Wavefunction Normalization

Equivalent forms

\langle\psi|\psi\rangle=1

The most basic sanity check in quantum mechanics: probabilities add to one, and that alone pins the amplitude's scale.

Symbol	Name	SI unit	Dimension	Range
$\psi$	Wavefunction 1D probability amplitude	m^{-1/2}	L^-1/2	—
$\|\psi\|^2$	Probability densityoutput Probability per unit length	m^-1	L^-1	—
$x$	Position Integration variable	m	L	-5 – 5
$A$	Normalization constant Overall amplitude fixed by the condition	m^{-1/2}	L^-1/2	—

Where it holds

Applies to bound (square-integrable) states. Momentum/position eigenstates are non-normalizable and use Dirac-delta normalization ⟨k|k'⟩

= \delta (k-k')

.

Dimensional analysis

\int |\psi |^{2} dx \to [L^-1][L]

= dimensionless

1 \to

dimensionless

A probability is a pure number; |

\psi |^{2}

must therefore carry 1/length in 1D.

\checkmark

Discovery

Max Born · 1926

Born's probabilistic interpretation of |ψ|² immediately demanded that the total probability be one. It converted Schrödinger's abstract wave into something measurable and imposed square-integrability as the criterion for a physical state — a footnote in Born's scattering paper that reshaped quantum mechanics.

Research status: stable

Real-world applications

Setting amplitudes of atomic and molecular orbitals
Monte-Carlo and variational quantum chemistry
Defining valid initial states for time evolution

Common misconceptions

Normalization fixes only the magnitude, not the phase — an overall $e^{i\theta }$ is still free
Plane waves cannot be normalized to 1 over infinite space
| $\psi |^{2} is$ a density (per length in 1D), not a probability itself

Experimental verification

Not measured directly; it is the consistency condition that makes measured relative frequencies sum to 100%. Time-independence

of \int |\psi |^{2}

(probability conservation) follows from the Schrödinger equation.

Derivation

Born: probability density

P(x) = |\psi (x)|^{2}

.

Total probability of the particle being somewhere

is \int P dx = 1

.

For a trial function

\psi = A \varphi (x)

, require |A|

^{2} \int |\varphi |^{2} dx = 1

, so

A = 1/\sqrt{\int |\varphi |^{2} dx}

.

The Schrödinger equation is linear, so multiplying

\psi by

this constant leaves it a valid solution while fixing the scale.

What if…

What

if \int |\psi |^{2}

diverges?

The state is non-normalizable and not a physical bound state — you need wave packets or delta-normalization.

What if you multiply

\psi by e^{i\theta }

?

Nothing observable changes: | $\psi |^{2} and all$ expectation values are unaffected by a global phase.

1

Normalize a Gaussian

Given ·

\psi (x) = A e^{-x^{2}/2\sigma ^{2}}

Find · A

Steps

$\int e^{-x^{2}/\sigma ^{2}} dx = \sqrt{\pi }\cdot \sigma$
|A| $^{2} \surd \pi \sigma = 1$
$A = 1/(\pi ^{1/4}\surd \sigma ) = (\pi \sigma ^{2})^{-1/4}$

Result ·

A = (\pi \sigma ^{2})^{-1/4}

Born Rule→Expectation Value of an Observable→The Schrödinger Equation→