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For a classical real scalar field $\phi(\vec{x},t)$ of the type:

$$\frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0$$ The modes $\phi(\vec{p},t)$ can be obtained by: $$\phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t)$$

Each mode satisfies the simple harmonic equation of motion: $$\frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi$$

Therefore each mode solves the equation of a harmonic oscillator vibrating in time at frequency: $$\omega_\vec{p}=\sqrt{\vec{p}^2+m^2}$$

Now, 2 statements:

  1. From the last equation it is evident that, for any given $\vec{p}$, we have a unique $\omega_\vec{p}$.
  2. This $\omega_\vec{p}$ is interpreted as the energy of the mode, respecting the energy-momentum relation.

I don't know how to reconcile these 2 statements with the following one:

  1. In this video by Sean Carroll about fields, when he is still talking about classical Klein-Gordon, he mentions that the energy of a mode is not unique! It depends on the amplitude of the mode, as in $E \propto (\phi_{max})^2$. The only way in which this statement can work, is if the two previous statements are mutually exclusive.

Option (a) to accept Statement 3: accept Statement 1, reject Statement 2.

Carroll is allowed to say that there are more energies possible for a given momentum-mode, but he cannot interprete the energy as the frequency of oscillation in time of the mode's amplitude, because omega is unique as $\omega_\vec{p}=\sqrt{\vec{p}^2+m^2}$.

Option (b) to accept Statement 3: accept Statement 2, reject Statement 1.

If we instead insist that the energy of a mode is equal to the frequency of oscillation in time of the mode's amplitude, and we agree with Carrol that there are more energies possible for a given mode, then by logic we cannot say that there is a unique frequency for a given momentum. We must reject $\omega_\vec{p}=\sqrt{\vec{p}^2+m^2}$ !

How do you solve this?

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  • $\begingroup$ The two modes $$\phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{\pm i\vec{p}\cdot \vec{x}}\phi(\vec{p},t)$$ verify the K-G equation,see:en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation $\endgroup$
    – The Tiler
    Commented Dec 10, 2023 at 17:48
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    $\begingroup$ Your "2 statements" are both wrong. You don't have to "reconcile" them with anything... $\endgroup$
    – hft
    Commented Dec 15, 2023 at 17:26

8 Answers 8

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A corrected version of Statement 2 is "$\omega_\vec{p}$ is the energy of one quantum of the mode with momentum $\vec{p}$." In a free quantum field theory, a stationary state is defined by the number of quanta in each mode (and their relative phases but that's a detail). Each mode then behaves like its own independent quantum harmonic oscillator. We can make three analogous statements about an individual harmonic oscillator, which may make it clear what is going wrong here:

  1. A harmonic oscillator has a definite, characteristic frequency $\omega$.
  2. This $\omega$ is interpreted as the energy of one quantum of the mode.
  3. The energy of a particular state is determined by the squared amplitude of the oscillation.

The important distinctions are between a "mode," which is a bit like it's own independent harmonic oscillator system that's tensor-producted with a bunch of other modes in the free field theory case, and a "state," which is a particular configuration of that oscillator. In quantum systems it's more convenient to describe a state with a quantum number $n$ such that $E_n = \hbar \omega (n + 1/2)$, but you could also say that there's a proportionality relationship between the squared-amplitude of the oscillation $\langle x^2 \rangle$ and the energy, which is $\langle x^2 \rangle = \frac{\hbar}{2m\omega} (2n+1) = E_n / m\omega^2$.

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Statement 2 is wrong. Both classically and in quantum field theory.

The energy of the (real) Klein-Gordon field is given by the expression \begin{equation} H = \frac{1}{2}\int \mathrm{d}^3 x \left[\pi^2 + (\nabla\phi)^2 + m^2\phi^2 \right] \end{equation} here $\pi(t, \vec{x})$ is the canonically conjugated momentum to $\phi(t, \vec{x})$, i.e. their Poisson bracket is $\{\phi(t, \vec{x}), \pi(t, \vec{y})\} = \delta^{(3)}(\vec{x}-\vec{y})$. The reason this is the energy is that it is the conserved charge associated to time-translations.

