Compton wavelength

The Compton wavelength is a quantum mechanical property of a particle, defined as the wavelength of a photon whose energy is the same as the rest energy of that particle (see mass–energy equivalence). It was introduced by Arthur Compton in 1923 in his explanation of the scattering of photons by electrons (a process known as Compton scattering).

The standard Compton wavelength $λ$ of a particle of mass $$m$$ is given by $$ \lambda = \frac{h}{m c}, $$ where $h$ is the Planck constant and $c$ is the speed of light. The corresponding frequency $f$ is given by $$f = \frac{m c^2}{h},$$ and the angular frequency $ω$ is given by $$ \omega = \frac{m c^2}{\hbar}.$$

The CODATA 2018 value for the Compton wavelength of the electron is $2.426 m$. Other particles have different Compton wavelengths.

Reduced Compton wavelength
The reduced Compton wavelength $ƛ$ (barred lambda, denoted below by $$\bar\lambda$$) is defined as the Compton wavelength divided by $2π$:
 * $$\bar\lambda = \frac{\lambda}{2 \pi} = \frac{\hbar}{m c},$$

where $ħ$ is the reduced Planck constant.

Role in equations for massive particles
The inverse reduced Compton wavelength is a natural representation for mass on the quantum scale, and as such, it appears in many of the fundamental equations of quantum mechanics. The reduced Compton wavelength appears in the relativistic Klein–Gordon equation for a free particle: $$ \mathbf{\nabla}^2\psi-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi = \left(\frac{m c}{\hbar} \right)^2 \psi.$$

It appears in the Dirac equation (the following is an explicitly covariant form employing the Einstein summation convention): $$-i \gamma^\mu \partial_\mu \psi + \left( \frac{m c}{\hbar} \right) \psi = 0.$$

The reduced Compton wavelength is also present in Schrödinger's equation, although this is not readily apparent in traditional representations of the equation. The following is the traditional representation of Schrödinger's equation for an electron in a hydrogen-like atom: $$ i\hbar\frac{\partial}{\partial t}\psi=-\frac{\hbar^2}{2m}\nabla^2\psi -\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r} \psi.$$

Dividing through by $$\hbar c$$ and rewriting in terms of the fine-structure constant, one obtains: $$\frac{i}{c}\frac{\partial}{\partial t}\psi=-\frac{\bar{\lambda}}{2} \nabla^2\psi - \frac{\alpha Z}{r} \psi.$$

Distinction between reduced and non-reduced
The reduced Compton wavelength is a natural representation of mass on the quantum scale and is used in equations that pertain to inertial mass, such as the Klein–Gordon and Schrödinger's equations.

Equations that pertain to the wavelengths of photons interacting with mass use the non-reduced Compton wavelength. A particle of mass $m$ has a rest energy of $E = mc^{2}$. The Compton wavelength for this particle is the wavelength of a photon of the same energy. For photons of frequency $f$, energy is given by $$ E = h f = \frac{h c}{\lambda} = m c^2, $$ which yields the Compton wavelength formula if solved for $λ$.

Limitation on measurement
The Compton wavelength expresses a fundamental limitation on measuring the position of a particle, taking into account quantum mechanics and special relativity.

This limitation depends on the mass $m$ of the particle. To see how, note that we can measure the position of a particle by bouncing light off it – but measuring the position accurately requires light of short wavelength. Light with a short wavelength consists of photons of high energy. If the energy of these photons exceeds $mc^{2}$, when one hits the particle whose position is being measured the collision may yield enough energy to create a new particle of the same type. This renders moot the question of the original particle's location.

