---
title: The Schrödinger Equation in One Dimension
module: Wave Mechanics in One Dimension
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  The wave equation for matter cannot be derived; it is postulated and judged by
  experiment. We build the time-dependent Schrödinger equation from the de Broglie
  relations, read Born's probability rule off the complex wave function, and separate
  the time and space dependence to get the time-independent equation whose bound-state
  solutions are the stationary states. The five acceptability conditions on the wave
  function are what force energy to be quantized.
topics: [Wave Mechanics in One Dimension]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 6 — The Schrödinger Equation; §6-1 The Schrödinger Equation in One Dimension"
---

The [de Broglie hypothesis](/quantum-mechanics/matter-waves/de-broglie-waves-and-electron-diffraction)
assigns a wavelength $\lambda = h/p$ to every particle, and the
[two-slit experiment](/quantum-mechanics/matter-waves/the-uncertainty-principle) shows
that electrons interfere. A wave that interferes must obey a wave equation. The
equation governing the matter wave of a nonrelativistic particle was found by
Erwin Schrödinger late in 1925, and it plays the role for atomic, molecular, and
solid-state physics that Newton's second law plays for classical mechanics.[^tl-61]

The Schrödinger equation is not derived. Like Newton's laws, it is a fundamental
postulate, and its authority rests entirely on agreement with experiment. What can
be derived is the _form_ the equation must take, by requiring that it reproduce the
de Broglie relations for a free particle and that it be linear in the wave function.
Schrödinger's equation is nonrelativistic; the relativistic wave equation for the
electron was supplied by Dirac in 1928. Within the nonrelativistic range it is
exact for practical purposes.

## The form dictated by de Broglie

Start from the classical wave equation for the electric field of light, which
propagates at $c$,

$$
\frac{\partial^2 \mathcal{E}}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 \mathcal{E}}{\partial t^2}.
$$

A harmonic solution $\mathcal{E}(x,t) = \mathcal{E}_0\cos(kx - \omega t)$
differentiated twice in space brings down $-k^2$ and twice in time brings down
$-\omega^2$, so substitution gives $k^2 = \omega^2/c^2$, i.e. $\omega = kc$. Using
$\omega = E/\hbar$ and $p = \hbar k$ this reads $E = pc$, the correct energy-momentum
relation for a photon. The classical wave equation encodes the photon dispersion
because it relates a second space derivative to a second time derivative.

For a massive particle the dispersion is different. The nonrelativistic energy is

$$
E = \frac{p^2}{2m} + V,
$$

and substituting the de Broglie relations $E = \hbar\omega$, $p = \hbar k$ gives

$$
\hbar\omega = \frac{\hbar^2 k^2}{2m} + V.
$$

This differs from the photon relation in two ways: it carries the potential energy
$V$, and $\omega$ is quadratic in $k$ rather than linear. Reading off the powers, one
factor of $\omega$ comes from a _first_ time derivative and the factor $k^2$ from a
_second_ space derivative. The wave equation for matter must therefore relate the
first time derivative to the second space derivative and must contain $V$ — the
opposite balance of derivatives from the classical wave equation.[^tl-61b]

A third requirement is **linearity**: every term must be linear in the wave function,
so that any linear combination of solutions for the same potential is again a
solution. Linearity is what lets matter waves add constructively and destructively,
the property interference demands.

> **Requirements on the matter-wave equation.** The equation must
> - reproduce $\hbar\omega = \hbar^2 k^2/2m + V$ for a free particle,
> - relate the first time derivative of the wave function to its second space
>   derivative,
> - contain the potential energy $V$,
> - be linear in the wave function, so superpositions are also solutions.

## The time-dependent equation

The equation meeting these requirements, postulated for a particle of mass $m$ in
one dimension, is the **time-dependent Schrödinger equation**:

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

> **Definition (Time-dependent Schrödinger equation).** The fundamental equation of
> nonrelativistic quantum mechanics. For a particle of mass $m$ in a potential
> $V(x,t)$, the complex wave function $\Psi(x,t)$ evolves so that the kinetic term
> $-\tfrac{\hbar^2}{2m}\partial_{xx}\Psi$ plus the potential term $V\Psi$ equals
> $i\hbar\,\partial_t\Psi$.

