computer Posted November 12, 2022 Share Posted November 12, 2022 It is not clear what complex wave function means in Schrödinger, Pauli, Dirac equations. Is it always two-component (complex), or can it be real, or are both variants possible in different situations? For example, how to understand: -i · h/(2·π) · ∂ψ/∂t = h^{2}/(8·π^{2}·m) · div grad ψ (for simplicity in absence of potential multiplied by function). The imaginary unit “i” simply shows that quantum operator is used instead of classical derivative, or function must be divided into two components: ψ = ψ_{1} + i · ψ_{2} and then in reality there are two equations ∂ψ_{1}/∂t ~ div grad ψ_{2} ∂ψ_{2}/∂t ~ div grad ψ_{1} (~ symbol means is proportional with a constant multiplier). In this case, the question arises how this relates to de Broglie equation, because it turns out to be ∂^{2}ψ_{1}/∂t^{2} ~ div grad (div grad ψ_{1}) ∂^{2}ψ_{2}/∂t^{2} ~ div grad (div grad ψ_{2}) instead of traditional ∂^{2}ψ/∂t^{2} ~ div grad ψ or ∂^{2}ψ/∂t^{2} ~ rot rot ψ for different kinds of waves. Or is function real (should be, or can be)? If Maxwell's equation is written as one formula, there are two components, electric field and magnetic, but instead of squared nabla single nabla (curl) is used, and this is consistent as de Broglie wave. Do Pauli and Dirac equations follow the same principle as Schrödinger equation with respect to the complexity of function, or there are differences? Link to comment Share on other sites More sharing options...

Markus Hanke Posted November 13, 2022 Share Posted November 13, 2022 12 hours ago, computer said: It is not clear what complex wave function means It’s a probability density. So, in order to extract actual probability distributions from it, you need to take its square norm and integrate this up over the region you are interested in - hence the result will be a real-valued probability distribution. 12 hours ago, computer said: Is it always two-component (complex), or can it be real, or are both variants possible in different situations? The wave function itself is always complex-valued, and it can be shown that it is fundamentally necessary for that to be the case: https://physicsworld.com/a/complex-numbers-are-essential-in-quantum-theory-experiments-reveal/ That being said, the observable quantities derived from the wave function are of course always real-valued. 12 hours ago, computer said: If Maxwell's equation is written as one formula, there are two components, electric field and magnetic Actually, Maxwell’s equations in their most general form involve only one entity, the electromagnetic field: \[dF=0\] \[d\star F=4\pi \star J\] The above equations are completely independent of choice of coordinate system, and are valid in all spacetimes. You can of course associate the 2-form above with a rank-2 tensor, and then express the components of that tensor as a mix of E and B fields. This will yield the usual four equations for the components of E and B. The problem is the form of these four equations depends on the observer. 12 hours ago, computer said: Do Pauli and Dirac equations follow the same principle as Schrödinger equation with respect to the complexity of function, or there are differences? The solutions to the Dirac equation are complex-valued bispinors. The Pauli equation is the non-relativistic version of the Dirac equation; its solutions are ordinary spinors that are also complex-valued. In general, quantum mechanical wave functions are always complex-valued, irrespective of exactly which representation of the Lorentz group you are dealing with. I suspect that’s because the complex numbers have a richer “structure” than can be represented by pairs of real numbers. Link to comment Share on other sites More sharing options...

computer Posted November 13, 2022 Author Share Posted November 13, 2022 7 hours ago, Markus Hanke said: The problem is the form of these four equations depends on the observer. I meant free electromagnetic field equations that we can write like this: ∂E/∂t = 1/ε_{0} · curl H ∂H/∂t = - 1/μ_{0} · curl E F = sqrt(ε_{0}) · E - i · sqrt(μ_{0}) · H - i · ∂F/∂t = c · curl F Thank you for you answer. I am still wondered why in quantum mechanics functions have prefix "wave". If we get double time derivative, it will be proportional to "nabla in the fourth power", when in typical de Broglie equation only second power (div grad or curl curl, depending on the type of waves, longitudinal or transverse). Has it to be so? Link to comment Share on other sites More sharing options...

