22 Jul 2025
Quantum Perturbation Theory and Transitions
First- and second-order stationary perturbation theory, degenerate levels, the Stark effect, time-dependent amplitudes, transition probabilities, and Fermi's golden rule.
Let
\[H=H_0+\lambda V, \qquad H_0\lvert n^{(0)}\rangle=E_n^{(0)}\lvert n^{(0)}\rangle,\]and expand
\[E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2E_n^{(2)}+\cdots,\] \[\lvert n\rangle=\lvert n^{(0)}\rangle+\lambda\lvert n^{(1)}\rangle +\lambda^2\lvert n^{(2)}\rangle+\cdots.\]Choose intermediate normalization, $\langle n^{(0)}\mid n\rangle=1$, so $\langle n^{(0)}\mid n^{(j)}\rangle=0$ for $j\ge1$.
Nondegenerate time-independent theory
Insert the expansions into $H\lvert n\rangle=E_n\lvert n\rangle$ and compare first powers of $\lambda$:
\[(H_0-E_n^{(0)})\lvert n^{(1)}\rangle =(E_n^{(1)}-V)\lvert n^{(0)}\rangle.\]Projection with $\langle n^{(0)}\rvert$ makes the left side zero and gives
\[\boxed{E_n^{(1)}=V_{nn},} \qquad V_{mn}=\langle m^{(0)}\rvert V\lvert n^{(0)}\rangle.\]For $m\ne n$,
\[(E_m^{(0)}-E_n^{(0)}) \langle m^{(0)}\mid n^{(1)}\rangle=-V_{mn},\]so
\[\boxed{ \lvert n^{(1)}\rangle =\sum_{m\ne n} \frac{V_{mn}}{E_n^{(0)}-E_m^{(0)}}\lvert m^{(0)}\rangle. }\]At second order,
\[(H_0-E_n^{(0)})\lvert n^{(2)}\rangle =(E_n^{(1)}-V)\lvert n^{(1)}\rangle +E_n^{(2)}\lvert n^{(0)}\rangle.\]Projection with $\langle n^{(0)}\rvert$ and intermediate normalization give
\[E_n^{(2)}=\langle n^{(0)}\rvert V\lvert n^{(1)}\rangle.\]Substituting the first-order ket yields
\[\boxed{ E_n^{(2)} =\sum_{m\ne n} \frac{\lvert V_{mn}\rvert^2}{E_n^{(0)}-E_m^{(0)}}. }\]These formulas require the perturbation matrix elements to be small compared with the relevant nonzero energy separations.
Degenerate case
Suppose $g$ orthonormal states $\lvert a\rangle$ share the unperturbed energy $E_d^{(0)}$. The zeroth-order ket inside this subspace is $\lvert\psi^{(0)}\rangle=\sum_{b=1}^gc_b\lvert b\rangle$. Projecting the first-order equation onto $\langle a\rvert$ gives
\[\sum_{b=1}^g(V_{ab}-E^{(1)}\delta_{ab})c_b=0.\]A nonzero coefficient vector exists only if
\[\boxed{\det(V_{ab}-E^{(1)}\delta_{ab})=0.}\]Thus the correct zeroth-order combinations are the eigenvectors of $V$ restricted to the degenerate subspace. Coupling to states outside that subspace then supplies the second-order correction through the nonzero energy denominators.
Linear Stark effect in hydrogen
Take a uniform electric field $\mathcal E\hat{\mathbf z}$. With scalar potential $\Phi=-\mathcal Ez$, the electron perturbation energy is
\[V=(-e)\Phi=e\mathcal Ez.\]Since
\[\cos\theta\,Y_l^m =A_{lm}Y_{l+1}^m+B_{lm}Y_{l-1}^m,\]the angular integral imposes the electric-dipole selection rules
\[\boxed{\Delta l=\pm1,\qquad \Delta m=0}\]for a field along $z$. In the $n=2$ manifold, only $\lvert200\rangle$ and $\lvert210\rangle$ mix. With $x=r/a_0$,
\[\psi_{200}=\frac{(2-x)e^{-x/2}}{4\sqrt{2\pi}\,a_0^{3/2}}, \qquad \psi_{210}=\frac{x e^{-x/2}\cos\theta} {4\sqrt{2\pi}\,a_0^{3/2}}.\]Parity makes the two diagonal matrix elements zero. The off-diagonal element is
\[\begin{aligned} \langle200\rvert z\lvert210\rangle &=\frac1{32\pi a_0^3} \int_0^\infty r^2dr\,(2-x)xe^{-x}r \int\cos^2\theta\,d\Omega\\ &=\frac1{32\pi a_0^3} \left[a_0^4\int_0^\infty x^4(2-x)e^{-x}dx\right] \left(\frac{4\pi}{3}\right)\\ &=\frac{a_0}{24}[2(4!)-5!] =-3a_0. \end{aligned}\]Therefore, in the ordered basis $(\lvert200\rangle,\lvert210\rangle)$,
\[V=e\mathcal E \begin{pmatrix}0&-3a_0\\-3a_0&0\end{pmatrix}.\]Its normalized eigenvectors and shifts are
\[\lvert\pm\rangle=\frac{\lvert200\rangle\mp\lvert210\rangle}{\sqrt2}, \qquad \boxed{E_\pm^{(1)}=\pm3ea_0\mathcal E.}\]The $\lvert21,\pm1\rangle$ states have zero first-order shift because $\Delta m=0$ forbids their coupling to the $2s$ state.
