Widdekind Posted February 1, 2012 Share Posted February 1, 2012 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). Link to comment Share on other sites More sharing options...

timo Posted February 2, 2012 Share Posted February 2, 2012 There is no time operator in mainstream quantum mechanics. Link to comment Share on other sites More sharing options...

ajb Posted February 2, 2012 Share Posted February 2, 2012 Time is not really a property of a particle and so one would not expect there to be a "time operator". There is of course the time evolution operator [math]U = e^{-\frac{i H t}{\hbar}}[/math]. Note this is not Hermitian, but rather unitary. Here [math]H[/math] is the Hamiltonian of the system. Assuming the Hamiltonian to be independent in time, (formal) solutions to the Schrödinger equation are [math]|\psi(t) \rangle = e^{-\frac{i H t}{\hbar}} |\psi(0) \rangle[/math]. Link to comment Share on other sites More sharing options...

Widdekind Posted February 2, 2012 Author Share Posted February 2, 2012 (edited) 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 ?? Edited February 2, 2012 by Widdekind Link to comment Share on other sites More sharing options...

ajb Posted February 3, 2012 Share Posted February 3, 2012 The Schrödinger wave equation is not invariant under the Lorentz transformations. In fact the solutions to the Schrödinger wave equation (and the equation itself) are not invariant under the Galilean group. You pick up a phase factor. It is not hard to see that under the Galilean transformations (in 1d) [math]x' = x + vt[/math] [math]p' = p + mv[/math] [math]t' = t [/math] that (free particle) states transform as [math]\psi'(x',t) = \psi(x+vt, t) e^{\frac{i}{\hbar}(px - Et)}[/math]. The phase factor is generally considered unphysical, but it may be important somewhere. This reminds me of the Aharonov–Bohm effect and the phase change is gauge transformation. So there maybe some physics in that phase, but I am not sure. Now, because all we pick up is a phase the probability density is invariant. This is really the physical thing, so we do maintain Galilean invariance at that level. The group that is important here is the Schrödinger group. The corresponding Lie algebra is the Galilean Lie algebra with a central extension [math]m[/math] which has the interpretation as the non-relativistic mass. The Schrödinger group looks very similar to the Galilean conformal group. This is probably why it is considered important, conformal field theories have a very special place in QFT. I am not familiar with the details of the Schrödinger group, other than it is regarded as the non-relativistic version of the conformal symmetry. You can find out details in [1,3]. (I too should read these a bit closer.) The earliest reference I am aware of on the topic of non-relativistic conformal symmetries and the Schrödinger group is [2]. References [1] Arjun Bagchi and Rajesh Gopakumar. Galilean Conformal Algebras and AdS/CFT. JHEP 0907:037,2009. Also available as arXiv:0902.1385v3 [hep-th]. [2]C. R. Hagen. Scale and conformal transformations in Galilean-covariant field theory. Phys. Rev. D 5, 377 (1972). [3] Yusuke Nishida and Dam T. Son. Nonrelativistic conformal field theories. Phys.Rev.D76:086004,2007. Also available as arXiv:0706.3746v2 [hep-th]. 1 Link to comment Share on other sites More sharing options...

Widdekind Posted February 4, 2012 Author Share Posted February 4, 2012 (edited) It is not hard to see that under the Galilean transformations (in 1d) [math]x' = x + vt[/math] [math]p' = p + mv[/math] [math]t' = t [/math] that (free particle) states transform as [math]\psi'(x',t) = \psi(x+vt, t) e^{\frac{i}{\hbar}(px - Et)}[/math]. The phase factor is generally considered unphysical, but it may be important somewhere. This reminds me of the Aharonov–Bohm effect and the phase change is gauge transformation. So there maybe some physics in that phase, but I am not sure. Now, because all we pick up is a phase the probability density is invariant. This is really the physical thing, so we do maintain Galilean invariance at that level. 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] ?? Edited February 4, 2012 by Widdekind Link to comment Share on other sites More sharing options...

Widdekind Posted February 4, 2012 Author Share Posted February 4, 2012 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" ?? Link to comment Share on other sites More sharing options...

ajb Posted February 4, 2012 Share Posted February 4, 2012 The Klein-Gordon equation is not really a single particle equation. You do not interpret the solutions as wave-functions, you need quantum field theory to make sense of this. Link to comment Share on other sites More sharing options...

Widdekind Posted February 4, 2012 Author Share Posted February 4, 2012 Is not the KG equation, generated by the same ansatz as the SWE, but applied to the full relativistic energy equation (E^{2} = p^{2} + m^{2}), instead of its classical approximation (E ~ p^{2}/2m) ? Why would one accept the ansatz, in the classical case; but then reject the same, in its natural relativistic extrapolation ? Link to comment Share on other sites More sharing options...

ajb Posted February 4, 2012 Share Posted February 4, 2012 (edited) Is not the KG equation, generated by the same ansatz as the SWE, but applied to the full relativistic energy equation (E^{2} = p^{2} + m^{2}), instead of its classical approximation (E ~ p^{2}/2m) ? Why would one accept the ansatz, in the classical case; but then reject the same, in its natural relativistic extrapolation ? I think you have already pointed to why the KG equation cannot really be an equation for single particle states. You cannot interpret the solutions as quantum wave functions of single particles. This is because you cannot build a probability density from the solutions. The way round this is to interpret the solutions as fields rather than quantum states. This is quite well explained in Ryder's book. In essence we have two problems: negative energy and negative probability if we take a naive interpretation of the solutions. Edited February 4, 2012 by ajb Link to comment Share on other sites More sharing options...