exchemist Posted November 13, 2022 Share Posted November 13, 2022 13 minutes ago, computer said: I meant free electromagnetic field equations that we can write like this: ∂E/∂t = 1/ε_{0} · curl H ∂H/∂t = - 1/μ_{0} · curl E F = sqrt(ε_{0}) · E - i · sqrt(μ_{0}) · H - i · ∂F/∂t = c · curl F Thank you for you answer. I am still wondered why in quantum mechanics functions have prefix "wave". If we get double time derivative, it will be proportional to "nabla in the fourth power", when in typical de Broglie equation only second power (div grad or curl curl, depending on the type of waves, longitudinal or transverse). Has it to be so? You have a point, I think. As I recall, Schrödinger's equation is not a true wave equation because it only has a single rather than double derivative with respect to time. I think I remember Peter Atkins telling us it is more properly a diffusion equation, rather than a strict wave equation. He went on say, rather enigmatically, that it might be thought more appropriate for a description of the behaviour of matter to be governed by a diffusion equation............. 2 Link to comment Share on other sites More sharing options...

Markus Hanke Posted November 14, 2022 Share Posted November 14, 2022 13 hours ago, computer said: I am still wondered why in quantum mechanics functions have prefix "wave". The terminology is mostly for historical reasons, I think, though of course (at least in the case of Schrödinger) many of the physically relevant solutions to these equations are wave-like. But, as exchemist has pointed out, technically speaking they are diffusion equations. 1 Link to comment Share on other sites More sharing options...

exchemist Posted November 14, 2022 Share Posted November 14, 2022 11 hours ago, Markus Hanke said: The terminology is mostly for historical reasons, I think, though of course (at least in the case of Schrödinger) many of the physically relevant solutions to these equations are wave-like. But, as exchemist has pointed out, technically speaking they are diffusion equations. In fact, is it not the case that the time-independent version of Schrödinger's equation is a wave equation, of the type appropriate to standing waves? I presume that would have been the first application of it (H𝚿 = E𝚿), back in the 1920s, as I think Schrödinger originally applied himself to the question of an electron in a bound state in an atom. If so, it would have been in its original form a special case kind of wave equation. When the time dependence was added to make it more general, that would have been the point at which it became a diffusion equation. Link to comment Share on other sites More sharing options...

studiot Posted November 14, 2022 Share Posted November 14, 2022 1 hour ago, exchemist said: In fact, is it not the case that the time-independent version of Schrödinger's equation is a wave equation, of the type appropriate to standing waves? I presume that would have been the first application of it (H𝚿 = E𝚿), back in the 1920s, as I think Schrödinger originally applied himself to the question of an electron in a bound state in an atom. If so, it would have been in its original form a special case kind of wave equation. When the time dependence was added to make it more general, that would have been the point at which it became a diffusion equation. Indeed so. In fact the naming of the equations is somewhat arbitrary and discipline dependent. Classically both the wave equation and the diffusion equation were derived from the more general telegraph equation by setting some of the coefficients to infinity or zero. 2 Link to comment Share on other sites More sharing options...

joigus Posted November 17, 2022 Share Posted November 17, 2022 On 11/13/2022 at 5:26 PM, exchemist said: You have a point, I think. As I recall, Schrödinger's equation is not a true wave equation because it only has a single rather than double derivative with respect to time. I think I remember Peter Atkins telling us it is more properly a diffusion equation, rather than a strict wave equation. He went on say, rather enigmatically, that it might be thought more appropriate for a description of the behaviour of matter to be governed by a diffusion equation............. You're absolutely right. +1. An image is worth a thousand words: The image is not mine, of course. It's from Wikipedia, and it represents the time evolution of a free quantum-mechanical wave packet. You can actually see how dispersive the non-relativistic regime is. The low-down is: Even empty space somehow operates as a dispersive medium for Schrödinger waves. Although you can get a similar behaviour for waves that are actually waves --same order in time and space derivatives-- by having them propagate through a dispersive material. The diffusion equation is, \[ \frac{\partial n}{\partial t}=-D\nabla^{2}n \] with D being what we call the diffusion coefficient. The Schrödinger equation, OTOH, is, \[ i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\psi \] So it's exactly mathematically equivalent to the evolution of a complex space-time valued function with complex values and purely imaginary diffusion coefficient, \[ D\rightarrow\frac{i\hbar}{2m} \] Whether something is a wave or not is, of course, a matter of definition. I would be happy enough with an equation that's linear in the field variable and admits travelling solutions being in some sense a wave. Travelling solutions meaning, \[ \psi\left(x,t\right)=u\left(\omega t-kx\right) \] If the equation is linear, we can do a Fourier analysis of the wave, and an arbitrary solution is a linear superposition of infinitely many travelling solutions like these. But the problem of whether our equation is dispersive or not is coded in the relation, \[ \omega\left(k\right) \] That's why it's called dispersion relation. Fourier components with different frequencies have different velocities. The velocity of propagation for each component of wave number k depends on that particular value of k. That's why the wave spreads out as it evolves. In the case of the Schrödinger equation, the dispersion relation is, \[ \hbar\omega=\frac{\left(\hbar k\right)^{2}}{2m}\Rightarrow \] so that, \[ \omega\left(k\right)=\frac{\hbar k^{2}}{2m} \] The phase velocity for Schrödinger waves being, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] And their group velocity being, \[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \] For light in a vacuum, there's no dispersion, or the dispersion relation is linear, so group velocity and phase velocity coincide. If we enter a medium, then we have dispersion. For relativistic (massive, matter) waves, the dispersion relation is very interesting, giving a group velocity that's subluminal, and a phase velocity that's superluminal, the product of both giving exactly c^{2}. The problem with relativistic equations is that they cannot be consistently interpreted in terms of one particle. They are multi-particle systems from the get go. 2 Link to comment Share on other sites More sharing options...

