A question from a mathematician far from physics :)

I have heard that Schrödinger equation for $n$ particles is hard in the following sense:

If $n$ is enough big then there is no computer which can give a numerical approximation of the solution to the corresponding Schrödinger equation.

Since the notion of "proof" in physics is intimately related to the experimental observations, I was wondering about the following statement "quantum mechanics is the most exact physical theory".

My question: how can we (objectively) claim such statements if there is no way to verify the numerical exactness of the Schrödinger equation for 100 particles?

Edit: let me be clear about what I am looking for: is there a concrete numerical simulation of Schrödinger equation with more than 100 particles interacting in not negligible way! And a real comparison with the experimental observations?

Edit2: everyone is free to downvote, but it will be nice if you provide the reason. But of course it is not obligatory:)

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    $\begingroup$ I disagree with the "opinion based" close votes. This is a fair question and "how do we know the Schrödinger equation for many particles is correct even though it's very hard or impossible to simulate and compare to experiment" is not up to opinion I think. $\endgroup$ Dec 10, 2020 at 7:53
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    $\begingroup$ We don't, just like we don't know if the moon is there when nobody's looking. There's nothing in principle or experiment preventing incredibly tiny deviations from Schrodinger's equation... $\endgroup$
    – knzhou
    Dec 16, 2020 at 0:11
  • $\begingroup$ hi @GSM, could you explain how the current answers do not address your question since you put a bounty? It would help people understand what you're looking for :) $\endgroup$ Dec 17, 2020 at 9:11
  • $\begingroup$ @user2723984, thanks for your comment. The point is that I'm not satisfied and if you read discussion in the comments it follows clearly that there is no consensus. Moreover it is not enough concrete to my taste. :) I am just an ignorant for sure:) $\endgroup$
    – GSM
    Dec 17, 2020 at 14:36
  • $\begingroup$ Can you clarify what you consider a 'concrete numerical simulation'. Directly treating it as a PDE in many variables is clearly not scalable but lots of approximate methods exist - eg Quantum Monte Carlo, DFT, DMRG, tensor networks etc. $\endgroup$
    – jacob1729
    Dec 17, 2020 at 16:29

7 Answers 7


Despite it being impossible to exactly simulate a large number of particles on a computer, physicists have found many tricks to approximately simulate it. From Quantum Monte Carlo to Density functional theory to the whole field of Tensor networks and many others, there is a plethora of methods to obtain accurate predictions for the physics of many particles that doesn't involve attacking the problem directly (which would be impossible).

As an example, a single particle is described by a $d$ dimensional vector space, so an $N$ particles state is described by an unmanageable $d^N$ dimensional vector. It turns out though that physically relevant states of $N$ particles (i.e. ground states of local Hamiltonians) have some nice properties that allows them to be tractable. Specifically, they have relatively low entanglement (i.e. since the physics is described by interaction between nearby particles, far away particles will not be very correlated). This allows to get an accurate description of such a state with a number of parameters that scales polynomially instead of exponentially in the system size. This is only possible because the physically relevant case is not the general case. This observation is where the techinques associated with tensor networks and matrix product states stem from.

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    $\begingroup$ This is another example of physics being largely unaware of the existence of quantum chemistry. $\endgroup$
    – my2cts
    Dec 12, 2020 at 7:45
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    $\begingroup$ @my2cts are you referring to the question or to my answer? $\endgroup$ Dec 12, 2020 at 9:52
  • $\begingroup$ We are not assuming that the interaction between particle is negligible! Imagine that we have a big number of particules in a very small box. $\endgroup$
    – GSM
    Dec 21, 2020 at 15:20

Statements such as

Quantum mechanics is the most exact physical theory

are usually based on precision tests of quantum electrodynamics, where QED predicts values for fundamental parameters such as the fine-structure constant that agree with experimentally measured values to a very high degree of precision - typically a few parts in a billion.


If n is enough big then there is no computer which can give a numerical approximation of the solution to the corresponding shrodinger equation.

Not true. In nuclear physics we routinely get good numerical approximations to the properties of nuclei that have 100 particles. In condensed matter physics, it's pretty routine to get good models of $10^{23}$ particles.

