Wave Functions and the Schrodinger Equation

Attempt these from memory before reading.

• What does $\int |\psi(\mathbf{r},t)|^2\,d^3\mathbf{r} = 1$ mean, and what does it say about the function space $\psi$ lives in?
• Write the time-dependent Schrodinger equation.
• Why can you separate variables when $V$ does not depend on time?
• State two properties of energy eigenstates, and say which fact about $\hat{H}$ each one depends on.
• What is probability current, and what does its integral over space equal?

Check your answers against the text. The second prompt is the one people find hardest, because the equation looks like it was derived from something. It was not, and saying where it comes from is the point of the page.

It is tempting to think of the Schrodinger equation as physics: some equation that happens to describe how $\psi$ moves. But it is really a restatement of the time-evolution postulate, and once you see that, several puzzles dissolve.

The postulate says time evolution is unitary: $U(t,0)$ is an operator with $U^\dagger U = I$. Every unitary operator is an exponential of something Hermitian, so write

$$U(t) = e^{-i\hat{H}t/\hbar}$$

for some Hermitian $\hat{H}$, and $\hbar$ is a choice of units. Acting on $\psi$ and differentiating at $t = 0$ gives

$$i\hbar\,\frac{\partial \psi}{\partial t} = \hat{H}\psi .$$

That is the Schrodinger equation, with no appeal to de Broglie, no Planck constant appearing from nowhere, and no assumption about trajectories. The only input was unitarity.

Why linearity is the whole ballgame. The equation is linear in $\psi$, so superpositions are solutions whenever their components are. That single fact produces interference, which is the experimental phenomenon that forced quantum mechanics to be linear in the first place. A nonlinear Schrodinger equation would not have interference, and with it gone there is little left that is recognisably quantum.

Why the Hamiltonian looks the way it does. It is not guessed from classical mechanics by replacing $p$ with $-i\hbar\partial_x$ — that is a mnemonic for finding the right answer, not a derivation. $\hat{H}$ is whatever generates time translation, and we identify it by reproducing two known results: free propagation needs $\hat{H} = \hat{p}^2/2m$, and a potential $V$ must enter additively as $\hat{H} = \hat{p}^2/2m + V(\mathbf{r})$. The non-commutativity of $\hat{x}$ and $\hat{p}$ is what forces the operator ordering to be fixed, and it is why $\hat{p}^2/2m$ must mean $-\hbar^2\nabla^2/2m$ rather than the other way round.

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 $\alpha \gt 0$.

Solution

$$\int_{-\infty}^{\infty} |A|^2 e^{-2\alpha|x|}\,dx = 2|A|^2 \int_0^{\infty} e^{-2\alpha x}\,dx = 2|A|^2 \cdot \frac{1}{2\alpha} = \frac{|A|^2}{\alpha} = 1$$

Therefore $|A| = \sqrt{\alpha}$ And we choose $A = \sqrt{\alpha}$:

$$\psi(x) = \sqrt{\alpha}\,e^{-\alpha|x|}$$

To find $\langle x \rangle$:

$$\langle x \rangle = \alpha \int_{-\infty}^{\infty} x\,e^{-2\alpha|x|}\,dx = 0$$

By symmetry (the integrand is odd). For $\langle x^2 \rangle$:

$$\langle x^2 \rangle = 2\alpha \int_0^{\infty} x^2 e^{-2\alpha x}\,dx = 2\alpha \cdot \frac{2}{(2\alpha)^3} = \frac{1}{2\alpha^2}$$

So $\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = 1/(\sqrt{2}\,\alpha)$.

The uncertainty product for this state is $\sigma_x\,\sigma_p = \hbar/(2\sqrt{2})$ Which is larger than the minimum $\hbar/2$ Showing this is not a minimum-uncertainty state.

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

Solution

$$\int_{-\infty}^{\infty} |A|^2 x^2 e^{-2\alpha x^2}\,dx = |A|^2 \cdot \frac{1}{4\alpha}\sqrt{\frac{\pi}{2\alpha}} = 1$$

Using the Gaussian integral $\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx = \frac{1}{2a}\sqrt{\frac{\pi}{a}}$. Therefore:

$$A = 2\sqrt{\alpha}\left(\frac{2\alpha}{\pi}\right)^{1/4}$$

Problem. A particle of mass $m$ is confined to a one-dimensional box of width $L$ (infinite square well). Find the energy eigenstates and eigenvalues.