Note that $H$ is precisely the Hamiltonian and using the Poisson structure written above you will recover the Klein-Gordon equation from the Hamilton's equations \begin{equation} \begin{cases} \partial_t \phi(t, \vec{x}) = \{\phi(t, \vec{x}), H\} = \pi(t, \vec{x})\\ \partial_t \pi(t, \vec{x}) = \{\pi(t, \vec{x}), H\} = \nabla^2 \phi(t, \vec{x}) - m^2 \phi(t, \vec{x}) \end{cases} \end{equation} of course the Hamilton equations are first order in the time derivative and you recover the Klein-Gordon equation by eliminating the momentum field $\pi(t, \vec{x})$.

Now let us consider a plane wave solution to the Klein-Gordon equation and compute its energy \begin{equation} \phi(t, \vec{x}) = A\exp(-i\omega_\vec{p}t + i \vec{p}\cdot\vec{x}) + \text{c.c.} \end{equation} the conjugated momentum is thus \begin{equation} \pi(t, \vec{x}) = -i\omega_\vec{p}A\exp(-i\omega_\vec{p}t + i \vec{p}\cdot\vec{x}) + \text{c.c.} \end{equation} Using this two expression we can compute the energy by simply substituting them in the formula. The energy density is given by \begin{equation} \mathcal{H} = 2 |A|^2 \omega_\vec{p}^2 - A^2 |\vec{p}|^2 e^{2 i \vec{p} \cdot \vec{x} - 2 i \omega_\vec{p} t} - \left(A^*\right)^2 |\vec{p}|^2 e^{2 i \omega_\vec{p} t - 2 i \vec{p} \cdot \vec{x}} \end{equation} now to take the integral notice that \begin{equation} \int \mathrm{d}^3 x \ e^{2 i \vec{p}\cdot\vec{x}} \propto \delta^{(3)}(\vec{p}) \end{equation} thus upon integration the only term surviving is (the other pieces would be proportional to $|\vec{p}|^2\delta^{(3)}(\vec{p})$ which is zero) \begin{equation} H = 2 |A|^2 \omega_\vec{p}^2 \int \mathrm{d}^3 x \end{equation} of course this is finite only in a finite volume. As you see the energy is proportional to the amplitude $A$ of the field and the frequency of the mode squared.

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    $\begingroup$ This answer makes valid points but it is not the whole story. I will update my answer with these additional physical considerations. $\endgroup$
    – my2cts
    Commented Dec 16, 2023 at 12:20
  • $\begingroup$ Your conclusion should be that E is proportional to the square of the amplitude. However, this is only possible if you drop the physical meaning of $m$. See my extended answer. $\endgroup$
    – my2cts
    Commented Dec 17, 2023 at 0:06
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  1. This $\omega_\vec{p}$ is interpreted as the energy of the mode, respecting the energy-momentum relation.

Not in the classical theory.

In the classical theory the total energy for all the modes that obey the classical equations of motion can be written as: $$ E = 2\int \frac{d^3k}{(2\pi)^3} \omega_{\vec k}^2|\bar\phi_{\vec k}|^2\;,\tag{1'} $$ where, yes, that is an omega squared in the integrand (explained further below!), and where $\omega_{\vec k}\equiv+\sqrt{k^2c^2+m^2}$, and where $\bar\phi_{\vec k}$ are the Fourier amplitudes of the real-space classical field modes with positive frequencies and $\bar\phi_{-\vec k}^*=\bar\chi_{\vec k}$ are the Fourier amplitudes of the real-space classical field modes with negative frequencies. I.e., $$ \phi(\vec x, t) \equiv \int \frac{d^3k}{(2\pi)^3}e^{i\vec k \cdot \vec x} \left(\bar\phi_{\vec k}e^{-i\omega_{\vec k} t} + \bar\chi_{\vec k}e^{+i\omega_{\vec k} t} \right)\;,\tag{2'} $$ where $\bar\chi_{\vec k} = \bar\phi_{-\vec k}^*$ to ensure that $\phi(\vec x, t)$ is always real (in addition to obeying the equations of motion). And further, where it is clear that $m$ must have units of frequency in the classical theory (since we do not have any way to turn units of energy like $mc^2$ into units of frequency in the classical theory).

