Plane Waves and Delta Normalization

Plane waves are the natural language of a free particle, but they are not ordinary square-normalizable wavefunctions on the full line. They are generalized eigenfunctions: basis objects used to build physical wave packets. Delta normalization is the bookkeeping that makes this continuum basis behave like an orthonormal basis.

This page uses the momentum convention

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}.$$

With this convention,

$$\langle p\vert p'\rangle = \delta(p-p').$$

The corresponding Fourier-transform conventions are summarized in Fourier Transform and Momentum-Space Representation.

Why Plane Waves Are Not Normalizable

For a nonzero constant amplitude $A$,

$$\psi_p(x)=A e^{ipx/\hbar}$$

has constant probability density:

$$\lvert\psi_p(x)\rvert^2 = \lvert A\rvert^2.$$

On the full real line,

$$\int_{-\infty}^{\infty} \lvert\psi_p(x)\rvert^2\,dx$$

diverges. A plane wave has perfectly definite momentum and is completely delocalized. It is therefore not a physical one-particle state with total probability one on the real line.

This does not make plane waves invalid. It means they are like Fourier basis functions: not themselves localized signals, but indispensable for decomposing localized states.

Delta Normalization

Momentum eigenkets are normalized by the Dirac delta distribution:

$$\langle p\vert p'\rangle = \delta(p-p').$$

In the position representation, this follows from

$$\begin{aligned} \langle p\vert p'\rangle &= \int_{-\infty}^{\infty} \langle p\vert x\rangle \langle x\vert p'\rangle\,dx \\ &= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{i(p'-p)x/\hbar}\,dx \\ &= \delta(p-p'). \end{aligned}$$

The last line is a distributional identity, not an ordinary convergent integral. It becomes meaningful when integrated against sufficiently well-behaved wave packets.

Completeness

The continuum analogue of an orthonormal basis resolution is

$$\int_{-\infty}^{\infty} dp\, \lvert p\rangle\langle p\rvert = \hat I.$$

In position space this gives

$$\int_{-\infty}^{\infty} dp\, \langle x\vert p\rangle \langle p\vert x'\rangle = \delta(x-x').$$

Using the plane-wave convention above,

$$\frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{ip(x-x')/\hbar}\,dp = \delta(x-x').$$

This is the statement that plane waves are complete as a continuum basis.

Momentum And Wavenumber Labels

One can label a plane wave either by momentum $p$ or by wavenumber $k$:

$$p=\hbar k.$$

The normalizations differ. A $p$-normalized plane wave is

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}.$$

A $k$-normalized plane wave is

$$\langle x\vert k\rangle = \frac{1}{\sqrt{2\pi}}e^{ikx},$$

with

$$\langle k\vert k'\rangle = \delta(k-k').$$

Because $p=\hbar k$,

$$\delta(p-p') = \frac{1}{\hbar}\delta(k-k').$$

Mixing $p$-normalization and $k$-normalization is one of the fastest ways to lose factors of $\hbar$.

Physical Wave Packets

A physical normalized state is built from a square-integrable momentum amplitude $\phi(p)$:

$$\lvert\psi\rangle = \int_{-\infty}^{\infty} dp\, \phi(p)\lvert p\rangle.$$

The position-space wavefunction is

$$\psi(x) = \langle x\vert\psi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p)e^{ipx/\hbar}\,dp.$$

The normalization condition is

$$\langle\psi\vert\psi\rangle = \int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp =1.$$

Thus $\lvert\phi(p)\rvert^2$ is the momentum probability density. A sharp momentum distribution gives a spatially broad packet; a sharply localized wavefunction requires a broad range of momenta. Gaussian packets make this tradeoff explicit.

Free Time Evolution

For a free particle,

$$E(p)=\frac{p^2}{2m}.$$

A momentum component evolves as

$$e^{-iE(p)t/\hbar} = e^{-ip^2t/(2m\hbar)}.$$

Therefore a free wave packet evolves by

$$\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p) e^{ipx/\hbar} e^{-ip^2t/(2m\hbar)} \,dp.$$

The packet spreads because the phase is quadratic in $p$. The detailed behavior is treated in Wave Packet Spreading.

