If $E=hf$ and the frequency of electromagnetic waves is continuous (i.e. you can have frequencies of $1.5\ Hz$ or $0.3\ Hz$ for example) then surely energy isn’t discrete or quantized into because one could simply have any multiple whatsoever of a place constant and so any value for energy. As an extension of this there would be no minimum value of the energy of a photon, so there would be no such thing as one quanta?
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$\begingroup$ There are two different things going on here. $f$ is the frequency of the EM wave. When I add energy to it, the energy of the wave increases in integral multiples of $f$. But the frequency of the wave itself is remains $f$. If you increase the frequency of the wave, then you are doing something else: changing the boundary conditions or the driving frequency. $\endgroup$– garypCommented Jun 14, 2020 at 14:36
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$\begingroup$ Exactly as @garyp. $E$ here is just the energy of any particular photon. That energy is indeed quantized by $h$. If you want to send a certain amount of energy at a frequency $f$, you'll have to do it in packets of energy $hf$ until that energy is sent. $\endgroup$– ChemiCalChemsCommented Jun 14, 2020 at 14:37
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$\begingroup$ Related: physics.stackexchange.com/q/52943/2451 , physics.stackexchange.com/q/73959/2451 , physics.stackexchange.com/q/130358/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 14, 2020 at 15:22
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$\begingroup$ @ChemiCalChems Just a minor thing: The photon is the quantum of transferred energy. It's the transferred energy that's quantized, not the energy of a single photon. $\endgroup$– Deschele SchilderCommented Jun 14, 2020 at 16:35
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$\begingroup$ @descheleschilder Sure, minor language thing. $\endgroup$– ChemiCalChemsCommented Jun 14, 2020 at 16:40
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3 Answers
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A photon, by itself, doesn't have a frequency or an energy because it has no rest frame. It just "is". When you pick a rest frame, then (ignoring polarization), it is described by a 4-wave vector:
$$k^{\mu}=(\omega/c, \vec k) $$
so that the frequency is $f = \omega/2\pi$, the direction is $\hat k$, and the wavelength is $\lambda = 2\pi/k$.
That's it.
The energy and momentum are then:
$$ p^{\mu} = \hbar k^{\mu} = (E/c, \vec p)$$
The only way to increase the energy is to add more photons to the mode. Since $\omega$ and $\lambda$ were fixed, you wind up with:
$$ p_n^{\mu} = np^{\mu}=n\hbar k^{\mu}$$
so energy is quantized for modes with frequency $\omega$:
$$ E_n = n\hbar\omega$$
while $\omega$ can have any value.
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It is related to the origin of the idea of energy quantization.
The function $E \times \lambda$ of blackbody radiation should follow a typical exponential curve, following the Boltzmann distribution of a gas of waves inside a cavity.
But the actual curves show a sharp decrease to zero for small wave lengths ($\lambda)$, below a peak value of energy.
Planck fixed the problem by imposing an energy threshold for emission proportional to $1/\lambda$ (or $\nu$). This way, all the emissions of very small wave lengths with energy below the threshold are eliminated.
The main point of quantization is not that energy must be multiple of some value, but that below a given value there is no emission for a given frequency.
answered Jun 14, 2020 at 22:19
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One should have clear the concept of quantization.
" Your question is analogous to asking : sound frequencies are continuous, why does the violin have specific harmonics?"
The answer for sound is that all sound frequencies can exist, and it is the boundary conditions on the interactions making the sound that define specific frequencies.
The same is true for quantization in quantum mechanics, (and it is not only for light). It is the boundary conditions imposed on the quantum mechanical equation solutions for a specific potential problem that generate specific frequencies in light emmision, which with the axiomatic assignment that the electromagnetic energy of a single photon is $E=hν$, defines quanta of energy.
See the solutions in the Coulomb potential for the hydrogen atom here.