Eq. (2') above is similar to your second equation, but I have provided the explicit time dependence that is required by the equations or motion. As you see, the energy per mode is not proportional to $\omega_{\vec k}$ when using the Fourier expansion of Eq. (2'), but rather is proportional to $\omega_{\vec k}^2$. This is because your Fourier transform convention for the Fourier amplitudes is not the typical one. See below for the more-conventional transformation, which includes an additional factor of $\frac{1}{\sqrt{2\omega_{\vec k}}}$.

It is more conventional to write the Fourier transform as: $$ \phi(\vec x, t) = \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec k}}}e^{i\vec k \cdot \vec x} \left(\tilde\phi_{\vec k}e^{-i\omega_{\vec k} t} + \tilde \chi_{\vec k}e^{+i\omega_{\vec k} t} \right)\;,\tag{2} $$ with a factor of $\sqrt{2\omega_{\vec k}}$ in the denominator, in which case the energy in terms of the modes is: $$ E = \int \frac{d^3k}{(2\pi)^3} \omega_{\vec k}|\tilde\phi_{\vec k}|^2\;,\tag{1} $$

Given the form of Eq. (1) above, the energy of a mode, $$ \omega_{\vec k}|\tilde \phi_{\vec k}|^2\;,$$ appears to be proportional to $\omega_{\vec k}$, but it is also clearly proportional to the Fourier amplitude squared!

And recall, as demonstrated above, that we are certainly free to alter our definition of the $\tilde\phi_{\vec k}$ to, say, the definition of $\bar \phi_{\vec k}$ and thereby absorb arbitrary powers of $\omega_{\vec k}$ if we so choose. Thus it is only the combination of $\omega_{\vec k}|\tilde \phi_{\vec k}|^2$, or equivalently $2\omega_{\vec k}^2|\bar \phi_{\vec k}|^2$, that should be called the energy per mode.


Our Eq. (1) for the classical field energy can be arrived at by considering the classical Lagrangian: $$ \mathcal{L} = \frac{1}{2}\dot \phi^2 - \frac{1}{2}(\nabla \phi)^2 - \frac{1}{2}m^2\phi^2\;,\tag{2} $$ where the units in Eq. (2) look pretty funny... Presumably we have set $c=1$, but as we mentioned above this "$m$" parameter must actually have units of frequency, since we have no $\hbar$ parameter in the classical theory that we can combine with $c$ and $m$ to get a frequency from a mass.

Given the classical Lagrangian density we can calculate the classical energy as: $$ E =\int d^3x \left(\frac{\partial \mathcal{L}}{\partial\partial_0\phi}\partial^0\phi - \mathcal{L}\right) $$ $$ = \int d^3x \left( \frac{1}{2}(\dot \phi)^2 + \frac{1}{2}c^2(\nabla \phi)^2 + \frac{1}{2}m^2 \phi^2 \right) $$ $$ =\int \frac{d^3k}{(2\pi)^3} \omega_{\vec k}|\tilde\phi_{\vec k}|^2\;, $$ where, again, we caution the reader that $m$ has units of frequency in the classical theory. Of course, we can still get an energy $E$ out on the LHS in the classical theory since the units of our $\phi$ function have not been specified yet. But apparently the units of $\phi$ must be $\sqrt{\frac{M}{L}}$, which, by the way, makes $\int d^3x m^2\phi^2$ have units of energy when $m$ has units of frequency (as it must in the classical theory). I also note here that the units of $\tilde \phi$ are $\sqrt{\frac{ML^5}{T}}$.

