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# Compton wavelength

The Compton wavelength $\lambda \$ of a particle is given by $\lambda = \frac{h}{m c} = 2 \pi \frac{\hbar}{m c} \$,

where $h \$ is the Planck constant, $m \$ is the particle's mass, $c \$ is the speed of light.

The CODATA 2002 value for the Compton wavelength of the electron is 2.4263102175×10-12 meter with a standard uncertainty of 0.0000000033×10-12 m. Other particles have different Compton wavelengths.

The Compton wavelength can be thought of as a fundamental limitation on measuring the position of a particle, taking quantum mechanics and special relativity into account. This depends on the mass $m \$ of the particle. To see this, 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 have 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 Compton wavelength is the cutoff below which quantum field theory– which can describe particle creation and annihilation – becomes important.

We can make the above argument a bit more precise as follows. Suppose we wish to measure the position of a particle to within an accuracy $\Delta x \$. Then the uncertainty relation for position and momentum says that $\Delta x\,\Delta p\ge \hbar/2$

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, 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. So, with a little algebra, we see there is a fundamental limitation $\Delta x \ge \frac{\hbar}{2mc}$

So, at least to within an order of magnitude, the uncertainty in position must be greater than the Compton wavelength $h/mc \$.

The Compton wavelength can be contrasted with the de Broglie wavelength, which depends on the momentum of a particle and determines the cutoff between particle and wave behavior in quantum mechanics.

For fermions, the 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 $(8\pi/3)\alpha^2\lambda_e^2$,

where $\alpha \$ is the fine-structure constant and $\lambda_e \$ is the Compton wavelength of the electron. For gauge bosons, the Compton wavelength sets the effective range of the Yukawa interaction: since the photon is massless, electromagnetism has infinite range.

The Compton wavelength of the electron is one of a trio of related units of length, the other two being the Bohr radius a0 and the classical electron radius re. The Compton wavelength is built from the electron mass me, Planck's constant h and the speed of light c. The Bohr radius is built from me, h and the electron charge e. The classical electron radius is built from me, c and e. Any one of these three lengths can be written in terms of any other using the fine structure constant α: $r_e = {\alpha \lambda_e \over 2\pi} = \alpha^2 a_0$

The Planck mass is special because ignoring factors of and the like, the Compton wavelength for this mass is equal to its Schwarzschild radius. This special distance is called the Planck length. This is a simple case of dimensional analysis: the Schwarzschild radius is proportional to the mass, whereas the Compton wavelength is proportional to the inverse of the mass.

## References

1. ^ CODATA 2002 value for Compton wavelength for the electron from NIST