27 Jun 2025
Energy-Eigenfunction Expansion and Time Evolution
Let a time-independent Hamiltonian have a complete orthonormal set of discrete energy eigenfunctions,
\[\hat H\phi_n(\mathbf r)=E_n\phi_n(\mathbf r), \qquad \int\phi_m^{\ast}(\mathbf r)\phi_n(\mathbf r)d^3r=\delta_{mn}.\]Completeness means that an arbitrary square-integrable initial state can be expanded as
\[\Psi(\mathbf r,0)=\sum_n c_n\phi_n(\mathbf r).\]Multiply by $\phi_m^{\ast}$ and integrate:
\[\int\phi_m^{\ast}\Psi(\mathbf r,0)d^3r =\sum_n c_n\int\phi_m^{\ast}\phi_n d^3r =\sum_n c_n\delta_{mn}=c_m.\]Therefore
\[\boxed{c_n=\int\phi_n^{\ast}(\mathbf r)\Psi(\mathbf r,0)d^3r}.\]If the initial state is normalized, orthonormality gives
\[\begin{aligned} 1&=\int\lvert\Psi(\mathbf r,0)\rvert^2d^3r\\ &=\sum_{m,n}c_m^{\ast}c_n \int\phi_m^{\ast}\phi_n d^3r =\sum_n\lvert c_n\rvert^2. \end{aligned}\]Thus $\lvert c_n\rvert^2$ is the probability of obtaining $E_n$ in an energy measurement.
General time-dependent solution
Each eigenfunction supplies a stationary solution $\phi_n e^{-iE_nt/\hbar}$. Linearity then gives
\[\boxed{ \Psi(\mathbf r,t)= \sum_n c_n\phi_n(\mathbf r)e^{-iE_nt/\hbar} }.\]The sign in the phase is fixed directly by substitution:
\[i\hbar\frac{\partial}{\partial t} e^{-iE_nt/\hbar}=E_ne^{-iE_nt/\hbar}.\]Consequently,
\[i\hbar\partial_t\Psi =\sum_nE_nc_n\phi_ne^{-iE_nt/\hbar} =\sum_nc_n(\hat H\phi_n)e^{-iE_nt/\hbar} =\hat H\Psi.\]At $t=0$ the expression returns the prescribed initial state, so it is the required solution. Only the phases change; hence the energy probabilities $\lvert c_n\rvert^2$ remain constant. The expectation value is
\[\boxed{\langle H\rangle=\sum_n\lvert c_n\rvert^2E_n},\]and is time independent for a time-independent Hamiltonian.
Density of a superposition
Although a single energy eigenstate has a time-independent density, a superposition need not. For two components,
\[\Psi=c_1\phi_1e^{-iE_1t/\hbar} +c_2\phi_2e^{-iE_2t/\hbar},\]so
\[\begin{aligned} \lvert\Psi\rvert^2={}&\lvert c_1\phi_1\rvert^2+\lvert c_2\phi_2\rvert^2\\ &+2\operatorname{Re}\!\left[ c_1c_2^{\ast}\phi_1\phi_2^{\ast} e^{-i(E_1-E_2)t/\hbar} \right]. \end{aligned}\]The interference term oscillates at angular frequency
\[\omega_{12}=\frac{\lvert E_1-E_2\rvert}{\hbar}.\]If $E_1=E_2$, the relative phase is constant. If the Hamiltonian also has a continuous spectrum, the corresponding part of the expansion is an integral,
\[\Psi(\mathbf r,t) =\sum_n c_n\phi_n e^{-iE_nt/\hbar} +\int c(E)\phi_E(\mathbf r)e^{-iEt/\hbar}dE,\]with Dirac-delta rather than Kronecker-delta normalization. This is the same spectral principle for both bound and free components.
Solved Problems
1. Energy statistics and time evolution of a three-state superposition
A normalized state has coefficients
\[c_1=\frac1{\sqrt6},\qquad c_2=\frac{i}{\sqrt3},\qquad c_3=-\frac1{\sqrt2},\]for energies $E_1=1.00\ \mathrm{eV}$, $E_2=2.00\ \mathrm{eV}$, and $E_3=5.00\ \mathrm{eV}$. Find the energy probabilities, mean energy, and standard deviation.
