Iyfjky

Question

This question already has an answer here:

What are the primary obstacles to solve the many-body problem in quantum mechanics?

5 answers

It is common knowledge that the SchrÃƒÂ¶dinger equation can be solved exactly only for the simplest of systems - such the so-called toy models (particle in a box, etc), and the Hydrogen atoms; and not for relatively complex systems, such as the Helium atom and other multielectron systems.

I have been trying to get to the reason for this for a long time, and it seems that it has something to do with one or more of these -

Correlation effects between the electrons as in the electrons trying to occupy positions opposite to each other with the nucleus between them.

Some kind of quantum corellation due to entanglement.

Appearance of inseparable 'cross-terms' in the expression for the Hamiltonian.

As in 3 above, we can't figure out to model the fact that we don't know (from our theory) the total energy of the system (while we are making the model), and the kinetic and poten tial energies depend upon each other, so we can't actually find any of these.

Some references for the above-

https://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics)#Many_particles

However, complications can arise in the many-body problem. Since the potential energy depends on the spatial arrangement of the particles, the kinetic energy will also depend on the spatial configuration to conserve energy. The motion due to any one particle will vary due to the motion of all the other particles in the system ... For N interacting particles, i.e. particles which interact mutually and constitute a many-body situation, the potential energy function V is not simply a sum of the separate potentials (and certainly not a product, as this is dimensionally incorrect). The potential energy function can only be written as above: a function of all the spatial positions of each particle.

https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/09._The_Electronic_States_of_the_Multielectron_Atoms/9.1%3A_The_SchrÃƒÂ¶dinger_Equation_For_Multi-Electron_Atoms

Unfortunately, the Coulomb repulsion terms make it impossible to find an exact solution to the SchrÃƒÂ¶dinger equation for many-electron atoms and molecules even if there are only two electrons. The most basic approximations to the exact solutions involve writing a multi-electron wavefunction as a simple product of single-electron wavefunctions, and obtaining the energy of the atom in the state described by that wavefunction as the sum of the energies of the one-electron components.

https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions

Very often, only numerical solutions to the SchrÃƒÂ¶dinger equation can be found for a given physical system and its associated potential energy. However, there exists a subset of physical systems for which the form of the eigenfunctions and their associated energies can be found.

https://en.wikipedia.org/wiki/Many-body_problem

The many-body problem is a general name for a vast category of physical problems pertaining to the properties of microscopic systems made of a large number of interacting particles ... In such a quantum system, the repeated interactions between particles create quantum correlations, or entanglement. As a consequence, the wave function of the system is a complicated object holding a large amount of information, which usually makes exact or analytical calculations impractical or even impossible.

I have been trying to figure out these things for years, but inspite of all the available information, I still have a question to which I cannot find a answer -

What is the fundamental reason behind our inability to solve the SchrÃƒÂ¶dinger equation for multielectron atoms - Is it actually imposibble to solve it (any proof?) or are we just not good enough at maths? Supposing that we kept trying g ti solve it, can someone clarify whether a solution could be found in principle?

Notes-

I know that there are other similar problems such as the 3 body problem in Classical mechancics and it would be great if the answer touches upon them too.

The reasons I listed are not exhaustive at any rate, they are just what came off the top of my head as some of the variations of "The math is too tough." that i have seen over the years.

Possible duplicate of What are the primary obstacles to solve the many-body problem in quantum mechanics? — Sep 9 at 14:55
@StephenG That comes quite close, but the answers there mostly deal with the constraints ( computational and physical-theoretical ) that make the solution difficult. I am interested in whether there is actually some solution which could be found in principle in addition to the difficulties in finding it. Si, I think my question is different. — Sep 9 at 15:12
I have a vague memory telling me that there is a solution for $mathrmH_2^+$ molecular ion (another three-body problem). Found it: arxiv.org/abs/1508.01359. This differs from the problem here in that two of the objects are heavy. Hmmm ... I can't find evidence of a peer reviewed publication for that. — Sep 9 at 17:22
I would recommend Landau-Lifschitz book on Quantum Mechanics as they explain these things, Landau was also very knowledgeable on molecular physics for obvious reasons. — Sep 9 at 18:08
Solving an n-body problem in classical mechanics is mathematically impossible, if by "solving" you mean "find a smooth function of initial positions, velocities and simulation time which outputs the system's evolution". I don't remember the details, but I believe V.Arnold's book "Mathematical methods of classical mechanics" discusses the main ideas of this theorem. I believe that the quantum n-body problem would be unsolvable for similar reasons, although I don't have a reference. If all particles are physically different,the $hbarto 0$ limit would give a classical solution, contradiction. — 2 days ago

