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Chapter 3 • Theory & Derivations

Schrödinger’s Equation, Probability Current & Ehrenfest Theorems

The dynamics of wave mechanics: derivation of the Time-Dependent and Time-Independent Schrödinger equations, stationary states, the conservation of probability, continuity equation, probability current density, time variation of observables, and Ehrenfest's theorem connecting quantum and classical trajectories.

§3.1 The Schrödinger Wave Equations

The Time-Dependent Schrödinger Equation (TDSE)

In 1926, Erwin Schrödinger formulated the fundamental wave equation for a non-relativistic particle of mass $m$ subjected to a potential energy $V(\mathbf{r}, t)$:

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

Where:

  • $i = \sqrt{-1}$ is the imaginary unit.
  • $\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ is the spatial Laplacian operator.
  • $\hat{H} = \frac{\hat{p}^2}{2m} + V(\mathbf{r})$ is the Hamiltonian operator.

Because the TDSE is first-order in time $t$, specifying the initial wavefunction $\Psi(\mathbf{r}, 0)$ uniquely determines $\Psi(\mathbf{r}, t)$ for all future times. Because it is linear in $\Psi$, it satisfies the superposition principle.

Separation of Variables: The Time-Independent Schrödinger Equation (TISE)

When the potential energy is independent of time ($V(\mathbf{r}, t) = V(\mathbf{r})$), we seek product solutions:

$$\Psi(\mathbf{r}, t) = \psi(\mathbf{r}) \phi(t)$$

Substituting into the TDSE and dividing both sides by $\psi(\mathbf{r}) \phi(t)$:

$$i\hbar \frac{1}{\phi(t)} \frac{d\phi(t)}{dt} = \frac{1}{\psi(\mathbf{r})} \left( -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}) \right) \psi(\mathbf{r})$$

The left side depends solely on $t$, while the right side depends solely on $\mathbf{r}$. Both sides must therefore equal a separation constant $E$ with dimensions of energy:

1. Temporal Differential Equation:

$$i\hbar \frac{d\phi(t)}{dt} = E \phi(t) \implies \phi(t) = e^{-i E t / \hbar}$$

2. Spatial Differential Equation (TISE):

$$\hat{H}\psi(\mathbf{r}) = E\psi(\mathbf{r}) \implies -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r}) + V(\mathbf{r})\psi(\mathbf{r}) = E\psi(\mathbf{r})$$
Stationary States and Their Properties

Solutions of the form $\Psi_n(\mathbf{r}, t) = \psi_n(\mathbf{r}) e^{-i E_n t / \hbar}$ are called stationary states because their physical properties are time-independent:

1. Probability Density:

$$\rho(\mathbf{r}, t) = |\Psi_n(\mathbf{r}, t)|^2 = \left| \psi_n(\mathbf{r}) e^{-i E_n t / \hbar} \right|^2 = |\psi_n(\mathbf{r})|^2$$

2. Expectation Values: For any time-independent observable $\hat{A}$:

$$\langle \hat{A} \rangle_t = \int \Psi_n^* \hat{A} \Psi_n d^3\mathbf{r} = \int \psi_n^* e^{i E_n t/\hbar} \hat{A} \psi_n e^{-i E_n t/\hbar} d^3\mathbf{r} = \langle \hat{A} \rangle_0 = \text{constant}$$

§3.2 Conservation of Probability and the Continuity Equation

Mathematical Derivation of the Continuity Equation

The total probability of finding a particle in all of space is:

$$P(t) = \int_{-\infty}^{+\infty} |\Psi(x,t)|^2 dx = \int_{-\infty}^{+\infty} \Psi^*(x,t) \Psi(x,t) dx$$

Taking the time derivative:

$$\frac{dP}{dt} = \int_{-\infty}^{+\infty} \frac{\partial}{\partial t}(\Psi^* \Psi) dx = \int_{-\infty}^{+\infty} \left( \frac{\partial \Psi^*}{\partial t} \Psi + \Psi^* \frac{\partial \Psi}{\partial t} \right) dx$$

From the TDSE:

$$\frac{\partial \Psi}{\partial t} = \frac{1}{i\hbar} \left( -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi \right) = \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi$$

Taking the complex conjugate:

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

Substituting these into the time derivative of the probability density $\rho = \Psi^* \Psi$:

$$\frac{\partial \rho}{\partial t} = \left( -\frac{i\hbar}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} V \Psi^* \right) \Psi + \Psi^* \left( \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi \right)$$

