Why can't a single photon produce an electron-positron pair?

Question

In reading through old course material, I found the assignment (my translation):

Show that a single photon cannot produce an electron-positron pair, but needs additional matter or light quanta.

My idea was to calculate the wavelength required to contain the required energy ($1.02$ MeV), which turned out to be $1.2\times 10^{-3}$ nm, but I don't know about any minimum wavelength of electromagnetic waves. I can't motivate it with the conservation laws for momentum or energy either.

How to solve this task?

Answer 1

Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy.

For example you've got a $5 MeV$ photon, so you think that there is plenty of energy to make $e^-e^+$ pair. Now you make a boost (a change by a constant velocity to another inertial reference frame) along the direction of the photon momentum with $v=0.99\,c$ and you get a $0.35 MeV$ photon. That is not enough even for one electron.

Answer 2

Other way to see why this is impossible is to look at inverse process: why annihilating positron and electron can't give up only one photon? Imagine these two particles at rest near each other (or look at center-of-mass system). They will annihilate giving 1MeV of energy, but single photon can't pick this energy up by itself because it would also have E/c of momentum and starting setup, the two charged particles, had non. You need two photons that move in opposite directions.

Answer 3

This has a very simple answer that also works the other way around (see my answer to Why during annihilation of an electron and positron 2 gamma rays are produced instead of 1?). The idea is that this process cannot satisfy both momentum and energy conservation simultaneously. Let's prove it.

Let us consider that the one photon in question is moving in the $z$ direction with energy $\hbar\omega$ and momentum $\hbar\omega/c$. That is, the $4$-vector describing the photon is $k=(\hbar\omega,0,0,\hbar\omega)$ in this frame (let's assume $c=1$). Now, the electron-positron system has $4$-vectors $p_1$ and $p_2$ describing their movement. Momentum energy conservation implies

$$k=p_1+p_2$$

If we square this equation (that is, take the inner product under the Lorenz signature), we have, noting $k^2=0$, $p_1^2=m^2$, and $p_2^2=m^2$, that

$$m^2+p_1\cdot p_2=0$$

Now, this equation is entirely frame independent. Thus, if we pick a frame in which the total momentum is zero, we have that $p_1=(m\gamma,m\beta\gamma\cos{\theta},0,m\beta\gamma\sin{\theta})$ and $p_2=(m\gamma,-m\beta\gamma\cos{\theta},0,-m\beta\gamma\sin{\theta})$ (where $\gamma$ is the Lorenz factor $1/\sqrt{1-\beta^2}$ and $\beta$ is the velocity in natural units). This gives

$$p_1\cdot p_2=m^2\gamma^2\left(1-\beta^2\right)=m^2$$

That is, the kinematic equation requires $2m^2=0$, which is not possible for an electron.

This is a lot of math to give not much intuition. The real intuition lies in the fact that there cannot exist a frame in which the photon has zero momentum, but there does exist a frame in which the electron-positron system has zero momentum. This is incompatible with relativity, and so this process is not kinematically possible.

This applies for the reverse process (where this argument becomes slightly more inherent). Like I said above, it may be helpful to check out my answer to a related question.

I hope this helped!

Answer 4

Very clear explanation by Griffiths:

The Feynman rules enforce conservation of energy and momentum at each vertex, and hence for the diagram as a whole. It follows that the primitive QED vertex (two electrons or a positron and an electron, coupled to a photon)by itself does not represent a possible physical process. We can draw the diagram, but calculation would assign to it the number zero. The reason is purely kinematical: $e^- \rightarrow e^- + \gamma$ would violate conservation of energy. (In the center-of-mass frame the electron is initially at rest, so its energy is $mc^2$. It cannot decay into a photon plus a recoiling electron because the latter alone would require an energy greater than $mc^2$.) Nor, for instance, is $e^+ +e^-\rightarrow \gamma$ possible, although it's easy enough to draw the diagram. In the center-of-mass system the electron and positron enter symmetrically with equal and opposite velocities, so the total momentum before the collision is obviously zero. But the final momentum cannot be zero, since photons always travel at the speed of light; an electron-positron pair can annihilate to make two photons, but not one.

Within a larger diagram, however, these figures are perfectly acceptable, because, although energy and momentum must be conserved at each vertex, a virtual particle does not carry the same mass as the corresponding free particle.

In fact, a virtual particle can have any mass, whatever the conservation laws require. The virtual particles do not lie on their mass shell. External lines, by contrast, represent real particles, and these do carry the “correct” mass.

Assume a photon can produce an electron and a positron. There exists a CM inertial frame for the electron and positron win which the electron and positron have opposite spatial momenta (the time part of the momentum is $mc$ for both the electron and the positron) so their total spatial momentum is zero. Now the total spatial momentum for the photon obviously can't be zero: it's traveling at the speed of light in all inertial frames. This contradicts the assumption.

Answer 5

Check if momentum can be conserved. That ought to do the trick.

Answer 6

If $e^+$ $e^+$ give photon From consevation of momuntum $$P_{before} = P_{after} $$ So, $$ p_e + p_{e^+} = p_{photon} \\ p_{photon} = 0 $$ And we know the momentum of photon can not be zero So , there must be two outgoing photons in opposite direction .

Answer 7

no. a single photon can decay to an electron-positron pair. It must do so however near a nucleus to conserve momentum. 2 gamma-ray reactions may be rare but pair production process dominates gamma-ray reactions as energy increases and also as nearby nucleus increases in mass.

Answer 8

A single electron sitting motionless in space cannot give off a photon and recoil otherwise we get energy for nothing. In another frame this electron is moving with momentum and energy ... even so it still cannot shoot off a single photon because the physics of the original centre of mass frame. Doing a spacetime “rotation” á la Feynman, we can therefore clearly neither have an electron coming in and an antielectron going out with a single photon, Nor a single photon in with an e- and e + going out as these would be equivalent to the previous impossibility.

However Compton scattering where a photon scatters from an electron and both alter course when spacetime “rotated” can look like e- e+ = ph ph and reverse. Just noticed this is basic version of asDeSchillers sophisticated answer.