The_Sympathizer 2,530520 · Answer 1 · 2018-09-09 14:43:18Z

None of those are really "good" answers. I will concur. Moreover, it is really more a question that ends up touching onto meta-mathematics, than physics.

I am not really sure one can levy a specific reason, other than the very general reason that suitably complex equations cannot be solved exactly as a general principle. But I believe the explanation below, while not a solid or firm answer, is nonetheless likely to suffice.

One way you can think of it is that there are only a finite number of operations we have for expressing solutions to equations, but an infinite number of possible actual solution functions - although I don't think that quite gets at it, either because you can combine the "expressing" functions in an infinite number of ways. Perhaps a better way to say it is that when we go to solve equations "exactly" what we are talking about is to express them in terms of the composition of some limited number of "base" functions we consider "acceptable" - almost always this includes and is at least the elementary functions, which are those composed of the constant functions, the four arithmetical operations (+, -, *, /), and the exponential ($exp$) and logarithmic ($log$) functions over the complex field ($mathbbC$), and the closure of this set is taken under functional composition ($circ$). We may also include a few non-elementary "special" functions as well like the gamma function ($Gamma$) or error function ($mathrmerf$, the integral of the Gaussian).

And effectively what happens is that these do not provide us with enough "degrees of freedom", so to speak, to express all functions of suitable generality. While I'm not sure of a result that proves this for the general composition operator, a very similar result along this line is the fact that the space of all differentiable (say) functions - like those that come out as solutions to a differential equation like Schrodinger - is algebraically infinite-dimensional, that is, the cardinal of any Hamel basis is infinite, meaning that if we want to express any differentiable function as a linear combination

$$f(x) = a_0 f_0(x) + a_1 f_1(x) + cdots + a_N f_N(x)$$

of basis functions $f_j$, then we actually need a basis set with infinitely many independent functions we can pull the $f_j$ from (i.e. they cannot be expressed in terms of each other) to be able to have enough expressing power to express every differentiable function $f$ in this fashion. (Such a set is called Hamel basis.) For the quantum theory of atoms, this is even more direct, it is the infinite-dimensionality of the Hilbert space.

Likewise, I'd strongly suspect something similar is happening when we allow general composition as well - it doesn't provide us with enough power, but I do not know of a proof.

However, this doesn't really answer the question as more specifically to why that a more restricted set of functions, like the solutions of a specific kind of differential equation - and in particular its eigenvectors, cannot be expressed with a suitable composition of a "nice" set of base functions. After all, linear constant-coefficient differential equations are famously known to have a completely-expressible solution set. Many other very specific differential equations have similar sets. Although, a key point may be that the Schrodinger equation offers effectively infinite freedom through the choice of the potential function $U(mathbfr)$, and more generally its Hamiltonian operator $hatH$.

That said, with a very restricted parameter set, say atoms very specifically, it may indeed be there is a finite or easily-described infinite set of functions that can provide a suitable basis set. We just don't, as of yet, know what the "atom functions" are or how to describe them.

So, is it correct to say that we haven't done enough math to know how to write down the solution, but since we have done a lot of math, we kinda assume that the written down thing would probably be too complicated too be useful(since we practical people haven't found it yet) and so stick with the approximations instead of wasting time on trying harder to solve it than we have done before. — Sep 9 at 14:59
Note that even the $exp$ and $log$ are effectively computed via their power series and you always get results up to a prechosen precision - except for maybe a few special cases. $erf$ can again not be expressed directly via $exp$ and $log$ but it turns out that it is useful in a lot of solutions that we define it as a special function. The solutions to the n-electron Hamiltonian could as well be defined to characterize some special functions but afaik they can't be reused for much else so we don't bother. — Sep 9 at 17:05
@DevashishKaushik: no, it's simpler than that. The solutions you are talking about are simply not common enough to deserve a name, as opposed to say the exponential, the trigonometric functions, or the error function. — Sep 9 at 19:42

Ponder Stibbons 43110 · Answer 2 · 2018-09-10 00:01:46Z

The problem is not at all unique to SchrÃƒÂ¶dinger's equation, it is a common feature of differential equations. The same thing happens all over the place in classical electro-magnetics. But, also to say there is no "exact" solution is misleading. It is more true to say that the solution cannot be expressed finitely in certain notations.

