Plane Waves

A plane wave is a generalized momentum eigenfunction and a character of the translation group. It is useful as a spectral component and as the carrier of a normalizable packet. A controlled approximation should compare such packets in Hilbert norm, rather than claim that a normalized state converges in norm to a plane wave.

Required background. Momentum Kets supplies the Fourier normalization. Continuous Spectrum Without Eigenvectors proves the free-particle approximate-eigenvector bounds; those spectral arguments are not repeated here. This page treats a free particle of mass $m>0$ on the line.

Translations multiply a plane wave by a phase

For the translation convention $(T(a)\psi)(x)=\psi(x-a)$, the generalized function $k_p(x)=(2\pi\hbar)^{-1/2}e^{ipx/\hbar}$ obeys

$$T(a)k_p=e^{-ipa/\hbar}k_p.$$

The scalar factors form a continuous character: the phase for $a+b$ is the product of the phases for $a$ and $b$. This identifies momentum as the label dual to spatial displacement.

The equality is meaningful on the generalized-function side. There is no normalizable whole-line vector whose probability density is a nonzero constant. On a periodic interval, the allowed characters instead have discrete momenta, and finite-volume modes can be normalized. Delta Normalization explains their relation to continuum factors.

A carrier and an envelope define an ordinary state

Choose a central momentum $p_0$ and a normalized momentum envelope $a(q)$, where $q=p-p_0$. The initial packet is

$$\psi(x,0)=e^{ip_0x/\hbar}g(x), \qquad g=\mathcal F_\hbar^{-1}a.$$

The plane-wave factor has modulus one, so $\|\psi(0)\|_2=\|g\|_2=\|a\|_2=1$. Normalization belongs to the envelope, not to the carrier. The central momentum is a reference parameter; it equals the mean only if the envelope has zero mean $q$.

For free evolution, expand the dispersion exactly:

$$\begin{gathered} E(p_0+q)=E_0+v_0q+\frac{q^2}{2m},\\ E_0=\frac{p_0^2}{2m},\quad v_0=\frac{p_0}{m}. \end{gathered}$$

Fourier evolution then gives

$$\psi(x,t) =e^{i(p_0x-E_0t)/\hbar}g_t(x-v_0t),$$

where $g_t=\mathcal F_\hbar^{-1}(e^{-itq^2/(2m\hbar)}a(q))$. This is an exact identity: the envelope translates with the carrier group velocity and also evolves dispersively through its quadratic phase.

A norm bound controls rigid-envelope propagation

The rigid-envelope approximation omits that quadratic phase:

$$\psi_{\rm rigid}(x,t) =e^{i(p_0x-E_0t)/\hbar}g(x-v_0t).$$

If $q^2a\in L^2$, Plancherel and $|e^{-iu}-1|\le|u|$ imply

$$\begin{aligned} \|\psi(t)-\psi_{\rm rigid}(t)\|_2 &=\|(e^{-itq^2/(2m\hbar)}-1)a\|_2\\ &\le\frac{|t|}{2m\hbar}\|q^2a\|_2. \end{aligned}$$

For an envelope supported in $|q|\le\delta p$, this becomes $|t|(\delta p)^2/(2m\hbar)$. The estimate is uniform over normalized envelopes with that support. It states both an accuracy and a time scale; “narrow band” alone is not a complete approximation claim.

The result compares two normalizable states, so it can control bounded-observable predictions. For an effect $0\le F\le I$, expanding the difference of the two expectation values bounds its magnitude by twice the vector-norm error. It does not automatically bound unbounded position or momentum moments.

Phase velocity and group velocity answer different questions

For nonzero $p_0$, a surface of constant carrier phase moves at $E_0/p_0=p_0/(2m)$. The translated envelope uses $dE/dp|_{p_0}=p_0/m$. These are different quantities. Neither should be interpreted as a universal signal speed, and the nonrelativistic free Schrödinger model is not a relativistic propagation theorem.

The plane wave can also carry a formal current density $j=(p_0/m)|A|^2$ for constant amplitude $A$. Its spatial integral is not the probability of a normalized single-particle state. Scattering theory uses such stationary currents only with specified incoming normalization or an underlying packet construction.

Finally, the two plane waves with momenta $p$ and $-p$ have equal free energy. Their superposition need not have definite momentum. A standing wave therefore illustrates energy multiplicity, not an exception to momentum’s spectral representation.

Exercise: a quantitative narrow-band requirement

For normalized envelopes supported in $|q|\le\delta p$, find a sufficient time range for the rigid approximation’s vector error to be at most $\eta>0$. What changes if $p_0$ is negative?

Solution

It suffices that $|t|\le2m\hbar\eta/(\delta p)^2$. A negative $p_0$ reverses the translation velocity $v_0$, but does not change the error bound, which depends on the curvature of the free dispersion and the envelope width. At long times the estimate can become uninformative even though the exact unitary packet evolution remains well defined.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — translation generators, Fourier transforms, and free propagation.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s edition — free Schrödinger dynamics and spectral calculus.