Negative probabilities in QFT

1. Sep 21, 2015

tzimie

Is there something wrong with negative probabilities per se? I don't what to hear that they are "unphysical" because it is not clear what reality is. Can negative probabilities (or probabilities > 1) be somehow consistent with macroscopic realism => not causing any macroscopic weirdness (no matter how weird it is in quantum world)

2. Sep 21, 2015

rootone

I'm not sure if the idea of a negative probability even makes logical sense.
A probability of zero means that whatever the subject of discussion is has no possibility to happen at all.
How can we sensibly talk of something happening which is less likely than being impossible?

3. Sep 21, 2015

tzimie

As you know, only in some cases particles are detected directly based on their tracks or based on calculations of a some specific run. Very often we can't point to an exact run where particle had manifested. Good example are resonances. Such cases could be compatible with negative probabilities.

4. Sep 21, 2015

mathman

Probability by definition must be in the range [0,1]. If the quantity you are talking about is negative or > 1, then it is not a probability. The physics should tell you what it is.

5. Sep 21, 2015

Vanadium 50

Staff Emeritus
You don't get to ask a question and decide the answer. They are unphysical. Indeed, I challenge you to post any (correct, obviously) calculation of a physical probability that comes out negative.

6. Sep 21, 2015

HomogenousCow

How does one even obtain a negative probability using Bohr's interpretation of the state vector?

7. Sep 21, 2015

atyy

8. Sep 21, 2015

kith

Feynman wrote a famous article about negative probabilities. John Baez talks about it and some other approaches in a blog post.

The bottom line seems to be that negative probabilities may be interesting and useful auxillary quantities. Of course, they can't occur as probabilities for physical outcomes.

9. Sep 21, 2015

Staff: Mentor

Negative probabilities contradict the Kolmogorov axioms and are not allowed - it by definition is a positive measure.

When they appeareed in relativistic QM it showed it was sick. Later it was shown when negative probabilities appeared in for example a probability current it wasn't the probability that turned negative, which of course it cant, but that something else like charge changed sign and was the first indication of the existence antiparticles.

I remember discussing negative probabilities with my statistical modelling professor who for some reason thought QM allowed them. I had to explain it didn't. I had a collection of essays at the time that included Feynman's famous essay on it and gave him the book. He assured me they are useful in certain more advanced areas of probability where you allow them to go infinitesimally negative then take a limit later - but its more of a trick than actually allowing it.

Thanks
Bill

10. Sep 22, 2015

Stephen Tashi

As a technicality, we must distinguish between "a probability" and "the value of a probability density function". A given value of a probability density function need not be a probability. Furthermore, there is no formalization in the Kolmogorov axioms of the concept that events that have probabilities can "actually happen".

11. Sep 22, 2015

Demystifier

Probabilities are useful in physics because you can measure them, provided that you adopt the frequency interpretation of probability. Other interpretations of probability also exist, but those cannot be directly measured.

Negative probability cannot be interpreted in terms of frequency. So whatever the interpretation of negative probability is, it cannot directly be measured. Therefore it is not so much useful in physics. Perhaps it can be useful as a mathematical tool in intermediate calculations, but not more than that.

12. Sep 22, 2015

Jazzdude

This is surely correct, but I think not really relevant for negative probabilities. A negative probability density will be able to produce a negative probability, if it is negative on any set of finite measure. If the set is of measure zero then there is an equivalent probability density without negative values on that set. So the non-negativity is sort of strict for the probability density too. This is different from for example the requirement that probabilities be less than or equal to 1, which doesn't hold for densities.

That said, there are negative "pseudo probability densities" if you try to construct a classical phase space from the quantum state space. The Wigner-Weyl function of a state gives a real density that is not necessarily non-negative as a function of the positions and momenta of the system. Integrating over that density will also violate non-negativity in general. However, this density function can still be interpreted as a probability density if you restrict the way of getting probabilities out of it by integration. Instead of directly integrating, we first multiply point-wise with a Wigner function of another state (or of any non-negative Hermitian operator in fact) and then integrate over the whole phase space. This is guaranteed to give a non-negative result and we can interpret it as the probability of finding the second state in the first.

So this quantum "phase space" representation is a very good example of how to utilise something potentially negative as a density function for probabilities without getting negative probabilities. But only with a non-standard definition of measures on that density.