There are theorems such as the Picard Lindelof theorem that ensure the existence of solutions to certain differential equations, but these theorems do not lead to finite expressions using common functions for these solutions. There are some equations that can be solved generically using polynomials. However, the simple equation $dotx=x$ is not one of them. This equation defines the exponential function (that can also be defined by an infinite sum). Effectively, when $exp(x)$ is used, it just means the solution to this equation.

More complicated differential equations require more new functions - the elliptical trigonometric functions, for example. This just keeps on going. There is no finite list of function, combinations of which will solve all differential equations. When a function defined as the solution to a differential equation is used commonly enough - we give it a name, and continue on.

This is a big area of study in formal solutions of any kind of equation:

http://mathworld.wolfram.com/Closed-FormSolution.html

aquirdturtle 1,324410 · Answer 3 · 2018-09-09 23:20:47Z

Why would you expect there to be a solution? If you just consider random complicated PDEs, most of them don't have "nice" solutions, so why would a multi-electron atom be so special? I suspect you're looking for a type of answer which may just not exist for the question at hand. The most literal answer is just the brute-force fact that it appears that nice solutions don't exist (although perhaps someday someone will find another nice solution).

If you want to get hand-wavy, you could say that the PDE is "too complicated", and point to the part which, if removed, would yield a solvable equation (the electron-electron interactions). However, there are also many "simple" equations which also don't have "nice" solutions, such as a simple linear potential $Vx=x$ (solutions being the not-nice airy functions), which is arguably much "simpler" than the harmonic oscillator or a single-atom potential. So, I would avoid the hand-wavy explanations here.

score 5 · Answer 4 · 2018-09-10 03:01:05Z

The question of whether an analytic solution exists for a given dynamical system or PDE goes under the name "integrability". This is a notoriously slippery concept, with precious few useful and rigorous ideas. However, there is one extremely important result (very relevant to the present question) in this area: The KAM Theorem.

One of the consequences of the KAM theorem is that, in general, a classical three-body problem can exhibit chaotic behavior. This can be interpreted as a sort of "proof" that the classical three-body problem has no general, analytic solution in any useful sense--i.e. it is not integrable. The reasoning behind this interpretation is that any formula you can write down must have "finite complexity", because the very fact that you can write it down requires this; however, to specify the motion of a chaotic system, you need "infinite information", because even an arbitrarily tiny perturbation to the state of the system produces wild deviations in the trajectory. This is not a "proof" in the usual rigorous sense, because nobody has a really good definition of what it means to be an "analytic solution". But you can be pretty sure that for whatever definition of "analytic solution" you could cook up, there would be no such solution for the general classical three-body problem$^*$.

Now the quantum case cannot possibly be any better than the classical one, because in the classical limit (i.e. large masses and quantum numbers, take expectation values, etc.), you recover the classical behavior. Hence, the "proof" that there is no analytic solution to the classical three-body problem implies the same for the quantum three-body problem, i.e. the Helium atom.

$*$ Note that there are some special cases (corresponding to particular mass ratios) of the three body problem which allow analytic solutions. See Goldstein's "Classical Mechanics" for details. I do not know whether there exist similar special cases of the Helium atom which are solvable, and I would be very interested to know if anyone has details.

The ground state of the helium atom is solvable in classical mechanics. — 2 days ago

Count Iblis 8,10911337 · Answer 5 · 2018-09-09 16:13:06Z

One way to address this problem is to consider high order series expansions and use that to study the analytical properties of the exact function using e.g. resummation methods. In case of the Helium atom, you multiply the interaction term by a factor $g$ and study the series expansion in powers of $g$.

