Widdekind

That sounds like Newton's law of motion, "a body in motion tends to stay in motion, unless acted upon by an outside force"; for the interiors of SCs. Er go, CPs in SCs "experience no outside forces", i.e. are non-interacting ?? So, the boson-like CPs form a BEC, of low (relative) entropy ?

Why is the definition, of the 4-current, not the more "physically intuitive" [math]j^{\mu} \equiv \frac{\imath \hbar}{2 m} \partial^{\mu} \left( \psi^{*} \psi \right)[/math] That would make all the "minus signs" into "plus signs" in eqs. 3.10 et seq., from which follows (?), via the KGE [math]\Box \psi \sim -m^2 \psi[/math]: [math]\partial_{\mu} j^{\mu} = \frac{\imath \hbar}{m} \left( \left( \partial_{\mu} \psi^{*} \right) \left( \partial^{\mu} \psi \right) - \frac{1}{\lambda_C^2} \psi^{*} \psi \right)[/math] That looks "tantalizing". Can one prove, that the RHS = 0, i.e (in normalized units) [math]\left( \hat{p}_{\mu} \psi \right)^{*} \left( \hat{p}^{\mu} \psi \right) = m^2 \left( \psi^{*} \psi \right)[/math]

I'm confused, about "negative probabilities". You give me a KGE WF [math]\phi[/math], for some particle. The WF associates, to every point in space, a complex number, representing so much "yin", and so much "yang", of "particle aether-field disturbance", at those points. I take [math]\phi^{*} \phi = |\phi|^2 > 0?[/math]; and then I integrate over all of space [math]\int d^3x |\phi|^2 >0?[/math]. How can I get a negative number, from summing such 'squares' ?? Please permit me to quickly compare, the KGE [math]j^0[/math] component, with the corresponding SWE component. According to this reference, the spatial "momentum-like" currents, clearly correspond, between the KGE and SWE. However, the time "energy-like" current, does not, immediately, clearly correspond. But, please watch what happens, if one "forces them to fit": [math]\rho_{KGE} \approx \rho_{SWE}[/math] [math] \frac{\imath \hbar}{2 m c^2} \left[ \phi^{*} \partial_t \phi - \partial_t \phi^{*} \phi \right] \approx \phi^{*} \phi [/math] [math]\frac{1}{2} \left[ \phi^{*} \left( \hat{E} \phi \right) + \left( \hat{E} \phi \right)^{*} \phi \right] \approx mc^2 \left( \phi^{*} \phi \right)[/math] [math]"E(x)" \approx mc^2 \rho(x)[/math] I wish to observe, that the SWE corresponds, to the classical limit, of the KGE -- but, normalized to the rest-mass energy, which is "dropped" from the equation, i.e. [math]E = \sqrt{p^2 + m^2} \approx m^2 + \frac{p^2}{2m} + ...[/math] and the SWE omits the mass term, "for simplicity". Thus, however, the SWE is not the "complete direct" low-energy limit, of the KGE; even as classical mechanics, by over-looking rest-mass energies, is not the "complete direct" low-energy limit, of the relativistic Einstein relation. Were you to put the mass term back into the SWE, then all classical QM WFs would acquire additional phase factors [math]e^{-\imath \frac{mc^2}{\hbar}t}[/math]. In practice, the SWE "filters out" that "carrier frequency", and represents merely the "frequency modulation" of the "signal" (likened to FM radio). So I'm trying to say, that, in the low-energy limit, [math]c p \rightarrow 0[/math], the KGE "says" that particle energies are dominated, by their rest-masses, cp. mec2 = 511KeV >> EH = 14eV. And that is why the "time derivative", from the KGE [math]j^0[/math], asymmetrically vanishes, from the SWE analog, e.g. "the 'carrier frequency' is 511,000eV; and the time-derivative is giving only 14eV, so forget-about-it". But, according to the source sited, the KGE satisfies a perfectly sensible continuity equation [math]\partial_{\mu} j^{\mu} = 0[/math], which vaguely amounts to saying that "energy flow" (from time derivative) = "momentum flow" (from space derivatives) [math]\partial_{ct} \left( \frac{1}{2} \left[ \phi^{*} \left( \hat{E} \phi \right) + \left( \hat{E} \phi \right)^{*} \phi \right] \right) = -\vec{\bigtriangledown} \cdot \left( \frac{1}{2} \left[ \phi^{*} \left( c\hat{p} \phi \right) + \left( c\hat{p} \phi \right)^{*} \phi \right] \right)[/math] "[math]\partial_{ct} \left( \hat{E} \left(\phi^{*} \phi\right) \right) = -\vec{\bigtriangledown} \cdot \left( c\hat{p} \left(\phi^{*} \phi\right) \right)[/math]" (vague simplified summary) Is that not, essentially, the correct QM probability continuity equation ? I.e. by implication, [math]\rho = \phi^{*} \phi[/math] is the particle probability density; and "[math]\hat{E} \rho[/math]" is the particle energy density; and "[math]c\hat{p} \rho[/math]" is the particle momentum density (please note the quotation marks, signifying vague simplified summary symbolism). Once you recognize, that the SWE "buries" or "hides" the mass term; then I cannot comprehend any inadequacy, in the KGE. For the KGE, [math]j^{\mu}[/math] is the "mom-energy" flow vector, obeying a sensible continuity equation; and, separately, [math]\rho = \phi^{*} \phi[/math] is the probability density; and, these terms, separate for the KGE, become "muddled" with the SWE, by its omission, of the mass term, causing confusion. Is this not so ? (According to this writer, at time of writing, eq. 3.52, in said cited source, is wrong; and is non-sensical.) If you write [math]\phi = \alpha + \imath \beta[/math]; then can you show, that [math]j^{\mu} = \partial^{\mu} \left( \alpha \beta \right)[/math] ? To my mind, a negative [math]j^0 < 0[/math] merely means, that energy density is evacuating some small place in space (along with momentum [math]\partial^i j^i = \vec{\bigtriangledown} \cdot \vec{j}_3 > 0[/math]), so that the WF, in that place in space, "red-shifts", i.e. phase-oscillates, at decreasing frequency. And conversely, when momentum flows into some small place in space; then local energy density increases; and the wave-function "blue-shifts", i.e. phase-oscillates, at increasing frequency. I'm trying not to second-guess eq. 3.52; but I do not perceive any unphysicality, with the KGE, i.e. [math]j^0[/math] is not a particle probability density, but a particle energy density measure, of the "red-shifting / blue-shifting", of the particle's WF, at that place in space. Perhaps, at low energies, in the SWE limit, due to some "deep connection", when momentum in-flows into some region; then the local energy density measure increases, i.e. the WF "blue-shifts", per KGE; and the WF "just so happens" to also build up, in said region ?? Comparing eqs. 3.50, 3.61, 3.65; in the low energy, SWE, limit; [math]j^0 \rightarrow \phi^{*} \phi[/math] (for mono-chromatic, free-particle, plane-wave, 'mom-energy' eigenstates).

