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I have been studying quantum mechanics and I came across Planck's relation which describes the energy $E$ of a photon as being directly proportional to its frequency $f$, with Planck's constant $h$ as the proportionality constant, i.e. $E=hf$.

My question arises from the observation that this linear relationship between energy and frequency seems to hold true specifically when considering sinusoidal bases for frequencies. Why is it that using a sinusoidal frequency base results in a linear relationship in Planck's relation? If we were to use a different basis for frequencies, would we not see a linear relation? What is it about the sinusoidal nature of frequencies that makes this linearity apparent? I'm trying to understand the fundamental reasons behind this and would appreciate any insights into why the energy-frequency relationship in Planck's formula would change, if at all, with a different frequency base.

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    $\begingroup$ What does "sinusoidal bases for frequencies" mean? $\endgroup$
    – Kyle Kanos
    Commented Mar 9 at 16:14
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    $\begingroup$ Well square and triangular waves aren't harmonic, so that kind of prohibits the whole idea that electrons oscillating can be treated as masses on (massless) springs. $\endgroup$
    – Kyle Kanos
    Commented Mar 9 at 16:58
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    $\begingroup$ Is the premise of this question true? If you used a different basis wouldn’t the relationship between energy and frequency still be linear? I mean, if you doubled the frequency of a triangular wave, I think it should still have double the energy. Other things might be weird, but I don’t think this is one of them $\endgroup$
    – Dale
    Commented Mar 9 at 18:00
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    $\begingroup$ I think it will still be linear. As you say, a triangular is simply a bunch of sinusoids. If you double the frequency of the triangular wave you will double the frequencies of all those sinusoids, and therefore the energy should be doubled. It will still be linear. $\endgroup$
    – Dale
    Commented Mar 9 at 18:48
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    $\begingroup$ I think everyone is missing the point here. The relationship $E = h f$ happens essentially because energy is the conserved quantity associated with time translation invariance and $\exp(i 2\pi f t)$ is precisely the function that is invariant (up to a scalar prefactor) under time translations. $\endgroup$
    – DanielSank
    Commented Mar 9 at 19:55

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For convenience I will be using the reduced Planck constant $\hbar=\frac{h}{2\pi}$ and angular frequency $\omega=2\pi f$ in this answer. Secondly, if you'll indulge me, it's easier to talk about complex exponentials (i.e. functions of the form $c(t)=e^{i\omega t}$ or $c(t)=e^{-i\omega t}$ ) instead of sines and cosines. One can show that these have a period $T=\frac{2\pi}{\omega}$. Of course, sines and cosines can be written in terms of these two functions via Euler's formula.

Thus the question becomes: Why does the Planck relation $E=\hbar \omega$ hold for particles (note: this relation holds for all particles, not just photons) with a $e^{-i\omega t}$ time dependence? Why doesn't a triangle or square wave with angular frequency $\omega$ have an energy $E=\hbar \omega$ as well?

In the Schrödinger picture, a quantum mechanical system with a Hamiltonian $\hat{H}$ and a state $|\psi(t)\rangle$ evolves over time according to the Schrödinger equation: $$\hat{H}|\psi(t)\rangle=i\hbar\frac{d}{dt}|\psi(t)\rangle $$ Furthermore, if a system has a well-defined energy $E$, then the state is an eigenstate of the Hamiltonian such that: $$\hat{H}|\psi (t)\rangle=E|\psi(t)\rangle = i\hbar\frac{d}{dt}|\psi(t)\rangle $$ This differential equation is solved by a state $|\psi(t)\rangle =e^{-i\frac{E}{\hbar}t}|\psi(0)\rangle$ (check for yourself!), where $|\psi(0)\rangle$ is the state of the system at time $t=0$. Thus, we can see that the state oscillates with an angular frequency $\omega=\frac{E}{\hbar}$, giving us Planck's famous relation. What happens when $|\psi(t)\rangle$ is not of this form? In that case $|\psi(t)\rangle$ is not an eigenstate: $$i\hbar\frac{d}{dt}|\psi(t)\rangle \neq E|\psi(t)\rangle$$ In other words, the system does not have a well defined energy. As you correctly point out in the comments, we can decompose any function into sines and cosines of different frequencies. In the same vein, we can decompose any state into eigenstates of the Hamiltonian: $$|\psi(t)\rangle=A_1e^{-i\frac{E_1}{\hbar}t}|\psi_1(0)\rangle + A_2e^{-i\frac{E_2}{\hbar}t}|\psi_2(0)\rangle + A_3e^{-i\frac{E_3}{\hbar}t}|\psi_3(0)\rangle + \dots$$ In this sense, even though the state doesn't have a single definite energy, the relation between the energy and the frequency is still 'linear' as you describe it. If you double the frequency of every eigenstate, then the energy of every eigenstate will also be doubled. But to reiterate: these states do not have a definite energy. I hope this answers your question.