With enough terms, the poles and branch points of the function in the complex plane can be found to a good approximation, this give you clues about whether the global structure of these singularities is compatible with a finite expression involving arbitrary analytic functions. Rodney Baxter has used such methods to argue that certain models in statistical mechanics are not exactly solvable.

Another way is to apply the differential approximant method to the series expansion. Here you write down a general inhomogeneous linear differential equation with polynomial coefficients for some finite order and and degrees of the polynomials with undetermined coefficients and solve for the coefficients that make the series expansion a solution. An exact solution then shows up as a differential equation of which the coefficients are determined by the first $N$ coefficients of the series expansion, but such that the series expansion with $M>N$ terms continues to satisfy that differential equation.

Functions that are solutions of finite order inhomogeneous linear differential equations with polynomial coefficients are called D-finite functions, but not all analytic functions are D-finite (e.g. the Gamma functions is not D-finite). So, this method is not as powerful as the previous method, but one can generalize this method to see if a function belongs to the larger class of differentially algebraic function, see e.g. here for such a study.

Joshua 709311 · Answer 6 · 2018-09-10 03:46:50Z

I'm certain your question is in fact more about math than physics.

The classical model of the Helium atom doesn't have a closed solution either. It is an instance of the three body problem. (This isn't quite true -- the minimum energy configuration can be solved in classical physics.)

The classical solutions of atom orbitals occasionally exist as series of powers, but the math works out such that the solvable orbitals have probability zero; however the quantum waveform interference may allow them to be the ones chosen anyway. The attack path to get closed form solutions to the orbitals would be to apply the SchrÃƒÂ¶dinger equation to the classical form of all the energy levels. (Most of the orbits don't repeat so you have to use the time-dependent SchrÃƒÂ¶dinger equation here.)

However you're going to have to integrate the waveform to get a probability function, and you're starting from a something that's already a power series. Mathematical theorem: if the function doesn't exist in closed form, neither does its integral.

I'm pretty sure somebody's going to tell me you can't use orbital mechanics to handle electrons around the atom. In fact you can, but there's a surprise. There are four orbits in any given plane for any given energy level except the lowest one for which there are only two. This is due to the orbital velocity energy becoming comparable to the energy well energy on the atomic scale. The center pole is at +INF rather than -INF for an orbiting object. We just can't normally observe this effect.

I haven't bothered to check whether the number of degrees of freedom allowed by the resonance orbits in three dimensions agrees with the number of degrees of freedom in the quantum solutions for the atom. I can't come up with a single reason to care.

score 1 · Answer 7 · 2018-09-09 14:49:51Z

The biggest problem for many electron systems is that you have to deal with correlation. This means that the behaviour of an electron cannot be captured with a single orbital. Instead many so called Slater determinants make up the wave function.

However quantum chemists have devised methods to solve the SchrÃƒÂ¶dinger and the Dirac equation with high accuracy for atoms and molecules. The solutions are not in closed form but are expansions of Slater determinants constructed from Gaussian basis functions.

Exact solutions are available only for the simplest systems. For atoms the reason is e-e correlation and also radiative effects, which require QFT. For more complex systems the complexity of nuclear motion is added.

Would've been a great answer but for the fact that it mainly reiterates, what is already written in a lots of places - it still does not address the question of whether it is fundamentally unsolvable or whether it's just a limitation of our current knowledge of maths. — Sep 9 at 14:38
That is a strange reaction: my answer should reiterate what us already written in many places. I will add my view on your second point. — Sep 9 at 14:41
Also, I don't see what you mean by it not being describable as a single orbital, unless you mean that the expression for one electron involves data for the other ones, si we can't separate them, in which case I get it but would need a clarification about whether that means that no exact solution exists ( and whether there is a proof for that). — Sep 9 at 14:43
Sorry if it seemed out of the blue, but the second point was the thrust of my question [ and thanks for taking the trouble to answer :-) ] — Sep 9 at 14:47
My pleasure. Look into quantum chemistry, you may be surprised! — Sep 9 at 14:54

The_Sympathizer 2,530520 · Answer 8 · 2018-09-09 14:43:18Z

None of those are really "good" answers. I will concur. Moreover, it is really more a question that ends up touching onto meta-mathematics, than physics.