The explicit factor $i = \sqrt{-1}$ is the sharpest break from the classical wave
equation. A real cosine cannot solve it: differentiating once in time turns a cosine
into a sine, while the second space derivative returns a cosine, so the two sides
cannot match. The exponential form does work. For a free particle, $V = V_0$
constant, try

$$
\Psi(x,t) = A e^{i(kx - \omega t)} = A\big[\cos(kx-\omega t) + i\sin(kx-\omega t)\big].
$$

Then $\partial_t \Psi = -i\omega\Psi$ and $\partial_{xx}\Psi = -k^2\Psi$, and
substitution reproduces exactly $\hbar\omega = \hbar^2 k^2/2m + V_0$. The free-particle
solution is intrinsically complex; there is no real wave function behind it.

## Born's probability interpretation

Because $\Psi$ is complex it is not itself a measurable field the way the classical
$\mathcal{E}(x,t)$ is — every measurement returns a real number. What is measurable is
the probability of finding the particle, and Max Born identified the rule connecting
$\Psi$ to that probability.[^tl-born]

> **Definition (Born rule).** The probability of finding the particle in the interval
> $\d x$ about $x$ at time $t$ is
> $$
> P(x,t)\,\d x = \Psi^\ast(x,t)\,\Psi(x,t)\,\d x = |\Psi(x,t)|^2\,\d x,
> $$
> where $\Psi^\ast$ is the complex conjugate, obtained by replacing $i$ with $-i$
> wherever it appears. The real quantity $|\Psi|^2$ is the **probability density**.

Physical meaning attaches not to $\Psi$ but to the product $\Psi^\ast\Psi = |\Psi|^2$.
The question "what is waving?" has no answer; $\Psi$ is a computational device whose
squared modulus is a probability distribution. This is the same reading of the wave
function introduced through
[wave packets](/quantum-mechanics/matter-waves/wave-packets-and-the-probability-interpretation), now attached
to the equation that produces $\Psi$.

Since the particle is certainly somewhere, the probabilities over all $x$ sum to one:

$$
\int_{-\infty}^{\infty} \Psi^\ast \Psi\,\d x = 1.
$$

> **Definition (Normalization condition).** A physically admissible wave function
> satisfies $\int_{-\infty}^{\infty} |\Psi|^2\,\d x = 1$. This requires $\Psi \to 0$
> fast enough as $x \to \pm\infty$ that the integral converges; a $\Psi$ that does
> not decay cannot be normalized and does not describe a bound particle.

The normalization requirement is not cosmetic. Together with boundary conditions at
finite $x$, it is what restricts the allowed solutions and forces the energy of a
bound particle to take discrete values, a result worked out in full for the
[square well](/quantum-mechanics/wave-mechanics-1d/particle-in-infinite-and-finite-square-wells).

## Separating time from space

Schrödinger's own applications — the hydrogen atom and the harmonic oscillator — have
potentials that do not depend on time, $V = V(x)$. For such potentials the space and
time dependence of $\Psi$ separate, and the equation collapses to something far
easier to solve. Assume a product form

$$
\Psi(x,t) = \psi(x)\,\phi(t).
$$

Substituting into the time-dependent equation and dividing through by
$\psi(x)\phi(t)$ gives

$$
-\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{\d^2\psi(x)}{\d x^2} + V(x) = i\hbar\,\frac{1}{\phi(t)}\frac{\d\phi(t)}{\d t}.
$$

The left side depends only on $x$ and the right side only on $t$. Two functions of
independent variables can be equal for all $x$ and $t$ only if both equal the same
constant. Call it the **separation constant** $C$. The single partial differential
equation splits into two ordinary ones:

$$
-\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{\d^2\psi(x)}{\d x^2} + V(x) = C,
\qquad
i\hbar\,\frac{1}{\phi(t)}\frac{\d\phi(t)}{\d t} = C.
$$

$$
% caption: Separation of variables factors the space-time wave function into a
% spatial shape fixed by V(x) and a universal phase that rotates at frequency E over
% hbar; the squared modulus of the phase is 1.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[draw, thick, minimum width=30mm, minimum height=12mm, align=center] (full) at (0,0) {full wave\\function};
  \node[draw, minimum width=26mm, minimum height=12mm, align=center] (sp) at (5.2,1.2) {spatial\\factor};
  \node[draw, minimum width=26mm, minimum height=12mm, align=center] (tp) at (5.2,-1.2) {temporal\\factor};
  \draw[->, thick] (full.east) -- (sp.west);
  \draw[->, thick] (full.east) -- (tp.west);
  \node[align=left, font=\scriptsize, text=black] at (5.35,2.35) {shape set by the potential};
  \node[align=left, font=\scriptsize, text=black] at (5.35,-2.35) {phase, same for every state};
\end{tikzpicture}
$$

