Schrödinger Equation

The Schrödinger equation is the closed-system dynamical law of nonrelativistic quantum mechanics. Once a Hamiltonian and an initial state are specified, it determines the state at later times.

In abstract Hilbert-space form,

$$i\hbar \frac{d}{dt} \lvert\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.$$

This equation is linear and first order in time. It preserves quantum superposition and, for a self-adjoint Hamiltonian, generates unitary evolution.

The equation does not construct the Hamiltonian, model a measurement outcome, or by itself describe dissipative open-system evolution. Those are separate layers of the theory.

Abstract Form

Let $\mathcal H$ be the system Hilbert space and let $H(t)$ be its Hamiltonian. The state curve

$$t \longmapsto \lvert\psi(t)\rangle \in \mathcal H$$

obeys

$$\lvert\dot\psi(t)\rangle = -\frac{i}{\hbar} H(t)\lvert\psi(t)\rangle.$$

The right side is tangent to unitary motion in Hilbert space. The factor $i$ is essential: replacing $-iH/\hbar$ by a real negative generator would generally produce decay rather than norm-preserving evolution.

Physical dimensions

The derivative has units of inverse time, so

$$\frac{H}{\hbar}$$

has units of angular frequency. Every term in $H$ must have units of energy.

Schrödinger picture

This page uses the Schrödinger picture: state vectors carry the time dependence, while an observable with no explicit control dependence is represented by a fixed operator. The Heisenberg and interaction pictures redistribute time dependence without changing measurable predictions; see Pictures of Motion Overview.

An Initial-Value Problem

The dynamical data are:

1. a Hamiltonian $H(t)$;
2. an initial time $t_0$;
3. a normalized initial state $\lvert\psi_0\rangle$.

The initial condition is

$$\lvert\psi(t_0)\rangle = \lvert\psi_0\rangle.$$

The solution can be written

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi_0\rangle,$$

where the propagator satisfies

$$i\hbar \frac{\partial U(t,t_0)}{\partial t} = H(t)U(t,t_0),$$

and

$$U(t_0,t_0) = I.$$

Because the equation is first order in time, one initial state is enough. There is no independent initial time derivative analogous to the second initial datum for a second-order wave equation.

For a time-independent self-adjoint $H$, Stone’s theorem gives the unique unitary solution

$$U(t,t_0) = \exp \left[ -\frac{i}{\hbar} H(t-t_0) \right].$$

Time-Evolution Operator develops composition, inverses, and time ordering.

Linearity and Superposition

Suppose $\lvert\psi_1(t)\rangle$ and $\lvert\psi_2(t)\rangle$ solve the same Schrödinger equation. For time-independent complex numbers $a$ and $b$,

$$\lvert\chi(t)\rangle = a\lvert\psi_1(t)\rangle + b\lvert\psi_2(t)\rangle$$

also solves it:

$$\begin{aligned} i\hbar \lvert\dot\chi\rangle &= aH\lvert\psi_1\rangle + bH\lvert\psi_2\rangle\\ &= H\lvert\chi\rangle. \end{aligned}$$

Linearity preserves coherent amplitudes, including their relative phases. It does not mean probabilities add for coherent alternatives; amplitudes must be combined before applying the Born rule.

Time-dependent coefficients $a(t)$ and $b(t)$ do not generally produce another solution, because their derivatives add extra terms.

Norm Preservation

Let $H(t)$ be self-adjoint and let $\lvert\psi(t)\rangle$ be a sufficiently regular solution. From

$$\lvert\dot\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle$$

and

$$\langle\dot\psi\rvert = \frac{i}{\hbar} \langle\psi\rvert H,$$

one obtains

$$\begin{aligned} \frac{d}{dt} \langle\psi\rvert\psi\rangle &= \langle\dot\psi\rvert\psi\rangle + \langle\psi\rvert\dot\psi\rangle\\ &= \frac{i}{\hbar} \langle\psi\rvert H\lvert\psi\rangle - \frac{i}{\hbar} \langle\psi\rvert H\lvert\psi\rangle\\ &= 0. \end{aligned}$$

An initially normalized state remains normalized.

The same calculation for two solutions gives

$$\frac{d}{dt} \langle\phi(t)\rvert\psi(t)\rangle = 0.$$

Thus all inner products are preserved. Orthogonal states remain orthogonal, and transition probabilities between coevolving states remain fixed.

