In Part 1, we argued that the Heisenberg uncertainty principle (standard deviation p) * (standard deviation x) >= hbar/2 only applies when a measurement (i.e. interaction) has occurred. Otherwise, the notion of standard deviation does not exist. Thus, if one has a Gaussian wavepacket, it is the measuring device which has created it and constrained is standard deviations of momentum and position. A free particle, which has not been measured, is described by exp(-iEt+ipx), as argued in Part 1, and not by a wavepacket. Here we ask: How does one interpret a wavepacket calculation? In (1), we argued that exp(-iEt) is a free particle energy conserving probability and exp(ipx), a momentum conserving one. Then, a sum of exp(ipx - iEt) represents an OR situation in probability theory. There is superposition and so one obtains a weighting of possible outcomes which is the probabilistic interpretation of the scenario. If one considers two slit interference, the viewpoint is the same. exp(ipx) is a probability which allows for momentum hits of p in a range of dx=hbar/p outside the center-of-mass, which follows x=vt. This built-in stochasticity of impulse delivery leads to the notion that probabilistically, a particle may interact with both slits if their separation is about hbar/p. One writes an OR probability statement: W(x)= exp(i p dot r1) + exp(i p dot r2), where r1 is the position vector from the center of slit 1 to a point far away on a screen, and r2, a vector from the center of slit 2 to the same point. The result, W*(x)W(x) is a probabilistic expression which describes the possible outcomes probabilistically. (1). Seen in this way, the Gaussian wavepacket should be interpreted in a similar manner and not as a physical description of the particle which moves freely. A single bound state in a potential V(x) is localized by a potential and one may consider only OR cases of exp(ipx), i.e W(x) = Sum over p a(p)exp(ipx). In the measurement case, there is no quasi-permanent localization and one uses a linear superposition of exp(-iEt+ipx)s. Nevertheless for a Gaussian wavepacket, two probability distributions describing possible outcomes are created. One has W*(x)W(x)= exp(- (x-u)(x-u) / (2 sigma(t) sigma(t))) where sigma is the x standard deviation. In addition, W(x) = Sum over p a(p) exp(ipx), leading to a momentum outcome distribution of a*(p)a(p). As written, sigma(t) for x which means that the Gaussian is expanding. We called this unphysical if one considers the wavefunction as representing a description of a single physical entity (2) at times beyond t=0. Like in the 2-slit case, we argue that it represents an outcome distribution. Thus, one would not consider any more time values than t=0, at which time the measurement is completed. An exp(-iEt+ipx), describing a free particle, is measured by a specific device which creates the Gaussian form and gives it a specific standard deviation in space and momentum. The measurement is then done and W*(x,t=0)W(x, t=0) represents the possible outcome positions of the particle at t=0 (i.e. probabilities for x portions of a phase shift). The momentum distribution yields possible p outcome values. As soon as the measurement is finished, the particle moves as exp(-iEt+ipx), with p being probabilistically linked to a*(p)a(p) (a Gaussian as well), where W(x,t) = Sum over p a(p)exp(-iEt+ipx).If one has a second measuring device which creates a tiny momentum standard deviation, one would be able to carry out experiments which would determine roughly the standard deviation of p created by the first measuring device. Furthermore, one would be able to measure phase shifts as well by comparing a free particle measured by a device with a tiny p standard deviation and a large x one with free particles which were not measured.