The time equation carries no potential, so it is solved once and for all. Writing
$\d\phi/\phi = (C/i\hbar)\,\d t = -(iC/\hbar)\,\d t$ and integrating,

$$
\phi(t) = e^{-iCt/\hbar}.
$$

This oscillates at frequency $f = C/h$. But de Broglie fixes the frequency of the
wave to $f = E/h$, so the separation constant is the total energy,

$$
C = E, \qquad \phi(t) = e^{-iEt/\hbar}.
$$

## The time-independent equation

With $C = E$, multiplying the spatial equation through by $\psi(x)$ yields the
**time-independent Schrödinger equation**:

$$
-\frac{\hbar^2}{2m}\frac{\d^2\psi(x)}{\d x^2} + V(x)\,\psi(x) = E\,\psi(x).
$$

> **Definition (Time-independent Schrödinger equation).** For a time-independent
> potential $V(x)$, the spatial factor $\psi(x)$ of a definite-energy state obeys
> $-\tfrac{\hbar^2}{2m}\psi'' + V(x)\psi = E\psi$. This is an ordinary differential
> equation in the single variable $x$; each solution belongs to one allowed energy
> $E$.

The full wave function of a definite-energy state is

$$
\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}.
$$

The time dependence is a pure phase of unit modulus. Its probability density is
therefore static:

$$
|\Psi(x,t)|^2 = \psi^\ast(x)\,e^{+iEt/\hbar}\,\psi(x)\,e^{-iEt/\hbar} = |\psi(x)|^2,
$$

independent of time. A state of definite energy is a **stationary state**: the phase
$e^{-iEt/\hbar}$ rotates in the complex plane, but nothing observable changes.

> **Definition (Stationary state).** A solution $\Psi(x,t) = \psi(x)e^{-iEt/\hbar}$ of
> definite energy $E$. Its probability density $|\Psi|^2 = |\psi|^2$ and every
> [expectation value](/quantum-mechanics/wave-mechanics-1d/operators-expectation-values-and-the-harmonic-oscillator)
> of a time-independent quantity are constant in time. These are the states Bohr
> called stationary orbits, now recovered as standing matter waves.

$$
% caption: In a stationary state the complex amplitude at a point traces a circle in
% the complex plane at angular rate E over hbar, while its squared length, the
% probability density, stays fixed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % complex plane
  \draw[->, black] (-2.1,0) -- (2.1,0) node[right, font=\scriptsize, text=black] {Re};
  \draw[->, black] (0,-2.1) -- (0,2.1) node[above, font=\scriptsize, text=black] {Im};
  \draw[thick] (0,0) circle (1.5);
  % phasor at some angle
  \draw[->, acc, very thick] (0,0) -- (60:1.5);
  \fill[acc] (60:1.5) circle (2pt);
  \node[acc, anchor=south west, font=\scriptsize] at (60:1.55) {state};
  % rotation arrow
  \draw[->, black] (20:1.85) arc[start angle=20, end angle=95, radius=1.85];
  \node[black!70, font=\scriptsize] at (58:2.15) {rotates};
  \node[align=center, font=\scriptsize, text=black] at (0,-2.6) {radius constant};
\end{tikzpicture}
$$

The normalization condition inherits the same simplification. Because the phase
cancels in $\Psi^\ast\Psi$,

$$
\int_{-\infty}^{\infty} \psi^\ast(x)\,\psi(x)\,\d x = 1,
$$

a condition on the spatial factor alone.

