Wave Functions and the Schrodinger Equation

In the position representation, the state is described by a wave function $\psi(\mathbf{r}, t)$ Where $|\psi(\mathbf{r}, t)|^2$ is the probability density:

$P(\mathbf{r} \in [\mathbf{r}, \mathbf{r} + d\mathbf{r}]) = |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r}$

Normalisation: $\int_{-\infty}^{\infty} |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r} = 1$.

$i\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi = \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}, t)\right)\psi$

For time-independent potentials $V(\mathbf{r})$Separate variables: $\psi(\mathbf{r}, t) = \phi(\mathbf{r}) e^{-iEt/\hbar}$:

$\hat{H}\phi = E\phi \quad \mathrm{i}.e., \quad -\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi$

This is an eigenvalue problem: $E$ is the energy eigenvalue, $\phi$ is the energy eigenstate.

Properties of energy eigenstates:

1. Orthogonality. If $\hat{H}\phi_n = E_n\phi_n$ and $\hat{H}\phi_m = E_m\phi_m$ with $E_n \neq E_m$ then $\int \phi_n^*\phi_m\,dx = 0$ (since $\hat{H}$ is Hermitian).

2. Completeness. The energy eigenstates form a complete basis: any state can be expanded as $\psi(x,0) = \sum_n c_n\phi_n(x)$ where $c_n = \int \phi_n^*(x)\psi(x,0)\,dx$.

3. Stationary states. If $\psi(x,0) = \phi_n(x)$Then $\psi(x,t) = \phi_n(x)e^{-iE_nt/\hbar}$. The probability density $|\psi|^2 = |\phi_n|^2$ is time-independent.

4. Reality of $\phi$. If $V(x)$ is real and there is no magnetic field, $\phi_n(x)$ can be chosen to be real. This is because if $\phi_n$ is a solution, so is $\phi_n^*$And degenerate solutions can be combined into real linear combinations.

The probability current density is

$\mathbf{J} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi \nabla\psi^*)$

It satisfies the continuity equation: $\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0$ Expressing conservation of probability.

Derivation of the continuity equation. Start with the Schrodinger equation and its complex conjugate:

$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi$

$-i\hbar \frac{\partial \psi^*}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi^* + V\psi^*$

Multiply the first by $\psi^*$ and the second by $\psi$Then subtract:

$i\hbar\!\left(\psi^*\frac{\partial \psi}{\partial t} + \psi\frac{\partial \psi^*}{\partial t}\right) = -\frac{\hbar^2}{2m}\!\left(\psi^*\nabla^2\psi - \psi\nabla^2\psi^*\right)$

The left-hand side is $i\hbar\,\partial|\psi|^2/\partial t$. The right-hand side is a divergence:

$\nabla \cdot (\psi^*\nabla\psi - \psi\nabla\psi^*) = \psi^*\nabla^2\psi - \psi\nabla^2\psi^*$

Therefore:

$i\hbar\frac{\partial |\psi|^2}{\partial t} = -\frac{\hbar^2}{2m}\nabla \cdot (\psi^*\nabla\psi - \psi\nabla\psi^*)$

Dividing by $i\hbar$:

$\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \left[\frac{\hbar}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)\right] = 0$

$\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0 \qquad \blacksquare$

Theorem 3.1. Time evolution governed by the Schrodinger equation with a Hermitian Hamiltonian Is unitary, and therefore preserves the norm of the state vector.

Proof. The time evolution operator $\hat{U}(t, t_0)$ is defined by:

$|\psi(t)\rangle = \hat{U}(t, t_0)|\psi(t_0)\rangle$

For a time-independent Hamiltonian:

$\hat{U}(t, t_0) = \exp\!\left(-\frac{i\hat{H}(t - t_0)}{\hbar}\right)$

To prove unitarity, we show $\hat{U}^\dagger \hat{U} = \hat{I}$:

$\hat{U}^\dagger = \exp\!\left(\frac{i\hat{H}^\dagger(t - t_0)}{\hbar}\right) = \exp\!\left(\frac{i\hat{H}(t - t_0)}{\hbar}\right)$

Since $\hat{H} = \hat{H}^\dagger$ (Hermitian). Therefore:

$\hat{U}^\dagger \hat{U} = \exp\!\left(\frac{i\hat{H}(t - t_0)}{\hbar}\right)\exp\!\left(-\frac{i\hat{H}(t - t_0)}{\hbar}\right) = \hat{I}$

Since commuting operators satisfy $e^A e^{-A} = I$.

Consequence. Norm preservation:

$\langle\psi(t)|\psi(t)\rangle = \langle\psi(t_0)|\hat{U}^\dagger\hat{U}|\psi(t_0)\rangle = \langle\psi(t_0)|\psi(t_0)\rangle$

Total probability is conserved under time evolution. $\blacksquare$

Composing evolutions. For successive time intervals, the evolution operator composes as:

$\hat{U}(t_2, t_0) = \hat{U}(t_2, t_1)\,\hat{U}(t_1, t_0)$

This composition law, combined with unitarity, is the group structure underlying quantum dynamics. For a time-dependent Hamiltonian, the evolution operator is given by Dyson”s time-ordered exponential:

$\hat{U}(t, t_0) = \mathcal{T}\exp\!\left(-\frac{i}{\hbar}\int_{t_0}^{t}\hat{H}(t')\,dt'\right)$

Where $\mathcal{T}$ denotes time ordering (later times appear to the left).