Solution. Phases and signs do not enter an energy probability:
\[P_1=\frac16,\qquad P_2=\frac13,\qquad P_3=\frac12,\]and $P_1+P_2+P_3=1$. Therefore
\[\begin{aligned} \langle E\rangle &=\frac16(1)+\frac13(2)+\frac12(5)\\ &=\boxed{\frac{10}{3}\ \mathrm{eV}=3.333\ \mathrm{eV}}. \end{aligned}\]The second moment is
\[\langle E^2\rangle =\frac16(1)^2+\frac13(2)^2+\frac12(5)^2 =14.00\ \mathrm{eV^2},\]so
\[\Delta E =\sqrt{\langle E^2\rangle-\langle E\rangle^2} =\boxed{\frac{\sqrt{26}}3\ \mathrm{eV} =1.700\ \mathrm{eV}}.\]The evolved state is
\[\lvert\Psi(t)\rangle =\sum_{n=1}^3c_ne^{-iE_nt/\hbar}\lvert n\rangle.\]Each phase has unit modulus, so the three probabilities, $\langle E\rangle$, and $\Delta E$ are time independent. The variance has units $\mathrm{eV^2}$ and its non-negative square root has energy units. If only one coefficient remains nonzero, the limiting spread is zero.
2. Beat period of an off-diagonal observable
Let $\lvert\Psi(0)\rangle=(\lvert1\rangle+\lvert2\rangle)/\sqrt2$, with $E_2-E_1=1.50\ \mathrm{eV}$. An observable has $A_{11}=A_{22}=0$ and $A_{12}=A_{21}=a=2.00$ in its stated units. Find $\langle A\rangle(t)$ and its period.
Solution. Time evolution gives
\[\lvert\Psi(t)\rangle =\frac1{\sqrt2}\left( e^{-iE_1t/\hbar}\lvert1\rangle+ e^{-iE_2t/\hbar}\lvert2\rangle\right).\]Only the two cross terms survive:
\[\begin{aligned} \langle A\rangle(t) &=\frac a2\left[ e^{-i(E_2-E_1)t/\hbar} +e^{i(E_2-E_1)t/\hbar}\right]\\ &=\boxed{a\cos\!\left(\frac{(E_2-E_1)t}{\hbar}\right)}. \end{aligned}\]Thus
\[T=\frac{2\pi\hbar}{E_2-E_1} =\frac{h}{1.50\ \mathrm{eV}} =\boxed{2.757\ \mathrm{fs}}.\]At $t=0$, $\langle A\rangle=+2.00$; after $T/2$ it is $-2.00$. The cosine argument is dimensionless and the sign of $E_2-E_1$ would not change the observable because cosine is even. In the degenerate limit $E_2-E_1\to0$, the relative phase stops evolving and the expectation becomes constant.
Descriptive Questions
- How are expansion coefficients obtained from an initial wavefunction, and why do their squared magnitudes sum to one?
- Why are energy probabilities constant for a time-independent Hamiltonian even when the position-space probability density oscillates?
- Explain how interference terms generate beat frequencies in a superposition of nondegenerate energy eigenstates.
- How is the spectral expansion modified when a Hamiltonian has both discrete bound states and a continuous spectrum?
Numerical Problems
- A two-state expansion has coefficients $3/5$ and $4i/5$. Find the two measurement probabilities.
Final answer: $\boxed{P_1=0.360,\quad P_2=0.640}$. - Find the beat period associated with an energy separation of $0.800\ \mathrm{eV}$.
Final answer: $\boxed{T=h/\Delta E=5.170\ \mathrm{fs}}$. - Energy values $2.00\ \mathrm{eV}$ and $8.00\ \mathrm{eV}$ occur with probabilities $0.250$ and $0.750$. Find the mean energy.
Final answer: $\boxed{\langle E\rangle=6.50\ \mathrm{eV}}$.
Every added probability, energy moment, and beat-period identity is checked in the MJ-11 problem-verification worksheet; every printed residual and check is zero.
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