## Effect of the potential on the wave function

Rearranged, the time-independent equation reads

$$
\frac{\d^2\psi}{\d x^2} = \frac{2m}{\hbar^2}\big[V(x) - E\big]\,\psi(x).
$$

The second derivative is the **curvature** of $\psi$. Its sign relative to $\psi$ is
fixed by whether the total energy exceeds the potential:

- **$E > V(x)$ (classically allowed).** The bracket is negative, so $\psi''$ and
  $\psi$ have opposite signs. Wherever $\psi$ is positive it curves down toward the
  axis, and wherever it is negative it curves up toward the axis. The wave function
  **oscillates**, like a sine or cosine. Larger $E - V$ means larger curvature, hence
  shorter wavelength — the classical statement that a faster particle has a shorter
  de Broglie wavelength.
- **$E < V(x)$ (classically forbidden).** The bracket is positive, so $\psi''$ and
  $\psi$ share a sign. A positive $\psi$ curves up, away from the axis; a negative
  $\psi$ curves down, away from the axis. The wave function grows or decays
  **exponentially** rather than oscillating.

$$
% caption: The sign of E minus V fixes the curvature: where the energy exceeds the
% potential the wave function bends toward the axis and oscillates; where the
% potential exceeds the energy it bends away and runs off exponentially.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % --- left panel: oscillatory ---
  \draw[->, black] (-0.3,0) -- (4.4,0) node[right, font=\scriptsize, text=black] {$x$};
  \draw[acc, very thick] (0,0) sin (0.8,1.1) cos (1.6,0) sin (2.4,-1.1) cos (3.2,0) sin (4.0,1.1);
  \node[align=center, font=\scriptsize, text=black] at (2.0,-1.75) {$E > V$: curves toward axis};
  \node[align=center, font=\scriptsize, text=acc] at (2.0,1.65) {oscillates};
  % --- right panel: exponential ---
  \begin{scope}[xshift=6.2cm]
    \draw[->, black] (-0.3,0) -- (4.4,0) node[right, font=\scriptsize, text=black] {$x$};
    \draw[acc, very thick] (0,0.15) .. controls (1.6,0.35) and (2.8,0.9) .. (4.0,2.0);
    \node[align=center, font=\scriptsize, text=black] at (2.0,-1.75) {$E < V$: curves away from axis};
    \node[align=center, font=\scriptsize, text=acc] at (1.5,1.6) {grows};
  \end{scope}
\end{tikzpicture}
$$

This curvature rule is the qualitative engine behind every problem in the chapter.
In a bound-state problem the particle is allowed in a central region and forbidden
outside it. Inside, $\psi$ oscillates; outside, it must decay. Only for special
values of $E$ do the oscillating interior and the decaying exterior join smoothly
with a solution that also vanishes at infinity. Those special energies are the
quantized levels.

## Acceptable wave functions

The form of $\psi(x)$ depends on $V(x)$, but every admissible solution obeys the same
regularity conditions. Where the potential jumps between regions, one solves the
equation separately in each region and joins the pieces.

Because the probability density cannot change discontinuously from point to point,
$\psi$ must be continuous. Because the equation involves $\psi'' = \d^2\psi/\d x^2$, the
first derivative $\psi'$ (the slope) must also be continuous, so the graph of $\psi$
is smooth. The one exception is where $V \to \infty$: no particle can have infinite
potential energy, so $\psi = 0$ there, and at the boundary of such a region $\psi'$
may be discontinuous.

$$
% caption: A continuous wave function joining two regions smoothly, versus rejected
% candidates with a kink in the slope or a jump in value; the physics admits only the
% smooth match except at an infinite wall.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % accepted
  \draw[->, black] (-0.2,0) -- (3.2,0) node[right, font=\scriptsize, text=black]{$x$};
  \draw[dashed, black] (1.5,-1.2) -- (1.5,1.3);
  \draw[acc, very thick] (0,-0.6) .. controls (0.8,0.6) and (1.2,1.0) .. (1.5,1.0) .. controls (1.9,1.0) and (2.4,0.2) .. (3.0,-0.7);
  \node[font=\scriptsize, text=black] at (1.5,-1.6) {smooth match: accepted};
  % kinked slope
  \begin{scope}[xshift=4.4cm]
    \draw[->, black] (-0.2,0) -- (3.2,0) node[right, font=\scriptsize, text=black]{$x$};
    \draw[dashed, black] (1.5,-1.2) -- (1.5,1.3);
    \draw[black, very thick] (0,-0.6) -- (1.5,1.0) -- (3.0,-0.7);
    \node[font=\scriptsize, text=black] at (1.5,-1.6) {kinked slope: rejected};
  \end{scope}
  % jump in value
  \begin{scope}[xshift=8.8cm]
    \draw[->, black] (-0.2,0) -- (3.2,0) node[right, font=\scriptsize, text=black]{$x$};
    \draw[dashed, black] (1.5,-1.2) -- (1.5,1.3);
    \draw[black, very thick] (0,-0.6) .. controls (0.8,0.5) and (1.2,0.8) .. (1.5,0.8);
    \draw[black, very thick] (1.5,-0.2) .. controls (1.9,-0.3) and (2.4,-0.6) .. (3.0,-0.8);
    \fill[black] (1.5,0.8) circle (1.6pt);
    \fill[black] (1.5,-0.2) circle (1.6pt);
    \node[font=\scriptsize, text=black] at (1.5,-1.6) {jump in value: rejected};
  \end{scope}
\end{tikzpicture}
$$

