The Maxwell-Boltzmann probability factor exp(-energy/T) / normalization may be applied to a quantum oscillator using energy levels given by hbar w (n+.5). This leads to the derivative in 1/T of a geometric series and yields an average energy: Eave = hbar w/2 + hbar w / (-1+exp(hbar w/T)). If one were to consider a semiclassical approach using p (momentum and x), energy would be described by pp/2 + xx/2 where for simplicity m=1 and k=1. This would appear in the MB factor with a temperature T and this expression would be expected to hold for high T or w/T small i.e. the quantum phonon jumps would be hardly noticeable compared to the average energy T in this limit. If one tries to write the classical energy, for w/T small, in the form of quantum energy, one has hbar w n in this limit, where n is the number of phonons (hbarw/2 is small in this limit) so: zz= {pp/2+kk/2}/w is the number of phonons which should be multiplied by hbar w/T. One may then ask: What function has the limit hbar w/T? A solution is: {1-exp(-hbarw/T)}. Thus the semiclassical normalized phase space probability: P(p,x) = {1-exp(-hbarw/T)}. Exp{ - zz {1-exp(-hbarw/T)} } yields the small hbar w/T limit quantum average energy. As a result, the average energy in this limit should match for both the classical phase space distribution and the quantum average energy expression. This classical distribution happens to be identical to the Husimi distribution. One may note (as is done in (1)), that {1-exp(-hbarw/T)} is the probability for the quantum oscillator particle to be in the ground state. In the limit of hbar w/T small, this probability is hbar w/T which is a measure of the effect of the quantum phonon leaps as well. Fisher’s information, as noted in (1), is equivalent to this ground state probability, but it is also the average energy which is proportional to the temperature type value, in this case {1-exp(-hbarw/T)}. This follows from the same kind of calculation done for a gas with particles with energy pp/2m, although instead of a factor of .5T one T because there is a contribution from pp/2 and xx/2.
The Maxwell-Boltzmann probability factor exp(-energy/T) / normalization may be applied to a quantum oscillator using energy levels given by hbar w (n+.5). This leads to the derivative in 1/T of a geometric series and yields an average energy: Eave = hbar w/2 + hbar w / (-1+exp(hbar w/T)). If one were to consider a semiclassical approach using p (momentum and x), energy would be described by pp/2 + xx/2 where for simplicity m=1 and k=1. This would appear in the MB factor with a temperature T and this expression would be expected to hold for high T or w/T small i.e. the quantum phonon jumps would be hardly noticeable compared to the average energy T in this limit. If one tries to write the classical energy, for w/T small, in the form of quantum energy, one has hbar w n in this limit, where n is the number of phonons (hbarw/2 is small in this limit) so: zz= {pp/2+kk/2}/w is the number of phonons which should be multiplied by hbar w/T. One may then ask: What function has the limit hbar w/T? A solution is: {1-exp(-hbarw/T)}. Thus the semiclassical normalized phase space probability: P(p,x) = {1-exp(-hbarw/T)}. Exp{ - zz {1-exp(-hbarw/T)} } yields the small hbar w/T limit quantum average energy. As a result, the average energy in this limit should match for both the classical phase space distribution and the quantum average energy expression. This classical distribution happens to be identical to the Husimi distribution. One may note (as is done in (1)), that {1-exp(-hbarw/T)} is the probability for the quantum oscillator particle to be in the ground state. In the limit of hbar w/T small, this probability is hbar w/T which is a measure of the effect of the quantum phonon leaps as well. Fisher’s information, as noted in (1), is equivalent to this ground state probability, but it is also the average energy which is proportional to the temperature type value, in this case {1-exp(-hbarw/T)}. This follows from the same kind of calculation done for a gas with particles with energy pp/2m, although instead of a factor of .5T one T because there is a contribution from pp/2 and xx/2.
In this note we consider, as in (1), finding an oscillator quantum state |z> (called the coherent state) such that <z| exp(-Hamiltonian/T |z> yields a classical-like phase space probability distribution in p and x. In (1) the approach is to use the known weights of <z|n> where Hamiltonian |n> = en |n>. Here we use instead the following ideas. First, we argue that for exp(-Hamiltonian/T) to be used at all, P(e1)P(e2)=P(e1+e2), but also P(z1*z1)P(z2*z2) = P(z1*z1+z2*z2) where z*z = pp/2+xx/2 with k=m=1. This yields a solution of exp(- b z*z) where b is unknown. In the limit of hbar w/T <<1, i.e. the classical limit, P(z*z) should become the Maxwell-Boltzmann factor exp(- (pp/2T + xx/2T). Writing |z> as a linear superposition of the energy eigenfunctions |n> one may find a solution for |z>, the coherent state which has an average of (pp/2+xx/2)/ hbar w phonons/photons. P(z*z) is called the Husimi distribution.