exchemist Posted November 17, 2022 Share Posted November 17, 2022 1 hour ago, joigus said: You're absolutely right. +1. An image is worth a thousand words: The image is not mine, of course. It's from Wikipedia, and it represents the time evolution of a free quantum-mechanical wave packet. You can actually see how dispersive the non-relativistic regime is. The low-down is: Even empty space somehow operates as a dispersive medium for Schrödinger waves. Although you can get a similar behaviour for waves that are actually waves --same order in time and space derivatives-- by having them propagate through a dispersive material. The diffusion equation is, ∂n∂t=−D∇2n with D being what we call the diffusion coefficient. The Schrödinger equation, OTOH, is, iℏ∂ψ∂t=−ℏ22m∇2ψ So it's exactly mathematically equivalent to the evolution of a complex space-time valued function with complex values and purely imaginary diffusion coefficient, D→iℏ2m Whether something is a wave or not is, of course, a matter of definition. I would be happy enough with an equation that's linear in the field variable and admits travelling solutions being in some sense a wave. Travelling solutions meaning, ψ(x,t)=u(ωt−kx) If the equation is linear, we can do a Fourier analysis of the wave, and an arbitrary solution is a linear superposition of infinitely many travelling solutions like these. But the problem of whether our equation is dispersive or not is coded in the relation, ω(k) That's why it's called dispersion relation. Fourier components with different frequencies have different velocities. The velocity of propagation for each component of wave number k depends on that particular value of k. That's why the wave spreads out as it evolves. In the case of the Schrödinger equation, the dispersion relation is, ℏω=(ℏk)22m⇒ so that, ω(k)=ℏk22m The phase velocity for Schrödinger waves being, vp=ωk=ℏk2m And their group velocity being, vg=dωdk=ℏkm For light in a vacuum, there's no dispersion, or the dispersion relation is linear, so group velocity and phase velocity coincide. If we enter a medium, then we have dispersion. For relativistic (massive, matter) waves, the dispersion relation is very interesting, giving a group velocity that's subluminal, and a phase velocity that's superluminal, the product of both giving exactly c^{2}. The problem with relativistic equations is that they cannot be consistently interpreted in terms of one particle. They are multi-particle systems from the get go. This is interesting. As mere chemists, at university we never delved into the significance of it being a diffusion equation. In most chemically relevant applications the time-independent version suffices, so there was perhaps little reason to do so. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted November 17, 2022 Share Posted November 17, 2022 (edited) 4 hours ago, joigus said: You can actually see how dispersive the non-relativistic regime is. The low-down is: Even empty space somehow operates as a dispersive medium for Schrödinger waves. Although you can get a similar behaviour for waves that are actually waves --same order in time and space derivatives-- by having them propagate through a dispersive material. I don't know what your definition of "wave" is (the definition above applies to equations, not their solutions), but eigenstates of the Schrödinger equation are purely oscillatory. The dispersion relation is the same as for spin waves in magnetic media. PS: Hello, Science Forums > Physics. New poster here. 🙂 Edited November 17, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

