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I was reading this thread and I'm a bit confused. The answer says negative probabilities can account for destructive wave interference and the events cancelling out. But if events just cancel out, shouldn't that make the probability zero? Why would it be negative?

Additionally, my (possibly incorrect) understanding has always been that we only get negative probabilities in QM in the context of "probability amplitudes", which is just a fancy name for the amplitudes of the wavefunctions. But it always seemed kind of weird to talk about probability amplitudes at all -- if we just square the normalized wavefunction and treat that as the PDF, like the Born rule says to do, everything is the same as in regular probability, right? Why even try to interpret the amplitudes themselves, as opposed to their squares, as probabilities?

Am I mistaken in the above paragraph, and we can't always get a regular PDF from squaring the normalized wavefunction and we instead sometimes get quasiprobability distributions that allow negative probabilities? Or, are there situations where we need to consider the wavefunctions themselves, as opposed to first normalzing and squaring them, in order to correctly compute some quantity?

I did also see this thread and the first answer basically says the math is just a lot simpler if we use probability amplitudes to explain interference patterns. I get that using actual wavefunctions to describe the very wave-like behavior of groups of particles, such as in the double slit experiment, is simpler and hence preferable; I wouldn't suggest that we do all the math using the PDFs.

But I don't see why that necesitates this idea of "probability amplitudes" and negative probability -- why can't we just call the wavefunctions wavefunctions and leave the talk of probability entirely to the actual PDF's we get by applying the Born rule? Or CAN we do that and calling the wavefunction amplitudes probability amplitudes is just one way of thinking about that math? In other words, is this just a question of language, of whether or not try to bring in the label of probability into more of the theory by generalizing the definition of probability, or is there more to it?

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    $\begingroup$ I would disregard what you have been reading there. Probabilities in quantum mechanics are what you get AFTER you apply the Born rule, i.e. after a projection and after taking the square of the modulus of the wave function. That is, by definition, a positive number. The wave function itself does not have a physical interpretation and it is complex, anyway, which means that it isn't "negative" in any meaningful mathematical sense. $\endgroup$ Mar 26, 2023 at 23:33
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    $\begingroup$ @Amit Where do you see a negative probability in that formula? All probabilities are positive. The only minus sign occurs in the formula itself, but the negative term alone does not represent a meaningful process. One can not undo the outcome of an experiment with another experiment. $\endgroup$ Mar 26, 2023 at 23:35
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    $\begingroup$ @Amit That has nothing at all to do with complex amplitudes or destructive interference in quantum mechanics. Moreover, it's a very misleading analogy. $\endgroup$
    – tparker
    Mar 26, 2023 at 23:38
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    $\begingroup$ @Amit Subtracting probabilities is not the same thing as negative probabilities. If you subtract two positive numbers, then they're still both positive. $\endgroup$
    – tparker
    Mar 27, 2023 at 0:12
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    $\begingroup$ On a lighter note: maybe (Jason) Bourne rules the screen, but it is (Max) Born who “rules” QM…. Hopefully I corrected all the occurrences of “Bourne”…. ;) $\endgroup$ Mar 27, 2023 at 0:31

4 Answers 4

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Not only do we not "need" negative probabilities in quantum mechanics, but in fact there are no negative probabilities in QM. All probabilities are real numbers between 0 and 1 by definition.

The answer says negative probabilities can account for destructive wave interference and the events cancelling out.

That is incorrect. Probability amplitudes can be negative and can experience destructive wave interfence, but probabilities cannot. Probability amplitudes are not probabilities.

My (possibly incorrect) understanding has always been that we only get negative probabilities in QM in the context of "probability amplitudes", which is just a fancy name for the amplitudes of the wavefunctions.

That is very close to correct; it's correct to a first approximation. 99% of the time that people talk about "negative probabilities" in QM, they really mean complex probability amplitudes. In very advanced applications, they might instead be referring to the Wigner quasiprobability distribution, which is a different notion that is loosely analogous to "negative probabilities" (but only analogous - actual probabilities are still always nonnegative). Until you become much more comfortable with QM, it's probably best to totally forget about the Wigner quasiprobability distribution for now.

But it always seemed kind of weird to talk about probability amplitudes at all -- if we just square the normalized wavefunction and treat that as the PDF, like the Bourne rule says to do, everything is the same as in regular probability, right?

Right (except that it's spelled "Born", not "Bourne").

Why even try to interpret the amplitudes themselves, as opposed to their squares, as probabilities?

We don't interpret them as probabilities. (At least, people who know what they are talking about don't.) They are closely related to probabilities, but they also have fundamental differences.