If CPs have zero spin, then they have zero magnetic moment, too ?

A quantum particle is represented by a wave-function, which extends throughout some finite volume of space. As such, a quantum-particle is like a "cloud of probability density". You can calculate the "center of probability" of the particle, i.e. its position expectation value: [math]\langle x \rangle = \int d^3x \; \psi^{*}(x) \; x \; \psi(x)[/math] That expectation value corresponds to the classical "position" of the particle. But, QM'ly, the particle has some statistical spread, in position, about that centroid. That statistical spread, is [math]\Delta x[/math]. Likewise, particles have momentum expectation values [math]\langle p \rangle[/math], which correspond, to the classical "momentum" of the particle; but also, QM'ly, have a statistical spread, of momentum components, [math]\Delta p[/math], about that centroid. I offer, that, in analogy, quantum particles are like "schools of fish", "herds of animals", or "flocks of birds". I.e. the "flock" is generally localized, to some vague region of space, with some semblance of a "center" [math]\langle x \rangle[/math], but also some spatial spread about that center [math]\Delta x[/math]; and, the "flock" is generally moving, in some general direction [math]\langle p \rangle[/math], but some "birds" are moving faster-or-slower, or this-way-and-that [math]\Delta p[/math]. As the particle interacts, with its environment, the "flock" scatters, gathers, roosts, lands, takes flight, swirls, etc.

