Correction: March 31, 2023 We remove the form D = 1-a(p)a(p) in the note as it does not equal the time evolution measure of (1). We wish to only point out that sinc(T/2 |(En-Em)|) in the expression for D has an analogue in the spatial behaviour of a particle in an infinite well where: sinc(T/2 |En-Em\) from D --> a(k) sqrt[3.14/(hbar L)] [n3.14 + kL] / (n3.14) i.e. is proportional to the momentum wavefunction a(k) multiplied by a linear function in k. We point out that even though this analogy exists, the summations involved in D do not map over to the single particle case. Thus there is not a complete analogy. In particular, one could sum over k, but would not generally sum over n, the energy level. --------------------------------------------------------------------------------------- In (1) a measure to determine time evolution of a quantum system is given as: D = 1- Integral dt (0,T) dt’ (0,T) <W(x,t)|W(x,t’)><W(x,t)|W(x,t’)>. Using W(x,t) = Sum over n cn exp(-iEnt) Wn(x) (1) shows: D= 1 - Sum (n,m) |cncm||cncm| sinc[wnm T/2] sinc[wnm T/2] The authors argue that this measure may be applied to qubit systems. Their result is based on the form exp(-iEnt)Wn(x) and the idea of a fixed time interval T. Here we argue that an identical form appears in the spatial realm for a system which instead of a finite time interval contains a finite length. The relevant system with a finite length is that of an infinite well potential with length L because the wavefunction is 0 outside this length. In such a case, instead of exp(-iEnt)Wn(x) we relate exp(-ipx)Wn(x) to a(p), the momentum wavefunction, and show that measure of time evolution is essentially of the same mathematical form as: 1- a(p)a(p) for an infinite well of length L. 1- a(p)a(p) has the interpretation of the probability not to find a particle in an infinite well in a momentum state of p for a given energy level with average momentum of n 3.14/L.
Correction: March 31, 2023 We remove the form D = 1-a(p)a(p) in the note as it does not equal the time evolution measure of (1). We wish to only point out that sinc(T/2 |(En-Em)|) in the expression for D has an analogue in the spatial behaviour of a particle in an infinite well where: sinc(T/2 |En-Em) from D --> a(k) sqrt[3.14/(hbar L)] [n3.14 + kL] / (n3.14) i.e. is proportional to the momentum wavefunction a(k) multiplied by a linear function in k. We point out that even though this analogy exists, the summations involved in D do not map over to the single particle case. Thus there is not a complete analogy. In particular, one could sum over k, but would not generally sum over n, the energy level. --------------------------------------------------------------------------------------- In (1) a measure to determine time evolution of a quantum system is given as: D = 1- Integral dt (0,T) dt’ (0,T) <W(x,t)|W(x,t’)><W(x,t)|W(x,t’)>. Using W(x,t) = Sum over n cn exp(-iEnt) Wn(x) (1) shows: D= 1 - Sum (n,m) |cncm||cncm| sinc[wnm T/2] sinc[wnm T/2] The authors argue that this measure may be applied to qubit systems. Their result is based on the form exp(-iEnt)Wn(x) and the idea of a fixed time interval T. Here we argue that an identical form appears in the spatial realm for a system which instead of a finite time interval contains a finite length. The relevant system with a finite length is that of an infinite well potential with length L because the wavefunction is 0 outside this length. In such a case, instead of exp(-iEnt)Wn(x) we relate exp(-ipx)Wn(x) to a(p), the momentum wavefunction, and show that measure of time evolution is essentially of the same mathematical form as: 1- a(p)a(p) for an infinite well of length L. 1- a(p)a(p) has the interpretation of the probability not to find a particle in an infinite well in a momentum state of p for a given energy level with average momentum of n 3.14/L.
