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I am trying to understand the photoelectric-effect deeply. My teacher used the Planck's law and integrated it to deduce the Stefan-Boltzmann law. He somehow showed some quantum-physical characteristic -- something that intensity did not increase the energy of photon as expected classically but the stopping voltage.

Now let's take a step back. He started with Planck's law and I want to understand how it is connected to other thermal equilibriums such as Bose-Einstein distribution, Fermi-Dirac distribution and Maxwell-Boltzmann distribution.

What are the thermal energy distributions? How to remember them? Some mnemonics? Are they somehow connected? I know BE and FD are the quantum-physical descriptions while MB is a classical approximation but I don't how Planck's law is related to them, how?

Wikipedia about Planck's law

As an energy distribution, it is one of a family of thermal equilibrium distributions which include the Bose–Einstein distribution, the Fermi–Dirac distribution and the Maxwell–Boltzmann distribution.

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...I understand from Wikipedia here that Planck's law is deduced from Bose-Einstein statistic so it is quantum-physical thing? I cannot yet understand what the teacher was trying to say: according QM something unlike classically. –  hhh Jan 4 '13 at 1:47
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I just remember $$ \frac{1}{\exp(\beta (E-\mu)) \pm 1}$$ You can work out the sign from the fact that Bose-Einstein distributions can diverge (so they go with the - sign), whereas Fermi-Dirac is bounded (so they go with the + sign). Maxwell-Boltzmann applies to classical systems, so quantum statistics don't matter, so take the limit that the two distributions are the same (so drop the $\pm1$).

These expressions represent the average number of particles occupying a state with energy $E$. The chemical potential $\mu$ is just a knob that lets you adjust the overall density. You can also think of it as (roughly) the energy it takes to add a particle to the system. To find the total number of particles in the system you have to sum this over all of the energy levels. You can use this information to find all kinds of thermal averages. For example:

$$ \text{total energy} = \sum_{\text{all energies}} \left(\text{distribution function}\times\text{number of states with a given energy}\right)$$

This is essentially what is going on in Planck's law, only the sum is left off. Planck's law is the Bose-Einstein distribution (with $\mu=0$ because photons can be freely created and destroyed) multiplied by the number of states with an energy in a small range about $E$. This tells you how much energy there is in the photons with energies in that range.

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@hhh Ahh, I see where you're getting confused. Planck's law - meaning the blackbody distribution function - is not directly connected to the photoelectric effect. Two different things. You're after "Planck's relation" en.wikipedia.org/wiki/Planck%27s_relation, the energy of photons is related to their frequency by $E = h \nu$. Planck proposed and used it to derive the blackbody distribution. Then Einstein used it to explain the photoelectric effect. –  Michael Brown Jan 4 '13 at 6:07
    
When a photon hits a surface, the surface may emit an electron or a phonon i.e. heat. I have heard this statement many times: higher intensity should classically result in larger energy per electron. But in reality, it does not and the energy of electron depends on the stopping voltage. I cannot understand how this is a proof of QM phenomenon -- which mathematical formulae contradicting...? –  hhh Jan 4 '13 at 6:26
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Light comes in discrete units - photons. Photons of a given color have a definite energy given by $E=h\nu$. If a photon strikes an electron, transferring its energy to the electron and ejecting it from the metal. The electron then has an energy $E=h\nu-W$, where $W$ is a characteristic of the metal that measures how much energy it takes to just barely remove an electron. The point is this energy doesn't depend on the intensity of the light, only the color. One photon -> one collision -> one electron ejected -> same energy every time. –  Michael Brown Jan 4 '13 at 6:32
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If you have a more intense light, that means more photons -> more electrons. But the electrons don't have any more energy. More electrons yields a greater current, but the voltage required to stop the current doesn't change. But the experiment also works with an extremely dim light, so low that you only get a single photon at a time. Then the current is so low that you can detect individual electrons getting kicked out, one at a time. But their individual energy is still the same. In this extreme situation it is very obvious that the energy is being transferred in discrete lumps - quanta. –  Michael Brown Jan 4 '13 at 6:34
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Classical here refers to the classical theory of electromagnetism, the complete description of which is found in Maxwell's equations: en.wikipedia.org/wiki/Maxwell%27s_equations This theory describes electromagnetic waves which can carry an arbitrarily small amount of energy in proportion to their intensity. There is no classical theory of light which involves photons. Photons are intrinsically quantum. –  Michael Brown Jan 5 '13 at 1:47
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The deduction of the thermal energy distributions are pretty much just Stirling approximation $\ln(x!)=x\ln(x)-x$, Lagrange-multipliers method and a lot of permutations/combinations. You can see it at the bottom.

Thermal energy distributions contain classical models such as Maxwell-Boltzmann statistics and quantum-physical models such as Bose-Einstein statistics and Fermi-Dirac statistics.

The "classical" term means models such as Maxwell equations, partial-derivations models, which do not contain the notion about discretization -- a big difference to QM models such as Planck's rule $E=hf$ where the energy of EM is quantified.

Light is an example of EM radiation. Maxwell realized this by analyzing earlier studies of Weber and Kohlrausch here and concluded $c_0=\frac 1 {\sqrt{\mu_0 \epsilon_0}}$. A more realistic model of light is a non-classical model here that cannot be described with classical mechanism but with quantified electromagnetic field and quantum mechanics. Photon is a boson so it obeys Bose-Einstein statistics, not the classical approximation ie Maxwell-Boltzmann statistics that is only realistic with extreme temperatures such as close to absolute zero or very high temperature.