joigus Posted November 17, 2022 Share Posted November 17, 2022 28 minutes ago, Lorentz Jr said: I don't know what your definition of "wave" is, but eigenstates of the Schrödinger equation are purely oscillatory. Eigenstates of what? Position eigenstates could not be farther from being oscillatory. Waves do not have to be oscillatory. But the question actually all depends on how you define a wave, and what you wish to include in the definition. Perhaps you should actually read what I posted, as I provided a somewhat restrictive one, although by no means necessarily unique: 3 hours ago, joigus said: Whether something is a wave or not is, of course, a matter of definition. I would be happy enough with an equation that's linear in the field variable and admits travelling solutions being in some sense a wave. Travelling solutions meaning, [...] You could, of course, weaken this definition and therefore expand the concept to include non-linear phenomena, like solitons or gravitational waves. In that case, it would be just any solution to a field equation that admits particular solutions that propagate as a travelling disturbance, \[ u\left( x-vt \right) \] Being "oscillatory" --most certainly-- is not a requisite in any definition I know. You are confusing the particular case with the general one. You are confusing the pieces into which we analyse waves with what is is an analysis of --the waves themselves. In fact, no realistic interesting solution of a wave equation is "oscillatory." Most waves suffer dispersion. What you are referring to is a monochromatic wave. The classical wave equation, the equation for the vibrating string, for example, has infinitely many oscillatory Fourier components, while the overall solution doesn't have to be oscillatory in any sense --even though every one of the pieces oscillate with its particular frequency. Welcome to the forums. 27 minutes ago, exchemist said: This is interesting. As mere chemists, at university we never delved into the significance of it being a diffusion equation. In most chemically relevant applications the time-independent version suffices, so there was perhaps little reason to do so. I don't think chemists are mere. Not anymore than isomers are mere "isos." I think in Chemistry you're mainly concerned with electrons being comfortably set in stationary states. Either atomic or molecular wave functions with a given energy. This dispersion is still going on, but due to the peculiarities of the function being complex, and the "diffusion coefficient" being imaginary, the stationary situation, when it's spatially confined, allows for states going back and forth withing a small volume. Like --another Wikipedia image--, Quote A harmonic oscillator in classical mechanics (A–B) and quantum mechanics (C–H). In (A–B), a ball, attached to a spring, oscillates back and forth. (C–H) are six solutions to the Schrödinger Equation for this situation. The horizontal axis is position, the vertical axis is the real part (blue) or imaginary part (red) of the wavefunction. (C, D, E, F), but not (G, H), are stationary states, or standing waves. The standing-wave oscillation frequency, times Planck's constant, is the energy of the state. Link to comment Share on other sites More sharing options...

exchemist Posted November 17, 2022 Share Posted November 17, 2022 1 hour ago, Lorentz Jr said: I don't know what your definition of "wave" is (the definition above applies to equations, not their solutions), but eigenstates of the Schrödinger equation are purely oscillatory. The dispersion relation is the same as for spin waves in magnetic media. PS: Hello, Science Forums > Physics. New poster here. 🙂 Hmm. I had understood it is the time-independent version that is an eigenvalue equation, which corresponds to a wave equation for standing waves, so it is not surprising that its eigenstates are purely oscillatory. Whereas it is the more general, time-dependent version that has the attributes of a diffusion equation. 13 minutes ago, joigus said: Eigenstates of what? Position eigenstates could not be farther from being oscillatory. Waves do not have to be oscillatory. But the question actually all depends on how you define a wave, and what you wish to include in the definition. Perhaps you should actually read what I posted, as I provided a somewhat restrictive one, although by no means necessarily unique: You could, of course, weaken this definition and therefore expand the concept to include non-linear phenomena, like solitons or gravitational waves. In that case, it would be just any solution to a field equation that admits particular solutions that propagate as a travelling disturbance, u(x−vt) Being "oscillatory" --most certainly-- is not a requisite in any definition I know. You are confusing the particular case with the general one. You are confusing the pieces into which we analyse waves with what is is an analysis of --the waves themselves. In fact, no realistic interesting solution of a wave equation is "oscillatory." Most waves suffer dispersion. What you are referring to is a monochromatic wave. The classical wave equation, the equation for the vibrating string, for example, has infinitely many oscillatory Fourier components, while the overall solution doesn't have to be oscillatory in any sense --even though every one of the pieces oscillate with its particular frequency. Welcome to the forums. I don't think chemists are mere. Not anymore than isomers are mere "isos." I think in Chemistry you're mainly concerned with electrons being comfortably set in stationary states. Either atomic or molecular wave functions with a given energy. This dispersion is still going on, but due to the peculiarities of the function being complex, and the "diffusion coefficient" being imaginary, the stationary situation, when it's spatially confined, allows for states going back and forth withing a small volume. Like --another Wikipedia image--, Well, not quite just electrons in atoms and molecules, but also molecules in vibration or rotation - and not forgetting the angular momentum of atomic nuclei, too. But mostly stationary states, it is true, though we do concern ourselves with the perturbation of the these states by EM fields and radiation, as in spectroscopic transitions and phenomena such as refractive index and magnetic properties of substances. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted November 17, 2022 Share Posted November 17, 2022 (edited) 1 hour ago, exchemist said: I had understood it is the time-independent version that is an eigenvalue equation, which corresponds to a wave equation for standing waves, so it is not surprising that its eigenstates are purely oscillatory. Whereas it is the more general, time-dependent version that has the attributes of a diffusion equation. Energy eigenstates are states where x and t can be separated in the time-dependent equation, so the solutions can be factored into two terms, as f(x)exp(iwt). Solutions with no potential-energy function (V(x,t) = 0) are traveling plane waves, and solutions with an infinite potential well are not dispersive. So calling the Schrödinger equation a "diffusion equation" seems misleading to me, and "wave equation" seems reasonable. I think the linear time derivative in the Schrödinger equation is misleading because its coefficient is imaginary. I think it has more in common with a real-valued second derivative from a physical or dynamical point of view, even though it superficially looks like a diffusion term. Traveling waves aren't defined in terms of oscillation, but standing waves are certainly associated with it, and classical diffusion is a completely different phenomenon, with no oscillation at all. PS: How do you edit equations here? Edited November 17, 2022 by Lorentz Jr 2 Link to comment Share on other sites More sharing options...