The potential terms cancel:

$$\frac{\partial \rho}{\partial t} = \frac{i\hbar}{2m} \left( \Psi^* \frac{\partial^2 \Psi}{\partial x^2} - \Psi \frac{\partial^2 \Psi^*}{\partial x^2} \right) = -\frac{\partial}{\partial x} \left[ \frac{\hbar}{2mi} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \right) \right]$$

Defining the Probability Current Density $J(x,t)$:

$$J(x,t) = \frac{\hbar}{2mi} \left( \Psi^* \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^*}{\partial x} \right) = \frac{\hbar}{m} \text{Im}\left( \Psi^* \frac{\partial \Psi}{\partial x} \right)$$

This yields the Quantum Continuity Equation:

$$\frac{\partial \rho(x,t)}{\partial t} + \frac{\partial J(x,t)}{\partial x} = 0 \qquad \left( \text{or in 3D: } \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0 \right)$$

Integrating over all space:

$$\frac{dP}{dt} = -\int_{-\infty}^{+\infty} \frac{\partial J}{\partial x} dx = -[J(\infty, t) - J(-\infty, t)] = 0$$

Since physical wavefunctions must vanish at infinity for square-integrability, $J(\pm \infty, t) = 0$. Consequently, total probability is conserved for all time: $\int_{-\infty}^{+\infty} |\Psi(x,t)|^2 dx = 1$.

§3.3 Time Evolution of Observables and Ehrenfest's Theorem

General Equation of Motion for Expectation Values

Let $\hat{A}$ be an arbitrary quantum observable. Its expectation value is $\langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle$. Taking the total time derivative:

$$\frac{d\langle A \rangle}{dt} = \frac{d}{dt} \langle \Psi | \hat{A} | \Psi \rangle = \left( \frac{d}{dt}\langle \Psi | \right) \hat{A} |\psi\rangle + \langle \Psi | \frac{\partial \hat{A}}{\partial t} | \Psi \rangle + \langle \Psi | \hat{A} \left( \frac{d}{dt}|\psi\rangle \right)$$

Using the TDSE $|\dot{\Psi}\rangle = \frac{1}{i\hbar}\hat{H}|\Psi\rangle$ and $\langle \dot{\Psi}| = -\frac{1}{i\hbar}\langle \Psi|\hat{H}$:

$$\frac{d\langle A \rangle}{dt} = -\frac{1}{i\hbar}\langle \Psi | \hat{H}\hat{A} | \Psi \rangle + \frac{1}{i\hbar}\langle \Psi | \hat{A}\hat{H} | \Psi \rangle + \left\langle \frac{\partial \hat{A}}{\partial t} \right\rangle$$
$$\frac{d\langle A \rangle}{dt} = \frac{1}{i\hbar} \langle [\hat{A}, \hat{H}] \rangle + \left\langle \frac{\partial \hat{A}}{\partial t} \right\rangle$$
Constants of Motion

If an observable $\hat{A}$ has no explicit time dependence ($\frac{\partial \hat{A}}{\partial t} = 0$) and commutes with the Hamiltonian ($[\hat{A}, \hat{H}] = 0$), then:

$$\frac{d\langle A \rangle}{dt} = 0$$

Such an observable is a constant of motion. Its expectation value is time-independent in any state.

Ehrenfest's Theorems (The Classical Limit)

Paul Ehrenfest (1927) showed that quantum expectation values obey classical equations of motion:

1. First Ehrenfest Theorem (Position):

Let $\hat{A} = \hat{x}$. Commuting with $\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})$:

$$[\hat{x}, \hat{H}] = \left[ \hat{x}, \frac{\hat{p}^2}{2m} \right] = \frac{1}{2m} (\hat{p}[\hat{x}, \hat{p}] + [\hat{x}, \hat{p}]\hat{p}) = \frac{i\hbar}{m}\hat{p}$$
$$\frac{d\langle x \rangle}{dt} = \frac{1}{i\hbar}\left( \frac{i\hbar}{m}\langle p \rangle \right) \implies \frac{d\langle x \rangle}{dt} = \frac{\langle p \rangle}{m}$$

2. Second Ehrenfest Theorem (Momentum):

Let $\hat{A} = \hat{p}$. Commuting with $\hat{H}$:

$$[\hat{p}, \hat{H}] = [\hat{p}, V(\hat{x})] = -i\hbar \frac{\partial V}{\partial x}$$
$$\frac{d\langle p \rangle}{dt} = \frac{1}{i\hbar}\left( -i\hbar \left\langle \frac{\partial V}{\partial x} \right\rangle \right) \implies \frac{d\langle p \rangle}{dt} = -\left\langle \frac{\partial V(\hat{x})}{\partial x} \right\rangle = \langle F(\hat{x}) \rangle$$

This reproduces Newton's Second Law ($mathbf{F} = mmathbf{a}$) for expectation values, showing how classical physics emerges from quantum mechanics in macroscopic systems.