Am I mistaken in the above paragraph, and we can't always get a regular PDF from squaring the normalized wavefunction and we instead sometimes get quasiprobability distributions that allow negative probabilities?

We always get a regular PDF from squaring the normalized wavefunction. We never get quasiprobability distributions; those come from a very different procedure, which it's probably best to complete ignore until you get much more familiar with QM.

Or, are there situations where we need to consider the wavefunctions themselves, as opposed to first normalzing and squaring them, in order to correctly compute some quantity?

Yes, there definitely are such situations. This is a deep and complicated subject. The fast and loose answer is that it's tremendously convenient to use the phase structure of amplitudes for practical calculations. The somewhat more complete answer is that we need to use the complex amplitudes in order to explain both time evolution and the possibility of changing the measurement basis. The full and deep answer is that the Kochen–Specker theorem and Bell's theorem demonstrate that we can't reproduce the predictions of the standard formalism of QM using only regular PDFs, at least not without making some extremely bizarre assumptions. The complex structure of the amplitudes is fundamentally necessary for reproducing the predictions of QM; it is not just a calculational convenience.

I mean this entirely respectfully, but these theorems are deep and complex, and you probably are not yet familiar enough with QM to fully understand them. But you can take a crack at it. You should pose any follow-up questions specific to these theorems in a separate Physics SE question.

But I don't see why that necesitates this idea of "probability amplitudes" and negative probability -- why can't we just call the wavefunctions wavefunctions and leave the talk of probability entirely to the actual PDF's we get by applying the Bourne rule?

"Negative probability" is just a misnomer. Most people who use that term are either just being sloppy and leaving off the word "amplitude", or they are just confused. "Probability amplitude" is correct terminology, but again, probability amplitudes are NOT probabilities. Their interpretation is very different. If you don't like using two similar terms for very different mathematical concepts, then for now it's fine to just stick to the term "wavefunction" instead (although there are few minor differences between the terms "wavefunction" and "probability amplitude" around the edges).

Or CAN we do that and calling the wavefunction amplitudes probability amplitudes is just one way of thinking about that math?

Correct.

In other words, is this just a question of language, of whether or not try to bring in the label of probability into more of the theory by generalizing the definition of probability, or is there more to it?

I don't quite understand this final question, but yes, it's basically just confusing terminology. Probability amplitudes are related to probabilities, but they are NOT probabilities. They are an intermediate tool that eventually get converted into true probabilities. There are no negative probabilities in QM. It's fine to drop the p-word and just call the components of the wavefunction "amplitudes" if you want - people will still understand what you mean.

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    $\begingroup$ "at least not without making some extremely bizarre assumptions" Are those the assumptions of pilot wave theory? The second answer I linked to mentioned the pilot wave formulation as being a more mathematically complicated model that doesn't use probability amplitudes. (Also, this answer is extremely thorough, well written, and did an excellent job explaining in terms I can understand. Thank you very much for such a good explanation!) $\endgroup$ Mar 27, 2023 at 1:09
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    $\begingroup$ @MikaylaEckelCifrese You're very welcome! No, I wasn't thinking about pilot-wave theory; I was thinking about something ever weirder called superdeterminism. Superdeterminism is pretty unpopular in the physics community; most physicists (though not all) find it to be philosophically even more unsatisfying and "unnatural" than standard quantum mechanics. $\endgroup$
    – tparker
    Mar 27, 2023 at 2:17
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    $\begingroup$ @MikaylaEckelCifrese But you're right that the pilot-wave interpretation is another way to get around Bell's theorem and make QM more "natural" in some ways, at the expense if making more "unnatural" in others (in this case, adding in explicit faster-than-light causal influences). $\endgroup$
    – tparker
    Mar 27, 2023 at 2:20
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    $\begingroup$ @tparker, ah gotcha. Yeah superdeterminism basically postulates that the initial conditions at the beginning of the universe were exactly right to make all the experiments testing Bell's theorem only appear to validate it, right? To me, that's more than bizarre, but outright absurd, and could be used to explain literally ANY experimental results. $\endgroup$ Mar 27, 2023 at 2:44
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    $\begingroup$ @MikaylaEckelCifrese Yep, that's right, and yep, many/most QM experts feel the same way about superdeterminism as you do. $\endgroup$
    – tparker
    Mar 27, 2023 at 2:46
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There are no negative probabilities.

There exists in the phase space formulation of QM “quasi-probability” distributions $W(x,p)$ that are somewhere negative, but the probabilities that occur as a result of integrating such distributions are always non-negative.