In a classical bound state, conservation of energy: p(x)p(x)/2m + V(x) = E holds for all x. Examined in a statistical sense, one may consider each constant dx to have a probability density proportional to dt(x) where v(x)=velocity =dx/ dt(x). Thus the particle is followed in time. In quantum mechanics, even for a free particle exp(ipx), there exists a wavelength hbar/p. Within this region one cannot follow the particle in time. In fact, there are two probability distributions cos(px) and the shifted sin(px). The two dimensional nature of exp(ipx) is required to describe the direction of motion. There exists a mathematical abstraction or average which holds for any x within the wavelength and yields the classical x= p/m t. Thus there is a duality, but within a wavelength one cannot follow the particle exactly using x=p/m t. Rather there is no time and one thinks of probabilities to be at one x or another. The operator -id/dx yields the abstracted or average momentum at any x and (-id/dx)(-id/dx) average pp if one divides by probability. In other words: (-id/dx) exp(ipx)/ exp(ipx) = p and (-id/dx) (-id/dx) exp(ipx) / exp(ipx) = pp. These then apply to any x point. For a bound state, exp(ipx) is replaced with W(x)= Sum over p a(p)exp(ipx). Thus p and -p are included together. From a statistical point of view this is fine because a quantum bound state is described in terms of energy conservation (time-independent Schrodinger equation). Unlike the classical case, the probabilities of p and -p differ, i.e. exp(ipx) and exp(-ipx). The sum exp(ipx)+exp(-ipx) yields cos(px) which is the resolution probability distribution. Taking -d/dx d/dx of W(x) and then dividing by W(x) yields the average kinetic energy at any x within the wavelength. In other words this is the average kinetic energy which matches the classical result for any energy level. As a result, prms(x)prms(x)/2m + V(x) = En holds for any quantum level, but prms(x) is an rms average value which bypasses wave structure. That is why the exp(ipx) for p,-p and -id/dx must be used to calculate average quantities within a wavelength. Many different p,-p combinations are needed so that at each x within a wavelength the correct average kinetic energy value appears. Now consider a high energy level and choose a fixed dx length to represent a tiny region in the system, a cutoff for which there is no further resolution. p(x) is constant within this region. If hbar/ prms(x) <= dx, the wave structure of quantum mechanics does not play a role. (Note that for an oscillator, W(x) is a set of crests and troughs with different heights and widths.) As a result, there is no need for d/dx acting on W(x) to describe average kinetic energy within dx. The average kinetic energy is prms(x)prms(x)/2m and dx is treated like a point. Only if exp(ipx) is used does -id/dx yield the proper classical result. (An alternative scenario is that d/dx acting on a function divided by the function yields the same result as acting on exp(ipx) divided by exp(ipx) e.g. -d/dx d/dx cos(px)/cos(px) = -d/dx d/dx exp(ipx) / exp(ipx). If one has the quantum kinetic energy expression -1/2m d/dx dW/dx / W ((1)) and considers one momentum, namely prms(x), and uses W(x)= cos( prms(x) x), then with prms(x) constant at x, ((1)) yields prms(x) prms(x)/ 2m. Thus d/dx has not broken down even though it does not describe motion within dx which is now a point. A problem arises, however, if one considers Wn(x)= Wn1(x)Wn2(x) as this yields an equation containing dW1/dx dW2/dx/ (W1W2). Then the above procedure yields: dW1/dx/W1 (prms(x) sin(prms(x) x) / cos(prms(x) x). This does not match the classical value. Furthermore, dW2/dx/ W2 contains +prms(x) and -prms(x) with different weights while a classical scenario is -pP(-p,x) + pP(p,x) = 0 because P(-p,x)=P(p,x) while quantum mechanically it doesn’t. We analyze this situation and conclude that d/dx breaks down/ does not apply in the classical limit because it is used to calculate average values within a wavelength only. It only “works” if the effect of the derivative(s) on a function divided by the function is the same as the effect on exp(ipx) divided by exp(ipx). If a wavelength fits within dx, there is no need to calculate values at any lower x level and d/dx is not needed. In fact, there is a single prms(x) value in dx and the average time links different dx’s so that one follows the particle (on average in the sense that one ignores structure within each x) by x(t). As a practical consequence, using -id/dx together with the wavefunction W(x) can lead to problems in the classical limit because d/dx does apply. In short, using -id/dx operating on Wn(x) divided by Wn(x) leads to issues unless it produces the same effect as acting on exp(ipx) divided by exp(ipx).
