Anonymous at Wed, 16 Oct 2024 21:23:45 UTC No. 16435235

>>16435152
>Why?
You can simply postulate it or derive it from other assumptions. One derivation method first defines something called a time-evolution operator which is postulated to have certain properties such as unitarity and the so-called composition property. Schwinger goes through the detailed derivation in his book. A more condense derivation is given in Sakurai's book.

Fundamentally, the time-evolution operator can be seen as a Lie exponent of the Hamiltonian operator. This is a consequence of Lorentzian group theory. The Hamiltonian acts as the generator of time translations.
>Does this imply that the probability of an object being found at various locations spreads out evenly over time, like heat does in matter?
Yes. See gif.
>That doesn't seem right, as everything doesn't seem to be approaching equal likelihood to be found anywhere in the universe.
That would be the case for a free particle in an infinite universe. On a finite region, free particles have the fundamental resonance mode as their ground state, so the probability has a peak in the middle. Our unbounded universe has potentials, so the ground states aren't trivial. For example, the ground state of a harmonic oscillator is a Gaussian. It has nice properties and we call it a coherent state. The process of measurement localizes the wavefunction, causing it to decohere. Once let go free, it goes back to the coherent state. This is what's shown in the gif.

Anonymous at Wed, 16 Oct 2024 21:35:43 UTC No. 16435263

>>16435152
It only looks like a heat equation superficially.

In practice, the i makes a significant difference in the actual behavior of the equation.

Anonymous at Wed, 16 Oct 2024 21:44:54 UTC No. 16435268

>>16435235
Thanks for the insight. I suppose I did not consider the influence of potential barriers.

Anonymous at Wed, 16 Oct 2024 21:54:56 UTC No. 16435281

>>16435152
Wave equations propagate
Heat equations dissipate toward equilibrium

Anonymous at Thu, 17 Oct 2024 05:41:55 UTC No. 16435814

>>16435152
shits vibrating yo

Anonymous at Thu, 17 Oct 2024 07:05:18 UTC No. 16435893

>>16435152
1/3
>Why does the Schrodinger equation take the form of the heat equation?
https://www.ams.org/notices/200902/rtx090200212p.pdf
>Starting from wave optics as a model, he wrote down a differential equation for a mechanical particle, but the equation made no sense. The equation looked like the equation of conduction of heat in a continuous medium. Heat conduction has no visible relevance to particle mechanics. Schrödinger’s idea seemed to be going nowhere. But then came the surprise. Schrödinger put the square root of minus one into the equation, and suddenly it made sense. Suddenly it became a wave equation instead of a heat conduction equation. And Schrödinger found to his delight that the equation has solutions corresponding to the quantized orbits in the Bohr model of the atom.
>This discovery came as a complete surprise, to Schrödinger as well as to everybody else. According to Schrödinger, his fourteen-year-old girlfriend Itha Junger said to him at the time, “Hey, you never even thought when you began that so much sensible stuff would come out of it.”

Anonymous at Thu, 17 Oct 2024 07:06:19 UTC No. 16435895

>>16435152
2/3
>Does this imply that the probability of an object being found at various locations spreads out evenly over time, like heat does in matter? That doesn't seem right, as everything doesn't seem to be approaching equal likelihood to be found anywhere in the universe.
1. Probability distributions can be "squeezed" provided they still obey the uncertainty principle. For macroscopic objects and units, they can be very localized (at macroscopic precision) and remain that way over extended periods of time.
2. Relative phase, interactions and conservation laws all constrain the support of probability distributions. The degree of spreading depends on the system and various details of the observables of interest: whether they are compatible, their dynamicity, compactness, etc. Particle position spreads more when there is an obstruction blocking trajectories that would otherwise interfere with each other. Measuring the spin of a particle along some axis over and over again without interposing measurements of incompatible observables reveals no spreading, the initially observed value is left untouched. The solution to the Mott problem isn't a spherically symmetric distribution of random bubbles but a spherically symmetric distribution of (almost) linear tracks of bubbles. The electron in a hydrogen atom is bound until ionization.
3. The universe has observers (or at least an observer, you, if you're a solipsist) that are making lots of observations. Observation, independently of the Schrödinger equation, irreversibly selects an element of the corresponding observable's spectrum weighted by spectral measure. So the probabilities are necessarily conditional and evolve only between observations. This allows manufacturing of "clean" initial states for experiments, the quantum Zeno effect, etc.

