26 Jun 2025

Schrödinger Equations and Stationary States

schrodinger-equation hamiltonian stationary-states energy-eigenvalues

For a non-relativistic particle in a scalar potential,

\[E=\frac{p^2}{2m}+V(\mathbf r,t).\]

The de Broglie plane wave $e^{i(\mathbf p\cdot\mathbf r-Et)/\hbar}$ is an eigenfunction of $i\hbar\partial_t$ with eigenvalue $E$ and of $-i\hbar\nabla$ with eigenvalue $\mathbf p$. This fixes the operator substitutions

\[E\longmapsto i\hbar\frac{\partial}{\partial t}, \qquad \mathbf p\longmapsto-i\hbar\nabla.\]

Applying the classical energy expression as an operator relation motivates the time-dependent Schrödinger equation; the equation itself is a postulate of non-relativistic quantum dynamics:

\[\boxed{i\hbar\frac{\partial\Psi}{\partial t}=\hat H\Psi}, \qquad \boxed{\hat H=-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r,t)}.\]

The minus sign in the kinetic term is $(-i)^2=-1$. Its units are those of energy because $\hbar^2\nabla^2/m$ has dimensions

\[\frac{(\mathrm{J\,s})^2}{\mathrm{kg\,m^2}} =\mathrm J.\]

The potential $V$ must have the same units. The equation is first order in time, so a specified initial wavefunction determines its later evolution; it is second order in space, so spatial boundary conditions are also required.

Conservation under dynamical evolution

Write the equation and its adjoint in state notation:

\[\lvert\dot\Psi\rangle=-\frac{i}{\hbar}\hat H\lvert\Psi\rangle, \qquad \langle\dot\Psi\rvert=\frac{i}{\hbar}\langle\Psi\rvert\hat H^\dagger.\]

For a self-adjoint Hamiltonian,

\[\begin{aligned} \frac d{dt}\langle\Psi\vert\Psi\rangle &=\langle\dot\Psi\vert\Psi\rangle +\langle\Psi\vert\dot\Psi\rangle\\ &=\frac{i}{\hbar}\langle\Psi\vert\hat H\vert\Psi\rangle -\frac{i}{\hbar}\langle\Psi\vert\hat H\vert\Psi\rangle=0. \end{aligned}\]

Thus normalization is preserved. If $\hat H$ is time independent, the evolution operator is

\[\boxed{U(t,t_0)=\exp\!\left[-\frac{i}{\hbar}\hat H(t-t_0)\right]},\] \[\lvert\Psi(t)\rangle=U(t,t_0)\lvert\Psi(t_0)\rangle.\]

Differentiation verifies $i\hbar\,\partial_tU=\hat HU$ and $U(t_0,t_0)=I$. Since $\hat H=\hat H^\dagger$,

\[U^\dagger U =e^{i\hat H(t-t_0)/\hbar}e^{-i\hat H(t-t_0)/\hbar}=I,\]

so the evolution is unitary.

Time-independent Schrödinger equation

When $V(\mathbf r)$ has no explicit time dependence, set

\[\Psi(\mathbf r,t)=\psi(\mathbf r)T(t).\]

Substitution gives

\[i\hbar\psi\frac{dT}{dt} =T\left[-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r)\right]\psi.\]

Divide by $\psi T$ wherever the product is non-zero:

\[i\hbar\frac1T\frac{dT}{dt} =\frac1\psi\left[-\frac{\hbar^2}{2m}\nabla^2+V\right]\psi.\]

The left side depends only on $t$ and the right side only on $\mathbf r$; both must equal a constant $E$. Hermiticity of $\hat H$ makes $E$ real. The separated equations are

\[i\hbar\frac{dT}{dt}=ET, \qquad \boxed{\hat H\psi=E\psi}.\]

Integrating $dT/T=-iE\,dt/\hbar$ gives

\[T(t)=T(0)e^{-iEt/\hbar}.\]

After absorbing $T(0)$ into $\psi$, an energy eigenfunction produces the stationary state

\[\boxed{\Psi_E(\mathbf r,t)=\psi_E(\mathbf r)e^{-iEt/\hbar}}.\]

Its probability density is stationary,

\[\lvert\Psi_E(\mathbf r,t)\rvert^2 =\lvert\psi_E(\mathbf r)\rvert^2,\]

because the time factor has unit modulus. The state vector still changes by a phase. Spatial boundary conditions and normalizability select the allowed eigenfunctions and hence the energy eigenvalues; this selection is derived for the stated potentials in Unit III.