I am not really sure one can levy a specific reason, other than the very general reason that suitably complex equations cannot be solved exactly as a general principle. But I believe the explanation below, while not a solid or firm answer, is nonetheless likely to suffice.

One way you can think of it is that there are only a finite number of operations we have for expressing solutions to equations, but an infinite number of possible actual solution functions - although I don't think that quite gets at it, either because you can combine the "expressing" functions in an infinite number of ways. Perhaps a better way to say it is that when we go to solve equations "exactly" what we are talking about is to express them in terms of the composition of some limited number of "base" functions we consider "acceptable" - almost always this includes and is at least the elementary functions, which are those composed of the constant functions, the four arithmetical operations (+, -, *, /), and the exponential ($exp$) and logarithmic ($log$) functions over the complex field ($mathbbC$), and the closure of this set is taken under functional composition ($circ$). We may also include a few non-elementary "special" functions as well like the gamma function ($Gamma$) or error function ($mathrmerf$, the integral of the Gaussian).

And effectively what happens is that these do not provide us with enough "degrees of freedom", so to speak, to express all functions of suitable generality. While I'm not sure of a result that proves this for the general composition operator, a very similar result along this line is the fact that the space of all differentiable (say) functions - like those that come out as solutions to a differential equation like Schrodinger - is algebraically infinite-dimensional, that is, the cardinal of any Hamel basis is infinite, meaning that if we want to express any differentiable function as a linear combination

$$f(x) = a_0 f_0(x) + a_1 f_1(x) + cdots + a_N f_N(x)$$

of basis functions $f_j$, then we actually need a basis set with infinitely many independent functions we can pull the $f_j$ from (i.e. they cannot be expressed in terms of each other) to be able to have enough expressing power to express every differentiable function $f$ in this fashion. (Such a set is called Hamel basis.) For the quantum theory of atoms, this is even more direct, it is the infinite-dimensionality of the Hilbert space.

Likewise, I'd strongly suspect something similar is happening when we allow general composition as well - it doesn't provide us with enough power, but I do not know of a proof.

However, this doesn't really answer the question as more specifically to why that a more restricted set of functions, like the solutions of a specific kind of differential equation - and in particular its eigenvectors, cannot be expressed with a suitable composition of a "nice" set of base functions. After all, linear constant-coefficient differential equations are famously known to have a completely-expressible solution set. Many other very specific differential equations have similar sets. Although, a key point may be that the Schrodinger equation offers effectively infinite freedom through the choice of the potential function $U(mathbfr)$, and more generally its Hamiltonian operator $hatH$.

That said, with a very restricted parameter set, say atoms very specifically, it may indeed be there is a finite or easily-described infinite set of functions that can provide a suitable basis set. We just don't, as of yet, know what the "atom functions" are or how to describe them.

So, is it correct to say that we haven't done enough math to know how to write down the solution, but since we have done a lot of math, we kinda assume that the written down thing would probably be too complicated too be useful(since we practical people haven't found it yet) and so stick with the approximations instead of wasting time on trying harder to solve it than we have done before. — Sep 9 at 14:59
Note that even the $exp$ and $log$ are effectively computed via their power series and you always get results up to a prechosen precision - except for maybe a few special cases. $erf$ can again not be expressed directly via $exp$ and $log$ but it turns out that it is useful in a lot of solutions that we define it as a special function. The solutions to the n-electron Hamiltonian could as well be defined to characterize some special functions but afaik they can't be reused for much else so we don't bother. — Sep 9 at 17:05
@DevashishKaushik: no, it's simpler than that. The solutions you are talking about are simply not common enough to deserve a name, as opposed to say the exponential, the trigonometric functions, or the error function. — Sep 9 at 19:42

Ponder Stibbons 43110 · Answer 9 · 2018-09-10 00:01:46Z

The problem is not at all unique to SchrÃƒÂ¶dinger's equation, it is a common feature of differential equations. The same thing happens all over the place in classical electro-magnetics. But, also to say there is no "exact" solution is misleading. It is more true to say that the solution cannot be expressed finitely in certain notations.

