The wavelength of visible light ranges from 750 - 400 nm, and so do the corresponding frequencies. However, a photon only has one frequency, given by $E =h\nu$, at a given time, and it can’t be changed unless the photon gets energy from somewhere, which isn’t possible in the vacuum; i.e. once it leaves the source it can’t possibly get enough packets of energy to become excited or even lose the energy to some other particle, as space is empty.

So, why does light come with so many wavelengths and frequencies if a single photon can only have one frequency at a time and are emitted from the same source?

A single photon goes with its individual oscillating electric and magnetic fields, right? So multiple photons mean multiple fields. Won’t these different fields affect the adjacent fields in any way and change their properties?

  • 21
    $\begingroup$ Why do pebbles come in so many sizes while a single pebble can only have one size? $\endgroup$
    – my2cts
    Oct 2, 2021 at 20:40
  • 1
    $\begingroup$ Unlike photons pebbles are NOT massless quantum particles that follow the laws of quantum mechanics and well there aren’t any extensive studies being conducted to find the true nature of a pebbles, currently. There’s that. $\endgroup$
    – A.M.
    Oct 2, 2021 at 21:18
  • $\begingroup$ Why does a photon have the frequency which it does? Why not a different frequency? $\endgroup$
    – Bill N
    Oct 2, 2021 at 21:47
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    $\begingroup$ No studies about the nature of pebbles? See petrology, the branch of geology dealing with the nature of rocks. $\endgroup$
    – nasu
    Oct 2, 2021 at 22:12
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    $\begingroup$ @BillN No, why different photons have different frequencies if emitted from the same source? $\endgroup$
    – A.M.
    Oct 3, 2021 at 9:10

5 Answers 5


Light comes with so many wavelengths because it is made of so many photons. A typical lightbulb puts out something on the order of $1\ \mathrm{W}$ of power in the visible spectrum, while individual photons in the visible spectrum each have an energy on the order of $10^{-19}\ \mathrm{J}$. Therefore, a typical lightbulb will produce something on the order of $10^{19}$ photons per second. While each photon has one specific frequency, there are so many of them that the ensemble of photons appears to have a continuous range of frequencies.

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    $\begingroup$ "While each photon has one specific frequency..." That's fine as an approximation, but I think it's worth pointing out that the OP's premise ("a single photon can only have one frequency at a time") is wrong. Any quantum superposition of single-photon states is still a single-photon state, even if it's a superposition of single photons with different frequencies. In the real world, that's unavoidable: each individual photon necessarily has nonzero bandwidth, at least as wide the inverse of the duration of the process that produced it. $\endgroup$ Oct 3, 2021 at 1:44
  • $\begingroup$ So you mean to say a single photon can have multiple frequencies like v1, v2, v3, simultaneously, corresponding to multiple Energy states E1, E2, E3 at the same time? $\endgroup$
    – A.M.
    Oct 3, 2021 at 8:58
  • $\begingroup$ @A.M. It can have multiple probabilistic frequencies. In quantum mechanics those probabilities are considered real so in theory it is considered to have multiple real frequencies until it interacts with something (your eye, a film, a sensor etc.) at the moment it is measure it can ONLY have one frequency. But as long as it is travelling in free space it "probably" has a range of frequencies (and as I said, in quantum physics those probabilities are considered real so it really has a range of frequencies) $\endgroup$
    – slebetman
    Oct 3, 2021 at 12:57
  • $\begingroup$ So when measured it can take up any random frequency or it is determined by any factors like temperature or the device with which it’s measured? And what happens to those frequencies when the photon interacts? And if these frequencies vary won’t the momentum of the photons vary too? $\endgroup$
    – A.M.
    Oct 3, 2021 at 21:24
  • $\begingroup$ @A.M. when measured the probability distribution for the frequency is given by the Born rule. $\endgroup$
    – Sandejo
    Oct 4, 2021 at 1:41

The question, as I understand it, is: why does a single body emit light of many frequencies?