There may be some sense in which this claim is true, but if so, the burden is on you or the person making this claim to define what they really mean.

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    $\begingroup$ In condensed matter physics you don't usually work with $10^{23}$ particles: instead you use mean field approximation for electrons, Born-Oppenheimer approximation for nuclear motion etc., infinite crystal approximation to avoid dealing with crystal surfaces, ... $\endgroup$
    – Ruslan
    Dec 15, 2020 at 23:10

Let me make a few comments from an ex mathematician (now physicist).

The first thing to understand is that the notion of "proof", namely a formal derivation with logical consistent steps to show that some statement is true, from purely theoretical grounds is not as central for physics as a science as opposed to mathematics.

Within science, specifically in physics, I would say the general view is definitely more pragmatic. Of course among all physicist you will find a whole spectrum of postures, were some will give more relevance to current theory or experiment, depending on where they themselves stand. Nonetheless, the ultimate test for a theory or a model will be experiment (that is science). So as theorists we must accept the fact that we should never confuse the model with reality. We can only produce models and they are just as good as their predictions.

Having said that we can discuss your concerns around the Schrödinger equation. As I explained this is a model of a piece (or a regime) of reality pertaining certain energy scales, lengths and time scales. As such it has been proven (verified) to work wonderfully for a specific regime, namely very small scales in length, and low energies (compared to rest masses for example). Under this view, the claim that the Schrödinger equation describes reality is just to broad and ignores the details. We know for example it breaks down at high energies (must be replaced with the Dirac equation). Alternatively, many argue that the higher the number of particles involved in your setup the closer you should be getting to the classical regime (high occupation numbers lead to classicalization...), however in the in between there is still a lot of phenomena and as physicist (some) we care about using what we would call the "current fundamental laws" to build up models that effectively describe the setup at hand. Thus, all the already mentioned, statistical mechanics, DFT, mean field treatments and so on, to be able to describe larger and larger numbers of particles.

Our job is precisely to keep testing and testing and testing... If we find that something disagrees with the Schrödinger equation, we change it, that is science. And that is what has been done in time. We know today that QM is way more than just the Schrödinger equation, its essence relies on non-commutativity of observables. This precise idea has been extended to a wider regime, such as high energy physics, also with excellent agreement. So objectively I can say, QM is the most fundamental description of the smallest scales of length we currently have.

I hope I have been objective enough.

P.D. By the way, modelling things fully classically with Newton's laws and the corresponding equations, faces the same issues when dealing with a large number of particles, so one goes again to effective theories derived from some averaging procedure. However, there is still little doubt that Newton's laws are correct (they predict up to our current experimental precision) at describing the mechanics of our daily life.


Experiments cannot be considered exact. There are many assumptions, models and uncertainties underlying any experimental result. Also the Schrödinger equation is not exact. One has to include special relativity, as the Dirac equation does. On top of this there are radiative corrections, which are treated as perturbations. "Experimental exactness" therefore can only mean that agreement with a experiment can be achieved to within the error boundaries of the experimental data.

Systems of 100 particles in general are hard to solve, in quantum mechanics but also in Newton gravity. One usually resorts to dynamical simulation, such molecular dynamics. Often one has to resort to serious approximations such as mean field or Hartree-Fock. Thus is because particle particle correlation is very difficult to treat.

To a physicist unfamiliar with quantum chemistry it may therefore come as a surprise that quantum chemists have no problem solving the Dirac equation for an atom of up to 100 electrons to very good accuracy. The results of such calculations can be compared to experiment and this constitutes a verification of the Dirac equation for n up to 100. One of the reasons is that electrons are so much lighter than nucleons so that for atoms the Born-Oppenheimer approximation, static nucleus, holds well. Another strong technique is Configuration Interaction, where the many particle wave function is expanded in Slater determinants. I would not be surprised if today QC can handle molecules of 100 atoms with the Schrödinger or the Dirac equation, again using BO or quasistatic nuclei. These approaches are based on Gaussian basis functions. You may prefer to think in terms of a grid in R$^{3n}$ space, but such a grid can also be seen as a set of basis functions, albeit a very inadequate one. In all of these cases, in principle one can increase accuracy to any level - but not in practice. NIST maintains a database of atomic spectra based on experimental and quantum chemical computations data. https://physics.nist.gov/PhysRefData/Elements/legend.html

There are also exactly soluble systems such the Heisenberg 1D antiferromagnet. This is a model system as only spin degrees of freedom are included.