There are theorems such as the Picard Lindelof theorem that ensure the existence of solutions to certain differential equations, but these theorems do not lead to finite expressions using common functions for these solutions. There are some equations that can be solved generically using polynomials. However, the simple equation $dotx=x$ is not one of them. This equation defines the exponential function (that can also be defined by an infinite sum). Effectively, when $exp(x)$ is used, it just means the solution to this equation.

More complicated differential equations require more new functions - the elliptical trigonometric functions, for example. This just keeps on going. There is no finite list of function, combinations of which will solve all differential equations. When a function defined as the solution to a differential equation is used commonly enough - we give it a name, and continue on.

This is a big area of study in formal solutions of any kind of equation:

http://mathworld.wolfram.com/Closed-FormSolution.html

aquirdturtle 1,324410 · Answer 10 · 2018-09-09 23:20:47Z

Why would you expect there to be a solution? If you just consider random complicated PDEs, most of them don't have "nice" solutions, so why would a multi-electron atom be so special? I suspect you're looking for a type of answer which may just not exist for the question at hand. The most literal answer is just the brute-force fact that it appears that nice solutions don't exist (although perhaps someday someone will find another nice solution).

If you want to get hand-wavy, you could say that the PDE is "too complicated", and point to the part which, if removed, would yield a solvable equation (the electron-electron interactions). However, there are also many "simple" equations which also don't have "nice" solutions, such as a simple linear potential $Vx=x$ (solutions being the not-nice airy functions), which is arguably much "simpler" than the harmonic oscillator or a single-atom potential. So, I would avoid the hand-wavy explanations here.

score 5 · Answer 11 · 2018-09-10 03:01:05Z

The question of whether an analytic solution exists for a given dynamical system or PDE goes under the name "integrability". This is a notoriously slippery concept, with precious few useful and rigorous ideas. However, there is one extremely important result (very relevant to the present question) in this area: The KAM Theorem.

One of the consequences of the KAM theorem is that, in general, a classical three-body problem can exhibit chaotic behavior. This can be interpreted as a sort of "proof" that the classical three-body problem has no general, analytic solution in any useful sense--i.e. it is not integrable. The reasoning behind this interpretation is that any formula you can write down must have "finite complexity", because the very fact that you can write it down requires this; however, to specify the motion of a chaotic system, you need "infinite information", because even an arbitrarily tiny perturbation to the state of the system produces wild deviations in the trajectory. This is not a "proof" in the usual rigorous sense, because nobody has a really good definition of what it means to be an "analytic solution". But you can be pretty sure that for whatever definition of "analytic solution" you could cook up, there would be no such solution for the general classical three-body problem$^*$.

Now the quantum case cannot possibly be any better than the classical one, because in the classical limit (i.e. large masses and quantum numbers, take expectation values, etc.), you recover the classical behavior. Hence, the "proof" that there is no analytic solution to the classical three-body problem implies the same for the quantum three-body problem, i.e. the Helium atom.

$*$ Note that there are some special cases (corresponding to particular mass ratios) of the three body problem which allow analytic solutions. See Goldstein's "Classical Mechanics" for details. I do not know whether there exist similar special cases of the Helium atom which are solvable, and I would be very interested to know if anyone has details.

The ground state of the helium atom is solvable in classical mechanics. — 2 days ago

Count Iblis 8,10911337 · Answer 12 · 2018-09-09 16:13:06Z

One way to address this problem is to consider high order series expansions and use that to study the analytical properties of the exact function using e.g. resummation methods. In case of the Helium atom, you multiply the interaction term by a factor $g$ and study the series expansion in powers of $g$.

With enough terms, the poles and branch points of the function in the complex plane can be found to a good approximation, this give you clues about whether the global structure of these singularities is compatible with a finite expression involving arbitrary analytic functions. Rodney Baxter has used such methods to argue that certain models in statistical mechanics are not exactly solvable.

Another way is to apply the differential approximant method to the series expansion. Here you write down a general inhomogeneous linear differential equation with polynomial coefficients for some finite order and and degrees of the polynomials with undetermined coefficients and solve for the coefficients that make the series expansion a solution. An exact solution then shows up as a differential equation of which the coefficients are determined by the first $N$ coefficients of the series expansion, but such that the series expansion with $M>N$ terms continues to satisfy that differential equation.