To be explicit, $$ \vert\psi(x)\vert^2=\int dp\, W(x,p) $$ (for instance) is obtained by integrating all momenta of the joint quasi-distribution $W(x,p)$, $\vert \psi(x)\vert^2$ agrees with the probability density computed from the Born rule, and is everywhere non-negative so that the probability of finding the system in some space interval is always non-negative.

To illustrate this, look at the following two figures:

enter image description here enter image description here

They show the same quasi-distribution function, which contains regions of negativity near the center, as illustrated in the right figure: the near the central "anti"-peak the quasidistribution is clearly negative. However, integrating this quasi distribution along the line $p$ at $x=-2$ (shown on the left figure by the red line) yields

$$ \int dp W(x=-2,p)=\vert\psi(x=-2)\vert^2 > 0 $$ so even if there are regions of negativity the probability of finding the system near $x=-2$, which is $\vert\psi(x=-2)\vert^2\,dx$, is of course positive. For completeness here $\psi(x)$ is in fact $\psi_1(x)$, the wavefunction of the $n=1$ harmonic oscillator state.

To make absolutely explicit the difference between a probability density and a probability, consider the following normalized Gaussian wavefunction:

$$ \psi(\sigma)=\frac{e^{-x^2/(2\sigma)}}{(\pi \sigma)^{1/4}} \tag{1} $$ for $\sigma=1/8$. $\psi(1/8)$ is a perfectly legitimate wavefunction for a particle in a harmonic oscillator. The resulting probability density $\vert \psi(1/8)\vert^2$ is $>1$ near the origin:

enter image description here

but of course this is not a problem. Since $\int_{-\infty}^{\infty} dx \vert \psi(1/8)\vert^2=1$ and $\vert \psi(1/8)\vert^2 \ge 0$ everywhere, it follows that, for any subinterval $[a,b]$: $$ \int_a^b dx \vert \psi(1/8)\vert^2 \le 1 $$ even if $\vert \psi(1/8)\vert^2$ can be greater than one somewhere. Nobody in their right mind would suggest that, because the probability density $\vert \psi(1/8)\vert^2$ is greater than 1 somewhere, probabilities greater than 1 are possible possible.

It's the same argument for negative regions of Wigner functions, which are (quasi)probability densities, with the distinction that these (quasi)probability densities can be negative.

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    $\begingroup$ This answer is completely correct, but (and I mean this entirely respectfully to both you and the OP) I think it assumes more prior background than the OP probably has, so they may not (yet) be equipped to fully understand it. $\endgroup$
    – tparker
    Mar 27, 2023 at 0:14
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    $\begingroup$ @tparker You are probably right but the first linked question in the OP does refer to quasi-distribution. I would think at least that part of the question is covered by my answer; I do agree my answer is sparse precisely because I think technical details would not be helpful. $\endgroup$ Mar 27, 2023 at 0:25
  • $\begingroup$ @tparker added to my answer to reinforce your comment on the difference between a probability density and a probability. $\endgroup$ Mar 28, 2023 at 17:46
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The other answers are correct, but I think their emphasis is pulling away from the question slightly. Statements like "there are no negative probabilities in quantum mechanics" are reasonable for an understanding of wavefunction picture quantum mechanics. But other ways of doing quantum physics exist, and some of them have negative quasiprobabilties - which are either negative probabilities or at least a bit like them depending on your interpretation.

Wavefunctions vs. Wigner World

There are several different ways of thinking about quantum mechanics. They all give the same experimental predictions, but the maths and the interpretations are very different. The most common is the wavefunction picture, where we have a complex field that when mod-squared gives the probability density function (PDF). This complex field (wavefunction) is not at all a probability field, only its mod-squares relates to probability. The parts of the field are sometimes called "probability amplitudes". In field physics fields waves have amplitude and intensity, with intensity the mod-squared of amplitude. So the phrase "probability amplitude" means "a thing that relates to probability in the same way amplitude does to intensity".

The reason this complex field is in the theory at all, instead of us just sticking to PDFs is because a locally evolving PDF cannot reproduce our experimental observations. For example in interference experiments with two slits (or two beam splitters) we find that additional paths to the same end point sometimes cancel one another out. Which a locally evolving PDF cannot do.

Another approach to quantum mechanics is with quasiprobability distributions, (things like Wigner functions). These are a different way of modelling the same physics. Quasiprobability distributions play the role of wavefunctions in this view. In some sense they are simpler, because they are real valued (but sometimes negative), instead of complex valued. Even with quasiprobabilities every probability you can ever measure will be positive, but negatives do appear for some joint measurements that cannot actually be done. (eg. "the quasiprobability it has exactly this momentum and is at exactly this position simultaneously is negative").