In Part 1, we argued that the Heisenberg uncertainty principle (standard deviation p) * (standard deviation x) >= hbar/2 only applies when a measurement (i.e. interaction) has occurred. Otherwise, the notion of standard deviation does not exist. Thus, if one has a Gaussian wavepacket, it is the measuring device which has created it and constrained is standard deviations of momentum and position. A free particle, which has not been measured, is described by exp(-iEt+ipx), as argued in Part 1, and not by a wavepacket. Here we ask: How does one interpret a wavepacket calculation? In (1), we argued that exp(-iEt) is a free particle energy conserving probability and exp(ipx), a momentum conserving one. Then, a sum of exp(ipx - iEt) represents an OR situation in probability theory. There is superposition and so one obtains a weighting of possible outcomes which is the probabilistic interpretation of the scenario. If one considers two slit interference, the viewpoint is the same. exp(ipx) is a probability which allows for momentum hits of p in a range of dx=hbar/p outside the center-of-mass, which follows x=vt. This built-in stochasticity of impulse delivery leads to the notion that probabilistically, a particle may interact with both slits if their separation is about hbar/p. One writes an OR probability statement: W(x)= exp(i p dot r1) + exp(i p dot r2), where r1 is the position vector from the center of slit 1 to a point far away on a screen, and r2, a vector from the center of slit 2 to the same point. The result, W*(x)W(x) is a probabilistic expression which describes the possible outcomes probabilistically. (1). Seen in this way, the Gaussian wavepacket should be interpreted in a similar manner and not as a physical description of the particle which moves freely. A single bound state in a potential V(x) is localized by a potential and one may consider only OR cases of exp(ipx), i.e W(x) = Sum over p a(p)exp(ipx). In the measurement case, there is no quasi-permanent localization and one uses a linear superposition of exp(-iEt+ipx)s. Nevertheless for a Gaussian wavepacket, two probability distributions describing possible outcomes are created. One has W*(x)W(x)= exp(- (x-u)(x-u) / (2 sigma(t) sigma(t))) where sigma is the x standard deviation. In addition, W(x) = Sum over p a(p) exp(ipx), leading to a momentum outcome distribution of a*(p)a(p). As written, sigma(t) for x which means that the Gaussian is expanding. We called this unphysical if one considers the wavefunction as representing a description of a single physical entity (2) at times beyond t=0. Like in the 2-slit case, we argue that it represents an outcome distribution. Thus, one would not consider any more time values than t=0, at which time the measurement is completed. An exp(-iEt+ipx), describing a free particle, is measured by a specific device which creates the Gaussian form and gives it a specific standard deviation in space and momentum. The measurement is then done and W*(x,t=0)W(x, t=0) represents the possible outcome positions of the particle at t=0 (i.e. probabilities for x portions of a phase shift). The momentum distribution yields possible p outcome values. As soon as the measurement is finished, the particle moves as exp(-iEt+ipx), with p being probabilistically linked to a*(p)a(p) (a Gaussian as well), where W(x,t) = Sum over p a(p)exp(-iEt+ipx).If one has a second measuring device which creates a tiny momentum standard deviation, one would be able to carry out experiments which would determine roughly the standard deviation of p created by the first measuring device. Furthermore, one would be able to measure phase shifts as well by comparing a free particle measured by a device with a tiny p standard deviation and a large x one with free particles which were not measured.
Traditional quantum mechanics calculates a probability in space through W*(x)W(x) for a time-independent wavefunction W(x) = Sum over p a(p)exp(ip dot r). We argue that this probability is created at the interaction region and indicates a probability for outcomes. In particular, given a 2-slit problem, one calculates W(r) = exp( i p dot r1) + exp(i p dot r2), where r1 and r2 are vectors from the center of each slit to the same point on a screen far away. W(r)*W(r) then represents the probability for particles to arrive at various points on the screen. This may be realized by sending many particles/photons through the 2-slit apparatus, one at a time. Given that exp(ip dot r) is an interaction probability (we argue), its relevance holds in the interaction region which is presumably of about a wavelength long or so. We argue that exp(i p dot r) exp(-i p dot r) = 1 does not mean that one should interpret the particle/photon as being possibly present at any r in space. The particle still has a center-of-mass which follows x=vt for a free particle. The only issue is that exp(ipx) implies that the particle does not interact at its center-of-mass point necessarily, but can interact with probability exp(ipx) at different points within .5 hbar/p of the center of mass on either side. As a result, we argue that there is no issue concerning localization of either a free particle or photon. The object is localized and moving at x=vt and it has a precise p which may render impulse hits at points about the center-of-mass point in keeping with exp(ipx). As a result, Heisenberg’s uncertainty relation (standard deviation p) (standard deviation x) >= hbar/2 does not hold for exp(ipx) because standard deviation p =0. exp(ipx) applies to a free particle which is not interacting. When the particle interacts (which is the way it is measured), then exp(ipx) becomes Sum over p a(p)exp(i p dot r). In such a case, a standard deviation of p and x arise. The Heisenberg uncertainty principle holds for interactions which are the only way the particle can be measured and so applies to W(x) and not a single exp(ipx). Given the above ideas, we suggest there is no reason to create a free particle wavepacket. Such a wavepacket seems to be created in the literature (1) in order to localize a particle in space and to ensure that Heisenberg’s uncertainty relationship is enforced. We have argued that there is no reason for this relationship if the particle is not interacting. As for localization, the free particle is localized by its center-of-mass in x=vt. It is just that one is not measuring this point. Furthermore, the increase in a Gaussian wavepacket’s x standard deviation in time is unphysical. Given that there is no reason for a wavepacket, we then question the notion of the wavepacket as being associated with a Bohmian potential which creates the wavepacket so to speak, i.e. -1/2m (d/dx dd/x A(x)) / A(x), where W(x,t) = A(x) exp(i S(x,t) ).