In the literature, stochastic approaches employed to derive the Schrodinger equation seem to focus on the idea of a classical spatial density which depends on time and space ie. density(x,t). As a result, d/dt (partial) density appears in various equations such as the continuity and Focker Planck equations. At one point in the derivation the quantum wavefunction is introduced “mathematically” as W(x,t)=sqrt(density(x,t)) exp(i S(x,t)) where S is related to a velocity field through u(x,t)=velocity= d/dx S. From quantum mechanics, -Et becomes a term of S the velocity field. The stochastic/hydrodynamical approaches employ the term du/dt partial = d/dt (partial) d/dx S and then “pull out”” d/dx leaving d/dt (partial) S which yields -E average energy from the -Et term. In this note we argue that for a bound quantum state density(x) so d/dt (partial) density = 0, yet the statistical ideas of quantum mechanics should be present in such a case. Furthermore for the free particle wavefunction exp(ipx-iEt), density=1. In fact, the Schrodinger equation should follow for a free particle even the case of density=1. Stochastic processes seem to be based on equations which consider a change in classical spatial density at a point x in time due to flow in or out or stochastic changes. Quantum processes seem to deal with periodic probability changes from one dimension to another exp(ipx) (2 dimensions) which are somehow related to a physical process of momentum flow. Ultimately one links this process to an energy conservation which does not seem to be the same as a classical stochastic/hydrodynamical approach. In this note we argue that a free particle with a constant momentum p has periodic stochastic processes associated with it through its complex probability change in the region of its wavelength, but that these are linked to creating the momentum p. Furthermore momentum is linked to energy so we argue that the Schrodinger equation is a stochastic equation linking energy and momentum and that time and space are separated i.e. there is no situation of following a change in spatial density in time for a free particle with momentum p. For a bound state, one has a distribution of free particles, but again time is removed from the problem in the sense that it only appears in exp(-iEt) in the wavefunction to again establish a relationship between energy and average momentum and potential. There is again no nonzero d/dt (partial) density term which seems to be key in stochastic equations. Only if one takes a linear combination of energy eigenstates Sum over En an(En) exp(-iEnt) Wn(x) does one obtain the form density(x,t), but we argue that quantum features should already appear in exp(ipx) and the bound state scenario with a zero d/dt (partial) density(x,t). We argue, however, that the Schrodinger equation and fundamental stochastic ideas of quantum mechanics should already be present at the level of exp(ipx).
Entropy in quantum mechanics is often based on von Neuman’s prescription (1) which yields zero for a pure eigenstate. Some alternative approaches also yield this result (2). In other words, it seems this type of entropy deals with transitions between different energy eigenstates and excludes the intrinsic entropy of a pure energy eigenstate. There is, however, literature (3) dealing with both spatial and momentum entropy of a single energy eigenstate. In particular, Shannon’s entropy equation yields - W*W ln[W*W] and -a(p)*a(p) ln[a*(p)a(p)] respectively where W, the wavefunction, is Sum over p a(p)exp(ipx). In a previous note (4), we argued that one may use Wn(x) a(p) exp(-ipx) as a probability in Shannon’s entropy equation to obtain an entropy which is the sum of Shannon’s spatial and momentum entropies for a single energy eigenstate. A third term proportional to ipx integrates to zero. In this note, we wish to extend this approach to a time-dependent wavefunction (for example: W(x,t)= Sum over n bn(En) exp(-iEn t) Wn(x).] We suggest W(x,t) bn(En)an(p) exp(iEn t) exp(-ipx) as a candidate probability for Shannon’s entropy and obtain: S= Sum over n bn(En)bn(En) ln(bn(En) + Sum over n bn(En)bn(En) {Sum over p an(p)an(p) ln(an(p))} Sum over x, t W*W ln(W(x,t)) + Sum over x,t,n,p i (Ent-px) W bn(En)an(p) For simplicity, we consider bn(En) and an(p) as being real. For the case of a single n present (i.e. a pure eigenstate) one obtains the entropy of such a state and not zero as in the von Neuman or other approaches. We argue that it might be good to have a single entropy formulation which works for both a single energy eigenstate and multiple states and not the two seemingly independent approaches which seem to be generally used. Furthermore, the approach here includes the intrinsic entropy of pure states together with that of transitions between states in the general W(x,t) (wavefunction) case.