joigus Posted November 17, 2022 Share Posted November 17, 2022 (edited) 28 minutes ago, Lorentz Jr said: Solutions with no potential-energy function (V(x,t) = 0) are traveling plane waves, This is incorrect. It's what some/many freshman or sophomore "introduction to quantum mechanics" books suggest say, and it's badly, badly wrong. Wanna know why? Edited November 17, 2022 by joigus minor correction Link to comment Share on other sites More sharing options...

exchemist Posted November 17, 2022 Share Posted November 17, 2022 (edited) 27 minutes ago, Lorentz Jr said: Energy eigenstates are states where x and t can be separated in the time-dependent equation, so the solutions can be factored into two terms, as f(x)exp(iwt). Solutions with no potential-energy function (V(x,t) = 0) are traveling plane waves, and solutions with an infinite potential well are not dispersive. So calling the Schrödinger equation a "diffusion equation" seems misleading to me, and "wave equation" seems reasonable. I think the linear time derivative in the Schrödinger equation is misleading because its coefficient is imaginary. I think it has more in common with a real-valued second derivative from a physical point of view, even though it superficially looks like a diffusion term. Traveling waves aren't defined in terms of oscillation, but standing waves are certainly associated with it, and classical diffusion is a completely different phenomenon, with no oscillation at all. PS: How do you edit equations here? Thanks for the explanation. I have to confess I have not used this stuff for about 40 years so I'm very rusty on the maths of it all, now, but the concepts intrigue me still. But I see @joigustakes issue with what you have said, so I'll sit back now and watch the debate. I don't know about editing equations on the forum. On the rare occasions I post one, I just do the best I can with regular text and the symbols library on my laptop. Edited November 17, 2022 by exchemist Link to comment Share on other sites More sharing options...

Lorentz Jr Posted November 17, 2022 Share Posted November 17, 2022 7 minutes ago, joigus said: This is incorrect. Wanna know why? Because they can't be normalized. Arguments about "waves" aside, my main complaint is that I don't think the word "diffusion" adequately describes the nature of quantum-mechanical phenomena. As I mentioned earlier, the Schrödinger equation has the same dispersion relation as spin waves in condensed matter, and I personally find that fascinating, even though most professional physicists apparently don't think about it much. I wonder what kind of spinning phenomenon might be going on in the vacuum that could implement matter and energy. 🤔 Link to comment Share on other sites More sharing options...

