I was thinking about this the other day after a quantum mechanics lecture (unrelated to the lecture I was taking) and pondered "Is there a minimum wavelength for a photon?", through searching online and some thought, there didn't seem to be many definitive answers and came across a few sources that said the Planck length, but didn't show why, just stated as such.

I believe that:



$$E=\frac{h c}{\lambda}$$

As when $$\lambda \to 0$$ $$E\to\infty$$

Which (assuming finite energy within the universe) isn't possible, however if energy within the universe isn't finite, I guess this could be an answer, that the wavelength of a photon could approach/reach 0.

But, upon further thought and inquiry, I considered the Schwarzschild Radius, where:

$$r_s=\frac{2 G M}{c^2}$$

Einstein's Mass energy equivalence formula can be used (I think, even though photons are massless):


Where $m=M$ gives:


And from above, $E=\frac{h c}{\lambda}$

$$Mc^2=\frac{h c}{\lambda}$$

Thus giving

$$M=\frac{h} {c\lambda}$$

Substituting $M$ into the Schwarzschild Radius equation, this yields:

$$r_s=\frac{2 G h}{c^3\lambda}$$

Assuming some of the online statements saying the minimum wavelength is the Planck length, substituting this into the equation (and all other values) yields:

$$r_s\approx 2.031\times 10^-34m$$

This means the Schwarzschild Radius of the photon with wavelength of Planck's length is larger than the Planck length? I may have assessed this result wrong but the diameter of this black hole is then $d_s\approx 4.062\times10^-34 m$ which is an order of magnitude larger than the Planck length - which doesn't seem to be possible in my mind. assuming my thought is correct this result (assuming I have thought this through correctly) means that $\lambda\ge 2\times r_s$?

If this is true, then the minimum wavelength of a photon is when $\lambda=2\times r_s$ we can denote $r_s$ and $\lambda$ as $\frac{z} {2}$ and $z$ respectively.

This rearranges to:

$$z^2=\frac{4 G h}{c^3}$$

Substituting values in and solving for $z$ when $z\ge0$, this gives the minimum wavelength of $\approx 8.1027\times10^-35 m$, this value of $\lambda=2r_s$ meaning the Schwarzschild Radius is $r_s\approx 4.0514\times 10^-35 m$

This means the Schwarzschild Radius is approximately 2.5066 times greater than the Planck length which is different to the answer scattered around the internet of "the minimum photon wavelength is the Planck length".

So I have arrived at two conclusions/answers to my question.

(1) The minimum wavelength of a photon is $\lambda\to 0$ (assuming the universe is infinite, but this violates the Schwarzschild Radius problem I considered in the second answer to this question)

(2) The minimum wavelength of a photon is $\lambda=2\times r_s$, for this case it is $\lambda\approx 8.1027\times10^-35 m$

Can someone with better understanding than me (as I am a 2nd year Physics student) please help me understand/explain if my reasoning is correct.


1 Answer 1


Existing comments on the question give links to answers. These are that (1) a wavelength can be arbitrarily small because for any given wavelength you can always get a smaller one by jumping to another inertial reference frame; and (2) nevertheless it is hard to maintain any claim that existing physical theories correctly treat interactions at Planck scale or below in the rest frame of the interacting systems.

The main thing wrong with the argument in the question is that the photon need not have its energy concentrated in a small region. A 'photon' is not necessarily a little thing spatially concentrated at a point or some small region. It can be spread out over huge volumes. Think of it like a plane wave having energy $E = h c/\lambda$, width $a$ and length $L$. So even if its energy is high, the energy per unit volume $E/(L a^2)$ need not be high, so no black hole will form. More generally, to form a black hole you need some stuff which is not moving at light speed.


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