Anonymous at Thu, 17 Oct 2024 07:07:31 UTC No. 16435896

>>16435152
3/3
>And would there then be a sort of "probability diffusivity coefficient"
Yes, the Schrödinger equation is just a diffusion equation where the main parameter is imaginary. The exact form will depend on the Hamiltonian and initial state, of course.

Anonymous at Thu, 17 Oct 2024 09:48:29 UTC No. 16436017

>>16435893
what a lucky bastard

Anonymous at Thu, 17 Oct 2024 12:51:43 UTC No. 16436179

>>16435152
It's only like a heat equation if you get rid of the imaginary i multiplying the time derivative. Theoretical physicists call this switching to "Euclidean time." Ordinarily an energy eigenstate fluctuates in time with a complex exponential e^{iEt}, but in Euclidean time it exponentially decays like e^{-Et}. This is why it is like a heat equation. Localizing an object involves higher energy levels, which decay in Euclidean time, and all that is left for late times is the ground state with lowest energy, which is very spread out.

Anonymous at Thu, 17 Oct 2024 12:58:25 UTC No. 16436186

>>16436179
>which decay
They can only decay if they’re coupled to something (eg the EM field). If you have a particle on its own in a state that is a linear combination of excited states, it will remain at these states for all time without decaying.

e^{iEt} is clearly not e^{-Et}. The exponential bit in decay comes from e^{-E/T}, where T is temperature.

Anonymous at Thu, 17 Oct 2024 13:24:06 UTC No. 16436238

>>16436186
That's why I said it is "Euclidean time" and not ordinary time. If you take ordinary time in the Schrodinger equation it is *not* like the heat equation, because of the i factor. Considering Euclidean time is an ordinary thing quantum field theorists do all the time to make the math nicer, it doesn't mean there is physical dissipation.

Anonymous at Thu, 17 Oct 2024 13:48:31 UTC No. 16436287

>>16436238
>to make the math nicer
That’s a very handwavy explanation. There are two different techniques used for two different purposes. Wick rotations are used when taking integrals over relativistic measures. They only work if at least two quarter-planes in complex space are absent of singularities. It’s basically a contour integration technique. The second technique is the one used by stat mech people because e^{E/beta} behaves like e^{iEt} if we substitute /beta = it and impose periodic boundary conditions. This allows one to use QFT techniques when working out thermodynamic properties of fields.

I say all this because there is no mapping between a Euclidean and a Minkowskian space that preserves topological structure. This is a basic result in pseudo-Riemannian geometry that has to do with metric signature. For example, SO(4) is compact, while SO(1,3) isn’t. So you have to be very careful when doing all those tricks.

Anonymous at Thu, 17 Oct 2024 14:03:53 UTC No. 16436314

>>16436287
You don't have to be very careful about doing it in this context. You are literally just changing the time evolution operator from e^{iHt} to e^{-H\tau}. Under Euclidean time evolution energy eigenstates exponentially decay, and my post was explaining how that is intuitively consistent with the Schrodinger equation acting like a heat equation if we just replace t with i\tau, with \tau real.

By the way, if you want to avoid the subtleties of Wick rotation you can literally just derive the Euclidean path integral directly from the Euclidean time translation operator e^{-H\tau} in a manner entirely analogous to the usual derivation from e^{iHt}. It is far clearer than first considering the path integral and then analytically continuing, which for some reason is usually what is presented.

Anonymous at Thu, 17 Oct 2024 14:08:06 UTC No. 16436321

>>16436314
>You don't have to be careful.
>You are changing the operator from a bounded to an unbounded one
Physicists really are something, huh?

Anonymous at Thu, 17 Oct 2024 14:11:29 UTC No. 16436326

>>16436321
In ordinary quantum mechanics we have countable energy eigenstates with the energy bounded from below. e^{-H\tau} is a bounded operator.

Anonymous at Thu, 17 Oct 2024 14:15:45 UTC No. 16436331

>>16436326
It's not. Does
[eqn]lim_{t\rightarrow -\infty} e^{-Ht}[/eqn]
exist?

Anonymous at Thu, 17 Oct 2024 14:21:43 UTC No. 16436340

>>16436331
Ok, whatever. If that's your point fine.

Anonymous at Thu, 17 Oct 2024 14:22:25 UTC No. 16436341

>>16436321
>noooooooo you have to rigorously prove every step!!!!
You're not going to make it far in physics

Anonymous at Thu, 17 Oct 2024 14:25:40 UTC No. 16436346

>>16436341
physicists be like
>if there is a mapping between the constant function and a parabola that makes our calculations easier, then it's ok
>wtf why does my shit diverge? This is so weird! Isn't nature wacky, guys?

