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Especially from this post I understand that the energy of a wave is directly proportional to the amplitude of that wave squared. Therefore, we can determine the total energy of a wave by summing the kinetic and potential energy which is briefly; $$ E=\Delta U+\Delta K=\frac12\omega^2y^2\,\mu\Delta x+\frac12v^2\,\mu \Delta x$$ where $y=A\sin\left(kx-\omega t\right)$ and $v=A\omega\cos(kx-\omega t)$

To this point I have understood the concept. However, Planck's formula confused me about how to determine the total energy of a light(EM) wave. Planck Formula also helps us to determine energy of an EM wave emission: $$ E=h \nu $$ where $h$ is Planck's constant and $\nu $ is the frequency of the wave. Thus, why doesn't Planck formula contain an element about amplitude of the wave if the energy is directly proportional to the amplitude squared? I know I am missing a fundamental knowledge here but I couldn't find a logical solution to that. Thus, I would be grateful if someone could help me on that.

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Consider the differences in the two equations describing the evolution of the wave.

  • In the classical case, the wave equation has the form $$ \frac{\partial^2u}{\partial t^2}+c\nabla^2u=0 $$
  • In the quantum case, the wave equation is the Schroedinger equation: $$ i\hbar\frac{\partial u}{\partial t}+c\nabla^2u=Eu $$

If you stick in a plane-wave, $u(x,t)\sim \exp\left(i(\mathbf k\cdot\mathbf x-\omega t)\right)$ into each of these, you'll find that the solution to the first has a factor of $A^2\omega^2$ in front of the time-derivative term while the latter has a factor of $\hbar\omega$.

So it seems that the difference in the power of the coefficient and the frequency is due to the fact that the two waves are evolved by different functions: one in which the time operator is first-order (quantum) and one in which the time operator is second-order (classical).

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The energy of an oscillator, as Planck defined, is quantized. It may be related to a harmonic oscillator, but it won't be simple to explain whose is the potential energy and the kinetic energy. We don't have here some sort of mass in some potential field, and the mass moves and has potential and kinetic energy. We have here the field itself and it is a thing that carries an energy of its own. So, the energy of the e.m. field is defined otherwise, not with potential and kinetic energy,

$ (\text I) \ H = \frac {1}{2} \int (\epsilon _0 |\vec E|^2 + \frac {1}{\mu _0} |\vec H|^2) \text d \vec r$.

where $H$ is the energy (the Hamiltonian). The vectors $\vec E$ and $\vec H$ also have an amplitude that oscillates in time.

Now, in the quantum theory, the fields $\vec E$ and $\vec H$, are replaced by the operators $\hat {\vec E}$ and $\hat {\vec H}$, (see D. F. Walls and G. J. Milburn, "Quantum Optics"). So, in the Hamiltonian $(\text I)$ the squared amplitudes $|\vec E|^2$ and $|\vec H|^2$ become operators. Next, to get the energy of the photon(s) in some state $|\psi\rangle$ of the e.m. field, we calculate the average of the Hamiltonian operator obtained, in that state. $ \langle \psi | \hat H|\psi \rangle$, and you will see that what we obtain is that the energy is given by Plank's formula.

I will show you a few steps of such a calculus.

a) In the classical electromagnetism $\vec E$ and $\vec B$ are obtained from the vector potential

$ (\text {II}) \hat E = - \frac {∂ \hat A}{∂t}, \ \ \ \hat B = \nabla \times \vec A .$

The same we do here, we define an operator vector potential

$ (\text {III}) \hat {\vec A(r,t)} = i \sum _k ( \frac {\hbar }{2 \omega _k \epsilon _0})^{1/2}[\hat a_k u_k(\vec r) e^{-i\omega _k t} + \hat a^{\dagger}_k u^*_k(\vec r) e^{i\omega _k t}]$,

where the index $k$ is for the $k$-th frequency (more exactly angular velocity) in the field, i.e. we number $\omega_1, \omega_2, ... \omega_k,...$; $C_k$ is a constant depending on $\omega_k$, and $a_k, a^{\dagger}_k$ are the annihilation and creation operators.

b) From $\hat {\vec A(r,t)} $ we calculate $\hat {\vec E}$ and $\hat {\vec B}$ applying the formulas in $ (\text {II})$. Then we take their absolute squares, i.e $\hat {\vec E}^{\dagger} \cdot \hat {\vec E}$ and $\hat {\vec B}^{\dagger} \cdot \hat {\vec B}$.

c) We introduce these vectors in the Hamiltonian, i.e.