Widdekind Posted February 4, 2012 Author Share Posted February 4, 2012 (edited) I understand, regarding the KGE, that: Unlike the Schrödinger equation, there are two values of ω for each k, one positive and one negative. Only by separating out the positive and negative frequency parts does the equation describe a relativistic wave-function [where] the particle propagates both forwards and backwards in time 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" in which an interference, between positive and negative energy states, produces what appears to be a fluctuation (at the speed of light) of the position of an electron, around the median, with a circular frequency of [math]2 m_e c^2 / \hbar [/math], or approximately 1.6×10^{21} Hz. 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: The position operator, at time t, consists of an initial position; a motion proportional to time; and an unexpected oscillation term, with an amplitude equal to the Compton wavelength. That oscillation term is the so-called "Zitterbewegung" [which] vanishes on taking expectation values, for wave-packets, that are made up entirely of positive- (or entirely of negative-) energy waves 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 ?? Edited February 5, 2012 by Widdekind Link to comment Share on other sites More sharing options...

DrRocket Posted February 5, 2012 Share Posted February 5, 2012 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" What part of "no" don't you understand ? -- Lorrie Morgan Link to comment Share on other sites More sharing options...

ajb Posted February 5, 2012 Share Posted February 5, 2012 For complex fields you reinterpret [math]J_{\mu}= i \phi^{\dag}\partial_{\mu}\phi - i \partial \phi^{\dag} \phi[/math] as the current associated with electric charge, rather than a probability density. Historically, the idea was that [math]J^{0}[/math] was the probability density density of the wave function. However the negative probabilities spoil this idea. Pauli and Weisskopf reinterpreted all this in terms of electric currents. The negative frequencies also need careful handling. Just as in the Dirac equation, these are reinterpreted as antiparticles. The important thing here is that we have to abandon the idea that the KG equation describes single particle states. The "states" appear to describe a mix of charged matter and antimatter. This leads to the idea of fields. 1 Link to comment Share on other sites More sharing options...

Widdekind Posted February 6, 2012 Author Share Posted February 6, 2012 (edited) 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. m_{e}c^{2} = 511KeV >> E_{H} = 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). Edited February 6, 2012 by Widdekind Link to comment Share on other sites More sharing options...

ajb Posted February 6, 2012 Share Posted February 6, 2012 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]. That is not the correct object to think about in this context. The complex KE equation has a U(1) symmetry [math]\phi \rightarrow e^{i \theta}\phi[/math] [math] \phi^{\dag} \rightarrow e^{-i\theta} \phi^{\dag}[/math] and it is the current associated with this that needs interpreting. Link to comment Share on other sites More sharing options...

Widdekind Posted February 6, 2012 Author Share Posted February 6, 2012 (edited) 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] Edited February 7, 2012 by Widdekind Link to comment Share on other sites More sharing options...

ajb Posted February 6, 2012 Share Posted February 6, 2012 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] What symmetry would that correspond to? Link to comment Share on other sites More sharing options...

Widdekind Posted February 7, 2012 Author Share Posted February 7, 2012 What symmetry would that correspond to? Wouldn't that definition of [math]j^{\mu}[/math] be invariant, under [math]\phi \rightarrow e^{\imath \theta} \phi[/math] ? I don't understand, how the conventional current, is invariant, under that transformation. Link to comment Share on other sites More sharing options...

ajb Posted February 7, 2012 Share Posted February 7, 2012 It maybe invariant, but it is not the current associated with that symmetry. Look up Noether's theorem. Link to comment Share on other sites More sharing options...

Widdekind Posted February 7, 2012 Author Share Posted February 7, 2012 (edited) If the Einstein "mom-energy" relation [math]p_{\mu} p^{\mu} = m^2[/math] implicitly involves the Minkowski metric matrix, [math]m^2 = \vec{p}^T \cdot \tilde{\eta} \cdot \vec{p}[/math]; and if, in the presence of matter-and-energy within the fabric of space-time, that Minkowski metric matrix "evolves" into the GR metric matrix, i.e. [math]\tilde{\eta} \rightarrow \tilde{g}[/math]; then could application of the QM operator ansatz, to [math]\vec{p}^T \cdot \tilde{g} \cdot \vec{p} = m^2[/math] represent relativistic QM, on a non-matter-free space-time ? If so, would the "mom-energy" operators act on the metric matrix elements; or would they "slide through" to act only on quantum particle WFs ? edit: According to Hyman 1989; and acknowledging one "minus sign" for reversing that author's choice of metric, from (-+++) to (+---); and acknowledging another "minus sign", arising from eq.11, when setting the source at the coordinate origin, so that [math]r^a \rightarrow -x^a[/math]; then the Schwarzschild metric, in cartesian-like coordinates, is: [math]g_{00} = 1 - \kappa[/math] [math]g_{0a} = \kappa (x^a/r)[/math] [math]g_{ab} = -\delta_{ab} - \kappa (x^a/r)(x^b/r)[/math] where [math]\kappa \equiv \frac{2GM}{c^2r} = \frac{R_S}{r}[/math]. The presence of such "cross terms" would make the KGE, in a Schwarschild space-time, much "messier". Edited February 8, 2012 by Widdekind Link to comment Share on other sites More sharing options...

juanrga Posted April 7, 2012 Share Posted April 7, 2012 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). Time is an evolution parameter, not an observable. Therefore there is not time operator in QM. Yes, in RQM one would try to treat time and space in the same footing, but since time is related to causality, it cannot be given by an operator. Precisely quantum field theory takes the contrary way, position is downgraded from observable to parameter and then both space and time are treated in the same footing. Link to comment Share on other sites More sharing options...

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