(\Rightarrow E_n^1=\dfrac{3b\hbar^2}{4m^2\omega^2}(2n^2+2n+1)\).
Example 12.6 Second order for \(\frac12bx^2\)
\(H=\frac{p^2}{2m}+\frac12m\omega^2x^2+\frac12bx^2\). Find \(E_n^2\).
Solution
Only \(\left|n\pm 2\right\rangle\) contribute: \(E_n^2=\dfrac{b^2}{4}\Big(\dfrac{\hbar}{2m\omega}\Big)^2\dfrac{n(n-1)-(n+1)(n+2)}{2\hbar\omega}=-\dfrac18\Big(n+\dfrac12\Big)\hbar\omega\dfrac{b^2}{k^2}\), \(k=m\omega^2\).
Check: exact \(\omega'=\omega\sqrt{1+b/k}\approx\omega\big(1+\frac{b}{2k}-\frac{b^2}{8k^2}\big)\).
