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I have read the following on Wikipedia but I can't understand it:

In quantum mechanics, the Schrödinger equation, which describes the continuous time evolution of a system's wave function, is deterministic. However, the relationship between a system's wave function and the observable properties of the system appears to be non-deterministic.

"Deterministic system", Wikipedia [links omitted]

How can a system be deterministic and not deterministic at the same time? Can anyone explain simply?

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    $\begingroup$ Possible duplicate of Why can't the outcome of a QM measurement be calculated a-priori? $\endgroup$ – Stéphane Rollandin Apr 16 '18 at 11:37
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    $\begingroup$ The time development of the probabilities to obtain certain measurement results is deterministic and is given by the solution of the time dependent Schrödinger equation. The measurements results are not deterministic. $\endgroup$ – freecharly Apr 16 '18 at 15:33
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    $\begingroup$ This is already a philosophical/interpretation issue. In principle, one would hope that all observable quantities are really just extracting deterministic information from some "universal wavefunction" with deterministic evolution. Whether this is strictly true or not, the fact is that you cannot isolate a quantum system from your measurement mechanism. This means that the wavefunction of the isolated system by itself cannot give you deterministic predictions for your observations. However, it can tell you what you should expect from those observations on a statistical basis. $\endgroup$ – Ian Apr 16 '18 at 17:03
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    $\begingroup$ IMHO the answers are all way too complicated. Here's an example of a probability distribution that changes deterministically. Suppose I have three cups, A, B, and C. I put a ball in either A or B, but don't tell you which. I now have a probability distribution: 1/2 probability that the ball is in A or B. Then I transfer whatever is in B to C. Now the probability distribution has deterministically changed so that there's 1/2 probability to be in A or C. $\endgroup$ – DanielSank Apr 16 '18 at 17:57
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    $\begingroup$ To me the question would be what is not deterministic in the Schrodinger equation. It seems more obvious to me what is deterministic. It's a differential equation, and it has a unique solution for a given set of initial conditions. $\endgroup$ – Ben Crowell Apr 17 '18 at 2:59
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In the standard interpretation of quantum mechanics the time-evolution of the system and what we observe are separated (unlike Newtonian mechanics). The system, while unobserved exists in a superposition of states (all the states that satisfy the Schrödinger equation). These states (while unobserved) evolve in time according to the Schrödinger equation. This part—all the way up to the point of observation—is completely deterministic.

If no observer comes along, then that is the end of the story. If, however, someone comes along and makes an observation, then the system, which prior to observation was in multiple states, is thrown into a single state, and only a single state is thus measured. How that state is chosen is posited by quantum mechanics to be completely non-deterministic.

Note that

  1. This interpretation of quantum mechanics is known as the Copenhagen interpretation and is by far the most common interpretation of quantum mechanics and is what’s in all the text books.
  2. This general problem is known as ‘the measurement problem’ and is by far the most controversial aspect of quantum mechanics
  3. Attempts have been made to add additional variables to the formalism to make the outcome of the measurement deterministic (see Einstein’s EPR paper, for example), however, these attempts cause inconsistencies in the theory of quantum mechanics. Therefor, the current view is that quantum mechanics is a complete theory—I.e we can’t add any more to it whitout generating contradictions.

It’s a very controversial part of the theory of quantum mechanics, but appears to be something we have to live with.

I hope this expiation helps.

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    $\begingroup$ +1 most accurate and complete answer yet $\endgroup$ – Thies Heidecke Apr 16 '18 at 23:54
  • $\begingroup$ +1, however there is also the open/closed system philosophical issue. I'd content that we, the observer, is within and part of the system, so from one viewpoint our system is fully computable, but unfortunately, a lot of the system is unobserved (no interaction) with/by us, because that part is behind the 'light cone', which is where 'random to us' part comes from. [aside: a similar 'unobserved but theoretically known' issue is noted in one Dempster-Shafer stats paper arxiv.org/abs/1011.0819 as the auxiliary variable] $\endgroup$ – Philip Oakley Apr 17 '18 at 15:07
  • $\begingroup$ Also have a look at Wavelets as a basis function that solves Schrodinger (square integral functions with compact support) Friendly Guide to, by Kaiser. $\endgroup$ – Philip Oakley Apr 17 '18 at 15:23
  • $\begingroup$ @ThiesHeidecke The very first sentence "In the standard interpretation of quantum mechanics the time-evolution of the system and what we observe are separated (unlike Newtonian mechanics). " refers to a mythical "standard interpretation of quantum mechanics". First, it seems reductionist to say that there even is a "standard" interpretation of quantum mechanics. Second, the story that is most commonly put forth in e.g. classrooms does not match with what the answer calls "standard". Furthermore, this answer only refers to strong projective measurement, which is not the whole story. $\endgroup$ – DanielSank Apr 17 '18 at 18:46
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    $\begingroup$ @DanielSank the standard interpretation of quantum mechanics is the Copenhagen interpretation, as I stated in the first item in the notes. It’s not the only interpretation of quantum mechanics, but it is by far the most widely held, and so it is referred to as such. $\endgroup$ – njspeer Apr 17 '18 at 19:16
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The fact is that there are two kind of things: 1) the wave function and 2) the physical observables.