Anonymous at Thu, 17 Oct 2024 14:28:18 UTC No. 16436355

>>16436346
Yeah they first do something that works. Only if in some special cases it doesn't work, do they bother to figure out the mathematical niceties. A much healthier approach to difficult physical problems then the culture of mathematics.

Anonymous at Thu, 17 Oct 2024 14:28:41 UTC No. 16436356

>>16436346
It just werks. Nature doesn't care about mathematical rigor and math autists can't get their heads around this

Anonymous at Thu, 17 Oct 2024 14:30:26 UTC No. 16436363

>>16436356
>nature doesn't care
I don't see infinities in nature. Yet I do see infinities in physics. Ever tried calculating the partition function of the hydrogen atom?

Anonymous at Thu, 17 Oct 2024 14:33:11 UTC No. 16436368

>>16436363
Did you calculate it taking into account that the electron can ionize and go into a scattering state?

Anonymous at Thu, 17 Oct 2024 14:33:52 UTC No. 16436369

>>16435893
>Anatomy of a Physicist

Anonymous at Thu, 17 Oct 2024 14:35:19 UTC No. 16436372

>>16436363
It doesn't matter if you get infinities in intermediate calculations. Only the physically observed quantities need to be finite.

Anonymous at Thu, 17 Oct 2024 14:38:08 UTC No. 16436377

>>16436368
>Did you calculate it taking into account that the electron can ionize and go into a scattering state?
It can, but why should it? And even if it does, you now have to calculate the partition function of an unbounded state which is even worse.
>>16436372
Riddle me this. How do you do algebraic manipulations with quantities that don't obey the rules of algebra? What's infinity - infinity?

Anonymous at Thu, 17 Oct 2024 14:40:09 UTC No. 16436378

>>16436377
>How do you do algebraic manipulations with quantities that don't obey the rules of algebra?
You use physical intuition

Anonymous at Thu, 17 Oct 2024 14:44:06 UTC No. 16436389

>>16436378
kek. Please use physical intuition to explain the following:
[eqn]lim_{x\rightarrow\infty} x = \infty[/eqn]
[eqn]lim_{x\rightarrow\infty} x^2 = \infty[/eqn]
[eqn]lim_{x\rightarrow\infty} (x - x) = 0[/eqn]
[eqn]lim_{x\rightarrow\infty} (x^2 - x) = \infty[/eqn]
[eqn]lim_{x\rightarrow\infty} (x - x^2) = -\infty[/eqn]
I've got two functions that evaluate to the same "quantitiy" (infinity). But if I subtract infinity from infinity I don't get 0 like the rules of algebra demand. Not only that, but if I get two entirely different answers depending on the "pre-image" of that infinity. How does le physical intuition get around this?

Anonymous at Thu, 17 Oct 2024 14:47:29 UTC No. 16436397

>>16436389
>But if I subtract infinity from infinity I don't get 0 like the rules of algebra demand.
That's not even a rule of algebra, you should brush up on your math.

Anonymous at Thu, 17 Oct 2024 14:48:16 UTC No. 16436399

>>16436397
Look up what a group inverse is.

Anonymous at Thu, 17 Oct 2024 14:50:54 UTC No. 16436401

>>16436399
??? Are you trolling?

Anonymous at Thu, 17 Oct 2024 14:54:18 UTC No. 16436405

>>16436401
The real line is an algebraic field ie it has well-defined operations of addition, subtraction, multiplication and division. When you try to “use physical intuition” and adjoin a new element called “infinity” to that set, you quickly find out that it doesn’t obey the field axioms. This is extremely basic shit if you’re not a brainlet who thinks math is le calculations. Someone came up with some dumb structure called a wheel to incorporate infinity, but it’s fucking useless.
https://en.wikipedia.org/wiki/Wheel_theory

Anonymous at Thu, 17 Oct 2024 14:56:52 UTC No. 16436408

>>16436389
>kek. Please use physical intuition to explain the following:
>trivial
>trivial
>trvial
>x(x-1)
>x(1-x)

Anonymous at Thu, 17 Oct 2024 14:58:24 UTC No. 16436412

>>16436405
Physical intuition just means using methods which have been proved to make correct experimental predictions. Making up random rules and definitions is not physical intuition.