Domain and Regularity Caveat

For an unbounded time-independent Hamiltonian, the unitary operator

$$e^{-iH(t-t_0)/\hbar}$$

acts on every vector in $\mathcal H$. The strong derivative

$$\frac{d}{dt} \lvert\psi(t)\rangle$$

need not exist as a Hilbert-space vector for every initial state. If

$$\lvert\psi_0\rangle \in \mathcal D(H),$$

then the orbit is a strong solution and the differential equation holds in the direct operator sense.

For less regular states, the unitary evolution remains well defined, while the equation may need a weak or integral interpretation. This is not a failure of dynamics; it is a reminder that an unbounded generator cannot act on every Hilbert-space vector.

For $H(t)$, existence of a propagator requires suitable assumptions on self-adjoint domains and time dependence. Finite-dimensional matrix models again hide these issues.

Matrix Form in a Fixed Basis

Choose a time-independent orthonormal basis $\{\lvert e_n\rangle\}$ and write

$$\lvert\psi(t)\rangle = \sum_n c_n(t)\lvert e_n\rangle.$$

Taking $\langle e_m\rvert$ of the abstract equation gives

$$i\hbar \dot c_m(t) = \sum_n H_{mn}(t)c_n(t),$$

where

$$H_{mn}(t) = \langle e_m\rvert H(t)\lvert e_n\rangle.$$

In vector notation,

$$i\hbar \dot{\mathbf c}(t) = H(t)\mathbf c(t).$$

This is a coupled system of first-order linear ordinary differential equations.

Time-dependent basis warning

If the basis vectors themselves depend on time,

$$\lvert\psi(t)\rangle = \sum_n c_n(t)\lvert e_n(t)\rangle,$$

then

$$\begin{aligned} i\hbar\dot c_m &= \sum_n \bigg[ H_{mn} - i\hbar \langle e_m\rvert\dot e_n\rangle \bigg] c_n. \end{aligned}$$

The additional connection term comes from differentiating the basis. Omitting it changes the dynamics.

Position Representation

The position-space wavefunction is

$$\psi(x,t) = \langle x\rvert\psi(t)\rangle.$$

For a general Hamiltonian with kernel

$$H(x,x';t) = \langle x\rvert H(t)\lvert x'\rangle,$$

the Schrödinger equation is

$$i\hbar \frac{\partial\psi(x,t)}{\partial t} = \int H(x,x';t) \psi(x',t)\,dx'.$$

The familiar partial differential equation is the local special case. For one particle in one dimension with

$$H(t) = \frac{P^2}{2m} + V(X,t),$$

one obtains

$$\begin{aligned} i\hbar \frac{\partial\psi(x,t)}{\partial t} &= \bigg[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\\ &\qquad + V(x,t) \bigg] \psi(x,t). \end{aligned}$$

This PDE must be accompanied by an initial wavefunction and the spatial domain or boundary conditions that define the Hamiltonian.

The wavefunction is a representation of the state, not an additional physical object alongside the ket. Wavefunctions as Representations develops this point.

Probability Conservation in Position Space

For a real local potential, define

$$\rho(x,t) = \lvert\psi(x,t)\rvert^2$$

and probability current

$$j(x,t) = \frac{\hbar}{m} \operatorname{Im} \left[ \psi^*(x,t) \frac{\partial\psi(x,t)}{\partial x} \right].$$

The Schrödinger equation and its complex conjugate imply the continuity equation

$$\frac{\partial\rho}{\partial t} + \frac{\partial j}{\partial x} = 0.$$

Integrating over a region gives

$$\frac{d}{dt} \int_a^b \rho(x,t)\,dx = j(a,t)-j(b,t).$$

Probability changes inside the region only through boundary flux. Boundary conditions that make the net flux vanish preserve total normalization.

This local statement is the position-space version of abstract norm preservation.

Time-Independent Hamiltonians

If $H$ has no explicit time dependence and

$$H\lvert E_n\rangle = E_n\lvert E_n\rangle,$$

then

$$\lvert\psi_n(t)\rangle = e^{-iE_n(t-t_0)/\hbar} \lvert E_n\rangle$$

solves the time-dependent equation.