Even though they are never seen the negatives can be used to "explain" interference. Imagine an interference experiment. We open a new slit, and see that fewer particles arrive at some places on the screen (the probability has dropped). But it feels like the new slit should only have added new options. So adding an option with negative probability makes it work. If this sounds fishy to you remember that it is exactly the same trick that is done with wavefunctions. Only frequencies (number of clicks) are actually measured. The quasiprobability distribution and wavefunction are never directly seen (but either can be worked out from the click distribution).

I take the view myself that probability is a subjective measure of how much some agent thinks they know about a system, and that if that agent wants to use negative probabilities to better describe the system then those are just as legitimate as positive ones. So in my writing I normally stop appending the word "quasi" after the introduction, and just talk about negative probabilities. The negatives are important after all: When we look at quantum mechanics using quasiprobability it is more or less true that the negatives are the thing that makes it quantum.

In summary, experiment tells us that a PDF doesn't quite work in quantum mechanics because of interference and locality issues. The basic way of dealing with this mathematically is to insert some (unseen) mathematical structure "underneath" the observed probabilities that can do interference. What kind of maths you put there is more or less a free choice. Two common options are wavefunctions and quasiprobabilty distributions. If you are using wavefunction theory the words "negative probability" are nonsensical. If you are using quasiprobability then the words "negative probability" are a perfectly acceptable shorthand for "negative quasiprobability", (where "quasiprobabiltiy" has the literal meaning "just like probability, but it can be negative sometimes").

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    $\begingroup$ Actually, nobody in a physics conference would seriously claim there are negative probabilities: statements of "negative probabilities" are provocative clickbaits. Some might talk of negativity of WF (which leads to cwey interesting consequences), and everyone agrees WF can have regions of negativity but the WF is a (quasi-)probability density so regions of negativity do not imply negative probabilities and is no more strange than regions where a PDF is larger than since what matters is their integral. $\endgroup$ Mar 27, 2023 at 13:01
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    $\begingroup$ @ZeroTheHero I have met a couple of them at conferences actually. Maybe I just go to weird conferences (quantum foundations people are strange). Although you are probably correct that provocative clickbaits are more common. $\endgroup$
    – Dast
    Mar 27, 2023 at 13:37
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    $\begingroup$ we have Feynman to blame for this, although he did like to be provocative and in that link he's really discussing negative probability regions in the quasi-probability distribution... I have myself never seen serious mainstream work where there are true negative probabilities. $\endgroup$ Mar 27, 2023 at 13:43
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    $\begingroup$ I agree with @ZeroTheHero that the distinction between probabilities and probability densities is important here. The fact that a quasiprobability distribution takes on negative values at certain points isn't so different from the fact that a regular PDF can be greater than 1 at certain points. In neither situation should any probability be thought of as either greater than 1 or less than 0, because those values aren't probabilities. In both cases, in order to get the actual probability, you need to integrate the (q-)PDF over a region such that the integral always lies in $[0,1]$. $\endgroup$
    – tparker
    Mar 27, 2023 at 23:47
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    $\begingroup$ So if you say that the existence of quasiprobability distributions should be interpreted as a negative probability, then you would also have to say the existence of a regular PDF should be interpreted as saying that probabilities can be greater than 1. But I don't think this perspective is useful. $\endgroup$
    – tparker
    Mar 27, 2023 at 23:49
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There are already some good answers here (I especially liked @Dast's answer). I would like just to add a couple of things.

  1. Let me refer you to Feynman's article: Simulating Physics with Computers. Int. J. Theor. Phys. 1982, 21, 467. The article mentions the Wigner function. One of the sections is called "Negative probabilities". I am not going to try to summarize this classical article. Just one quote:

The only difference between a probabilistic classical world and the equations of the quantum world is that somehow or other it appears as if the probabilities would have to go negative, and that we do not know, as far as I know, how to simulate.

  1. It seems to me now that it is possible to simulate quantum world using classical models and make sense of negative probabilities. Previously (Eur. Phys. J. C (2013) 73:2371), I proposed modeling quantum particles by a large number of particles and antiparticles (for example, an electron can be modeled by a collection of N+1 electrons and N positrons). Later, I found out (Entropy 2022, 24(2), 261) how to approximate a smooth charge distribution (with an integral total charge) by such collections of discrete quantized charges. Recently, I proved (Quantum Rep. 2022, 4(4), 486-508) that such approximation can be arbitrarily precise in the one-dimensional case (see the exact wording in the article).

This approach suggests that negative probabilities (for example, for the Wigner function) can be explained by existence of antiparticles.

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