In Part 1, we argued that the consideration of all possible outcomes is a defining idea of a probabilistic calculation. We then argued that in free particle quantum mechanics there were two sets of outcomes, one driving the other. In particular, a single particle moves as x=vt, but at each x,t point, one has interaction uncertainty/fluctuations of dt and dx linked with exp(-iEt) and exp(ipx). As a result, in a 2-slit interference problem, exp(ipx) (a probability linked with possible positional fluctuations about x) leads to a further probability of interacting with both slits. The two sets of outcomes must be considered, i.e. exp(i p dot r1) + exp(i p dot r2), where r1 is a vector from the center of the first slit to a point on a screen far away, and r2, a vector from the center of the second slit to the same screen point. We also considered the case of reflection-refraction from an n1-n2 index of refraction junction. The point is that these two cases are clearly probabilistic, i.e. not deterministic at all. Here we consider two different scenarios which have the appearance of a deterministic result, but are actually probabilistic and so linked with all possible outcomes, i.e. an average over these which must be understood as an average (as in classical probability). The first is a single particle bound state and the second, transmission through a potential wall. In the bound state case, the analogue is a Newtonian bound state and in fact between the classical turning points KEave(x) + V(x) = E in the probabilistic case which exactly matches the classical one. One is, however, dealing with an average KEave(x) = -1/2m d/dx d/dx W/ W, where W=Sum over p a(p)exp(ipx), it involves all possible outcomes and is not a physical state. Given the similarity to a deterministic problem there is a danger of not considering the quantum bound calculation as a probabilistic one. In the quantum single particle case, one may be surprised by tunneling, but the quantum probabilistic scenario is one in which all possible interaction outcomes are considered together (as is required by probability) and so one cannot possibly ignore interactions linked to a particle entering or leaving the potential region. Furthermore, it is only KEave(x) which equals the classical result and so E which is single value in the classical case, is linked to an average value in the quantum case, i.e. a particle may have a p such that pp/2m is not zero at the classical turning point. Only KEave(x) =0 at this point. An interesting feature is that a probabilistic calculation applies to a single particle run (which only produces one outcome). One, however, cannot measure the full probability associated with a problem without making a very large number of measurements.
Quantum mechanics is largely based on a wavefunction, which we consider to be a probability. The theory is, however, not solely probabilistic because if an interaction does not occur, x=vt holds as does the classical notion of impulse hit and kinetic energy = pp/2m (nonrelativistic case). The probability exp(-iEt+ipx) describes probabilistic interactions in time and space (we argue) and so one does not think of x and t as in x=vt. As a result, there is a probability for a quantum particle to interact with both slits of a two slit apparatus if they are about hbar/p apart. This leads us to the notion that W*(x)W(x) (W(x)=Sum over p a(p)exp(ipx)=wavefunction) is not spatial density, but rather an interaction profile. A particle can still exist at W(x)=0 points, it just has very little to no probability of interacting at these points on average, with the average considering both exp(ipx) and exp(-ipx) in a bound case scenario. Given that E=kinetic energy = pp/2m (nonrelativistic case only) and W*(x)W(x) not= spatial density, we suggest that one must be careful with interpreting d/dt partial W*(x)W(x) = constant grad dot Im (W* (-id/dx) W) (from (1)) as a continuity equation which represents hydrodynamic motion. We suggest that the quantum features are governed by exp(-iEt+ipx) and that any derivatives d/dx and d/dt are related to changes in phase and not hydrodynamic flow (in a classical sense). We note that for a free particle d/dt partial density = 0 = constant grad dot Im (W* (-id/dx) W). The same zero equation holds for a bound state (even though there is physical motion to the right and then left) , but if one creates a superposition of bound states (which is a probability statement), then the same equation is not 0. We note that for a classical bound state there exists a vrms(x), but this is an average quantity. It does not show any resolution pattern which is represented by W(x) and so one must be careful when interpreting it, we argue.