studiot Posted November 17, 2022 Share Posted November 17, 2022 (edited) 1 hour ago, Lorentz Jr said: and classical diffusion is a completely different phenomenon, with no oscillation at all. I'm sorry Have you come across reaction-diffusion dynamics ? https://pubs.acs.org/doi/10.1021/acsomega.1c04269 However thank you for the other worthwhile contributions to the discussion +1 Welcome to SF 1 hour ago, Lorentz Jr said: PS: How do you edit equations here? Please note you have 1 hour to edit any posting. Additionally new members have a limit of 5 posts in thier first 24 hour, which is a seriously good spam limiting precaution. If by editing equations you actually mean writing them, this is the only website (since the demise of the old all-about-circuits) that offers direct super and subscript. This is really useful. You can also use Charmap.exe (Arial) to get greek and other alphabets and quite a few maths characters such as pi, square root and arrows. The site also parses MATHML quite well enclose your code in "[math} and [/math]" tags. You can use commercial or some free online equation editors such as and copy/paste them into SF. https://latex.codecogs.com/eqneditor/editor.php or http://www.sciweavers.org/free-online-latex-equation-editor The equation editor in MS Word does not work here. Edited November 17, 2022 by studiot 1 Link to comment Share on other sites More sharing options...

joigus Posted November 17, 2022 Share Posted November 17, 2022 (edited) 1 hour ago, Lorentz Jr said: Because they can't be normalized. No. That's one good reason, and a pretty important one, but not the only reason. Any initial-condition wave function of any shape you like --not necessarily a function for which Ax+By+Cz=K (plane) is a surface of constant phase, and let it propagate freely. Eventually, it will get close to a plane wave if you leave it alone, but it never reaches that profile. It's curved and contorted for a long, long while, ever so slightly less so as time goes by, but never totally plane. It takes infinite time to do so, and then the multiplicative constant must become zero. "Plane" is what they tend to be, given enough time, but not what they are. Plane waves are extreme simplifications. Their localisation probabilities produce an infinity, so they're not the actual representation of a physical state. They're toy models. Plane waves are, eg, what the amplitude looks like in some region when you prepare the state having it go through infinitely many collimating screens, and then let it "relax" until it reaches this situation in some region of interest. OTOH, there has been extensive study of states which propagate in one direction, but package orbital angular momentum in the directions perpendicular to the propagation direction, so they're not plane waves. Look up for Bessel and Airy packets. They're very interesting, and quite a surprise when you're used to this simplifying idea that free waves are plane waves. Many people say it, but it's very old, sloppy, non-rigorous QM. We understand it better now. Another more realistic approach to a free Schrödinger wave is a Gaussian wave packet. Another one is the wave function of a particle coming out of a slit. It's never plane, although once it's got out of the slit, it's totally free. So V=0. But even more simply. Take the free Schrödinger equation: \[ i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\psi \] Now suppose you know, for some reason, that the momentum is in the z-direction. So you can do the separation \( \psi\left(x,y,z,t\right)=e^{-iEt/\hbar}e^{ip_{z}z/\hbar}\varphi\left(x,y\right) \). Now plug it into the time-independent Schrödinger equation: \[ -\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)\psi=\frac{p_{z}^{2}}{2m}\psi \] So your Schrödinger equation splits into, \[ \frac{\hbar}{i}\frac{\partial}{\partial z}\psi=p_{z}\psi \] and, \[ \left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)\varphi=0 \] The second one is the Laplace equation, so any harmonic function in the variables perpendicular to the selected momentum will do as a perfectly valid --and actually much more realistic-- solution to the Schrödinger equation. This is why people have been studying for some time now these very interesting states with orbital angular momentum packaged in them that I like to call --privately-- fusilli or tagliatelle electrons. They are free particles, and they are not plane waves. Edited November 17, 2022 by joigus Addition 2 Link to comment Share on other sites More sharing options...

studiot Posted November 17, 2022 Share Posted November 17, 2022 12 minutes ago, joigus said: I like to call --privately-- fusilli or tagliatelle electrons. They are free particles, and they are not plane waves. Now you are making me hungry. +1 Link to comment Share on other sites More sharing options...

joigus Posted November 17, 2022 Share Posted November 17, 2022 9 minutes ago, studiot said: Now you are making me hungry. +1 I love pasta. Link to comment Share on other sites More sharing options...