In (1), it is argued that if the classical energy of an oscillator E = pp/2 + wwxx/2 (m=1) is written as E = (p+iwx)(p-iwx) and this in turn written using the Moyal star product { d/dp left d/dx right - d/dx left d/dp right } in terms of a parameter ( i hbar /2), one obtains: E = pp/2 + wwxx/2 + hbar w/2 i.e. a semiclassical expression for the quantum result with the ground state energy hbar w/2. In Part I, we showed that this result follows from combining a time-independent Schrodinger equation in x-space with one in p (momentum) - space. The two second order equations yield second order derivatives composed of two parts. One part returns the form pp/2m and k/2 xx and the other returns a constant which is the ground state energy if one considers the ground state p and x wavefunctions which are of the form exp(- (pp/2 + wwxx/2) /b) where b is the ground state energy.. The first part cancels with kinetic energy from the p-space equation and .5kxx from the x-space equation. Thus, this combination of two Schrodinger equations really involves three pieces: ((A)) an energy term in p,x, ((B)) a cancellation term of this energy and ((C)) a constant ground state energy term. In (2) it is stated that the Moyal star product ** has the property: Integral dx dp f(p,x) g(p,x) = Integral dx dp f(p,x) ** g(p,x) This is then used to argue that: H(p,x) ** f(p,x) = E f(p,x) ((1)). For the ground state oscillator case, only the first and second order operator terms contribute. The first order operator term is zero, while the second order term behaves in exactly the same manner as the combined x-space and p-space time-independent Schrodinger equations used in Part I. Thus the combined p and x space time-independent Schrodinger equations for the oscillator Hamiltonian pp/2 +wwxx/2 is identical to the Moyal approach.
In Part III we argued that a Schrodinger equation of the form: a d/dx d/dx W(x) + bxx W(x) = E W(x) may be scaled x→y such that yy = xx sqrt(ba) and E→ E sqrt(a/b) so that E is proportional to sqrt(b/a). We applied this idea to the quantum oscillator and noted that classically w, the angular frequency, for the same equation is sqrt(b/a). We further argued that this implies the momentum operator in quantum mechanics should scale as 1/xx as d/dx d/dx does. In this note we use a slightly different argument to analyze this same problem. Instead of treating the classical oscillator and time-independent Schrodinger equation as separate, we link them and show that scaling must occur with E being proportional to w. Furthermore we argue that if quantum mechanically one wishes to use probabilities f(p,x) to calculate average kinetic energy at x i.e. KE(at x) = {Sum over p pp/2m f(p,x) } / {Sum over p f(p,x)}, then f(p,x) = f(px) and pp should scale as 1/xx. Thus comparison of the classical oscillator equation directly with a statistical average type of quantum equation which uses V(x), E and KE(at x) yields information about E, f(p,x) and p i.e. introduces the idea of wavelength. Furthermore the discrete quantum energy values E=hbar w (n+.5) are dependent on w not through any boundary value considerations.
Classical statistical mechanical density maximizes entropy subject to constraints i.e. one has the Maxwell-Boltzmann (MB) distribution P(x)P(p) proportional to exp(- pp/2mT - V(x)/T). Quantum mechanics is also a statistical theory, but it is not based on the maximization of entropy. Rather it is based on a conditional probability P(p/x) = a(p)exp(ipx)/W(x) where W(x) =wavefunction = Sum over p a(p)exp(ipx) such that: Sum over p pp/2m P(p/x) + V(x) = E (bound states). As a result, it has classical densities P(x)=W(x)W(x) and P(p)=a(p)a(p), but these in general do not maximize entropy and so do not match the MB distribution. There is a special case, however, namely the ground state oscillator for which the quantum mechanical results P(x) and P(p) are both Gaussians. It is shown in (1) that for a fixed variance, a Gaussian maximizes entropy. Thus, the MB scenario must be a Gaussian for an oscillator as it is, but the quantum result is also a Gaussian and so matches the classical statistical mechanical result. In this note, we show that the time-independent Schrodinger equation collapses to the equality situation in Heisenberg’s uncertainty principle, suggesting that the solution also represents maximum entropy.
In (1) the high energy limit of the Wigner transform W1(p,x,n) is investigated for the quantum oscillator. The WKB semiclassical approximation is used and the high energy wavefunction approximated by: 1/ p(x) { exp(i p(x) x) + exp(-i p(x) x) } leading to the result of delta ( PP/2m+kXX/2 - E) where p(x)= sqrt(2m(E-.5kxx)), P,X are classical momentum and position and PP/2m +kXX/2 energy. In this note, we first observe that the WKB approximation may be used for any potential V(x) to find a similar result. We then focus on obtaining the result without using the WKB approximation, but instead utilize: H(p,x) ** W1(p,x,n) = En W1(p,x,n) (from (2)) where H(p,x) is the classical Hamiltonian, ** the Moyal star product involving exp(ihbar/2 Operator) and W(p,x,n) the Wigner function (as used in (2)). The main idea is that in the classical limit, hbar—>0 so exp(ihbar/2 Operator) —>1. Thus H(p,x)W1(p,x,n→large) = En W1n(p,x,n -> large). In (2) it is shown that Integral dp dx W1(p,x,n) = 1, so Integral H(p,x) W1(p,x,n->large) = E. As a result, W1(p,x,n->large) acts as a delta function enforcing pp/2m + V(x)=E.