We can also calculate other conserved quantities like the total momentum of the classical field $$ \vec P = -\int d^3x \left(\frac{\partial \mathcal{L}}{\partial\partial_0\phi}\vec{\nabla}\phi \right) $$ $$ =\int \frac{d^3k}{(2\pi)^3}\frac{\vec k}{2}\left(|\phi_{\vec k}|^2-|\tilde\phi_{-\vec k}|^2\right)\;, $$ which again depends on the squared Fourier amplitudes of the classical field.


When we transition to the quantum theory in the canonical way, we quantize the real Klein-Gordon field in exact analogy to quantizing a (set of) simple harmonic oscillator(s).

That is, we recall the canonical quantization rules (with $\hbar=1$): $$ [q_i, p_j] = i\delta_{ij} $$ and extend these to fields as: $$ [\hat \phi(\vec x), \hat \pi(\vec y)] = i\delta(\vec x - \vec y)\;, $$ where now $\hat \phi$ and $\hat \pi$ are quantum fields (hence the "hat" notation to help us remember the difference.)

The quantum field $\hat \phi(\vec x)$ (now in the Heisenberg picture) just so happens to obey the same equation of motion as the classical field: $$ \partial_t^2\hat\phi(\vec x, t) =(\vec \nabla^2 - m^2)\hat\phi(\vec x, t)\;, $$ or, Fourier transforming in space, $$ \partial_t^2 \hat{\tilde{\phi}}_{\vec p}(t) =(-\vec p^2 - m^2)\hat{\tilde{\phi}}_{\vec p}(t)\;, $$ which is exactly the same equation of motion as a quantum simple harmonic oscillator with frequency $\omega= \omega_{\vec p}$.

In fact, each quantum mode just acts like a completely independent quantum harmonic oscillator.

We already know that the Hamiltonian for a single oscillator is $\hat H_{SHO}=\omega\hat a^\dagger a$. (You are free to add a constant like $\frac{\omega}{2}$ to this Hamiltonian if you would like, but I won't.)

So the Hamiltonian for our real Klein-Gordon field is just a sum (integral) of individual SHO Hamiltonians (one for each mode): $$ \hat H_{KG} = \int \frac{d^3p}{(2\pi)^3}\omega_{\vec p}\hat a_{\vec p}^\dagger a_{\vec p}\;, $$ where $$ \hat \phi(\vec x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec p}}} \left( a_{\vec p}e^{i\vec p\cdot\vec x} + a^\dagger_{\vec p}e^{-i\vec p\cdot \vec x} \right) $$ and $$ \hat \pi(\vec x) = i\int \frac{d^3p}{(2\pi)^3}\sqrt{\frac{\omega_{\vec p}}{2}} \left( a^\dagger_{\vec p}e^{-i\vec p\cdot \vec x} -a_{\vec p}e^{i\vec p\cdot\vec x} \right)\;, $$ which are analogous to the quantum simple harmonic oscillator ladder operator relations: $$ \hat x = \sqrt{\frac{\hbar}{2m\omega}}(a^\dagger + a) $$ and $$ \hat p = i\sqrt{\frac{\hbar m\omega}{2}} (a^\dagger - a) \;. $$

As with the quantum simple harmonic oscillator, we say that each of the $a^\dagger_{\vec p}$ creates a quantum excitation in mode $\vec p$ and the energy of such a single quantum excitation is $\omega_{\vec p}$. (Recall, we still have $\hbar=1$.)

Note that, unlike in the classical theory, we have little choice but to insert the appropriate factors of $\sqrt{2\omega_{\vec p}}$ if we want our ladder operators $a_{\vec p}$ and $a_{\vec p}^\dagger$ to combine to give a number density (in momentum space). Thus we are forced to arrive at the expression $$ \hat H_{KG} = \int \frac{d^3p}{(2\pi)^3}\hbar\omega_{\vec p}\hat a_{\vec p}^\dagger a_{\vec p} = \int \frac{d^3p}{(2\pi)^3}\hbar\omega_{\vec p}\hat n_{\vec p}\;, $$ with a single power of $\hbar \omega_{\vec p}$ that can now be interpreted as the energy of a single mode excitation.