Some final notes:

  • It is conceivable to construct a universe in which the energy eigenstates don't depend on time according to $e^{-i\omega t}$. In that case, the Schrödinger equation would be of a different form, perhaps with triangle or square waves as solutions like you propose. However, this is not the universe we live in.
  • This of course does just kick the can further down the road: why is the Schrödinger equation the way it is? It is possible to motivate the form from certain symmetries (i.e. the Hamiltonian, and thus the energy, is related to time translation symmetry. Thus, it shouldn't be too surprising that there is a relation between energy and frequency), but that goes beyond the scope of this answer.
  • It is often convenient, especially when talking about photons specifically, to use the Heisenberg picture, in which the operators (the fields, the Hamiltonian, the momenta) change over time and the state remains constant. I used the Schrödinger picture mainly for clarity, but suffice it to say the physics are the same.
  • It should go without saying, but historically the relation $E=\hbar \omega$ was found before the Schrödinger equation. In this case, what we mean by $\omega$ is 'the angular frequency of a sine/cosine/complex exponential' and not 'the angular frequency of a triangle/square/other wave'. As it happens, this is consistent with reality (and thus subsequent developments in quantum mechanics).
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  • $\begingroup$ Thanks for your insightful answer. But here I don't get what does it mean for an state to not have definite energy, do those kind of states exist? Do they have multiple energies at the same time? What is an example? Also, I've never seen a proof showing that the state you showed as solution is the only solution of that equation. $\endgroup$ Commented Mar 9 at 23:50
  • $\begingroup$ It's true that the statement 'the state does not have a definite energy' is a bit vague. I'll try to elaborate. Let's say that our system allows for a solution $|\psi_1 \rangle$ with energy $E_1$ and another solution $|\psi_2 \rangle$ with energy $E_2$. Then indeed $|\psi \rangle = C_1 |\psi_1 \rangle + C_2 |\psi_2 \rangle$ is also an allowed solution. Can a real system be in such a state? Absolutely! This is easy to show mathematically (any linear combination of solutions to the Schrödinger equation is also a solution), but somewhat harder to interpret physically. $\endgroup$
    – Sturrum
    Commented Mar 10 at 0:58
  • $\begingroup$ One can indeed say that a particle has 'two energies at once', but that raises more questions. What happens when I measure the energy of the particle? Obviously, my instrument will only read one number, not two at once. More specifically, I'll measure either an energy $E_1$ or an energy $E_2$ with a probability of $|C_1|^2$ and $|C_2|^2$ respectively. $\endgroup$
    – Sturrum
    Commented Mar 10 at 0:58
  • $\begingroup$ This is the core of quantum mechanics: A system isn't in one particular eigenstate but a linear combination (also called superposition) of many different eigenstates. This isn't unique to the energy of the system either. Just as the Hamiltonian operator has energy eigenstates, there is an operator for the position/momentum/electric field/any other physical quantity. Only when we measure a particular physical quantity of the system does it "choose" one value. $\endgroup$
    – Sturrum
    Commented Mar 10 at 0:58
  • $\begingroup$ As for the uniqueness of the solution to $i \hbar \frac{d}{dt} |\psi(t) \rangle = E|\psi(t) \rangle$, you can essentially think of $|\psi(t) \rangle$ as one big vector, and each component $f_i$ of the vector most satisfy $i \hbar f_i' = Ef_i$ and I hope it's clear why the only solution to this ODE is an exponential function. If not, you could probably find a more rigorous explanation on MathSE. $\endgroup$
    – Sturrum
    Commented Mar 10 at 0:59
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You should look at this way: if you have an atom where an electron goes from an upper (excited) energy level to a lower level, with the energy difference between the two levels being $E$, this will lead to the emission of radiation with some sharp frequency $f$ that turns to be proportional to the energy $E$ i.e. $f=E/h$ where h is the constant of proportionality as determined by the observations. The fact that the observed spectral line is a sharp line implies that the oscillation is sinusoidal. If the oscillation is not sinusoidal you would not see a sharp line but a broadened one and/or multiple lines.