Please ponder an electron, propagating towards a proton. Now, "before the electron reaches the proton", the electron's incident wave-function will not appreciably over-lap, with any of the potentially available bound hydrogenic states. And "after the electron has been there and gone", the electron's scattered wave-function will not appreciably over-lap, with any of those bound states, either. However, "in between", when the wave-function of the electron has "enveloped" the proton, i.e. as the electron is "on top of" the proton; then the over-lap integrals will be non-zero. To make an investment analogy, each available bound H state, could be likened, to a "stock". And, the spectrum, of available bound H states, could be likened, to a "portfolio of stocks". Now, as the electron impinges upon the proton, the "stocks" in its "portfolio" will, at first, "rise in value". Then, as the electron scatters away from the proton, those "stocks" will "fall in value". Is it not true, that the electron does not directly "perceive" the proton, but only the "virtual photons" emanated by the proton, i.e. the proton's electrical potential well ?? And so, is it not true, that the electron's only possible "perception", of its "interaction", with the proton, would be its "stock values", in its "investment portfolio" ?? And so, are not those "stock values", qualitatively visualized, in the following figure, the only possible information, which could possibly "trigger" a wave-function "collapse", of the electron, into some captured, bound, H state, i.e. a QM detection / measurement / observation ?? If so, then something, seemingly, must happen, between "the stocks are all making money", and "the stocks are all losing money", to trigger the "selling off of some stocks", and the "buying more of (one) other stock" ?? I want to ask, if wave-function "collapse" is, ultimately, a smooth & gradual process, via which a wave-function is "con-formed" into a "desired shape"; then would that immediately imply, that the "collapse" process, involves a "positive feedback" process, via which the electron (say) "perceives an error", between its current wave-function; and a "desired" bound H (say) wave-function ?? I want to ask, what other known physical process, besides feed-back control, can "conform" things, into "desired" homeostases ?? Would not the electron (say) need to "perceive" the "wave-function discrepancy" [math]\Delta \psi \equiv \psi - \phi_0[/math], as some sort of "phantom potential" in the SWE, which "drives" the electron into the "desired" distribution, i.e. "shoves the electron into shape" ?? Perhaps there are published articles, detailing such scenarios ??

I understand, that when two particles are entangled; then their wave-functions are spatially inter-mingled. And therefore, when you separate "the two particles"; then each particle, is partially present, in both locations. Each "partial particle" is also in the super-position state. So, in a coin analogy, each coin is half on one side, half on the other; and on each side, is partially H, and partially T. I read that in a book. I.e. "H+T" is conserved; and "L+R" is conserved: [math]|\psi\rangle \sim |H,L\rangle_1 |T,R\rangle_2 + |T,L\rangle_1 |H,R\rangle_2 + |H,R\rangle_1 |T,L\rangle_2 + |T,R\rangle_1 |H,L\rangle_2[/math]

I understand, regarding the KGE, that: Vaguely, such smacks of the particles-and-anti-particles interpretation, for the Dirac equation. Strikingly, the KGE "does not admit a positive definite probability". If physicists have no qualms, with interpreting negative frequencies, as anti-particles; then negative probabilities could possibly be interpreted, as corresponding anti-particle "anti-probabilities" ?? Perhaps, such could possibly be the KGE equivalent, of the DE "zitterbewegung" The factor of "two", in front of the electron mass, suggests some sort of "pair production process" -- as if relativistic particles "ablate" a "retinue" of particle-anti-particle pairs, a little like a satellite fragmenting as it burns up in atmospheric re-entry ?? Note that: and that, if KGE WFs are naively interpreted, as probability densities, occasionally integrating to a total probability P<1; then the KGE could possibly imply, that relativistic particles have proportionately reduced force charges, e.g. mass, electrical charge ??

B.R.Martin's Particle Physics; and K.Ford's 101 Quantum Questions; give me the impression, that "wave-function collapse" is spontaneous, when exo-thermic, in a manner akin to chemistry, i.e. "chemical reactions are QM wave-function collapses". Vaguely, wave-functions, of quantum particles, are spatially extended; and so, quantum particles "feel" a "variety" of potential energies, as they "reach out" through space. And meanwhile, wave-functions have "multiple personalities", some of which have higher momenta-and-energies; whilst others have lower momenta-and-energies. And, in regions of space, where the particle's local energy, is less than the local potential, there the WF decays away. Conversely, in regions of space, where the particle's local energy, is higher than the local potential, there the WF is reinforced. In such a way, the energy environment, of a particle's WF, can "cultivate" that WF, demoting the WF in some regions, whilst promoting it in others -- so that WF "collapse" is, at sufficient time resolution, a gradual process of "pruning & cultivating" the WF. Physically analogous to chemical reactions, spontaneous wave-function "collapses" are "driven" by energy, i.e. when one "personality" of the wave-function finds an energetically advantageous "fit", that "persona" is gradually-but-quickly spontaneously "promoted" -- perhaps as, somehow, energy is transferred into the "promotion" of virtual force carriers, per G.Zukav's Dancing Wu Li Masters ??