See Simulation 3.1 below for a real-time visualization of a free Gaussian wavepacket spreading and dispersing over time.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Hard Example 3.1: Rigorous Calculation of Gaussian Wave Packet Dispersion Rate

A free particle ($V(x)=0$) is initialized at $t=0$ as a Gaussian wave packet $\psi(x,0) = (2\pi\sigma_0^2)^{-1/4} e^{-x^2 / 4\sigma_0^2}$. Solve the time-dependent Schrödinger equation to find $\psi(x,t)$, and prove that the packet width broadens according to $\sigma(t) = \sigma_0 \sqrt{1 + \left( \frac{\hbar t}{2m\sigma_0^2} \right)^2}$.

Step 1: Fourier Transform to Momentum Space
$$\phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \psi(x,0) e^{-i k x} dx = \left( \frac{2\sigma_0^2}{\pi} \right)^{1/4} e^{-\sigma_0^2 k^2}$$

In momentum space, each plane wave component evolves with a simple phase factor: $e^{-i E_k t / \hbar} = e^{-i \frac{\hbar k^2}{2m} t}$.

Step 2: Inverse Fourier Transform at time t
$$\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{i(kx - \omega_k t)} dk = \frac{(2\sigma_0^2 / \pi)^{1/4}}{\sqrt{2\sigma_0^2 + i \frac{\hbar t}{m}}} \exp\left( -\frac{x^2}{4\sigma_0^2 + 2i \frac{\hbar t}{m}} \right) $$ $$|\psi(x,t)|^2 = \frac{1}{\sqrt{2\pi} \sigma(t)} \exp\left( -\frac{x^2}{2\sigma(t)^2} \right), \quad \text{where } \sigma(t) = \sigma_0 \sqrt{1 + \left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}$$

Because different momentum components travel at different group velocities ($v_g = \hbar k / m$), the wave packet broadens over time. This dispersion is purely quantum mechanical.

An electron is accelerated from rest through an electrostatic potential difference of $V = 150 \text{ V}$.\n(a) Determine its de Broglie wavelength using non-relativistic mechanics.\n(b) At what accelerating potential does the relativistic correction to the de Broglie wavelength exceed $1\%$?
Step 1: Non-relativistic Calculation
$$K = e V = 150 \text{ eV} = 150 \times 1.602 \times 10^{-19} \text{ J} = 2.403 \times 10^{-17} \text{ J} $$ $$p = \sqrt{2 m_e K} = \sqrt{2(9.109 \times 10^{-31} \text{ kg})(2.403 \times 10^{-17} \text{ J})} = 6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s} $$ $$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.001 \times 10^{-10} \text{ m} = 1.001 \text{ Å}$$
For quick calculations, note that $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$. For $V = 150 \text{ V}$, $\lambda = 1.00 \text{ Å}$, which corresponds to typical atomic crystal lattice spacings.
Step 2: Relativistic Condition
$$E^2 = p^2 c^2 + m_0^2 c^4 \implies p = \frac{1}{c}\sqrt{K(K + 2m_0 c^2)} $$ $$\lambda_{\text{rel}} = \frac{h c}{\sqrt{K(K + 2m_0 c^2)}} = \frac{\lambda_{\text{class}}}{\sqrt{1 + \frac{K}{2m_0 c^2}}} \approx \lambda_{\text{class}}\left(1 - \frac{K}{4 m_0 c^2}\right) $$ $$\frac{\Delta \lambda}{\lambda} \approx \frac{K}{4 m_0 c^2} \ge 0.01 \implies K \ge 0.04 m_0 c^2 = 0.04 (511 \text{ keV}) \approx 20.44 \text{ keV}$$
When accelerating potentials exceed roughly $20 \text{ kV}$ (typical in transmission electron microscopes), relativistic momentum corrections become necessary.