For an initial state expanded in a discrete energy basis,

$$\lvert\psi(t_0)\rangle = \sum_n c_n\lvert E_n\rangle,$$

the solution is

$$\lvert\psi(t)\rangle = \sum_n c_n e^{-iE_n(t-t_0)/\hbar} \lvert E_n\rangle.$$

The energy probabilities $\lvert c_n\rvert^2$ remain constant. Relative phases between different energies evolve and can make other observables time dependent.

Time-Dependent Versus Time-Independent Equations

The time-dependent Schrödinger equation

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

is the dynamical law.

The time-independent Schrödinger equation

$$H\phi = E\phi$$

is an eigenvalue problem. It does not contain a time derivative and is not a second competing law of motion.

For time-independent $H$, a separated solution

$$\psi(x,t) = \phi(x)T(t)$$

leads to

$$T(t) = e^{-iE(t-t_0)/\hbar}$$

and

$$H\phi = E\phi.$$

The spatial eigenvalue problem supplies modes; the time-dependent equation evolves their amplitudes and phases. Time-Independent Schrödinger Equation owns boundary-value methods and spectral examples.

Time-Dependent Hamiltonians

When $H(t)$ depends explicitly on time, the equation remains

$$i\hbar \lvert\dot\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.$$

The formal solution is

$$\lvert\psi(t)\rangle = \mathcal T \exp \left[ -\frac{i}{\hbar} \int_{t_0}^{t} H(s)\,ds \right] \lvert\psi(t_0)\rangle,$$

where $\mathcal T$ orders later Hamiltonians to the left. Time ordering is unnecessary when

$$[H(t),H(t')] = 0$$

for all relevant times.

The detailed Dyson expansion and driven examples belong in Time-Dependent Hamiltonians.

Density-Operator Form

If

$$\rho(t) = \sum_k w_k \lvert\psi_k(t)\rangle \langle\psi_k(t)\rvert$$

with fixed ensemble weights, applying the Schrödinger equation to every ket and bra gives

$$i\hbar \frac{d\rho}{dt} = [H(t),\rho(t)].$$

The solution is

$$\rho(t) = U(t,t_0) \rho(t_0) U^\dagger(t,t_0).$$

Unitary evolution preserves the eigenvalues of $\rho$, and therefore preserves purity and von Neumann entropy. A changing mixedness requires averaging, measurement, discarded degrees of freedom, or other open-system structure beyond the closed-system equation.

Example: Relative Phase in a Two-Level System

Let

$$H = \frac{\hbar\omega}{2} \sigma_z$$

and prepare

$$\lvert\psi(0)\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

Because the energies are $\pm\hbar\omega/2$,

$$\lvert\psi(t)\rangle = \frac{ e^{-i\omega t/2}\lvert0\rangle + e^{i\omega t/2}\lvert1\rangle }{\sqrt2}.$$

Computational-basis probabilities remain $1/2$. In the Hadamard basis,

$$\begin{aligned} p(+;t) &= \cos^2\frac{\omega t}{2},\\ p(-;t) &= \sin^2\frac{\omega t}{2}. \end{aligned}$$

The Schrödinger equation changes relative phase, which a rotated measurement converts into oscillating probabilities.

Closed-System Boundary

The Schrödinger equation describes deterministic, norm-preserving evolution for a closed system between interventions. It should not be used alone to model:

• a selected measurement outcome;
• discarded environmental degrees of freedom;
• irreversible relaxation or dephasing;
• stochastic classical control noise after averaging;
• particle creation in a fixed-particle nonrelativistic Hilbert space;
• relativistic quantum fields.

A larger closed system can obey a Schrödinger equation while a subsystem obeys a nonunitary reduced equation. The boundary depends on which degrees of freedom are retained.

The equation also does not derive $H$. Choosing the Hamiltonian remains a modeling task constrained by symmetry, experiment, and approximation scales.

A Reliable Solution Workflow

1. Declare the Hilbert space and representation.
2. Specify $H(t)$, including its domain or boundary conditions when needed.
3. State the normalized initial condition at $t_0$.
4. Decide whether $H$ is time independent, commuting at different times, or genuinely time ordered.
5. Use a spectral expansion, matrix ODE, PDE method, or propagator suited to the model.
6. Verify the initial condition.
7. Substitute the result back into the differential equation.
8. Check normalization and boundary flux.
9. Test dimensions and limiting cases.
10. Compute observables from the evolved state rather than interpreting amplitudes before specifying a measurement.