Time-dependent perturbation theory
Let $H(t)=H_0+V(t)$ and expand an exact state as
\[\lvert\Psi(t)\rangle=\sum_n c_n(t)e^{-iE_nt/\hbar}\lvert n\rangle.\]Insert this in $i\hbar\partial_t\lvert\Psi\rangle=H\lvert\Psi\rangle$. The $H_0$ terms cancel, leaving
\[i\hbar\dot c_f(t) =\sum_nV_{fn}(t)e^{i\omega_{fn}t}c_n(t), \qquad \omega_{fn}=\frac{E_f-E_n}{\hbar}.\]If the system begins in $\lvert i\rangle$, first order replaces $c_n(t)$ on the right by $\delta_{ni}$:
\[\boxed{ c_f^{(1)}(t) =-\frac{i}{\hbar}\int_0^t V_{fi}(t^{\prime})e^{i\omega_{fi}t^{\prime}}dt^{\prime}, \qquad f\ne i. }\]The transition probability to this order is
\[\boxed{P_{i\to f}(t)=\lvert c_f^{(1)}(t)\rvert^2.}\]Constant perturbation
For $V_{fi}(t)=V_{fi}$ during $0<t<T$,
\[\begin{aligned} c_f^{(1)}(T) &=-\frac{iV_{fi}}{\hbar} \frac{e^{i\omega_{fi}T}-1}{i\omega_{fi}}\\ &=-\frac{2iV_{fi}}{\hbar} e^{i\omega_{fi}T/2} \frac{\sin(\omega_{fi}T/2)}{\omega_{fi}}. \end{aligned}\]Hence
\[\boxed{ P_{i\to f}(T) =\frac{4\lvert V_{fi}\rvert^2}{\hbar^2} \frac{\sin^2(\omega_{fi}T/2)}{\omega_{fi}^2}. }\]At $\omega_{fi}=0$, the limiting probability is $\lvert V_{fi}\rvert^2T^2/\hbar^2$.
Harmonic perturbation and Fermi’s golden rule
Write a real harmonic perturbation as
\[V(t)=We^{-i\omega t}+W^\dagger e^{i\omega t}.\]The absorption part of the first-order amplitude is
\[c_{f,\rm abs}^{(1)}(T) =-\frac{iW_{fi}}{\hbar} e^{i(\omega_{fi}-\omega)T/2} \frac{2\sin[(\omega_{fi}-\omega)T/2]} {\omega_{fi}-\omega}.\]It is sharply peaked at the energy-selection condition
\[\boxed{E_f-E_i=\hbar\omega.}\]The conjugate term produces the stimulated-emission amplitude
\[c_{f,\rm em}^{(1)}(T) =-\frac{i(W^\dagger)_{fi}}{\hbar} e^{i(\omega_{fi}+\omega)T/2} \frac{2\sin[(\omega_{fi}+\omega)T/2]} {\omega_{fi}+\omega},\]which is resonant when
\[\boxed{E_f-E_i=-\hbar\omega.}\]For final states forming a continuum with density $\rho(E_f)$, use
\[\lim_{T\to\infty} \frac1T\frac{4\sin^2[(E_f-E_i-\hbar\omega)T/(2\hbar)]} {(E_f-E_i-\hbar\omega)^2} =\frac{2\pi}{\hbar}\delta(E_f-E_i-\hbar\omega).\]Integration over final energies gives Fermi’s golden rule,
\[\boxed{ \Gamma_{i\to f} =\frac{2\pi}{\hbar}\lvert W_{fi}\rvert^2\rho(E_f) \bigg\rvert_{E_f=E_i+\hbar\omega}. }\]If the perturbation is written $V_0\cos\omega t$, then $W=V_0/2$ for the absorption component. Any additional selection rules follow from a vanishing matrix element. For an electric dipole, $\Delta l=\pm1$. Polarization along $z$ selects $\Delta m=0$; transverse linear polarization is a superposition of $\Delta m=+1$ and $-1$, while a fixed circular helicity selects one of these two signs. An arbitrary polarization can contain all three spherical components $\Delta m=0,\pm1$.