The problem lies in the question itself. The sources we are dealing with, such as stars or light bulbs, are single macroscopic sources. They are composed of countless particles, each producing light because of its thermal motion.

We know from thermodynamics that a body of a certain temperature T is composed of particles whose energy is not uniform. I have to admit it's been a while since the last time I had to do with thermodynamics, and I don't want to give you wrong information, but I think a look into Boltzmann distribution may be... illuminating.

Once you see that what you call "single source" is actually a group of sources with different energies, it should be easy to see why the body emits photons of different energies.

As for your last question, photons are electromagnetic particles, that's true, but at the same time they have no electric charge, so they can't change the properties of other photons. Only a charged particle can interact with the electromagnetic field.

  • $\begingroup$ Yeah I had a question regarding that too, how are photon charge less if they’re nothing but packets of Energy propagating through alternating Electric and Magnetic fields? If there’s an electric field there must be an electric charge producing it right? $\endgroup$
    – A.M.
    Oct 3, 2021 at 11:30
  • $\begingroup$ That's how photons start their existence, but once they are created, they are self-sustaining. If you take the Maxwell's equations and set the electric charge density and electric current density to zero, you can still find a non-zero solution, one where the electric and magnetic field oscillate perpendicular to one another and to the direction of motion. Light, for short. $\endgroup$
    – GRB
    Oct 3, 2021 at 14:10

Light is made up of individual photons and like you said photons get their energy at the source. Depending on the amount of energy it receives, a photon can have any frequency.


Once it is emitted, the wavelength of a photon does not change (if we ignore cosmological redshift due to the expansion of space). But different photons can be emitted with different wavelengths.

  • $\begingroup$ A single photon goes with its individual oscillating electric and magnetic fields right? So multiple photons means multiple fields. So won’t these different fields affect the adjacent fields in any way and change their properties? $\endgroup$
    – A.M.
    Oct 2, 2021 at 21:23
  • $\begingroup$ No, they won't. $\endgroup$
    – nasu
    Oct 2, 2021 at 22:13
  • $\begingroup$ Technically photons do interact with other photons but for all everyday purposes and naïve understanding it's so oh insignificant they even teach you in schools they don't. $\endgroup$
    – Lodinn
    Oct 2, 2021 at 23:04
  • $\begingroup$ @A.M. This is the right answer. There is no magic to it other than "they get emitted at different frequencies". At least the classical Electrodynamic equations follow the superposition principle, i.e. 'neighbouring' photons do not change their energies. $\endgroup$ Oct 4, 2021 at 0:45

Planck's law predicts the spectral energy density $u$ of a radiating blackbody and is given by $$ u(\omega) = \frac{\omega^2}{\pi^2 c^3} \hbar\omega \frac{1}{e^{\frac{\hbar\omega}{k_B T}}-1} $$ A typical derivation involves the canonical partition function from statistical mechanics to derive the average energy of the system. In statistical mechanics every quantity has a certain distribution and isn't exact anymore, because while we could know the exact position and momentum of every constituent this is an unreasonable ask. To get around this problem we discard a lot of the information and ask for simpler quantities such as the pressure or energy of the system instead of asking about every single position. As such we can only ascribe a probability to the system to be in a specific state and this leads to a continuous distribution of the various quantities of interest.

Another thing to note is that as ChiralAnomaly points out in the comments is that on a quantum mechanical level we need to consider the Heisenberg Uncertainty principle. Despite time not being an operator we can derive $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$ So if you consider an atom in an excited state it will necessarily decay into a lower energy state with a certain half life $\Delta t$ and as such the emitted photon won't have an exact energy. If you're doing spectroscopy on real gases it's even worse because the lines you observe will be broadened by the Doppler Effect due to the thermal motion of the atoms and also pressure broadened due to collisions between the atoms.


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