# On the Discretization of Energy Levels

We consider a system of "n" particles whose total energy E and net momentum $\vec{P}$ are fixed are fixed.There no net force on the system(assumed)

$$\Sigma \epsilon_i= E$$ $$\Sigma\vec{p_i}=\vec{P}$$ For an individual particle its momentnum and energy remain constant for the time $\tau$,the relaxation time(average time between successive collisions----a constant). That's an extra constraint for each particle [Radiational energy density at some point is assumed to be constant for some physically small time interval]

All that seem to indicate that classical theories favor the discretization of energy levels in the equilibrium state. It is important to note that for each particle to satisfy on the mass shell condition we should have: $$\epsilon_i^2-p_i^2=m_0^2c^4$$ The above equation is frame invariant.Any theoretical speculation considering energy and momentum independent of each other should correspond to what we understand by "off the mass shell" situation An orbiting electron in an atom ,so far as the classical theories are concerned , should radiate energy. But it can also receive energy form other particles.The net radiation from a block of iron at constant temperature is zero.This should favor the discretization of the electronic orbits in the atoms in the expected manner.

[For a cluster of charged particles one should take into account the electromagnetic potential energy of the system( a closed system ) due to the presence of charges and currents.This energy for the system should be considered constant for our model]

Should we use intuition(commonsense classical theories ) to interpret QM with the understanding that the results of observation should not change in view of the system interacting with the measuring instrument and that QM will be no less useful to human activity?

[Some points to examine have been placed on the following link:

The document may be downloaded though you get a message that conversion is going on]]

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Just think of a pair of particles participating in an elastic collision.You don't get arbitrary solutions satisfying the equations involved.Now you may extend your thinking to ten particles and then to ten million particles involved in an elastic collision(net force on the system being zero). The relaxation time and its constant nature become important when you are considering a huge number of particles. – Anamitra Palit Aug 18 '12 at 3:21
$F(E,p,x,t)=\int_{-\infty}^{x}\int_{-\infty}^{t}f(x,t)e^{ia(Et-px)}dxdt$----(A).‌​$\frac{\partial F}{\partial x}=\int_{-\infty}^{x}\int_{-\infty}^{t}(\frac{\partial f}{\partial x}-ipxf)e^{ia(Et-px)}$ Again $\frac{\partial F}{\partial x}=\int_{-\infty}^{x}\int_{-\infty}^{t}(\frac{\partial f}{\partial t}-iEf)e^{ia(Et-px)}$. If $\frac{\partial F}{\partial x}=0$ favors $\frac{\partial f}{\partial x}-iap=0$ =>$(x,t)=Ae^{iapx}$. Again, $\frac{\partial F}{\partial x}=0$ should favor $f(x,t)=Be^{-iaEt}$. Finally we obtain $f(x,t)=Ae^{-ia(Et-px)}$--(B)which is a solution of the Klien Gordon equation. – Anamitra Palit Aug 26 '12 at 8:11
Relation (A) becomes the Fourier transform when both x and t tend to infinity.Using (B) in (A) and allowing x and t to tend to $\infty$,we obtain:$F(E,p)=Const\times \delta(E-E_0)\delta(p-p_0)$ Integration of the last formula on the (E,p) domain counts the number of (E0,p0) modes present.If we divide this by the total number of possible modes we obtain a probability picture – Anamitra Palit Aug 26 '12 at 8:16
Regarding Relation B in the scond last comment:(1) It satisfies the Klein Gordon Relation. (2)For the invariance of the exponential part the Lorentz transformations are a suitable candidate provided "a" is a universal constant.(3)Energy and momentum are suitable choices for E and p if Et and px are dimensionally identical.(4)The psi indicated by relation (B)is periodic nature associated with a probability picture. – Anamitra Palit Aug 26 '12 at 8:18

This is nonsense--- the "quantization of energy" you are referring to in classical theories is not a quantization at all, it is equipartition. It is only true that independent degrees of freedom have an average energy which is roughly quantized in classical mechanics.

You can't make a classical equilibrium between a field and an atom, because the field always has infinitely many degrees of freedom, and the atom finitely many. So the atom goes to the classical ground state, the electrons sit on top of the nucleus, and the field jitters thermally infinitesimally (because the finite energy is all sucked to the smallest wavelength modes).

This is called the ultraviolet catastrophe, and it dooms attempt to explain quantum behavior from a classical theory.

The simplest argument against a modified classical mechanics reproducing quantum mechanics is that quantum mechanics introduces a dimensional constant $\hbar$ which is not there in the classical limit. So if you start with the classical limit, you have to explain what sets the scale $\hbar$. This was debated in the early quarter of the twentieth century, and these ideas do not work.

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Maxwell' Boltzman's Distribution is a unimodal bell shaped one assigning the highest probability to the mean speed of the entire system of oscillators at least in an approximate way.The mean speed of the highest frequency oscillator is much greater than the said mean value pertaining to the entire mass of oscillators.Any effort to assign the largest strength to the oscillators on the higher frequency side would be a big mistake even from the classical point of view. Maxwellian distribution of momenta,incidentally, remain valid even in a potential field. – Anamitra Palit Aug 18 '12 at 5:45
Let's go into a simple treatment:The number of modes of frequency $\nu_n$ for the first octant of a sphere of radius "n":$N=\frac{1}{8}\frac{4\pi}{3} n^3=\frac{\pi}{6}(\frac{2L}{c})^3 \nu_n^2$---(1)Therefore number of modes on the interval$(\nu_n,\nu_n+d \nu_n)$:$dN=\frac{4\pi}{c^3}L^2{\mu_n}^2 d\mu_n$. N and dN correspond to possible modes and not to the actual modes that may be realized consistently if the energy of the system ,its total momentum etc remain unchanged.These factors asre considered in the derivation of the Maxwell -Boltzman distribution. – Anamitra Palit Aug 18 '12 at 6:38
@AnamitraPalit: The number of modes is infinite you are finding density of modes. The infinite number of modes means an infinite energy. This is the ultraviolet catastrophe, there is nothing more to say. – Ron Maimon Aug 18 '12 at 7:15
Let's consider the example of a string stretched between a pair of fixed points. You may pluck the string at some point suppressing a multitude of modes. Still you have an infinite number of modes left in the oscillation. Do the highest or rather the higher frequency modes have as large an amplitude as the fundamental? – Anamitra Palit Aug 18 '12 at 7:33
@AnamitraPalit: Yes--- the string analogy is perfect. When you pluck the string, it will eventually decay so that all the frequency modes have the same amplitude on average--- this is the thermal state. All other states are unstable to this. If your string has random density perturbations, or is attached to thermal walls, then it will eventually decay to be stationary, and then all modes will have the same (infinitesimal) energy on average. The series will blow up for a continuous string, it will not blow up for a string made of atoms. This is the ultraviolet catastrophe. – Ron Maimon Aug 18 '12 at 8:42