$ (\text IV) \ \hat H = \frac {1}{2} \int (\epsilon _0 \hat {\vec E}^{\dagger} \hat {\vec E} + \frac {1}{\mu _0} \hat {\vec B}^{\dagger} \hat {\vec B} ) \text d \vec r$.

Performing the integral and applying different rules of the annihilation and creation operators, we get

$ (\text {V}) \ \hat H = \sum _k \hbar \omega _k (\hat a^{\dagger} \hat a + \frac{1}{2})$.

d) Now, as I said, we calculate the average for the state given for the e.m. field. Let's take a simple state, $|n\rangle$, which means that we have $n$ photons of the same frequency $\nu = \omega/2\pi$. We get the energy

$ (\text {VI}) \ \mathscr E = \langle \hat {H} \rangle = (n + \frac {1}{2})\hbar \omega$

in agreement with Planck's formula up to the term $\hbar \omega /2$ that I won't discuss here.

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  • $\begingroup$ Thanks for your answer. I think I have understood the main idea but I will probably go on and look at "Quantum Optics" for more detailed description. One more question: Planck's formula is valid for calculating the energy of a light beam(visible light). Right? $\endgroup$ – Starior Feb 7 '15 at 22:23
  • $\begingroup$ @Starior , Planck's formula is valid for any frequency of electromagnetic wave. (It is used also for other types of waves, as you will learn e.g. in quantum mechanics, but before you learn that topic, we speak only of electromagnetic waves.) $\endgroup$ – Sofia Feb 7 '15 at 22:51
  • $\begingroup$ You explain the orthodox point of view well, but you cannot attribute it to Planck. He did not support such point of view in his papers; his quanta $h\nu$ were a mathematical device to calculate entropy of set of material oscillators interacting via ordinary EM waves. $\endgroup$ – Ján Lalinský Feb 8 '15 at 3:37
  • $\begingroup$ The question was what is the connection between Planck's formula and the absolute square of the amplitude. $\endgroup$ – Sofia Feb 8 '15 at 9:13
  • $\begingroup$ That's right. I was addressing your first sentence. $\endgroup$ – Ján Lalinský Feb 8 '15 at 11:53
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Planck's expression

$$ h\nu $$

is not meant to give energy in EM wave, it was introduced as the smallest amount of energy an harmonic oscillator (as a model of matter) may give away when inside a reflective cavity containing matter in thermodynamic equilibrium.

This quantity of energy $h\nu$ may be interpreted in different ways, but usual EM waves have immensely much greater energy than that.

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  • $\begingroup$ That's not true. When introducing the field operators, the energy of the EM wave has to give us a multiple of the Planck's expression. $\endgroup$ – Sofia Feb 7 '15 at 21:05
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    $\begingroup$ What's not true? Which part? $\endgroup$ – Ján Lalinský Feb 8 '15 at 3:39
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The formula $E=h\nu$ is not the most general formula, it's the formula for a single quanta of light. Often times light has many quanta so the change in the total energy is $\Delta E=nh\nu$ where $n$ is an integer representing how many quanta you removed or added.

If you have a wave with a particular amplitude and frequency you could find the total energy $E$, then divide it by $h\nu$ to find $n=E/h\nu$, the number of quanta in the light.

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So let's turn that around. To make a wave with a large amplitude it turns out you must use many quanta to make the wave, so it just means there are many photons involved in a uniform wave with a large amplitude. This is OK because photons do not obey the Pauli exclusion principle so you can have many photons in the exact same state, in fact it becomes easier to get a photon into a particular state when there is another photon already in that state, it's like an anti-exclusion principle.

So, a big classical wave can and does have many photons in it, the amplitude of the big classical wave can tell you how many photons are in it. And the Planck formula can tell you how the energy (and thus the amplitude) changes when some photons are removed or added to the photons already there.

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  • $\begingroup$ Yes that is true but I am asking that why doesn't Planck's formula contain any element for amplitude if the energy of a wave is related with its amplitude? $\endgroup$ – Starior Feb 8 '15 at 8:45
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Planks constant has units of Energy/Hz. Also, the energy of a wave is proportional to the square of its amplitude. Therefore, the square root of Planks constant IS the amplitude of the wave. All light waves have the same amplitude, they differ only in frequency.

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  • $\begingroup$ This answer is patently, obviously incorrect. "All light waves have the same amplitude"? Presumably you've only ever seen light from a single source ever. $\endgroup$ – Emilio Pisanty Feb 14 '17 at 15:14
  • $\begingroup$ (for the record, though the answer is indeed wrong, that's not my downvote, btw) $\endgroup$ – Emilio Pisanty Apr 17 '17 at 15:48

protected by Qmechanic Jan 28 '17 at 15:33

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