Collecting the requirements gives the checklist every wave function must pass.[^tl-accept]

> **Conditions for an acceptable wave function.** The spatial factor $\psi(x)$ must:
> 1. exist and satisfy the Schrödinger equation;
> 2. have $\psi$ and $\d\psi/\d x$ continuous (except $\d\psi/\d x$ at an infinite-potential
>    wall);
> 3. have $\psi$ and $\d\psi/\d x$ finite;
> 4. have $\psi$ and $\d\psi/\d x$ single-valued;
> 5. satisfy $\psi \to 0$ fast enough as $x \to \pm\infty$ that the normalization
>    integral stays bounded.

Conditions 3 and 4 exist because measurable quantities such as position and momentum
are always finite and single-valued; if $\psi$ or $\psi'$ failed either, the
predictions built from them would not be. Condition 5 is normalizability. Together
these conditions are the mathematical content of "bound state": a particle trapped in
a region must have a wave function that decays to nothing far away, and only special
energies produce such a function.

## A free-particle solution

For a free particle $V(x) = 0$, so the time-independent equation is

$$
-\frac{\hbar^2}{2m}\frac{\d^2\psi}{\d x^2} = E\psi
\quad\Longrightarrow\quad
\frac{\d^2\psi}{\d x^2} = -k^2\psi,
\qquad k^2 = \frac{2mE}{\hbar^2}.
$$

Any combination $\psi(x) = A\sin kx + B\cos kx$ solves this. Differentiating twice,

$$
\frac{\d^2\psi}{\d x^2} = -k^2 A\sin kx - k^2 B\cos kx = -k^2\psi(x),
$$

and since $\hbar^2 k^2 = 2mE$, substitution gives $E\psi = E\psi$, an identity, so the
combination is a solution for any $A$, $B$.[^tl-ex61] With no boundary to confine it,
$k$ is unrestricted and $E = \hbar^2 k^2/2m$ takes any nonnegative value: the free
particle has a continuous energy spectrum. Confinement is what changes this. Impose a
boundary and only discrete $k$ survive, which is the subject of the
[infinite and finite square wells](/quantum-mechanics/wave-mechanics-1d/particle-in-infinite-and-finite-square-wells).
The continuum itself, and how genuinely localized states are assembled from it, is
taken up next for the
[free particle and its wave packets](/quantum-mechanics/wave-mechanics-1d/the-free-particle-and-wave-packet-dynamics).

[^tl-61]: **Tipler & Llewellyn**, _Modern Physics_, §6-1 — the Schrödinger equation is postulated, not derived, and validated by experiment; nonrelativistic, superseded at high speed by the Dirac equation.
[^tl-61b]: **Tipler & Llewellyn**, _Modern Physics_, §6-1 — building the equation from the photon dispersion and the de Broglie relations, showing it must relate the first time derivative to the second space derivative and carry $V$.
[^tl-born]: **Tipler & Llewellyn**, _Modern Physics_, §6-1 — Born's probabilistic interpretation of the complex wave function, $P(x,t)\,\d x = \Psi^\ast\Psi\,\d x = |\Psi|^2\,\d x$, and the normalization condition.
[^tl-accept]: **Tipler & Llewellyn**, _Modern Physics_, §6-1 — the five conditions for an acceptable wave function and the continuity of $\psi$ and $\d\psi/\d x$ at a potential discontinuity.
[^tl-ex61]: **Tipler & Llewellyn**, _Modern Physics_, §6-1, Example 6-1 — verification that $A\sin kx + B\cos kx$ solves the free-particle time-independent equation.