Solution

The potential is:

$$V(x) = \begin{cases} 0 & 0 < x < L \\ \infty & \text{otherwise} \end{cases}$$

Inside the box ($0 < x < L$), the time-independent Schrodinger equation is:

$$-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi$$ $$\frac{d^2\phi}{dx^2} = -k^2\phi, \quad k = \frac{\sqrt{2mE}}{\hbar}$$

General solution: $\phi(x) = A\sin(kx) + B\cos(kx)$

Boundary conditions: $\phi(0) = 0 \implies B = 0$

$\phi(L) = 0 \implies \sin(kL) = 0 \implies kL = n\pi, \quad n = 1, 2, 3, ...$

Energy eigenvalues:

$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$$

Eigenstates: $\phi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$

The normalization constant $A = \sqrt{2/L}$ ensures $\int_0^L |\phi_n|^2 dx = 1$.

$\blacksquare$

Intuition. The energy levels scale as $n^2$, so the spacing between levels increases with $n$. The ground state energy $E_1$ is non-zero — this is the zero-point energy, a direct consequence of the uncertainty principle. A particle confined to a region of size $L$ must have momentum $\sim \hbar/L$, giving kinetic energy $\sim \hbar^2/(2mL^2)$.

Problem. For the ground state of the infinite square well, find $\langle x \rangle$, $\langle x^2 \rangle$, $\langle p \rangle$, and $\langle p^2 \rangle$. Verify the uncertainty principle.

Solution

Ground state: $\phi_1(x) = \sqrt{2/L}\sin(\pi x/L)$

$$\langle x \rangle = \frac{2}{L}\int_0^L x\sin^2\left(\frac{\pi x}{L}\right)dx = \frac{L}{2}$$

By symmetry, the average position is at the center of the well.

$$\langle x^2 \rangle = \frac{2}{L}\int_0^L x^2\sin^2\left(\frac{\pi x}{L}\right)dx = \frac{L^2}{3} - \frac{L^2}{2\pi^2}$$

For momentum, use $\hat{p} = -i\hbar\frac{d}{dx}$:

$$\langle p \rangle = \frac{2}{L}\int_0^L \sin\left(\frac{\pi x}{L}\right)\left(-i\hbar\frac{d}{dx}\right)\sin\left(\frac{\pi x}{L}\right)dx = 0$$

By symmetry, the average momentum is zero.

$$\langle p^2 \rangle = \frac{2}{L}\int_0^L \sin\left(\frac{\pi x}{L}\right)\left(-\hbar^2\frac{d^2}{dx^2}\right)\sin\left(\frac{\pi x}{L}\right)dx = \frac{\pi^2\hbar^2}{L^2}$$

Uncertainties:

$$\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = L\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} \approx 0.18L$$ $$\sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} = \frac{\pi\hbar}{L}$$

Uncertainty product:

$$\sigma_x\sigma_p = \frac{\pi\hbar}{L} \cdot L\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} = \pi\hbar\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}} \approx 1.14\hbar > \frac{\hbar}{2}$$

The uncertainty principle is satisfied.

$\blacksquare$

Common mistake. Forgetting that $\langle p \rangle = 0$ for a standing wave. The momentum expectation value is zero because the particle is equally likely to be moving left or right.

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 $V_{fi} \to \langle f|\hat{V}_1|i\rangle$.

The Schrodinger equation is the quantum version of Newton’s second law: it tells you how a quantum state evolves in time. The wave function is like a ghostly cloud that describes where a particle might be found, with the square of its height giving the probability. A particle in a box is like a vibrating string fixed at both ends: only certain wavelengths fit, so only certain energies are allowed. The zero-point energy is the quantum version of the jitterbug: you cannot pin a particle down completely because the uncertainty principle demands it keep moving. Probability current is like water flow: the continuity equation ensures that probability is never created or destroyed, only moved around. Unitarity guarantees that quantum mechanics is reversible: if you run time backward, you recover the initial state, just like rewinding a video.

The first row is the most important, and the global-phase version is worth sitting with. Multiply $\psi$ by $e^{i\theta}$ and nothing measurable changes, so $\psi$ and $e^{i\theta}\psi$ are the same state. Multiply one component of a superposition by a phase and everything measurable changes. The phase is individually unobservable and collectively decisive — which is why interference is not optional.

The last row is the one that is usually taught as a consequence rather than a theorem. Normality is a property of a state at a moment; conservation of probability is a statement about the dynamics, and it is exactly what fails when the potential is complex.