Scientific American has published a simplified explanation, with 10 minute video, of QE.

Is not the KG equation, generated by the same ansatz as the SWE, but applied to the full relativistic energy equation (E2 = p2 + m2), instead of its classical approximation (E ~ p2/2m) ? Why would one accept the ansatz, in the classical case; but then reject the same, in its natural relativistic extrapolation ?

Per the "point-by-point translation" procedure, perhaps transformation, of point-like position eigenstates, of the Klein-Gordon equation [math]\left(E_0 = mc^2, E' = \gamma E_0, cp' = \gamma \beta E_0 \right)[/math]: [math]\delta(x') e^{- \imath \frac{E_0}{\hbar c} t'} \longrightarrow \delta\left( \gamma \left( x - \beta ct \right) \right) e^{- \imath \frac{\gamma E_0}{\hbar c} \left( ct - \beta x \right) } = \frac{1}{\gamma} \delta\left( x - \beta ct \right) e^{- \imath \frac{\gamma E_0}{\hbar c} \left( ct - \beta x \right) }[/math] could effect a relativistic transformation, of particle wave-functions ?? Vaguely, I understand, that the KG equation does not preserve probability; and, indeed, I observe, that a wave-function normalized in one reference frame, would be "all mixed up" in another relativistic reference frame, so that normalization, i.e. probability, would not be preserved. Vaguely, therefore, I intuit, that "all of that probability non-conservation originates from the same (mathematical) source". Vaguely, a relativistic observer "integrates through a longer diagonal slice" of the "wave-function world 'swath'" through a space-time diagram. Their "slice" is longer by the factor [math]\sim \sqrt{1 + \beta^2}[/math]. In crude combination, with the Lorentz-contraction factor, in front of the delta-function; and squared, to cancel out phase factors, in favor of potentially observable probabilities, [math]P' \sim \left(1 - \beta^4 \right) P = \left(1 - \beta^4 \right)[/math] If so, then (highly) relativistic observers perceive particles to be only "partially present" ??

Is it not true, that predicted stellar lifetimes, are based upon human computer simulations ?? Surely, no human has ever "sat and watched the star grass grow" for millions to billions of years, and "stopped their stop-watch" when the star exits MS ??

I understand, that multiple particles are only "entangled", when they form a single "system", with a single quantum state, which must be a super-position, of several "pure" product states. Now, that is not to say, that super-positions -- specifically factorable super-positions -- must represent entanglements. For counter example, consider two separate electrons, each in the ground state, of two separate hydrogen atoms. Each of their individual states, can be written in any basis, e.g. position basis [math]\psi_0(x) = \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ...[/math]. And, if so, then their composite state would involve cross terms: [math]\psi_{0,1} \otimes \psi_{0,2} = \left( \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ... \right)_1 \otimes \left( \psi_0(x_1) \delta(x-x_1) + \psi_0(x_2) \delta(x-x_2) + ... \right)_2[/math] However, upon a position measurement, the location "chosen" by one particle, would in no way constrain or limit, the position "chosen" by the other particle, i.e. "each could appear 'here' or 'there', without care, as to what the other was doing". Such "care free attitude" defines non-entangled states. Intuitively, the particles are non-entangled, b/c they behave, upon measurement, as if they "never noticed each other". Conversely, entangled particles must be represented, by non-factorable super-positions, of pure product states, embodying some "mutually exclusive behaviors", i.e. some "either-or" quality, according to which the "behavior" of one particle, upon measurement, necessarily limits & constrains, the "behavior" of their partner particle, as if "there is only so much, of some limited quantity, to be shared between them". Er go, a system state such as [math]|0\rangle_1 |1\rangle_2 + |1\rangle_1 |0\rangle_2[/math], embodying the possible "behaviors" of "either the first particle is in the ground state, and the second particle is in an excited state; or vice versa". Such entangled states are due to interactions, which impose the familiar Conservation Laws, for Conserved Quantities, e.g. "conservation of energy". I.e. particles are entangled, when they behave, upon measurement, as if "they negotiated with the other, on how to divvy up the earnings" of their interaction, e.g. "particle one takes 'spin up', particle two takes 'spin down'". I understand all of this, from Compendium of Quantum Physics (Springer, 2009); and from 101 Quantum Questions by Kenneth Ford.