Quantum mechanics utilizes a wavefunction W(x)=Sum over p a(p)exp(ipx) (in time independent cases) with exp(ipx) being what we call a free particle complex probability. In such a case, W(x) is an OR scenario of these various probabilities. We have argued in previous notes that exp(ipx) is the probability for a p impulse hit at x, even if the center-of-mass of the particle is not at x. This means that impulse hits may occur probabilistically in space as in the 2-slit example which requires adding two probabilities as an OR case for each slit. The point then seems to be that W(x) (or even its momentum frozen form W*(x)W(x) gives a profile in space of the number of interactions (various p impulse hits) at each x. Classically, the kinetic energy (nonrelativistic) is pp/2m. In a classical bound problem this kinetic energy is considered to exist (in calculations) at the center-of-mass, but forces are also said to be rendered at this point so pp/2m, the center-of-mass and the force hit point all coincide. In quantum mechanics, a p impulse hit and the center-of-mass do not coincide. If p is said to hit at a probabilistic x, then it seems that pp/2m should also be seen to exist at this same hit point. A physical interaction point verifies in a sense the location of pp/2m in a probabilistic manner, i.e. a(p) exp(ipx)/W(x) is the probability used to calculate KEaverage and this is based on impulse hit locations. By normalizing using W(x), one removes the relative number of average hits at each x and obtains a pure energy value at x. (This may then be used in a KEave(x)+V(x)=E equation.) On average W(x) has zero points, which means that if one considers W*(x)W(x) to represent the motion-frozen-out spatial distribution of hits, then no hits are rendered at certain x points (on average). One might argue then that the particle cannot exist at x, but the center-of-mass can exist anywhere. One might next argue that there is no kinetic energy at x, but that cannot be true because 1/2m d/dx d/dx W has the same 0 points as W and the two divide. Thus, there is average kinetic energy at all bound points and this represents an average based on impulse hits (exp(ipx)). Classically, if one has many particles in a potential, one may write density(x)V(x) to find the potential energy of the particles. This implies that the particles interact with V(x), e.g. neutrinos usually do not interact. Density(x) in a classical sense means that the center-of-mass and force hit point are both at x. In quantum mechanics, exp(ipx) represents the probability for a force hit point, W(x) represents an average and W*(x)W(x) represents a motion-frozen average hit point. W*(x)W(x)V(x) seems to represent the average potential energy of the particle at x because W*(x)W(x) is linked with interaction. Even if a particle is not present, however, there is energy V(x) in space. It seems that if W*(x)W(x)=0 at some x, then there is no average particle interaction at this point, but there is still average kinetic energy Sum over p a(p)pp/2m exp(ipx) / W(x) based on the probability for average probability hits. The key idea is that one normalizes probability by dividing by W(x) in the average kinetic energy calculation. W*(x)W(x) is normalized not at x, but integrating over dx and setting this value to 1. One has relative hit at x values represented by W*(x)W(x). For exp(-ipx)exp(ipx)=1, each dx has the same average hit value (with one not being able to distinguish details in hbar/p, i.e. this is a motion frozen result. The question then becomes: Why would one write: W*(x)W(x) KEave(x)? One might argue that this term occurs if one takes the quantum mechanical equation: KEave(x) +V(x) = En and multiplies by W*(x)W(x). One may ask: Does V(x)W*(x)W(x) have physical meaning? V(x) is the potential energy at x. The average interaction probability with V(x) at x should be W(x)V(x) and not W*(x)W(x)V(x). The latter is a classical mechanical idealization. If W*(x)W(x) is 3 at x1 and 2 at x2, then it is as if the particle interacts 1.5 times more at x1 and so if V(x1)=V(x2), then this approach assumes 1.5 the interaction energy at x1. KEave(x), however, is simply free particle energy (average) at x and one does not need to take this average value and multiply it by the relative number of interactions at x. W(x) has zero points meaning that there is no interaction at such a point on average. KEave(x) on the other hand is the average energy at x based on the particle kinetic energy presence pp/2m being at the point of an impulse hit. This does not mean that an interaction is occurring with V(x) at this point in this formalism. Even a free particle has an exp(ipx) probability when it is not interacting. The sum Sum over p a(p)exp(ipx) means that all of the interactions linked with x have occurred and one is making an OR calculation. KEave(x) is free particle kinetic energy at x and V(x) is potential energy (not interaction potential energy with the