# Compton Wavelength

I have the formula for Compton wavelength: $$\lambda_{c}= \frac{h}{m_{0}c}$$ In this equation, is $m_0$ the mass of the electron that the photon hit?

I got online that this might be the photon rest mass, but it is basically 0, and it is impossible to do the calculation to get $\lambda_c$ if it is equal to 0.

• Since photons do not have mass, then process of elimination says that you are correct: it is the mass of the electron (particle, in the general case) Commented May 13, 2014 at 19:49
• so, it would be $9.1\times10^-31$?
– TYZ
Commented May 13, 2014 at 19:51
• Assuming you are using SI, yes. Commented May 13, 2014 at 19:52

The Compton wavelength given by,

$$\lambda=\frac{\hbar}{mc}$$

is a natural length scale associated to any particle with mass $m\neq 0$. As Professor Tong, states:

At distances shorter than this [Compton wavelength], there is a high probability that we will detect particle-anti-particle pairs swarming around the original particle that we put in. [The wavelength] is always smaller than the de Broglie wavelength $\lambda_{dB}=h/|\vec{p}|$. If you like, the de Broglie wavelength is the distance at which the wavelike nature of particles becomes apparent; the Compton wavelength is the distance at which the concept of a single pointlike particle breaks down completely.

• In first equation shouldn't instead of $\hbar$ be $h$?, or in first equation you mean reduced compton wavelength?
– user21420
Commented May 13, 2014 at 20:14
• @GigiButbaia: I took the definition directly from Prof. Tong, it's $\hbar$, but it's really a matter of convention because they both have the same dimensions, and hence one can construct a natural length scale from either. Commented May 13, 2014 at 20:15
• @JamalS Can you share the link where Prof. Tong states this? Commented Oct 18, 2017 at 7:13
• @abstract damtp.cam.ac.uk/user/tong/qft/qft.pdf (Page 3 in the document; page 9 in PDF viewer.) Commented Oct 18, 2017 at 13:47

The Compton wavelength is a characteristic scale in Quantum Electrodynamics.

the photon rest mass, but it is basically 0

we should force ourselves to only state what we can observe. In free space, photon can never be at rest according special relativity. Therefore, we cannot talk about the rest mass.

$$\lambda_c$$ should be interpreted through the energy of a photon $$E = hc/\lambda_c$$. Twice this value should be enough energy to create a pair electron-positron.

When the frequency spectrum of light contains non-negligible components in the gamma domain, with wavelength below $$\lambda_c$$ then the standard formalism of quantum optics fails as the fundamental superposition principle in electromagnetism does not hold any more as light can interact with light through the creation of electron-position pairs.

See. "The quantum theory of radiation" Heitler - 1938.

If it is an electron wavelength, the mass to insert is the electron mass. If you speak about photons and you wanna define their wavelengths, then you cannot just insert the electron mass. The Compton wavelength of a photon is just the photon wavelength, which is simply $$\lambda=c/\nu=ch/E=2\pi\hbar/p$$ where $E$ and $p$ are the energy and linear momentum of the photon.

Two more things:
1- Compton postulated its Compton wavelength for electrons since they were hitting on photons like waves. Therefore in its case you will have to insert the electron mass.
2- You may force your formula also for photons. Folk, do not get me wrong here, please. You may define the energy of the photon by $E=h \nu$ (which is the standard form) $\equiv m c^2$. It is just a definition of your new parameter $m$. Then you will obtain $\lambda =h/mc$ also for photons, where now $m$ is your new parameter. I owe to add that this definition of $m$ is not so bad, since photons at rest have mass 0, but for running photons you may define a mass. The photon mass in this case would be the product $m_0 \,\gamma$, where $m_0$ is the rest mass (which is zero) and $\gamma$ is its relativistic gamma (which is infinite). Of course, that is a product $0\cdot(+\infty)$ and is not defined. But as long as you know that the full consistent definition is $m=E/c^2$, you may say that the product $0\cdot(+\infty)$ makes the finite number $E/c^2$, and you are safe. After all, if you try to estimate how much photons are attracted by gravitational masses (gravitational lensing) and you postulate that their mass is your parameter $m$, you do not get far from the exact calculation in General Relativity.