That is a very clear & cogent explanation. I observe, that those transformed states have the expected Classically-corresponding expectation values, [math]\langle p \rangle' = \langle p \rangle + mv[/math]. Vaguely, the "core concept", of the frame-to-frame conversion, is a "point-by-point", i.e. "spacetime-event-by-spacetime-event" transfer, of [math]\psi(x,t) \longrightarrow \psi'(x',t')[/math]. Vaguely, such transfers preserve wave-function phase values (albeit modified by relative motion), which are, therefore, "mathematically real" field values, similar to EM fields. Vaguely, some sort of "Faraday tensor" could be constructed, to represent a relativistically transformable "wave-function field" ?? Vaguely, were one to consider a hypothetical particle, confined to a "box" of finite size L, in its own rest-frame, such that its wave-function had constant value, within that region, i.e. [math]\psi \approx \frac{1}{\sqrt{L}}[/math]; then, in another frame, moving relativistically swiftly, w.r.t. the particle rest frame, the relativistic observer would "slice" the wave-function differently, and so observe a different probability density, which would not necessarily integrate to one, in the relativstic coordinate frame ?? Vaguely, the relativistic observer would "slice through" the box wave-function, along the line ct = vx/c, so "slicing through" more wave-function, by an amount [math]\propto \sqrt{1 + \beta^2}[/math]. Yet, the relativistic observer would perceive, that that "slice", was concentrated, by the factor [math]\gamma^{-1}[/math]. If so, then the relativistic observer's "time slice normalization integral", integrated along the t'=0 path through spacetime, would be calculated to be [math]\propto \frac{ \sqrt{1 + \beta^2} }{\sqrt{1 - \beta^2}} > 1[/math] ??

I did not derive, from those words, any implication, that "observer induced uncertainty" is currently conceived to be the same, as "intrinsic Heisenberg uncertainty". I favor the popularization of important science, e.g. QM, given its immense potential power, e.g. for improving human life, on earth, so I'm partial to sites like SD.

Clarifying, in the CoM frame, each particle would veer to "its own right", yes ? But I may have mis-stated -- all virtual particle "pushes or pulls" could plausibly be modeled, with virtual particles, having momenta, either parallel to, or anti-parallel to, their direction of motion. All other angular scattering variabilities, could plausibly be accounted for, by effective "impact parameters", i.e. "off axis" collisions.

Could DE be "Hot Dark Matter", whilst 'DM' is "Cold Dark Matter" ? I.e. I understand computer simulations are run, with either CDM, or HDM; but what about both, in the same simulation ?? Perhaps 'DM' = CDM ("brown dwarves" ?); and 'DE' = HDM ("primordial neutrinos" ?).

Due to the relativistic "rotation", of a fast-moving observer's "sense of simultaneity"; would not a relativist observer integrate over a different "slice" of space-time, in computing the normalization of wave-functions [math]\int d^3x \Psi(x)[/math] ?? How, then, is the spatial normalization of a wave-function relativistically invariant ? Vaguely, aren't all relativistic invariants 4D, i.e. [math]\sim \int dt d^3x \Psi(t,x) \rightarrow 1[/math] ?? Vaguely, the wave-functions, deriving from the Classical Schrodinger wave-equation, seem to be relativistically variable. Given the Lorentz transformation, does not the space-time 4-volume element [math]d\tau dx \longrightarrow \gamma^2 \left( d\tau dx \left( 1+\beta^2 \right) - \beta \left( d\tau^2 + dx^2 \right) \right)[/math] Would relativistically-invariant wave-functions need to be invariant, under such "messy" transformations ?? How do conventional matter-and-energy densities [math]\rho(x)[/math] transform, relativistically ??