  1. From the last equation it is evident that, for any given $\vec{p}$, we have a unique $\omega_\vec{p}$.

(Emphasis added.)

This assertion that "for any" $\vec p$, we have "a unique" $\omega_{\vec p}$ is wrong. (At least, given my interpretation of your meaning here.)

For example, any other $\vec p'$ where $\vec p' = R\vec p$, where $R$ is any rotation matrix, will have the same $\omega_{\vec p}$ value. So the $\omega_{\vec p}$ is not "unique" for "any" $\vec p$.

Similarly, we have $$ \omega_{\vec p} = \omega_{-\vec p}\;, $$ which is sort of important for a lot of the formal manipulations, so I just want to stress that these two frequencies are the same (i.e., not "unique").

Anyways, the frequency $\omega_{\vec p}$ is a function of the $\vec p$, this does not mean it is a one-to-one function, which OP seems to imply by saying "for any given $\vec p$... a unique $\omega_{\vec p}$ value."


I don't know how to reconcile these 2 statements with the following one:

You don't have to reconcile those two statements, since they are either wrong or irrelevant.

  1. In this video by Sean Carroll about fields, when he is still talking about classical Klein-Gordon, he mentions that the energy of a mode is not unique! It depends on the amplitude of the mode, as in $E \propto (\phi_{max})^2$.

OK. This seems fine (although it is not entirely clear what your $\phi_{max}$ is supposed to mean).

We have seen explicitly that the classical mode energy for a set of real classical waves that obey the equations of motion is equal to the square of the Fourier amplitude $|\tilde\phi_{\vec k}|^2$ times the frequency $\omega_{\vec k}$.

The only way in which this statement can work, is if the two previous statements are mutually exclusive.

No, that is not the only way this statement can work, since the previous two statements were both wrong or irrelevant and thus don't have to be mutually exclusive of each other or any other statement.

Anyways, at the end of the day, you can't interpret a classical frequency such as $\omega_{\vec p}$ as an "energy" in the classical theory. In the classical theory, just like for a classical oscillator, you need the combination of the frequency and the amplitude (squared) to get the classical mode energy. To interpret $\hbar \omega_{\vec p}$ as an energy you have to wait until you transition to the quantum theory, where there are no classical amplitudes, but rather discrete (i.e., "quantum") excitations. Note that the transition back to the classical theory can also be considered, and one typically considers states with very large numbers of excitations in the quantum modes.

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    $\begingroup$ he said that for every $\vec{p}$ there is one $\omega_\vec{p}$, no the opposite. what he said does not exclude that multiple $\vec{p}$ can have the same $\omega_\vec{p}$, just that there is a unique frequency associated to each wavenumber $\vec{p}$. it is not wrong (well exept for the fact that $\omega_\vec{p}$ is a non monodromous function) $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 17:50
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    $\begingroup$ He said that for every $\vec p$ there is "a unique $\omega_{\vec p}$." (Emphasis added.) @ekardnam_ $\endgroup$
    – hft
    Commented Dec 15, 2023 at 20:05
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    $\begingroup$ What's written is basically the definition of function, for every x there is a unique y such that y = f(x). It doesn't imply the function is injective $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 21:00
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    $\begingroup$ it is. Please see the definition of function as a total univalent relation on wikipedia en.wikipedia.org/wiki/Function_(mathematics), univalent or right unique is precisely what we are discussing right now. For each $x$ there is a unique output which is $f(x)$, if the output was not unique then $f(x)$ would be a set of values. This doesn't mean that the unique outputs of two different $x_1$ and $x_2$, which are $f(x_1)$ and $f(x_2)$ cant be the same $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 22:02
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    $\begingroup$ to me saying "for any given" still sounds more like a "for each", anyhow, we both clearly understand that it is a function but not a injective (1-1) function and your answer also states it clearly which doesn't hurt $\endgroup$
    – lucabtz
    Commented Dec 16, 2023 at 23:03
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I think you've confused classic waves and quantum "waves".