This argument also shows that the reduced Compton wavelength is the cutoff below which quantum field theory – which can describe particle creation and annihilation – becomes important. The above argument can be made a bit more precise as follows. Suppose we wish to measure the position of a particle to within an accuracy $Δx$. Then the uncertainty relation for position and momentum says that $$\Delta x\,\Delta p\ge \frac{\hbar}{2},$$ so the uncertainty in the particle's momentum satisfies $$\Delta p \ge \frac{\hbar}{2\Delta x}.$$

Using the relativistic relation between momentum and energy $E^{2} = (pc)^{2} + (mc^{2})^{2}$, when $Δp$ exceeds $mc$ then the uncertainty in energy is greater than $mc^{2}$, which is enough energy to create another particle of the same type. But we must exclude this greater energy uncertainty. Physically, this is excluded by the creation of one or more additional particles to keep the momentum uncertainty of each particle at or below $mc$. In particular the minimum uncertainty is when the scattered photon has limit energy equal to the incident observing energy. It follows that there is a fundamental minimum for $Δx$: $$\Delta x \ge \frac{1}{2} \left(\frac{\hbar}{mc} \right).$$

Thus the uncertainty in position must be greater than half of the reduced Compton wavelength $ħ/mc$.

Relationship to other constants
Typical atomic lengths, wave numbers, and areas in physics can be related to the reduced Compton wavelength for the electron ($\bar{\lambda}_\text{e} \equiv \tfrac{\lambda_\text{e}}{2\pi}\simeq 386~\textrm{fm}$ )|undefined and the electromagnetic fine-structure constant ($\alpha\simeq\tfrac{1}{137}$ ).

The Bohr radius is related to the Compton wavelength by: $$a_0 = \frac{1}{\alpha}\left(\frac{\lambda_\text{e}}{2\pi}\right) = \frac{\bar{\lambda}_\text{e}}{\alpha} \simeq 137\times\bar{\lambda}_\text{e}\simeq 5.29\times 10^4~\textrm{fm} $$

The classical electron radius is about 3 times larger than the proton radius, and is written: $$r_\text{e} = \alpha\left(\frac{\lambda_\text{e}}{2\pi}\right) = \alpha\bar{\lambda}_\text{e} \simeq\frac{\bar{\lambda}_\text{e}}{137}\simeq 2.82~\textrm{fm}$$

The Rydberg constant, having dimensions of linear wavenumber, is written: $$\frac{1}{R_\infty}=\frac{2\lambda_\text{e}}{\alpha^2} \simeq 91.1~\textrm{nm}$$ $$\frac{1}{2\pi R_\infty} = \frac{2}{\alpha^2}\left(\frac{\lambda_\text{e}}{2\pi}\right) = 2 \frac{\bar{\lambda}_\text{e}}{\alpha^2} \simeq 14.5~\textrm{nm}$$

This yields the sequence: $$r_{\text{e}} = \alpha \bar{\lambda}_{\text{e}} = \alpha^2 a_0 = \alpha^3 \frac{1}{4\pi R_\infty}.$$

For fermions, the reduced Compton wavelength sets the cross-section of interactions. For example, the cross-section for Thomson scattering of a photon from an electron is equal to $$\sigma_\mathrm{T} = \frac{8\pi}{3}\alpha^2\bar{\lambda}_\text{e}^2 \simeq 66.5~\textrm{fm}^2 ,$$ which is roughly the same as the cross-sectional area of an iron-56 nucleus. For gauge bosons, the Compton wavelength sets the effective range of the Yukawa interaction: since the photon has no mass, electromagnetism has infinite range.

The Planck mass is the order of mass for which the Compton wavelength and the Schwarzschild radius $$ r_{\rm S} = 2 G M /c^2 $$ are the same, when their value is close to the Planck length ($$l_{\rm P}$$). The Schwarzschild radius is proportional to the mass, whereas the Compton wavelength is proportional to the inverse of the mass. The Planck mass and length are defined by:

$$m_{\rm P} = \sqrt{\hbar c/G}$$ $$l_{\rm P} = \sqrt{\hbar G /c^3}.$$

Geometrical interpretation
A geometrical origin of the Compton wavelength has been demonstrated using semiclassical equations describing the motion of a wavepacket. In this case, the Compton wavelength is equal to the square root of the quantum metric, a metric describing the quantum space: $$\sqrt{g_{kk}}=\lambda_\mathrm{C}$$.