Before our ultra-early universe cooled, down to temperatures ~2 GeV, pair-production, of nucleons-with-anti-nucleons, would have kept nucleons & anti-nucleons in relative equilibrium ? I.e. our observed matter-over-anti-matter dominance, only "froze in", after nucleon/anti-nucleons "solidified", at T < 2 GeV ? Meanwhile, electrons/anti-electrons would still have been pair-produced, down to temperatures ~1 MeV, at which time, they would also have "solidified" & "frozen in". Er go, are there not two matter-over-anti-matter asymmetries, the first for baryons ~2 GeV; and the second for leptons ~0.001 GeV ? I'm confused, for several reasons. First, pions would have been produced, prodigiously, down to temperatures ~0.14 GeV. And, pions can inter-convert [math]n^0 \leftrightarrow p^{+}[/math]. So, even if matter baryons, and anti-matter anti-baryons, "froze out", in fixed-or-decaying numbers, back at ~2 GeV; then any surviving baryons / anti-baryons could have kept inter-converting, down to ~1/7 GeV. And, during that epoch (2 GeV --> 1/7 GeV), protons & anti-productions would possibly have interacted more strongly, than neutrons & anti-neutrons. Also, if the observed matter nucleon ratio, of 7p:1n, is attributed to a Boltzmann factor [math]e^{-\frac{\Delta E}{k_B T}} \sim \frac{1}{7}[/math], then [math]\Delta E \approx 1.3 \; MeV \implies k_B T \approx 2.5 \; MeV[/math]. But why would p,n have been able to so "freely" inter-convert, down to such "cold" temperatures, when pion production would plausibly have ceased, back at temperatures nearly a hundred times higher ?? And, if pions / anti-pions were produced, prodigiously, down to ~140 MeV; could their interactions, have generated additional nucleons / anti-nucleons ?? Vaguely, several temperature regimes would possibly have been identifiable: T > 2 GeV -- pair-production generates nucleons, anti-nucleons T > 1/7 GeV -- pair-production generates pions, anti-pions; nucleons, anti-nucleons freely inter-convert T > 1/1000 GeV -- pair-production generates electrons, anti-electrons How would the "bare" quark / anti-quark masses, ~few MeV, affect these phenomena ? The n:p number ratio is comparable to the nucleon:pion mass ratio, ~1/7, a coincidence; but meaningful, or a fluke ?

Reportedly, "entangelment" decoheres "gradually", as interactions with the environment "whittle away" at the entanglement relationship:

Reportedly, So, as molecules cool down, and "vibrate less", they "settle into" some stable cold condition, wherewithin they are "more receptive" to incident light ?? If so, then conversely, as molecules heat up, they "vibrate more", and their wave-functions "blur", "smearing out" in momentum-and-energy. Does that mean, that at higher temperatures, wave-functions are "constantly perturbed" into "blurred semi-excited states", which become only "partially present", at any single given "mono-chromatic" energy/frequency ?? Is that why high-T molecules "see less of the light" incident upon them, i.e. the molecules become "energetically blurred" and "detuned" from any given incident frequency of light ??

I understand, that in the Feynmann path integral approach, all possible paths are attributed phase factors [math]e^{\frac{1}{\hbar}\int p dx}[/math]. Relativistically, [math]p \rightarrow \infty[/math] as [math]v \rightarrow c[/math], which drives the phase factors to oscillate "wildly", and so tend to cancel out, i.e. destructively interfere. Does that, then, limit possible paths, to the future light-cone, of some starting "event" (t,x) ?? (I.e. the Classical limit corresponds to [math]c \rightarrow \infty[/math], so that the future light-cone "opens out" until the cone "lies flat", and admits paths over arbitrary distances, at arbitrary velocities, throughout the "entire future" of that even (t>0, x="anything"), in Classical QM ??)

Is not a Cooper Pair a spinless boson ? If so, then CPs have no magnetic moment, i.e. "the electrons are spinning oppositely, and so their spins, and magnetic moments, cancel" ?? Would "magnetic neutrality" help CPs propagate, without resistance ??

By comparison to position, and momentum (proportional to the derivative of position); and the parallel to time, and energy (as the time derivative); is there a "time operator" [math]\hat{t} \equiv t \times[/math], having eigenstates [math]|t_0\rangle = \delta(t-t_0)[/math] ? If particles must be normalized w.r.t. space, s.t. [math]\int d^3x \Psi(x) = 1[/math]; then why aren't wave-functions normalized, w.r.t. time, s.t. [math]\int dt \Psi = 1[/math] ? Vaguely, to tie in to relativity, would seemingly require combining time & position, into "quantum events" (t,x).