particle). Multiplying by W*(x)W(x) is a classical minded approach to creating a classical equation: density(x) (KEave(x)+V(x)) = E density(x) ((1)) Here KEave(x) would represent the average kinetic energy of a single particle and density(x) would be the average number of particles at x. density(x)KE(x) has a clear meaning. W*(x)W(x) KEave(x) on the other hand is different. KEave(x) is actually the classical kinetic energy value at x and so it is not clear why one would multiply it by W*(x)W(x). W*(x)W(x) represents the probability for an impulse hit at x in a p-frozen out picture. The notion of exp(ipx) (i.e. probability for a p hit at x) is already buried within the quantum calculation of KEave(x) (which yields the classical value). In order to find the physical meaning of W*(x)W(x) multiplied by KEave(x) or V(x) it seems that one has broken dx into equally sized segments which correspond to different detection hits of the particle. This is then somehow equated with the presence of the particle per unit time in dx which is then somehow rendered as the average energy at x multiplied by the detected presence at x through a steady rate of measurement/time. Thus, multiplying by W*(x)W(x) is not multiplying by a physical density.
Classically, probability is considered in terms of the number of possible outcomes. For example, a die toss has six possible outcomes and a coin toss, two. Probability calculations make use of probabilities and hence of the total number of possible outcomes. Thus, they are not deterministic calculations which follow “what happens” as in Newtonian mechanics. We suggest that if free particle quantum mechanics is probabilistic, it must also involve the notion of possible outcomes in an experiment and not represent a deterministic approach. We suggest, however, that something interesting applies to quantum mechanics because we have suggested that free particle quantum mechanics arises from the Lorentz invariant A = -Et+px. We note that there exist x,t trajectory points which follow x=vt. We also note, however, that one may introduce the idea of intervals dx, dt (because Newton’s dx→0, dt→0 is an idealization) and that A is unchanged for x,t and x+hbar/p , t+hbar/E. In other words, we suggest that there are intervals with probability about each x,t trajectory point. A trajectory point is associated with deterministic physics, but probability about this point is not. In other words, one has a kind of hybrid probability scheme. Thus, exp(ipx) and exp(-iEt) really represent fluctuation probabilities about a trajectory point x and t. These ranges of possible x and t fluctuations about a trajectory point x,t allow for further probability to occur in certain interactions. We consider two examples. The first is 2-slit interference which involves a deterministic particle moving along a straight line towards a 2-slit mechanism perpendicular to its path. At the 2-slit apparatus, one considers a probabilistic calculation as the uncertainty in x, i.e. dx=hbar/p allows for possible interaction with both slits. This involves considering all possible outcomes (i.e. motion through one slit or the other which is a classical situation), but exp(i p dot r1) and exp(i p dot r2) “sit on top” of these trajectories at the 2-slit apparatus and one has interference of the probabilities linked with each classical possible outcome. Thus, there are two sets of probabilities which are intertwined, the probability range hbar/p for each path and the two possible paths which have equal probability. One must consider the possible outcomes for both sets together. We suggest that a similar analysis applies to the case of 1-D reflection-refraction for x=0 at an n1-n2 index of refraction junction. Thus, we suggest that a deterministic free particle with x=vt may encounter probabilistic interaction situations, such as a two-slit apparatus or an n1-n2 index of refraction junction. These are probabilistic only because there is a probability fluctuation scheme in x associated with each deterministic x point. It is this probability which allows for a second probability to arise due to interaction with the slits or at the 1-D n1-n2 junction. In such a case, one must consider all possible outcomes which means that one does not only consider the exp(ipx) type probabilities, but also those linked with the physical interaction, i.e. two possible paths, one from each slit, or an incident, reflected and refracted photon at x=0 (n1-n2 junction). It is a combination of all possible paths which must be used in calculations. We thus suggest that even though there is a sense of determinism, i..e free particles move as x=vt and these exist in the 2-slit case and the 1-D reflection-refraction case, in a probabilistic calculation one must consider all probabilistic outcomes consistent with these. Thus, quantum mechanical calculations are true probabilistic calculations in the sense that they involve all possible outcomes in an analysis.