Box Normalization As A Regulator

Delta normalization can be understood by first putting the particle in a large periodic box of length $L$:

$$\psi_n(x) = \frac{1}{\sqrt L}e^{ik_nx}, \qquad k_n=\frac{2\pi n}{L}, \qquad n\in\mathbb Z.$$

Then

$$\int_0^L \psi_m^*(x)\psi_n(x)\,dx = \delta_{mn}.$$

The allowed wavenumbers are separated by

$$\Delta k=\frac{2\pi}{L}.$$

As $L\to\infty$, the spectrum becomes dense and sums become integrals:

$$\sum_n \longrightarrow \frac{L}{2\pi}\int dk = \frac{L}{2\pi\hbar}\int dp.$$

The box is a regulator, not a claim that space is literally periodic. It is useful when deriving density-of-states factors, handling intermediate infinities, or comparing continuum formulas with numerical calculations. The boundary-condition details are collected in Periodic Boundary Conditions.

Relating Box And Delta Normalization

Box-normalized and delta-normalized states have different dimensions. A box-normalized plane wave has amplitude $1/\sqrt L$. A $k$-delta-normalized plane wave has amplitude $1/\sqrt{2\pi}$.

The replacement

$$\delta_{mn} \longleftrightarrow \Delta k\,\delta(k-k')$$

is a compact way to remember the continuum limit. Since $\Delta k=2\pi/L$, the discrete Kronecker delta turns into a Dirac delta together with the density of states.

For physical predictions, the artificial factors of $L$ cancel once states, sums, and densities are converted consistently.

Probability Current

A plane wave

$$\psi(x)=A e^{ikx}$$

has probability current

$$j=\frac{\hbar k}{m}\lvert A\rvert^2.$$

For a box-normalized state, this is

$$j=\frac{\hbar k}{mL}.$$

For a scattering calculation, one often works instead with amplitudes whose incident or outgoing waves carry a convenient flux. Current normalization is a separate convention from delta normalization. The common rule is the same: state the convention before interpreting amplitudes.

Common Mistakes

• Trying to force $\int\lvert e^{ikx}\rvert^2dx=1$ on the full line.
• Treating $\delta(p-p')$ as an ordinary function rather than a distribution.
• Mixing $p$-normalized and $k$-normalized states without the factor of $\hbar$.
• Forgetting that a wave packet, not a single plane wave, represents a localized free particle.
• Keeping box factors after taking the continuum limit.
• Comparing scattering amplitudes without checking whether the normalization is square, delta, box, or flux normalization.

Where This Is Used

• Free Particle gives the Hamiltonian, dispersion relation, and current of plane waves.
• Free Particle in Three Dimensions extends the same normalization ideas to vector momentum and energy shells.
• Density of States: First Encounter uses the box-to-continuum conversion to count three-dimensional free-particle states.
• Periodic Boundary Conditions makes the finite-volume regulator and density-of-states conversion explicit.
• Momentum Eigenstates derives plane waves from the momentum operator.
• Normalization Conventions compares bound-state, box, delta, flux, radial, and numerical conventions.
• Delta Function explains the distribution used in continuum normalization.
• Position and Momentum Representations gives the Hilbert-space basis viewpoint.
• Gaussian Wave Packets builds normalizable packets from plane-wave components.
• Reflection and Transmission Coefficients uses plane waves with current ratios.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Verify the delta normalization of $\langle x\vert p\rangle$ using the Fourier representation of the delta function.

Solution

Using

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar},$$

one finds

$$\begin{aligned} \langle p\vert p'\rangle &= \int_{-\infty}^{\infty} \langle p\vert x\rangle \langle x\vert p'\rangle\,dx \\ &= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{i(p'-p)x/\hbar}\,dx \\ &= \delta(p-p'). \end{aligned}$$

The final equality is understood distributionally.

2. Convert the sum over periodic-box wavevectors into an integral as $L\to\infty$.

Solution

Periodic boundary conditions give

$$k_n=\frac{2\pi n}{L},$$

so adjacent wavenumbers are separated by

$$\Delta k=\frac{2\pi}{L}.$$

For a smooth test function $f(k)$,

$$\sum_n f(k_n) \approx \frac{1}{\Delta k} \int f(k)\,dk = \frac{L}{2\pi} \int f(k)\,dk.$$

3. If $\phi(p)$ is normalized by $\int\lvert\phi(p)\rvert^2dp=1$, show that the corresponding $\psi(x)$ is normalized.

Solution

The momentum kets obey completeness and delta normalization. Therefore

$$\begin{aligned} \langle\psi\vert\psi\rangle &= \int dp \int dp'\, \phi^*(p)\phi(p') \langle p\vert p'\rangle \\ &= \int dp \int dp'\, \phi^*(p)\phi(p') \delta(p-p') \\ &= \int dp\,\lvert\phi(p)\rvert^2 =1. \end{aligned}$$