Why is the oscillation sinusoidal? The reason is the fact that the Coulomb field of the atomic nucleus is spherically symmetric, which, in the semi-classical picture of the atom, implies circular electron orbits, and a sinusoidal is just the projection of a circular motion.

Note that one can not really explain the sinusoidal through the form of the Schrödinger equation (as done in some of the other answers) as this behaviour was implicitly put in in the first place when postulating the equation by assuming it to be a wave equation, and the solutions of a wave equation are just sinusoidal functions (see here for a good account of this).

Having said this, the radiation emitted in an atomic transition is actually not strictly sinusoidal but the amplitude is damped by some amount as shown schematically below (in reality this damping will usually be much slower). Otherwise the electron would never get to the lower level and keep on emitting radiation forever, which would violate energy conservation

enter image description here

Mathematically, this damped oscillation is described by a sinusoidal multiplied by an exponential damping term i.e.

$$ x(t)=x_0\cdot sin(2\pi ft)\cdot e^{-\gamma t} $$

This damping then also means that the associated spectral line is actually not infinitely sharp but broadened by some amount, which results in a so-called Lorentz line shape as shown below

enter image description here

The width of this curve is directly given by the damping term $\gamma$ in the equation for the damped oscillator above (again, in reality the broadening will usually be much smaller).

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  • $\begingroup$ Yes but, why is it sinusoidal? It is a bit strange right? What in the atom makes an energy transition (from one orbital to another) a sinusoidal oscilation? $\endgroup$ Commented Mar 9 at 22:31
  • $\begingroup$ @CuriousMind Please see my edited answer $\endgroup$
    – Thomas
    Commented Mar 10 at 10:32
  • $\begingroup$ thanks but that is a semi classical picture, I would like to see the real quantum reason for it. $\endgroup$ Commented Mar 10 at 15:00
  • $\begingroup$ @CuriousMind There is no real quantum reason as such. Even the Schrödinger equation mentioned in one of the other answers is actually constructed by combining two classical concepts, 1) classical mechanics and 2) the wave equation from classical electrodynamics. It is just so that the solutions of the wave equation (and thus the Schrödinger equation) are sinusoidal functions (see chem.libretexts.org/Courses/Pacific_Union_College/… ) and this appears to match observations $\endgroup$
    – Thomas
    Commented Mar 10 at 16:03
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It does not. It's linear in other cases too. Square wave can be approximated by pushing sinusoidal function to some transformation $g(f)$, like : $$y(t)=\tanh\left(k\sin\left(\omega t\right)\right) \tag 1$$

Triangle wave similarly can be achieved by filtering sinusoidal by: $$ y(t) = \arcsin(\sin(\omega t)) \tag 2 $$

Nor hiperbolic tangent, neither $\sin^{-1}$ are periodic functions, only $\sin()$ is, so in both cases $E= \hbar \omega$ linear relationship also holds as long as custom wave can be represented by sinusoidal function $g(f)$ transformations.