Facts such as photoelectric effect, X-emission (opposite to photoelectric effect) and Compton-scattering prove the discretization of the EM that QM describes. Wave-particle-dualism explains events where light acts like a wave and like a particle. This is impossible to explain with Maxwell equations. Examples of such events are double-slit-experiment and single slit expriment.

Now the double-slit expriment lead into the realization of uncertainty. You cannot see the wave nature at the same time as the particle nature. An example of this is Heisenberg's uncertainty principle $\Delta p\Delta x \geq \frac{h}{4\pi}$ that means you cannot know the location of physical object and and its momentum at the same time -- if the $\Delta p$ is close to zero, you have a particle -- and if the $\Delta p$ is close to zero, you have a wave. Bohr generalized this concept of complementary events from mere waves and particles in his complementarity here where he realized

"It is impossible to design a measuring device that demonstrates both phenomena simultaneously not because of lack of creativity on the part of the experimenter, but simply because such a device is literally inconceivable." (Sentence in the Wikipedia about complementarity)

which is actually quite thought-provoking statement. For example, I understand this so that you cannot have a camera that minimizes all types of noises. The QM models infer a new type of noises such as quantum noise aka shot noise that dominates low signal-to-noise-ratio noises in certain situations.

My lecture documents here at the end are confusing in this point. It mentions "You cannot force wave nature into particle nature without losing interference." after mentioning "You will lose interference pattern on the left if you try to find out from which hole the photon went by filling the other hole one-by-one" (not word-to-word translation) but the meaning should be the same.

enter image description here

Now back to the statistics 'why are the statistics called "thermal" or "energy"?'

QM models such as Bose-Einstein and Fermi-Dirac describe bosons and fermions, respectively. Classical models (ambiguous term but meaning now Maxwell equations) are energy equations in a way: you need energy to see their working. Thermal prepending is a bit odd but perhaps it wants to stress the association of energy and temperature. The word "distribution" stresses the statistical connotation.

I hope someone more experienced can explain what the "thermal energy distributions" really are! I feel my explanation is not thorough.

Mathematical formalism

Bose-Einstein

We have particles with states $N_i$ and walls $M$ where particles can have the same quantum state, a big difference to fermions where $(n,s,l,m_l)$ cannot be the same. So horizontal alignment

$$W_h^i=\frac{(N_i+M-1)!}{N_i!(M-1)!}$$

where the total alignment is the product of all horizontal alignment so the probability function $P=\Pi_i w_h^i=\Pi_i\frac{(N_i+M_i-1)!}{N_i!(M_i-1)!}$ so

$$\ln(P)=\sum_i \ln\left[(N_i+M-1)!-\ln(N_i!)-\ln((M-1)!)\right]$$

Now we use Lagrange-multiplier method so the F-function is

$$F=\ln(P)+\alpha (N-\sum_i n_i) + \beta (MU-\sum_i n_i u_i)$$

where the first $\alpha$ restriction means the amount of particles is the sum of all particles in the states $N=\sum_i n_i$ and the second $\beta$ condition means the system energy is the sum of all energies in states.

Now we derivative this one with respect to the states variable $n_i$ where we need to use the Stirling approximation $\ln(x!)\approx x\ln(x)-x$ because of large number of particles (small amount of particles requires an extra term here). So

$$\ln(M)-\ln(N_i)-\alpha-\beta E_i=0$$

$$N_i=Me^{-\alpha-\beta E_i}$$

Fermi-Dirac

Pauli's exclusion principle is the key difference. It is otherwise the same deduction as with Bose-Einstein but $W_h^i=\frac{M!}{N!(M-N_i)!}$ where $N!$ is for "miehitetty" manned states and $(M-N_i)!$ for un-manned states due to Pauli's exclusion principle -- you cannot have same two Q-states with fermions!

Maxwell Boltzmann

I use now lectures 2061 here pages 63-65. I am not sure of this because the two teachers use a slightly different notations but I understand it this way

$$W_h^i =\frac{g_i^{n_i}}{n_i!}$$

where $g_i$ is the degenerazy, $n_i$ is the amount of state so the probability $P_{MB}=\Pi_i W_h^i.$ And we will get the statistics but taking the logarithm and using Lagrange multipliers. Our conditions are $N=\sum_i n_i$ and $E=\sum_i n_i E_i$.

SUMMARY

Most states are with Maxwell-Boltzman then Bose-Eistein and least states with Fermi-Dirac because of Walls and Pauli's exclusion principle. Please, note that there are no "walls" with Maxwell-Boltzmann where systems such as ideal-gas particles can occupy the same quantum state -- perhaps related to superfluidity phenomenon. Horizontal occupation formulae for Bose-Einstein, Fermi-Dirac and Maxwell-Boltzman:

$$W_h^i(BE)=\frac{(N_i+M-1)!}{N_i!(M-1)!}$$

$$W_h^i(FD)=\frac{M!}{N_i!(M-N_i)!}$$

$$W_h^i(MB) =\frac{G_i^{N_i}}{N_i!}$$

Study Questions

  1. Do Fermions and Bosons have degenerazy like a Maxwell-Boltzmann system?

  2. In other words, why no $G_i$ with BE and FD formulae?

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