Anonymous at Thu, 17 Oct 2024 14:58:52 UTC No. 16436413

>>16436408
>doesn’t understand the question
I’ll rephrase it for idiots. Why does the same operation result in different answers? Operations are functions, but here we have the same argument mapping to three different arguments.

Anonymous at Thu, 17 Oct 2024 15:02:47 UTC No. 16436417

>>16436412
>I have shit methodology
>but look, it works when I do some voodoo magic on it
>Wtf do you mean I can do this correctly? I don't understand consistency. I just want to publish more garbage for grants.

Anonymous at Thu, 17 Oct 2024 15:06:38 UTC No. 16436424

>>16436417
>I have shit methodology
You just don't understand how science works
>I just want to publish more garbage for grants.
The irony

Anonymous at Thu, 17 Oct 2024 15:08:25 UTC No. 16436426

>>16436424
I understand that theoretical physics uses math. If you use your tool incorrectly, you’re a retard. Imagine someone hammering a nail with a sledgehammer. When you come up to them and say “you know you can use a nail hammer, right? You won’t make a mess and it’s much easier.” they respond with “I don’t need your fancy hammers. I just use physical intuition.” Like lol. Lmao even.

Anonymous at Thu, 17 Oct 2024 15:15:10 UTC No. 16436432

>>16436426
You're free to come up with a mathematically rigorous formulation of all of physics whenever you want.

Anonymous at Thu, 17 Oct 2024 15:20:58 UTC No. 16436439

>>16436432
Thank you. Look up quotient spaces btw. That’s what you’re doing when you “remove infinites” from non-invariant quantities to obtain invariant answers. Very elegant alternative to physicists’ niggerlicious approach.

Anonymous at Thu, 17 Oct 2024 15:23:31 UTC No. 16436441

>>16436439
What a dumb fucking post

Anonymous at Thu, 17 Oct 2024 15:37:33 UTC No. 16436461

>>16436441
QFT guy deriving the Sackur-Tetrode equation be like
>the answer is infinite! What do we do?
>let’s make the volume into a constant and take it to infinity at the end
>See? We get the Sackur-Tetrode formula just fine! Now even though both the entropy and the volume are infinite according to our own assumptions, we can just subtract lnV from S to get a finite entropy difference.
>this entropy difference contains all relevant observables, so it’s totally fine
>le observations confirm my nonsense!
meanwhile, a normal person
>oh, we expect a translation-invariant answer even though we have coordinates in the equation. We obviously need to renormalize by quotioning the function space by R^3.

Anonymous at Thu, 17 Oct 2024 16:25:21 UTC No. 16436517

It is not heat equation, in real solution. Remeber that cucklex numbers are not behaving like real numbers

Anonymous at Fri, 18 Oct 2024 16:16:37 UTC No. 16438154

>>16435152
wouldn't you like to know.

Anonymous at Fri, 18 Oct 2024 17:23:32 UTC No. 16438283

>>16435152
You can get it pretty easily starting from the integral ∫ |Ψ|^2 dx = 1.
Differentiate this with respect to t.
You get some dΨ/dt terms under the integral that you can't really do anything with at this level of generality.
Pick the simplest nontrivial pde for Ψ to satisfy that is "most convenient" to substitute for dΨ/dt.
Suppose dΨ/dt = AΨ + BΨ_x + CΨ_xx where A,B,C are functions of x and t (no dependence on Ψ).
After substituting then using integration by parts, you can constrain A,B,C enough to get something resembling the SE.

Anonymous at Sat, 19 Oct 2024 05:43:05 UTC No. 16439316

>>16436331
Sure, that limit exists in finite-dimensional systems. The ground state is the limit of the quantity you wrote divided by its trace; that rigorously exists when the spectrum is bounded from below.
The limit of the operator e^{-beta H}/{tr e^{-beta H}} with beta to infinity is the ground state or degenerate ground space (as a density operator).
So when field theorists consider this operator, it's a way of expressing the vacuum. They are often confused and start with exp(i t H) and then say some nonsense, but whether they know it or not, they are just trying to get to an expression for the vacuum.

Anonymous at Sat, 19 Oct 2024 10:17:31 UTC No. 16439506

>>16439316
ah, so you do quotient spaces in disguise

Anonymous at Sat, 19 Oct 2024 10:20:45 UTC No. 16439511

>>16436413
It is not the same operation.

Anonymous at Sat, 19 Oct 2024 10:22:00 UTC No. 16439512

>>16435235
This guy will stop posting on /sci/ within a year tops