In addition I believe that almost all "convoluted" signals can be decomposed into series of sinusoidal functions by Fourier transform.

Also sometimes (usually ?), the REAL signal used in telecommunications and other areas, does not have well-defined "single frequency". Lets look at this type of signal :

enter image description here

It can be shown that it's addition of two sinusoidal waves, by using trigonometric identity we see that

$$ \sin (\omega_1 t) + \sin (\omega_2 t) = 2 \sin \left( \frac{\omega_1+\omega_2}{2} t \right) \cos \left( \frac{\omega_1-\omega_2}{2}t \right) \tag 3$$

where in the right side of equation there's beat and modulation frequencies. Hence we can define separate beat and modulation quanta energies :

$$ E= \hbar \frac{\omega_1 \pm \omega_2}{2} = \frac {E_1}{2} \pm \frac{E_2}{2} \tag 4$$

So it is seen that beat and modulation frequencies are linear combination of some base sinusoidal frequencies $\omega_1,\omega_2$ for whom linear combination of energies can also be stated.

Overall conclusion is that Planck linear relation is very universal and fits to all signals, because they can be decomposed into summation of sinusoidal components or simply transforming sinusoidal functions.

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  • $\begingroup$ Yes but those are not periodic functions as you say. I am talking about why in E=h*f the frequency is of sinusoidal signals, instead of square periodic or triangular signals. Like any of these (look how they repeat constantly with a specific frequency). upload.wikimedia.org/wikipedia/commons/7/77/Waveforms.svg $\endgroup$ Commented Mar 9 at 22:39
  • $\begingroup$ No, you misunderstood something. (1) and (2) expressions ARE periodic, which represents square and triangular wave approximations. Try to draw their charts to see that they are periodic. What I mean is that non-sinusoidal waves can be approximated by putting sinusoidal function in some $g(f)$ transformation(s) like I gave you in (1),(2). And so linear relationship $E=h\nu$ still holds for the non-sinusoidal waves, because they are same sinusoidal waves "under the hood". In general all signals can be decomposed into sinusoidal ones by Fourier transform. $\endgroup$ Commented Mar 9 at 23:30
  • $\begingroup$ Here's take a look. $\endgroup$ Commented Mar 9 at 23:37
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  1. As for the photon, this linear relationship is verified in experiment (as Thomas said, we measure the energy of the electronic transition and divide it by the frequency of the emitted light). But before that, we must agree in advance that the states of the electron that emits this light are discrete. That is, we must assume that any energy (not only light, but also electrons and any other objects) changes in portions (quanta) and attribute to these energy portions some unknown oscillatory process, the frequency of which is determined by Planck’s relation. Here it (Planck's relation) becomes not an experimental law (as for light), but a universal postulate (axiom) of the new quantum mechanics, its decisive difference from classical mechanics.

  2. Why is it that using a sinusoidal frequency base results in a linear relationship in Planck's relation? Because this is a very successful axiom (or postulate, or rule of primary quantization) - a free material point corresponds to a plane wave with precisely such a linear relationship between energy and frequency. Here, the frequency of a sinusoidal process (and not a rectangular one, for example) is taken because this is the simplest process of uniform periodic motion (circular rotation at a constant speed). All other periodic processes turn out to be simply more complicated (their speeds are not constant) and several sinusoids (and even many) are needed to describe them.

  3. If we were to use a different basis for frequencies, would we not see a linear relation? The linear relationship is the main postulate of quantum mechanics, that is, a proposal accepted without proof precisely in order to deduce all other observable phenomena from it. For example, if instead of a sine wave you take a square wave, then it can be represented as a sum of sinusoids, each of which will have its own energy. Such an object is called a wave packet and behaves (for example, gradually spreads out) depending on another postulated relationship - the relationship between frequency and wavelength (or between energy and momentum), called the "dispersion law".