Classical waves have momentum and energy. They're continuum, and generally depend on the amplitude, not the wave number/frequency. Think about Poynting vector $\vec{S}=\vec{E}\times\vec{H}$ - it is about momentum, but it has nothing to do with $\vec{k}$ (apart from the fact that in linear media they point in the same direction). Similarly the EM energy density $u_{EM}=\frac{E^2}{2\varepsilon_0} + \frac{\mu_0}{2} B^2$.

Quantum "waves" or excitations have discrete energy and momentum, which are obtained by de-Broglie: $\vec{p}=\hbar \vec{k}$ and $E = \hbar \omega$. These relations are not relevant at all in the case of the classical Klein-Gordon equation you're describing. The energy-momentum relation $E^2=p^2 + m^2$ is relevant for scalar Bosons, but Bosons are only relevant for describing this field if it is quantized.

So, your (2) is irrelevant in the quantum case; but moreover, as to (1), discussing $\omega_{\vec{p}}$ is misleading - what you have is a dispersion relation $\omega(\vec{k})$, and as long as you're analyzing classical wave, this is the right point of view, and not energy/momentum relations.

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It is somewhat unfortunate that the OP is using natural units ($\hbar = 1 = c$) throughout (without explicitly saying so) when the issue is a comparison of the classical and quantum theories. I will thus retain both $\hbar$ and $c$ in what follows. For simplicity I will assume $d=1+1$. The first thing to notice is that, strictly speaking, there is no "classical Klein-Gordon (KG) equation". Recall that this equation was Schrödinger's first attempt of a matter wave equation. He abandoned it as it did not yield the correct hydrogen spectrum. Using the relativistic energy momentum relation, $E^2/c^2 - p^2 = m^2 c^2$, and the heuristic quantisation rules, $E \to i\hbar \partial_t$, $p \to -i\hbar \partial_x$, acting on a wave function $\phi$, the KG equation becomes (upon dividing by $\hbar^2$), $$ \left( \frac{1}{c^2} \partial_t^2 - \partial_x^2 + \frac{m^2c^2}{\hbar^2} \right) \phi \equiv \left( \Box + \frac{m^2c^2}{\hbar^2} \right) \phi = 0 \; . $$ This is of course consistent with dimensional analysis as $mc/\hbar$ is the inverse Compton wavelength, hence a wave number. The appearance of $c$ signals that we are working in a relativistic context. More importantly, we see $\hbar$ appearing as it must for a matter wave equation.

In the classical theory, there obviously is no $\hbar$, and all we can do is replace $mc/\hbar$ by an unspecified wave number $K$. This leads to the equation $$ (\Box + K^2) \phi = 0 \; , $$ which formally looks like the KG equation, but has a rather different interpretation. It is closely related to the telegraph equation, now almost forgotten in physics, but discussed e.g. in Courant and Hilbert, Vol. II. As the name says, it describes dispersive waves moving along wires. Solutions are of the standard form $$ \phi(t,x) = A \, e^{-i\omega t + ikx} \; , \qquad (1) $$
with amplitude $A$, angular frequency $\omega$, wave number $k$ and dispersion relation $\omega^2 = k^2 + K^2$. There is no $\hbar$ involved whatsoever.

It was the great achievement of Planck and de Broglie to note that one needs $\hbar$ to turn angular frequencies into energies ($E = \hbar \omega$) and wave numbers into momenta ($p = \hbar k$). This can be seen by applying the energy and momentum operators, $i\hbar \partial_t$ and $-i\hbar \partial_x$, to the wave solution (1). Thus, after quantisation, (1) can be interpreted as a matter wave with Energy $E = \hbar \omega = \hbar (k^2 + K^2)^{1/2}$ and momentum $p = \hbar k$, identifying $K \equiv mc/\hbar$. One can then write $$ \phi(t,x) = A \, e^{-iE t/\hbar + ipx/\hbar} \; , $$ where the actions $Et$ and $px$ are measured in units of $\hbar$, the 'quantum of action'. Thus, in the OP, all stated equations are missing factors of $\hbar$. Reinstating them would have revealed the quantum nature of the equations.