Functions that are solutions of finite order inhomogeneous linear differential equations with polynomial coefficients are called D-finite functions, but not all analytic functions are D-finite (e.g. the Gamma functions is not D-finite). So, this method is not as powerful as the previous method, but one can generalize this method to see if a function belongs to the larger class of differentially algebraic function, see e.g. here for such a study.

Joshua 709311 · Answer 13 · 2018-09-10 03:46:50Z

I'm certain your question is in fact more about math than physics.

The classical model of the Helium atom doesn't have a closed solution either. It is an instance of the three body problem. (This isn't quite true -- the minimum energy configuration can be solved in classical physics.)

The classical solutions of atom orbitals occasionally exist as series of powers, but the math works out such that the solvable orbitals have probability zero; however the quantum waveform interference may allow them to be the ones chosen anyway. The attack path to get closed form solutions to the orbitals would be to apply the SchrÃƒÂ¶dinger equation to the classical form of all the energy levels. (Most of the orbits don't repeat so you have to use the time-dependent SchrÃƒÂ¶dinger equation here.)

However you're going to have to integrate the waveform to get a probability function, and you're starting from a something that's already a power series. Mathematical theorem: if the function doesn't exist in closed form, neither does its integral.

I'm pretty sure somebody's going to tell me you can't use orbital mechanics to handle electrons around the atom. In fact you can, but there's a surprise. There are four orbits in any given plane for any given energy level except the lowest one for which there are only two. This is due to the orbital velocity energy becoming comparable to the energy well energy on the atomic scale. The center pole is at +INF rather than -INF for an orbiting object. We just can't normally observe this effect.

I haven't bothered to check whether the number of degrees of freedom allowed by the resonance orbits in three dimensions agrees with the number of degrees of freedom in the quantum solutions for the atom. I can't come up with a single reason to care.

score 1 · Answer 14 · 2018-09-09 14:49:51Z

The biggest problem for many electron systems is that you have to deal with correlation. This means that the behaviour of an electron cannot be captured with a single orbital. Instead many so called Slater determinants make up the wave function.

However quantum chemists have devised methods to solve the SchrÃƒÂ¶dinger and the Dirac equation with high accuracy for atoms and molecules. The solutions are not in closed form but are expansions of Slater determinants constructed from Gaussian basis functions.

Exact solutions are available only for the simplest systems. For atoms the reason is e-e correlation and also radiative effects, which require QFT. For more complex systems the complexity of nuclear motion is added.

Would've been a great answer but for the fact that it mainly reiterates, what is already written in a lots of places - it still does not address the question of whether it is fundamentally unsolvable or whether it's just a limitation of our current knowledge of maths. — Sep 9 at 14:38
That is a strange reaction: my answer should reiterate what us already written in many places. I will add my view on your second point. — Sep 9 at 14:41
Also, I don't see what you mean by it not being describable as a single orbital, unless you mean that the expression for one electron involves data for the other ones, si we can't separate them, in which case I get it but would need a clarification about whether that means that no exact solution exists ( and whether there is a proof for that). — Sep 9 at 14:43
Sorry if it seemed out of the blue, but the second point was the thrust of my question [ and thanks for taking the trouble to answer :-) ] — Sep 9 at 14:47
My pleasure. Look into quantum chemistry, you may be surprised! — Sep 9 at 14:54

Search This Blog

Iyfjky

Why can't the SchrÃ¶dinger equation be solved exactly for multi-electron atoms? Does some solution exist even in principle? [duplicate]

Notes-

Notes-

Notes-

Notes-

7 Answers
7

7 Answers
7

7 Answers
7

Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

List of Gilmore Girls characters

Confectionery

Category

Random preview

Why can't the SchrÃ¶dinger equation be solved exactly for multi-electron atoms? Does some solution exist even in principle? [duplicate]

Notes-

Notes-

Notes-

Notes-

7 Answers 7

7 Answers 7

7 Answers 7

Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

List of Gilmore Girls characters

Confectionery

7 Answers
7

7 Answers
7

7 Answers
7