3.How is the linear relantioship a postulate (axiom)? It is real right? It matches with experimentation. The slope of that relation experimentally is h, or not? This relationship is precisely an assumption (postulate, axiom) about the existence of quanta (pieces) of energy. This refers to quanta not of matter (the usual particles in classical physics), but of energy. In relation to a free electron, this means that you determine (postulate) its frequency from this ratio (divide the measured energy by Planck's constant). There is no way you can measure the frequency in an experiment, since the expression “the frequency of a stone falling into a pond” simply does not make sense (unlike the waves that it generates).

4.A century after the invention of quantum mechanics, we believe that the frequency of any object with a given energy is the frequency of oscillations of its (harmonic) wave function (see Sturrum's answer). If these oscillations are not harmonic (not sinusoidal, but for example triangular), then the object has not one specific energy, but a whole spectrum (coinciding with the Fourier frequency spectrum of the wave function expansion). We can say that such an object has many energies at once. Or we can say that the object somehow, unnoticed by us and randomly jumps between states with different energies. Or we can even say that the very concept of energy here loses its classical meaning. All this is connected with the fundamental unresolved problems of entangled states and quantum causality (the rejection of the classical causality through local interactions that we are familiar with).

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  • $\begingroup$ 1. Well a cosine is the projection along the x axis of that circular movement, it isn't really the circular movement itself. 2. How is the linear relantioship a postulate (axiom)? It is real right? It matches with experimentation. The slope of that relation experimentally is h, or not? $\endgroup$ Commented Mar 9 at 18:14
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Mar 9 at 20:09
  • $\begingroup$ With respect to point 3, of course we can measure the frequency right? We just have to make the Fourier Transform of the measured radiation emitted by the atom. What is strange is that in the spectrum there are lines instead of continous regions. Meaning that they are purely sinusoidal. $\endgroup$ Commented Mar 9 at 22:34
  • $\begingroup$ Of course, CuriousMind, we can measure the frequency of light. But we cannot measure the frequency of a free electron simply because its “frequency” does not make sense, because a free electron flies straight and uniformly. $\endgroup$ Commented Mar 9 at 22:53
  • $\begingroup$ @Ванек Огонек I think in this case you are wrong. I think the frequency of any particle is the frequency of its wavefunction. In that case, the frequency of a free electron would be the frequency of its wavefunction, in the case of a free electron it would be the frequency of a sine if I am not mistaken. $\endgroup$ Commented Mar 9 at 23:47
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The quantization of a single-mode EM field results in a Hamiltonian of a harmonic oscillator, given by $H=\hbar \omega (a^\dagger a + \frac{1}{2})$, where $a^\dagger a$ can be understood as the "photon number" operator. Therefore, each photon has an energy of $\hbar \omega$. The result stems directly from the formulation of the Maxwell equations, where your propagating field can be represented by plane waves of a given frequency, which is also emphasized by the fact that we are calling that field a single-mode field.

By a single-mode field we understand a solution, that represents only a single frequency of light manifested in the temporal dependency of our solution. The spatial part does not matter when obtaining the energy formula. In the case of a harmonic oscillator that models our field the $\exp(i\omega t)$ (or a sine wave) is a natural choice for a frequency basis.

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  • $\begingroup$ We could also represent the propagating field with any other basis different of plane waves, like triangular or square waves. The strange thing is that we choose plane waves for simplicity and that matches with reality by pure coincidence? $\endgroup$ Commented Mar 9 at 22:41
  • $\begingroup$ Hi, then you are asking question: why we want solutions that are harmonic in time. When you assume that, you get the energy as described. However, spatial form of the solution does not matter. $\endgroup$
    – Noct
    Commented Mar 10 at 8:25
  • $\begingroup$ And that's because $\exp(i\omega)$ is the most physical representation of a pure frequency when your systems can be understood as some variation of a harmonic oscillator (physics.stackexchange.com/questions/716271/…). $\endgroup$
    – Noct
    Commented Mar 10 at 11:17

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