Common Mistakes

• Confusing the time-dependent equation with the time-independent eigenvalue problem.
• Forgetting the factor of $i$ or $\hbar$.
• Supplying an independent initial time derivative to a first-order equation.
• Treating the position-space wavefunction as different from the abstract state.
• Forgetting spatial boundary conditions or the Hamiltonian domain.
• Assuming every formal Hilbert-space orbit has a strong derivative under an unbounded $H$.
• Applying a time-independent exponential to noncommuting $H(t)$.
• Omitting the connection term in a time-dependent basis.
• Adding probabilities rather than amplitudes for coherent solutions.
• Calling every superposition of energy eigenstates stationary.
• Using nonunitary phenomenological decay inside a ket equation without specifying the effective or conditional interpretation.
• Treating the closed-system equation as a measurement-update rule.

References

• E. Schrödinger, “Quantisierung als Eigenwertproblem,” Annalen der Physik 384, 361–376, 1926.
• E. Schrödinger, “An Undulatory Theory of the Mechanics of Atoms and Molecules,” Physical Review 28, 1049–1070, 1926.
• M. H. Stone, “On one-parameter unitary groups in Hilbert space,” Annals of Mathematics 33, 643–648, 1932.
• A. Messiah, Quantum Mechanics, Dover Publications, 1999.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

Exercises

1. Check dimensions

Use dimensional analysis to show that both sides of

$$i\hbar \frac{d}{dt} \lvert\psi\rangle = H\lvert\psi\rangle$$

have the same units.

Solution

The state vector carries the same units on both sides and can be treated abstractly as normalized. Since

$$[\hbar] = \text{energy}\times\text{time},$$

the factor $\hbar\,d/dt$ has units of energy. Therefore the left side has units of energy times state, matching $H\lvert\psi\rangle$.

2. Prove norm preservation

Assume $H(t)=H^\dagger(t)$ and that $\lvert\psi(t)\rangle$ is a strong solution. Derive

$$\frac{d}{dt} \langle\psi(t)\rvert\psi(t)\rangle = 0.$$

Solution

The equation and its adjoint give

$$\lvert\dot\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle,$$

and

$$\langle\dot\psi\rvert = \frac{i}{\hbar} \langle\psi\rvert H.$$

Hence

$$\begin{aligned} \frac{d}{dt} \langle\psi\rvert\psi\rangle &= \langle\dot\psi\rvert\psi\rangle + \langle\psi\rvert\dot\psi\rangle\\ &= \frac{i}{\hbar}\langle\psi\rvert H\lvert\psi\rangle - \frac{i}{\hbar}\langle\psi\rvert H\lvert\psi\rangle\\ &= 0. \end{aligned}$$

3. Test superposition

Let $\lvert\psi_1(t)\rangle$ and $\lvert\psi_2(t)\rangle$ solve the same equation with Hamiltonian $H(t)$. Show that

$$a\lvert\psi_1(t)\rangle + b\lvert\psi_2(t)\rangle$$

is also a solution for constant $a,b\in\mathbb C$. Explain why $a(t)$ and $b(t)$ generally fail.

Solution

For constant coefficients, define

$$\lvert\chi\rangle = a\lvert\psi_1\rangle + b\lvert\psi_2\rangle.$$

Linearity then gives

$$\begin{aligned} i\hbar\frac{d}{dt}\lvert\chi\rangle &= aH\lvert\psi_1\rangle + bH\lvert\psi_2\rangle\\ &= H\lvert\chi\rangle. \end{aligned}$$

If the coefficients depend on time, differentiation adds

$$i\hbar \left( \dot a\lvert\psi_1\rangle + \dot b\lvert\psi_2\rangle \right),$$

which is not present on the right side in general.

4. Verify an energy eigenstate solution

If

$$H\lvert E\rangle = E\lvert E\rangle,$$

verify that

$$\lvert\psi(t)\rangle = e^{-iE(t-t_0)/\hbar} \lvert E\rangle$$

solves the time-dependent equation and the initial condition $\lvert\psi(t_0)\rangle=\lvert E\rangle$.

Solution

The time derivative is

$$\lvert\dot\psi(t)\rangle = -\frac{iE}{\hbar} e^{-iE(t-t_0)/\hbar} \lvert E\rangle.$$

Therefore

$$i\hbar \lvert\dot\psi(t)\rangle = E e^{-iE(t-t_0)/\hbar} \lvert E\rangle.$$

The Hamiltonian gives the same result:

$$H\lvert\psi(t)\rangle = E e^{-iE(t-t_0)/\hbar} \lvert E\rangle.$$

At $t=t_0$, the exponential equals one.