Lorentz Jr Posted November 19, 2022 Share Posted November 19, 2022 (edited) On 11/17/2022 at 4:41 PM, studiot said: Quote classical diffusion is a completely different phenomenon, with no oscillation at all. Have you come across reaction-diffusion dynamics ? Sorry, I meant heat diffusion. Maybe there should be a third term for "oscillatory equations". Quote However thank you for the other worthwhile contributions to the discussion +1 Welcome to SF Thanks studiot. I'll try not to post too much nonsense. 🤔😶😏 I hope I don't wear out my welcome here with too much speculation and criticism of mainstream theory. I've been reading books by Peter Woit, Sabine Hossenfelder, Lee Smolin, and Jim Baggott, that are critical of modern physics, and I have some similar ideas of my own. I'll try not to be too argumentative. I can also start a "this is me" thread in the lounge if that would help. Thanks for the editing tips too. I'll look into them as needed. +1 😋 Edited November 19, 2022 by Lorentz Jr Link to comment Share on other sites More sharing options...

studiot Posted November 19, 2022 Share Posted November 19, 2022 2 hours ago, Lorentz Jr said: Sorry, I meant heat diffusion. Maybe there should be a third term for "oscillatory equations". Thanks studiot. I'll try not to post too much nonsense. 🤔😶😏 I hope I don't wear out my welcome here with too much speculation and criticism of mainstream theory. I've been reading books by Peter Woit, Sabine Hossenfelder, Lee Smolin, and Jim Baggott, that are critical of modern physics, and I have some similar ideas of my own. I'll try not to be too argumentative. I can also start a "this is me" thread in the lounge if that would help. Thanks for the editing tips too. I'll look into them as needed. +1 😋 You are very welcome here and as I sais before offer some penetrating questions. Many folks, not excluding many famous Physicists heavily criticise the way modern Physics seems to be working out. For myself I have seen too many bright ideas come and go and be replaced by others, although I am always open to think about new ideas. I don't know much about the authors you mention, I like Chad Orzel and Frank Wilczek, though Frank has got rather whimiscal of late. Another approach is counterfactuals, I posted this thread about them some months ago. Link to comment Share on other sites More sharing options...

computer Posted November 21, 2022 Author Share Posted November 21, 2022 On 11/17/2022 at 6:24 PM, joigus said: The image is not mine, of course. It's from Wikipedia, and it represents the time evolution of a free quantum-mechanical wave packet. 1-dimensional and even 2-dimensional models are deceptive. If you try to find out explicit 3-dimensional description of some wave process, infinite in space, not inside of conditional "pit with vertical walls", it will not be easy. With finite integrals of energy and charge density. On 11/17/2022 at 6:24 PM, joigus said: The phase velocity for Schrödinger waves being, It seems Schrödinger equation is not useful at all in this case (directional motion of compact soliton). Only for some spheric waves in 3-dimensional space. And I do not understand how it can be adapted to describe "free particle". Statistical and diffusive by its nature Schrödinger equation works only in presence of attraction centres (atomic nuclei). Otherways electron cloud will expand to the sizes of whole Universe. Theoretically this is "correct" also, but unusable practically, especially in computer simulation. Link to comment Share on other sites More sharing options...

joigus Posted November 21, 2022 Share Posted November 21, 2022 4 hours ago, computer said: 1-dimensional and even 2-dimensional models are deceptive. If you try to find out explicit 3-dimensional description of some wave process, infinite in space, not inside of conditional "pit with vertical walls", it will not be easy. With finite integrals of energy and charge density. It seems Schrödinger equation is not useful at all in this case (directional motion of compact soliton). Only for some spheric waves in 3-dimensional space. And I do not understand how it can be adapted to describe "free particle". Statistical and diffusive by its nature Schrödinger equation works only in presence of attraction centres (atomic nuclei). Otherways electron cloud will expand to the sizes of whole Universe. Theoretically this is "correct" also, but unusable practically, especially in computer simulation. The non-relativistic Schrödinger equation in thre dimensions is exactly solvable. It's not deceptive at all what happens in one direction, as it's completely separable. The propagator can be obtained exactly and shows equal dispersion --in empty space-- in every direction, as couldn't be otherwise, because it's completely symmetric under rotations. Because the Schrödinger equation is linear, it has no soliton solutions. You are right though in that other very different things are very different things than the thing I was talking about. 1 Link to comment Share on other sites More sharing options...

joigus Posted November 22, 2022 Share Posted November 22, 2022 5 hours ago, joigus said: The non-relativistic Schrödinger equation Sorry: I meant "The free non-relativistic Schrödiger equation," of course. Link to comment Share on other sites More sharing options...

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