In classical physics, a particle traveling to the right is the mirror image of one traveling to the left. In terms of a spatial density (the time spent in each constant dx), the value is the same, as is the energy. A similar notion of such a spatial invariance is in the classical conservation of energy equation for a bound particle: p(x)p(x)/2m + V(x) = E. Quantum mechanics seems to be different because exp(ipx) and exp(-ipx) are distinctly different spatial probabilities, i.e. they are dynamic. It is not just that case that a particle moves to the right for exp(ipx) and to the left for exp(-ipx). The impulse hit probabilities (represented by exp(ipx) and exp(-ipx)) are distinctly different. If one wishes to create an OR situation and write: W(x) = Sum over p a(p)exp(ipx) and use this in a conservation of energy equation, then simply including only positive or negative p values in W(x) leads to a complex average kinetic energy -1/2m d/dx dW/dx / W which cannot be used with V(x) and E. Furthermore, W_ (only negative p) and W+ (only positive p) values do not yield the same average kinetic energy. Fortunately, given that this is a statistical average for a localized object (somewhat analogous to a gas), one may combine W+ + W- and obtain a real kinetic energy. We have argued in previous notes that there exists a quantum-classical duality. For example, for a free particle one has two independent notions. One is x=vt which describes the center-of-mass of the particle. We take this to represent a state description which treats v and -v as mirror images. exp(ipx), on the other hand, represents probabilistic impulse hits and does not define the state by itself, we argue, only its interactions. Interactions are physical and define what happens in an interacting system. One may, however, measure a system, i.e interact with exp(ipx), and this should define a measurement state ( a second state ensemble average). For example, exp(ipx)exp(-ipx) = 1 which is equivalent to the constant dt in each dx associated with x=vt. Thus, the two states coincide, but exp(ipx)exp(-ipx) means that one may place the 2-slit apparatus at any x point and is not about the center-of-mass, as x=vt is. In some cases, the two states are quite different. W*(x)W(x) represents an interaction state, but vrms(x) from .5mvrms(x)vrms(x) = -1/2m d/dx dW/dx / W represents a center-of-mass state description. (Note: in these cases a state means an average state, i.e. an average over many ensemble entries.) If one takes W*(x)W(x) and integrates over x, then interference terms vanish and one obtains a classical counting of distinct p value states with their weight a*(p)a(p).This, we argue, leads to the notion of orthogonality of probabilities. In classical probability, there is no such notion, but there is the notion of AND (multiply) probabilities. Given exp(ipx) and exp(-ipx), if one multiplies one removes p, i.e. the interactions and only sees the state left. W*(x)W(x) yields a set of peaks which represent a pattern of probabilistic hits in space which define a state which one would measure in space. The interference pattern, however, must be one such that if one integrates over x, one only obtains a classical sum of probabilities a*(p)a(p)’s as mentioned above. This leads to the notion of orthogonal exp(ipx)’s (which follows mathematically anyway). If one considers different bound En states, then these two are orthogonal because W*n1(x)Wn2(x) mean n1 and n2 are associated with different energies and integrating over x cannot show a result which looks like a state result, where W*n(x)Wn(x) shows a state result. In order to achieve this orthogonality (linked directly to the notion of a state) one must have that Wn(2) usually has 0 points and is positive and negative with the possible exception of a ground state. In other words, if Wn1*(x)Wn2(x) integrated over x is not 0, one violates conservation of energy. As a result, we argue an average probability for a bound state must statistically include +p and -p contributions and must create a function which has positive and negative peaks (except possibly for the ground state), i.e. W(x) must have 0 points. This W(x) is linked to a measurement state W*(x)W(x), but this is strictly considering exp(ipx)s. We argue that there are two pieces of information for a free particle, x=vt and exp(ipx). There must also be an average state linked to x=vt and this given by KEave(x) = .5m vrms(x)vrms(x) + V(x) = E which shows that the particle may exist and have average kinetic energy even at W(x)=0 points. When measuring a system, one interacts with the exp(ipx)s and so the W*(x)W(x) state is the one which is traditionally presented, but whether one measures the center-of-mass motion or not, one also has the notion of the x=vt, i.e. vrms(x) state (average) as well, we argue.