The evolution of the wave function is dictated by the Schrödinger equation and is deterministic meaning that if you know the wave function at some time, then you know it at any time just using Schrödinger equation $$ i \hbar \frac{d}{dt} \left \lvert \Psi (t) \right \rangle = H(t) \left \lvert \Psi (t) \right \rangle \, .$$

On the other hand any observable (e.g. position) is not deterministic in the sense that if you know observables at some time, in general you cannot say anything of the particle at some future time. There exist no equation for observables themselves.

Coming back to the wave function, I have said that it is deterministic. This is true unless someone measures an observable. In that case the wave function collapses in a non deterministic way. Next to the measurement the evolution becomes deterministic again, but just at the instant of the measurement it does not.

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    $\begingroup$ Your use of the word "observable" and e.g. the statement "There exist no equation for observables themselves" may lead to confusion as in QM it's used for the operators. The "measurement outcome" is more appropriate. $\endgroup$ – OON Apr 17 '18 at 5:02
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In quantum mechanics, the solution of the equations (Schrodinger, Dirac...), called wave functions are deterministic, at each $\left(x,\,y,\,z,\,t\right)$ point, but the only prediction they give is a probability distribution, which depends on the boundary conditions of the problem. $Ψ$ is a complex valued function, and measurements are real numbers and this is the distribution of $Ψ^*Ψ$ (the absolute square or squared norm of $ψ$) which gives the probability of finding a particle at a given space time point.

Probabilities by definition means that many measurements in the same boundary conditions have to be carried out, and a comparison made between the predicted probability distribution and the measured one. So even though the distribution is strictly deterministic, its comparison with one datum is probabilistic.

Note the same boundary conditions statement. Once a measurement is carried out, the boundary conditions are different, a different $Ψ$ is needed for the system after the measurement, which is called a "collapse of the wavefunction". In experiments one does not observe the same particle scattering or decaying, but a large number of same boundary condition set ups to accumulate the probability distribution to compare.

Edit after comment:

By boundary conditions I mean the real numbers that have to be introduced so that the mathematical formula will give predictions for the specific observables of the experiment. For example the energy and momentum for getting the cross section from a scattering experiment of two protons, as in the LHC.

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    $\begingroup$ they only prediction they give is a probability distribution, which depends on the boundary conditions of the problem This doesn't make much sense. What do boundary conditions have to do with this? Do you mean boundary conditions in the sense in which the phrase is normally used in mathematics, or in some looser sense? $\endgroup$ – Ben Crowell Apr 16 '18 at 20:03
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    $\begingroup$ @BenCrowell I mean the exact real numbers that have to be introduced in the mathematical formula so that the results can be compared with data from an experiment, as, for example, dimensions and distance between slits, in the double slit experiment and energy of the photons. Input energies and momenta in scattering etc. $\endgroup$ – anna v Apr 17 '18 at 4:03
  • $\begingroup$ Isn't it $\overbarΨ∗Ψ$? $\endgroup$ – Acccumulation Apr 17 '18 at 22:03
  • $\begingroup$ @Acccumulation it is the complex conjugate squared $\endgroup$ – anna v Apr 18 '18 at 3:44
  • $\begingroup$ It's the absolute square, not the complex conjugate squared. The complex conjugate squared is just the conjugate of it squared. $\endgroup$ – Acccumulation Apr 18 '18 at 14:38
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Although there are other complete answers, I'd like to put down a one sentence summary, since this point is not emphasized enough. To your title question:

What exactly is deterministic in Schrödinger's equation?

The answer is: Everything. Absolutely, unambiguously, vollständig, del tutto, complètement, whollus bolus, bet-yer-life-on-it Everything.

The time evolution of a quantum state is utterly deterministic and Schrödinger's equation would be something that Laplace himself, with his daemon familiar, would have been altogether comfortable with. The complex valued system system state might have looked a bit weird to him, but, with his mathematical background, he'd have had no trouble grasping the idea and fitting it seamlessly into his philosophy of determinism. Aside from its special properties of being unitary and complex, it is exactly like any other linear system state transition.

The nondeterministic (in most interpretations) part of quantum mechanics and what is different from Laplace's idea is how we infer values of measurements from the system state - the so called measurement problem. In classical physics, in theory there is a one to one mapping from the system state to any measurement we make on that system, and the only limit to this notion is practicalities like measurement noise and the in-theory-invertible observer effect. In quantum mechanics, the quantum state defines the statistical distributions of measurements made on an ensemble of quantum systems with the state in question. I say more about this in my answer about quantum measurement here.