For a time-dependent Hamiltonian, the evolution operator satisfies $i\hbar\,\partial\hat{U}/\partial t = \hat{H}(t)\hat{U}$ With $\hat{U}(t_0, t_0) = \hat{I}$. Unitarity still holds: $d(\hat{U}^\dagger\hat{U})/dt = 0$ since $\hat{H}(t) = \hat{H}^\dagger(t)$.

A physically valid wave function must satisfy $\int |\psi|^2\,dx = 1$. This determines the normalisation Constant.

Example 3.1. Normalise the wave function $\psi(x) = Ae^{-\alpha|x|}$ for $-\infty \lt x \lt \infty$ Where $\alpha \gt 0$.

Example 3.2. Normalise $\psi(x) = Axe^{-\alpha x^2}$ for $-\infty \lt x \lt \infty$.

When the Hamiltonian has a time-dependent perturbation, $\hat{H}(t) = \hat{H}_0 + \hat{V}(t)$The Transition probability from initial state $|i\rangle$ to final state $|f\rangle$ (with $E_i \neq E_f$) is computed in the interaction picture.

First-order transition amplitude. If the system starts in $|i\rangle$ at $t = 0$The probability Amplitude for being in $|f\rangle$ at time $t$ is, to first order:

$c_f(t) = -\frac{i}{\hbar}\int_0^t \langle f | \hat{V}(t') | i \rangle\, e^{i\omega_{fi}t'}\,dt'$

Where $\omega_{fi} = (E_f - E_i)/\hbar$ is the Bohr frequency.

Constant perturbation. If $\hat{V}(t) = \hat{V}_0$ (constant) for $0 \lt t \lt T$:

$c_f(T) = -\frac{i}{\hbar}V_{fi}\int_0^T e^{i\omega_{fi}t'}\,dt' = -\frac{V_{fi}}{\hbar\omega_{fi}}\!\left(e^{i\omega_{fi}T} - 1\right)$

The transition probability is:

$P_{i \to f}(T) = \frac{|V_{fi}|^2}{\hbar^2}\,\frac{\sin^2(\omega_{fi}T/2)}{(\omega_{fi}/2)^2}$

This function is sharply peaked around $\omega_{fi} = 0$ (resonance), with width $\Delta\omega \sim 2\pi/T$.

Interpretation. As $T \to \infty$The function $\sin^2(\omega_{fi}T/2)/(\omega_{fi}/2)^2 \to 2\pi T\,\delta(\omega_{fi})$ So transitions occur only when energy is conserved ($E_f = E_i$). For finite $T$Energy conservation Is approximate to within $\Delta E \sim \hbar/T$A manifestation of the time-energy uncertainty Relation.

Fermi’s Golden Rule. For a transition to a continuum of final states with density of states $\rho(E_f)$The transition rate (probability per unit time) is:

$\Gamma_{i \to f} = \frac{2\pi}{\hbar}|\langle f | \hat{V} | i \rangle|^2\,\rho(E_f)$

This is one of the most important results in quantum mechanics, with applications to spontaneous Emission, scattering theory, and condensed matter physics.

Sudden and adiabatic approximations.

• Sudden approximation. If the Hamiltonian changes rapidly compared to the system’s natural timescale $\sim \hbar/\Delta E$The state does not have time to adjust: $|\psi_{\mathrm{after}\rangle = |\psi_{\mathrm{before}\rangle}}$. The probability of finding the system in the new $n$-th eigenstate is $P_n = |\langle n_{\mathrm{new}|\psi_{\mathrm{before}\rangle|^2}}$.

• Adiabatic theorem. If the Hamiltonian changes slowly enough (specifically, if $|\langle m|\partial\hat{H}/\partial t|n\rangle|/(\hbar\omega_{mn}^2) \ll 1$ for all $m \neq n$), the system remains in an instantaneous eigenstate without transitions. The adiabatic condition requires the rate of change to be much slower than the energy gap divided by $\hbar$.

Harmonic perturbation. For a sinusoidal perturbation $\hat{V}(t) = \hat{V}_1\,e^{-i\omega t} + \hat{V}_1^\dagger\,e^{i\omega t}$ The first-order transition rate from $|i\rangle$ to $|f\rangle$ is significant only when $\omega \approx \omega_{fi}$ (absorption) or $\omega \approx -\omega_{fi}$ (stimulated emission). The transition probability for Resonant absorption ($\omega \approx \omega_{fi}$) is:

$P_{i\to f}(t) = \frac{|\langle f|\hat{V}_1|i\rangle|^2}{\hbar^2}\,\frac{\sin^2((\omega - \omega_{fi})t/2)}{(\omega - \omega_{fi})^2/4}$

In the long-time limit, this reduces to Fermi’s Golden Rule with the replacement $V_{fi} \to \langle f|\hat{V}_1|i\rangle$.