The energy of the solution, both in the classical and quantum theories, is given by the integral $H$ of the Hamiltonian density, $$ \mathcal{H} = T^{00} = \frac{1}{2} (\Pi^2 + \phi'^2 + K^2 \phi^2) \; , $$ where one should take the real parts of $\phi$ and $\Pi = \dot{\phi}/c$. Evaluation in the classical theory yields the energy $$ H = \frac{2}{c^2} \omega^2 |A|^2 \qquad (\text{no} \; \hbar) \; , $$ proportional to the amplitude mod-squared. Quantisation is done by introducing the Fock operator $$ \hat{a} \equiv \frac{1}{c} \left( \frac{2\omega}{\hbar} \right)^{1/2} \hat{A} \; , $$ which yields the standard result $$ \hat{H} = \hbar \omega \, \hat{a}^\dagger \hat{a} \; , \qquad [\hat{a}, \hat{a}^\dagger] = 1 $$ Note that the Fock operator $\hat{a}$ is dimensionless, unlike the amplitude $A$. In summary then Statement 2 is wrong as $\omega$ is an angular frequency and thus very distinct from an energy in the classical theory.

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  • $\begingroup$ i think $\mathcal{H} = \frac{2}{c^2} \omega^2 |A|^2$ is the energy density no, not the energy, right? anyhow this is a great answer $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 17:59
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    $\begingroup$ Thanks! I've used a single mode and box normalisation without making it explicit, so $A = A'/\sqrt{2\omega L}$ and factors of box size $L$ cancel upon integration. $\endgroup$
    – Tom Heinzl
    Commented Dec 15, 2023 at 18:07
  • $\begingroup$ oh i see now, maybe it could be stated explicitly for clarity? $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 18:09
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Statement 2 should be rejected. It is difficult to explain why until you provide the source for this statement. So far, it has no basis. The argument about the energy-momentum relation does not seem relevant because it only works for a single quantum in the mode, not for classical theory (note that you need $\hbar$ to have correct dimensions). It is obvious that the energy in the mode depends on the amplitude.

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    $\begingroup$ the $\hbar$ serves to connect the wave number $\vec{k}$ to the momentum $\vec{p}$, of course classically it would be more proper to write $\omega_\vec{k} = \sqrt{|\vec{k}|^2+m^2}$ which is just the dispersion relation, but then you can just call the wave number $\vec{p}$ and it is the same, it just doesn't have the interpretation of momentum $\endgroup$
    – lucabtz
    Commented Dec 14, 2023 at 14:45
  • $\begingroup$ @ekardnam_ : I don't understand this. Even if you "adjust" the dimension of one term under the square root, you still need $\hbar$ to adjust the dimension of mass $m$. $\endgroup$
    – akhmeteli
    Commented Dec 15, 2023 at 4:26
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    $\begingroup$ see @Tom Heinzl answer for a reply to your doubt $\endgroup$
    – lucabtz
    Commented Dec 15, 2023 at 17:55
  • $\begingroup$ @ekardnam_ : I believe my answer is correct, OP's Statement 2 has no basis, and, with all due respect, your comment seems worthless for the reason I provided. I don't see how Tom Heinzl's answer changes that. $\omega$ is in no way the energy of the mode, it can only be a quantum of energy in the mode. Note that Tom Heinzl also believes Statement 2 is wrong. $\endgroup$
    – akhmeteli
    Commented Dec 16, 2023 at 4:09
  • $\begingroup$ I also think statement 2 is wrong. Actually it isn't even an opinion, it is wrong. I'm just saying the dispersion relation is a classical thing, and Tom's answer tackles that $\endgroup$
    – lucabtz
    Commented Dec 16, 2023 at 9:09
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Quantized modes have amplitudes normalized to unity in Hilbert space. Classical modes have energies composed of kinetic and stress energy, dependent on amplitude and frequency. Quantum waves squared represent position densities. There are no similarities besides mathematical formulas.