So in principle we can push the accuracy of theoretical predictions very far but in practice often this would be required unlimited resources, with a few beautiful exceptions.

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    $\begingroup$ This is not true, solving e.g. dissociation energies even of relatively simple molecules can be rather hard. $\endgroup$ Dec 15, 2020 at 22:51
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    $\begingroup$ I agree with @NorbertSchuch that this answer is misleading. There are certainly difficulties in modeling large systems in quantum chemistry, that's why it's an active area of research. If you can make your statements more precise, then I'll remove my downvote. $\endgroup$ Dec 17, 2020 at 12:34
  • $\begingroup$ I've removed some back-and-forth comments that were degenerating into an unproductive discussion. Stay nice, folks. $\endgroup$
    – rob
    Dec 18, 2020 at 1:30

Quantum mechanics is an exact theory if the value of an observable seen in experiment $\langle X \rangle_\mathrm{observed}$ matches that calculated by solving the Schrödinger equation $\langle X \rangle_\mathrm{calculated}$. $$ \langle X \rangle_\mathrm{observed} = \langle X \rangle_\mathrm{calculated} $$ Thus for an approximate solution $\langle X \rangle_\mathrm{approx}$ the following holds $$ |\langle X \rangle_\mathrm{approx} - \langle X \rangle_\mathrm{calculated}| \leq \epsilon \implies |\langle X \rangle_\mathrm{approx} - \langle X \rangle_\mathrm{observed}| \leq \epsilon. $$ As approximated solutions can be obtained, and errors can be bounded, this statement can be verified.

By extension utility of the Schrödinger equation is not limited to the cases in which it may be solved exactly.

In reality there are also experimental and statistical errors to contend with, but this is outside the scope of the question.


You are right there is no proof like that. However, we can check that QM works for 2, 3 and say 10 particles (to be honest I do not know what the state of the art is at present for many body simulations) and there is no reason to think that something changes when at specific $N$ that all of a sudden (or smoothly) QM stops working. Moreover, sometimes systems with many fundamental degrees of freedom can be described by just a handful of effective ones. In those cases we still have the rules of QM in full action.

Btw you can ask the same question about any classical system with many constituents.

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    $\begingroup$ "We can check classical mechanics at speeds of 10, 100, and 1000 m/s, and there is no reason to think that something changes at some specific speed." $\endgroup$ Dec 15, 2020 at 22:50
  • $\begingroup$ @NorbertSchuch for an equally glib summary of the problem of induction: the laws of thermodynamics held yesterday and the day before, and there is no reason to think that something will change tomorrow. $\endgroup$ Dec 15, 2020 at 23:45
  • $\begingroup$ @Norbert Schuch what's your point? How is what I am saying wrong? Note that I am not saying that QM does not break down, I just have no reasons to think that it does. Is it more reasonable to think that QM is not applicable to describe many body systems, although we know that for extremely large number of constituents it does work? $\endgroup$
    – nwolijin
    Dec 17, 2020 at 10:11
  • $\begingroup$ @ComptonScattering how is what I am saying glib? Note that I am not saying that QM does not break down, I just have no reasons to think that it does. Is it more reasonable to think that QM is not applicable to describe many body systems, although we know that for extremely large number of constituents it does work? $\endgroup$
    – nwolijin
    Dec 17, 2020 at 10:13
  • $\begingroup$ That QM can also describe large systems: That's not what your answer is saying. Your answer is saying: 2,3,10 -> many. (And it is also only partly true that we know that QM works perfectly well for very large systems: Only certain things can be tested well for large systems. Otherwise, what is the point of e.g. doing interference experiments with large objects?) $\endgroup$ Dec 17, 2020 at 12:43

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