Solved Problems
1. Corrections to a three-level nondegenerate state
Let the unperturbed energies be $E_0^{(0)}=0$, $E_1^{(0)}=\Delta$, and $E_2^{(0)}=3\Delta$, with $\Delta>0$. Suppose the real perturbation elements connected to $\lvert0\rangle$ are
\[V_{00}=\alpha, \qquad V_{10}=g, \qquad V_{20}=2g.\]The first-order shift is immediately
\[\boxed{E_0^{(1)}=\alpha.}\]Both excited levels lie above the state of interest, so both denominators in the second-order shift are negative:
\[\begin{aligned} E_0^{(2)} &=\frac{g^2}{0-\Delta} +\frac{(2g)^2}{0-3\Delta}\\ &=-\frac{g^2}{\Delta}-\frac{4g^2}{3\Delta} =\boxed{-\frac{7g^2}{3\Delta}}. \end{aligned}\]The first-order ket correction is
\[\boxed{ \lvert0^{(1)}\rangle =-\frac g\Delta\lvert1\rangle -\frac{2g}{3\Delta}\lvert2\rangle.}\]$\alpha$, $g$, and $\Delta$ have units of energy, so the ket coefficients are dimensionless and $g^2/\Delta$ has units of energy. The negative second-order sign reflects level repulsion from states above $E_0^{(0)}$; the expansion requires $\lvert g/\Delta\rvert\ll1$.
2. A twofold degenerate perturbation
Within a degenerate subspace, let
\[V_d=\begin{pmatrix}3&4\\4&-3\end{pmatrix}\mathrm{meV}.\]The first-order shifts satisfy
\[\det(V_d-\varepsilon I) =(3-\varepsilon)(-3-\varepsilon)-16 =\varepsilon^2-25=0.\]Thus $\varepsilon_+=5\,\mathrm{meV}$ and $\varepsilon_-=-5\,\mathrm{meV}$. For the positive shift, $-2c_1+4c_2=0$, so $c_1=2c_2$; for the negative shift, $8c_1+4c_2=0$, so $c_2=-2c_1$. After normalization,
\[\boxed{ \lvert+\rangle=\frac{2\lvert1\rangle+\lvert2\rangle}{\sqrt5}, \qquad \lvert-\rangle=\frac{\lvert1\rangle-2\lvert2\rangle}{\sqrt5}.}\]The states are orthonormal. The two shifts sum to zero, equal to $\operatorname{tr}V_d$, and their separation is $10\,\mathrm{meV}$.
3. Transition probability under a constant pulse
A constant matrix element $\lvert V_{fi}\rvert=0.0200\,\mathrm{eV}$ acts between levels separated by $\Delta E=E_f-E_i=0.100\,\mathrm{eV}$. Choose
\[T=\frac{\pi\hbar}{\Delta E}=20.68\,\mathrm{fs}.\]Since $\omega_{fi}=\Delta E/\hbar$, the phase in the probability is
\[\frac{\omega_{fi}T}{2}=\frac{\Delta E}{2\hbar} \frac{\pi\hbar}{\Delta E}=\frac\pi2.\]Therefore
\[\begin{aligned} P_{i\to f}(T) &=\frac{4\lvert V_{fi}\rvert^2}{\Delta E^2} \sin^2\left(\frac{\Delta E T}{2\hbar}\right)\\ &=4\left(\frac{0.0200}{0.100}\right)^2 =\boxed{0.160}. \end{aligned}\]The ratio of energies and the sine are dimensionless. The probability is below unity and tends to zero with $V_{fi}$, as first-order perturbation theory requires.
Descriptive Questions
- Derive the first- and second-order energy corrections for a nondegenerate level, stating the normalization convention and the small-denominator limitation.
- Explain why the perturbation must be diagonalized within a degenerate subspace before ordinary nondegenerate formulas can be used outside that subspace.
- Derive the linear Stark matrix for the hydrogen $n=2$ manifold, including the sign of the electron’s potential energy and all angular selection rules.
- Starting from the first-order time-dependent amplitude, obtain the constant- and harmonic-perturbation probabilities and the continuum limit leading to Fermi’s golden rule.
Numerical Problems
- Calculate the first-order $n=2$ hydrogen Stark shifts and their separation for $\mathcal E=5.00\times10^4\,\mathrm{V\,m^{-1}}$, using $a_0=5.29177\times10^{-11}\,\mathrm m$.
- For a $z$-polarized electric-dipole perturbation, identify which of the final states $\lvert3,2,m^{\prime}\rangle$ can couple to an initial $\lvert2,1,0\rangle$ state at first order.
- Find the resonant angular frequency and ordinary frequency for absorption across a $2.00\,\mathrm{eV}$ gap. Use $\hbar=6.58212\times10^{-16}\,\mathrm{eV\,s}$.
- A continuum has $\rho(E_f)=5.00\,\mathrm{eV^{-1}}$ and $\lvert W_{fi}\rvert=2.00\,\mathrm{meV}$. Calculate the golden-rule transition rate and lifetime.
Final answers: 1. shifts $\pm7.93766\,\mu\mathrm{eV}$ and separation $15.8753\,\mu\mathrm{eV}$; 2. only $m^{\prime}=0$; 3. $\omega=3.03853\times10^{15}\,\mathrm{s^{-1}}$ and $f=4.83598\times10^{14}\,\mathrm{Hz}$; 4. $\Gamma=1.90917\times10^{11}\,\mathrm{s^{-1}}$ and $\tau=5.23788\,\mathrm{ps}$.
The core derivations and all problem answers are checked in the original Maxima worksheet and the problems worksheet; every printed residual is zero.
References
- Perturbation theory in quantum mechanics.
- J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Chapter 5, “Approximation Methods.”
- D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Chapters 6 and 9.
Discussion