The time-independent Schrodinger equation: En= -1/2m [d/dx dWn/dx] / Wn(x) + V(x) ((1)) matches a classical conservation of energy equation where -1/2m [d/dx dWn/dx] / Wn(x) = KEclassical(x). In classical physics, however, one has x(t) so each x point is associated with a specific t for a given energy En. In the quantum time-independent equation, one has exp(-iEnt) as the time dependent factor and does not consider any link with x although if one is comparing ((1)) to a classical conservation equation then a link between x and t could be made to complete the analogy. This is usually not considered, but we examine this point as we argue it becomes important in the time-dependent situation. For the time-dependent Schrodinger case we consider two approaches. First, a linear combination of time independent solutions is taken i.e. W(x,t) = Sum over n an(En) exp(-iEnt) Wn(x) ((2)) Again each Wn(x) has an association with KEclassical(x) and so one may think in terms of (x,t) classical pairs for a given En. The second approach is to consider the function: exp(-iEnt + m Integral (0,x) v(y) dy) ((3)). This satisfies the time-dependent Schrodinger equation and yields a classical conservation of energy equation with v(x) being the classical velocity. Noting that En= .5m v(x)v(x) + V(x), ((3)) may be rewritten as: exp(i Integral L dt) where L = T-V is the Langrangian with T=.5mv(x)v(x). At this point En has been removed from the picture, but is replaced with: x(t)= X g(t)/g(T) ((4)) where (X,T) are a space-time pair corresponding to a particular energy E value. Using exp(i Action) ((5)) in the time-dependent Schrodinger equation yields a conservation equation which holds for a particular classical energy as long as (x,t) values in the equation are classically linked to the appropriate classical (X,T) values i.e. for a classical trajectory there is a specific family of X at time T for a given energy. For another energy, there is another family of (X,T) pairs. Thus the equation obtained represents energy conservation for many different energy values depending on which simultaneous values one uses for x and t. There is, however, no overlap as a specific (X,T) family corresponds to an energy E and another nonoverlapping family of (X,T) to another energy. This, however, is the same situation as ((2)) for which there are also families of (X,T) corresponding to each En. The only difference is that the En are quantized i.e. discrete. ((3)) and ((5)) are both solutions to the time-dependent Schrodinger equation which represent a variety of energies and families of (X,T) associated to the classical solution of the energy in question. We thus suggest that ((2)) and ((4)) are equivalent for this reason without using the idea of a complete set of basis functions represented by Wn(x). We consider these ideas as the Schrodinger propagator is given by ((5)), but with ((4)) generalized to: x(t)= Xi g(tf-t)/g(tf-ti) + Xf g(t-ti)/g(tf-ti) for the case that g(0)=0 (2) so that Xi,Xf, tf and ti enter the picture. This then should be equivalent to: Sum over n exp(-iEntf) Wn(xf) Wn* (xi) ((6)) In this note, we suggest that one may link ((5)) and ((6)) based on ideas of space-time families associated with different energies as well as the two being solutions of the time-dependent Schrodinger equation.
Quantum mechanics is concerned with measurement and we suggest that in a bound problem impulses due to momentum are the form of measurement which occurs in x space. The Lorentz invariant A= -Et+px, which is also the free particle relativistic or nonrelativistic action with v=x/t, associates energy = pp/2m with time and p with x. Creating eigenvalue equations yields id/dt exp(-iEt) = E exp(-iEt) and -id/dx exp(ipx) = p exp(ipx). P (momentum) is associated with physical impulses and with its own two dimensional probability distribution exp(ipx). E is associated with t and has its own distribution in time exp(-iEt) which implies that the same E holds for each x point without any notion of x probability associated with it. In classical physics, a momentum p exists in a tiny dx (dx→0). For exp(ipx), a p impulse may occur within hbar/p with different probabilities, so one cannot restrict the measurement of p to a tiny dx unless dx is of the order of hbar/p. Nevertheless energy exists at all x points. Given that p measurements occur through specific impulses at different x points with exp(ipx) probabilities, one must calculate an average energy using {Sum over p pp/2m a(p) exp(ipx)} / {Sum over p a(p)exp(ipx)} using the probability associated with the measurement of p in space i.e. exp(ipx). This adds to V(x) to yield a constant E at each x point. We suggest that weighted exp(ipx)s are used to calculate average energy at each x, because it is through impulse measurements that one determines p at different x values and hence the average energy. One may note that one does not calculate {Sum over p a(p)exp(ipx) p } / {Sum over p a(p)exp(ipx)}. An impulse hit is associated with a certain probability and this measurement in x yields information about the corresponding energy pp/2m. Each impulse hit also yields information about a time probability exp(-i ppt/2m), but only if the particle is knocked out of the bound system i.e. only if pp/2m is the only energy present, which is not the case for a single particle bound system.
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).
We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.