5. Observe a changing relative phase

For the two-level example on this page, derive $p(+;t)$, where

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

Solution

The amplitude is

$$\begin{aligned} \langle+\rvert\psi(t)\rangle &= \frac12 \left( e^{-i\omega t/2} + e^{i\omega t/2} \right)\\ &= \cos\frac{\omega t}{2}. \end{aligned}$$

Thus

$$p(+;t) = \left\lvert \langle+\rvert\psi(t)\rangle \right\rvert^2 = \cos^2\frac{\omega t}{2}.$$

6. Separate variables carefully

For time-independent

$$H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x),$$

insert $\psi(x,t)=\phi(x)T(t)$ into the time-dependent equation and derive both separated equations.

Solution

Substitution gives

$$i\hbar \phi(x)\frac{dT}{dt} = T(t)H\phi(x).$$

After dividing by $\phi T$ where nonzero,

$$i\hbar \frac{1}{T} \frac{dT}{dt} = \frac{1}{\phi} H\phi.$$

The left side depends only on $t$ and the right side only on $x$, so both equal a separation constant $E$. Therefore

$$H\phi = E\phi,$$

and

$$i\hbar \frac{dT}{dt} = ET.$$

The temporal solution is

$$T(t) = T(t_0) e^{-iE(t-t_0)/\hbar}.$$

7. Derive the continuity equation

For

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

with real $V$, derive

$$\frac{\partial}{\partial t} \lvert\psi\rvert^2 + \frac{\partial j}{\partial x} = 0$$

and identify $j$.

Solution

Multiply the Schrödinger equation by $\psi^*$ and its complex conjugate by $\psi$, then subtract. The real potential terms cancel:

$$i\hbar \frac{\partial}{\partial t} \lvert\psi\rvert^2 = -\frac{\hbar^2}{2m} \left( \psi^*\frac{\partial^2\psi}{\partial x^2} - \psi\frac{\partial^2\psi^*}{\partial x^2} \right).$$

Writing $\partial_x=\partial/\partial x$ and using c.c. for the complex conjugate of the immediately preceding term, use

$$\psi^*\partial_x^2\psi -\mathrm{c.c.} = \partial_x \left( \psi^*\partial_x\psi -\mathrm{c.c.} \right).$$

Then

$$\frac{\partial}{\partial t} \lvert\psi\rvert^2 + \frac{\partial j}{\partial x} = 0,$$

with

$$\begin{aligned} j &= \frac{\hbar}{2mi} \left( \psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^*}{\partial x} \right)\\ &= \frac{\hbar}{m} \operatorname{Im} \left( \psi^*\frac{\partial\psi}{\partial x} \right). \end{aligned}$$

8. Derive density-operator evolution

Let

$$\rho(t) = \sum_k w_k \lvert\psi_k(t)\rangle \langle\psi_k(t)\rvert,$$

where every $\lvert\psi_k(t)\rangle$ obeys the same Schrödinger equation and the weights are constant. Derive the von Neumann equation.

Solution

For $P_k=\lvert\psi_k\rangle\langle\psi_k\rvert$, differentiation gives

$$\begin{aligned} \dot P_k &= \lvert\dot\psi_k\rangle\langle\psi_k\rvert + \lvert\psi_k\rangle\langle\dot\psi_k\rvert\\ &= -\frac{i}{\hbar} HP_k + \frac{i}{\hbar} P_kH. \end{aligned}$$

Summing with weights gives

$$\dot\rho = -\frac{i}{\hbar} [H,\rho].$$

Equivalently,

$$i\hbar\dot\rho = [H,\rho].$$

Summary

The Schrödinger equation is a linear, first-order initial-value equation for closed-system quantum states:

$$i\hbar \lvert\dot\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.$$

Self-adjoint Hamiltonians generate unitary evolution, preserving norms and inner products. Matrix ODEs, position-space PDEs, spectral phase evolution, and the von Neumann equation are representations or consequences of the same abstract law.

The time-independent equation is an energy eigenvalue problem used to construct modes; it is not a separate dynamical postulate. Domains, boundary conditions, time ordering, and the closed-system boundary must be stated whenever they matter.