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  • $\begingroup$ Hi, thanks for your answer, but unfortunately I am not talking about quantum field theory. My question is formulated entirely in the realm of classical field theory $\endgroup$
    – TrentKent6
    Commented Dec 8, 2023 at 20:16
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    $\begingroup$ @TrentKent6 Yes, you are talking about quantum fields. Statement 2 is a quantum statement, although you dropped $\hbar$ from it. $\endgroup$
    – my2cts
    Commented Dec 12, 2023 at 8:24
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    $\begingroup$ Precisely; in the classical world simple harmonic oscillators have a well-defined frequency, but the oscillations can have arbitrarily high energy. The identification between frequency and energy is essentially a quantum-mechanical construct. $\endgroup$
    – printf
    Commented Dec 12, 2023 at 12:43
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The answer is: it depends on the physical meaning of the constant $m$. If $m$ represents the total mass of the wave, then 2) applies and not 3). If $m$ represents some unit mass of which the wave can contain a, not necessarily integral, quantity, then statement 3) applies and not 2).

Explanation. Using the Klein-Gordon lagrangian one finds the expression given in @Ekardnam_'s answer: \begin{equation} E = 2 \omega^2 |A|^2 \,, \end{equation} with $\omega^2=k^2+m^2$ ($c=1$ units). Let's consider such a wave at rest. Now $\omega=m$ so $E=2A^2 m^2$. If the parameter $m$ indeed represents the mass of the wave it follows that in the rest frame, where $E=m$, $$A^2=\frac{1}{2m} \,.$$ The charge of this wave is proportional to $A^2\omega$, to $qA^22m$, Since the charge should be the same in all reference frames we find that in any reference frame $$A^2=\frac{1}{2\omega}\,.$$ so that $$E=\omega\,.$$ This means that statement 2 would apply. However one can be interpret $m$ as some unit mass. Then the total mass will be, say, $M=fm$. The energy is then proportional to the amplitude factor f, which favors statement 3) over 2).

Original (incomplete) answer: Statement 2 is incorrect. Your fourth equation should read $$A\omega_\vec{p}=\sqrt{\vec{p}^2+m^2}$$ where $A$ is the amplitude of the wave.

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  • $\begingroup$ Hi, please help me understand E=Aω. In natural units, [E] has mass dimension 1, [A] mass dimension -1, and [ω] mass dimension 1, so 1 = 0 does not add up. Don't we need other terms? $\endgroup$
    – TrentKent6
    Commented Dec 13, 2023 at 6:52
  • $\begingroup$ @TrentKent6 You just have $$A^2=\int d^3x |\psi|^2 \,.$$ $A$ is dimensionless and $|\psi|^2$ has dimension $m^{-3}$. $\endgroup$
    – my2cts
    Commented Dec 13, 2023 at 9:22
  • $\begingroup$ how would you classically express the energy as a function of the angular frequency? $\endgroup$
    – TrentKent6
    Commented Dec 13, 2023 at 11:13
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    $\begingroup$ the equation you write is just wrong, in the litterature $\omega_\vec{p} = \sqrt{|\vec{p}|^2+m^2}$, this is just the dispersion relation of the equation, there is no amplitude in there $\endgroup$
    – lucabtz
    Commented Dec 14, 2023 at 14:42
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    $\begingroup$ The energy is not independent of the amplitude. The formula $\omega = \sqrt{p^2 + m^2}$ is the dispersion relation for the Klein-Gordon equation, there is no amplitude dependence. The energy density of a plane wave with wave number $\vec{p}$ does definitely depend on the amplitude, but not in the way you have written, see my reply if you want the correct expression $\endgroup$
    – lucabtz
    Commented Dec 14, 2023 at 16:26

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