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 time-independent self-adjoint Hamiltonian, generates unitary evolution. Time-dependent unbounded Hamiltonians require additional domain and regularity assumptions.

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.

Helpful background. State Vectors supplies ket notation; Hamiltonians explains the generator of motion.

Schrödinger Dynamics in Hilbert Space

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

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

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 develops 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;

The abstract Schrödinger equation can also generate dynamics in quantum field theory; the required Hilbert space and Hamiltonian differ from the fixed-particle models discussed here.

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.

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.

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,

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

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• A. Messiah, Quantum Mechanics, Dover Publications, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. H. Stone, “On one-parameter unitary groups in Hilbert space,” Annals of Mathematics 33, 643–648, 1932.