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Oversimplified, since OP asked for a simple explanation:

QM deterministically describes the time evolution of a system which is not a model for "the observed universe" but for "all possible universes". (In reality that's not quite right for any particular real-world problem you look at, because it's already been simplified down to a minimal closed-system model that's tractable, but the concept still makes sense to talk about.)

Unfortunately (from the standpoint of someone who wants to use a model to make predictions) there's no way to measure or even approximate the "state of all possible universes" at a given time, and even if you could somehow do that to run the evolution as a simulation, your results would only tell you about all possible universes, not the real world you live in.

So with this in mind, in order to use that to make meaningful predictions, you have to interpret what are essentially conditional probabilities, conditioned on observations: given that I measured [something] at time $t_0$, the chance of finding [some other measurement] in [some neighborhood/interval] at time $t_1$ is [some probability].

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Mathematically speaking, the Schrodinger equation is a type of linear PDE known as the wave equation. The wave equation, which you may be familiar with from other studies of PDEs, is deterministic, because it evolves in the sense that if you know the initial conditions at an initial time, you know exactly what the wave described by this equation will be at a later time. The state at a later time is determined by the initial conditions and time-evolves toward that state according to the wave equation: you freeze time at an instant and you 'know' what it will be.

In the case of the Schrodinger equation, the wave (as it was original conceived of) is a probability distribution (this is the origin of the term 'wave function'). The probability distribution evolves in a deterministic way according to a linear PDE (probability distributions can do this in general), but the results of measurements are obviously 'not deterministic', because you 'don't know what you are getting' when you dip your hand in the bag, although you do know the probabilities of what you can get.

There are more complicated ways of coming to the same conclusion, but I think is the simplest way (without getting into the foundations of QM).

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"Can anyone explain simply?"

The outcome of a measurement of an observable for some system in some state is, in general, not determined by QM.

The (in general, time varying as determined by the SE) expectation value of the measurement of an observable for some system is determined by QM.

That is to say, if you have an ensemble of identically prepared systems, QM predicts the ensemble mean of the measurement of an observable as it evolves with time. QM does not (cannot) predict, in general, the outcome of the measurement for any member of the ensemble.

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TL;DR Schrödinger's equation determines the wave function. The wave function determines probabilities. But a particle's position, for example, is not determined by a probability.


If you know the wave function $\Psi(x,t)$ at some time $t_0$, Schrödinger's equation will give you the wave function for all $t>t_0$. This means that the equation determines the evolution of the wave function$\dagger$.

If we want to know, say, the position of a particle, QM won't give us a function $x(t)$ defined for all $t$. What QM can give us is the probability of the particle being in some interval $(a,b)$ at time $t$. But a probability alone doesn't determines the position of a particle. The particle may as well be outside $(a,b)$. Thus the theory isn't deterministic.


$\dagger$ Note however that measurements will collapse the wave function in a non-deterministic way.

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The standard answer to this question is that a system evolves deterministically according to the Schrodinger equation except when it is measured. When it is measured the wave function collapses into one of the possible measurement results. There are many problems with this way of understanding the world. For example, unless you specify how the wave function collapses the whole idea of collapse is vague and ad hoc. If you do specify how it collapses you have to modify quantum theoretical equations of motion and all existing modifications create a lot of problems.

So it is worth considering whether you need collapse at all. What would the world look like if collapse never happened? Physical reality would evolve entirely according to some quantum mechanical equation of motion. And as Everett pointed out in 1957, multiple versions of each physical system would exist after a measurement or any other interaction that transfers information between systems. This is not widely accepted by physicists for reasons that seem unclear to me, but it has the advantage of being consistent and clear while explaining experimental results. A system might evolve so that $$|x\rangle\to\tfrac{1}{\sqrt{2}}(|x\rangle+|y\rangle)$$.

So a system that started out in one state could differentiate into multiple versions over time. If there is only a single version of each system and that version evolves into two different states, that's non-deterministic. If there are multiple versions then how can they evolve deterministically into different versions without being different to start with? The process is deterministic because the vector $|x\rangle$ can be partitioned into sets each with the same value $x$ of some observable because $|x\rangle = \sum_i \alpha_i|x\rangle$ for any set of real numbers $\alpha_i$ such that $\sum_i \alpha_i=1$. So the vector represents a continuous set of identical versions of a system. They are identical in the sense that no measurement could distinguish them. As such, the evolution given above describes a situation in which some versions of the system are different after the evolution compared to their state before the evolution but there is no factor of the matter about which of the versions of the system before the evolution correspond to a particular state after the evolution. the rule specifies the amplitude of each successor state but it doesn't say that you will see a single version of the system after the measurement. Different versions of you will see different versions of the outcome and there is no single fact of the matter about which outcome you will see. For more explanation of this and other issues see "The Beginning of Infinity" by David Deutsch chapter 11.

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protected by Qmechanic Apr 17 '18 at 18:32

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