• The many-body Schrodinger equation. Solving it exactly is exponential in the number of particles, and this is a statement about the problem, not about our algorithms. Quantum Monte Carlo works because the sign problem can be sidestepped for fermions with nodes, at a cost that grows with system size.
• Real-time electron dynamics. For large molecules, propagating the wave function directly (TDCI) is prohibitive, so TDDFT and related approaches propagate an orbital basis instead. The accuracy of those methods near conical intersections and in strong fields is an active question.
• Stochastic Schrodinger equations. Replacing the evolution by a complex noise turns the problem into a diffusion in Hilbert space. Trajectories reproduce the Born statistics for a single outcome, but the time step must shrink with $\sqrt{\mathrm{d}t}$ because of the quadratic variation, which is exactly the Itô correction on the stochastic calculus page.
• Discontinuous Galerkin methods. Error analysis for the Schrodinger equation remains active because the time propagator is skew-Hermitian and not energy-conserving at finite step, which matters for long-time simulations.
• Schrodinger operators on graphs and quantum walks. The spectral structure — and whether the continuum picture survives — is still being worked out, with direct consequences for quantum search algorithms.
• Tensor networks. Matrix product states make the many-body problem tractable in one dimension, and the bond dimension needed grows exponentially with entanglement, which is the obstruction being worked around.

Mix these with the postulates, Hamiltonian Mechanics, Angular Momentum, Linear Algebra and Measure Theory.

1. (Postulates.) Starting only from “time evolution is unitary”, prove that $U(t) = e^{-i\hat{H}t/\hbar}$, and show that $\hat{H}$ must be Hermitian. Then explain why this makes the Schrodinger equation a statement about time rather than a law about particles.
2. (Linear algebra.) Prove that eigenstates of a Hermitian operator with distinct eigenvalues are orthogonal, by taking the inner product of $\hat{H}\phi_n = E_n\phi_n$ with $\phi_m$ and using $\langle\phi_n, \hat{H}\phi_m\rangle = \langle \hat{H}\phi_n, \phi_m\rangle$. Identify which step fails when the eigenvalues coincide, and what replaces it.
3. (Measure theory.) Normalisability is the statement that $\psi \in L^2$. Explain why $L^2$ rather than $L^1$ is the right space, and why $\int|\psi|^2 = 1$ fixes the scale of the state rather than adding a constraint. What would go wrong if we used $L^1$?
4. (Classical mechanics.) Show that $\hat{p} = -i\hbar\partial_x$ and $[\hat{x}, \hat{p}] = i\hbar$ reproduce the canonical Poisson bracket of classical mechanics. Then say precisely why this is a consistency requirement and not a derivation of the Schrodinger equation.
5. (Stochastic analysis.) Write the stochastic Schrodinger equation with a complex noise $dW = dW_1 + i\,dW_2$. Show that unitarity requires the noise to be complex, and explain why a real noise cannot preserve the norm. What does this say about the cost of simulating a wave function stochastically?
6. (Research.) Derive the Hartree equations from the variational principle and show that the resulting energy is an upper bound on the exact ground-state energy. Then explain why that bound does not converge as the basis is improved — and what the missing piece is called.

• Postulates of Quantum Mechanics: The postulates provide the mathematical framework for wave functions and the Schrodinger equation.

• Operators and Observables: Operators represent physical observables and act on wave functions to extract measurable quantities.

• One-Dimensional Problems: The infinite square well and other 1D problems illustrate the application of the Schrodinger equation.

• Calculus

• Linear Algebra

• Vector Calculus

• Quantum Computing

Mistake 1: Forgetting to normalise the wave function A physically valid wave function must satisfy $\int |\psi|^2\,dx = 1$. Students often solve the Schrodinger equation and present the unnormalised solution. The normalisation constant is determined by this integral and affects all expectation values and probabilities. Always normalise before computing physical quantities.

Mistake 2: Applying incorrect boundary conditions For the infinite square well, $\psi = 0$ at the walls because the potential is infinite there. For the finite square well, $\psi$ and $d\psi/dx$ must be continuous at the boundary. For the delta function potential, $\psi$ is continuous but $d\psi/dx$ has a discontinuity. Using the wrong matching conditions gives incorrect energy eigenvalues.

Mistake 3: Assuming all wave functions are normalisable Plane waves $e^{ikx}$ are not normalisable over all space and cannot represent physical states by themselves. Physical states are wave packets constructed by superposing plane waves. The energy spectrum for free particles is continuous, and the eigenfunctions form a continuum rather than a discrete set.

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• Approximation Methods
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• Historical Motivation