Cheers,

Jazz

Last edited: Sep 22, 2015
13. Sep 22, 2015

Jazzdude

In the title of this thread you refer to QFT. Where exactly in QFT do you think you encounter negative probabilities? Negative probabilities arose in relativistic quantum mechanics before the idea of field quantisation was established. And at that time those negative probabilities were pretty much a deal breaker that led to the search of better descriptions.

Cheers,

Jazz

14. Sep 22, 2015

tzimie

What's about the following example. Particle-resonance is detected by observing a positive spike, a higher frequency for some energy. If such spike is observed we conclude there is a particle with very short lifetime and with positive probability of creation. What if probability is negative? Then we should detect a negative spike, some energy will be avoided. As such hypothetical particle has extremely short lifetime, it does not cause any macroscopic weirdness. Just a model.

15. Sep 22, 2015

vanhees71

As all other posters have pointed out, negative probabilities are a contractio in adjecto. They don't make sense in mathematics and thus also not in physics. The good news is that there are no negative probabilities in QFT; only wrong attempts to quantize gauge fields. So you don't even need to think about them. Either you use the path-integral formalism with the Faddeev-Popov technique (this you should do first) and/or the covariant operator-quantization formalism. For the first topic I recommend

Bailin, Love, Gauge Theories

and for the second the review paper by Kugo and Ojima:

Taichiro Kugo and Izumi Ojima. Local covariant operator formalism of nonabelian gauge theories and quark confinement problem. Prog. Theor. Phys. Suppl., 66:1 (1979)
http://dx.doi.org/10.1143/PTPS.66.1

16. Sep 22, 2015

Demystifier

A typical spike you are talking about looks like this:
https://qph.is.quoracdn.net/main-qimg-ac61f5ba8ae784a7bc31952ca51caf2f?convert_to_webp=true [Broken]
You have a large background and a small spike slightly above the background. In principle you may also have a spike below the background, but the value (on the y-axis) cannot be negative. You cannot have less than zero number of events at a given energy. In other words, the probability of given energy cannot be less than zero.

Last edited by a moderator: May 7, 2017
17. Sep 22, 2015

tzimie

By negative spike I mean it goes below the green line.

18. Sep 22, 2015

Demystifier

I know, the green line is the background I was talking about. As you can see, below the green line the probability is still positive.

19. Sep 22, 2015

martinbn

They can be interpreted as less often than never.

20. Sep 22, 2015

Demystifier

I am more than convinced (more than 100%) that it doesn't make sense.

Which leads us to an interesting question: Does Bayesian interpretation of probability allow probabilities larger than 1 or smaller than 0?
(Hint: of course not.)

Last edited: Sep 22, 2015
21. Sep 22, 2015

Staff: Mentor

Yes.

And via the strong law of large numbers leads to measuring them in a frequentest way which makes it utterly obvious it cant.

Thanks
Bill

22. Sep 24, 2015

stevendaryl

Staff Emeritus
Just another point about negative probabilities. There can be a quantity that can be interpreted as a probability in many cases, but in extreme cases, the probability interpretation breaks down, and a different interpretation is needed.

Here's an example that isn't really physics, but I think it relates to the appearance of negative values for the "current" in solutions to Klein-Gordon equation:

Suppose I have a string in the x-y plane that runs basically in the x-direction, as shown in the top figure below. You can imagine that the exact shape of the string is nondeterministic. So instead of describing it by $y = f(x)$, a function, we might instead have a probability density function $P(y,x)$: the probability of the string passing through the point $(x,y)$. If the shape of the string is well-behaved in a certain sense, you would expect the density function to satisfy:

$\int_{-\infty}^{+\infty} P(y,x) dy = 1$

That is, as long as the string is running basically in the x-direction, there will only be a single value of $y$ for any fixed $x$.

If you knew something about how the probabilities for various shapes were computed, you could come up with a kind of "Green's function" $G(x_1, y_1, x_2, y_2)$, which would be the probability of the string passing through $(x_1, y_1)$ given that it passed through $(x_2, y_2)$.

Now, imagine a wilder shape for the string, as shown in the middle figure. In this case, the string doubles-back at several places, so that $y$ is no longer a single-valued function of $x$. At point $\mathcal{A}$, there is only one value of $y$ that the string passes through. At point $\mathcal{B}$, there are three values of $y$. At point $\mathcal{C}$, there are five values. So it is no longer the case that $P(y,x)$ must add up to 1 when integrated over all $y$ (with x held fixed). However, what we can do is this: Let $x(s), y(s)$ be a parametrization of the path of the string. In terms of this parametrization, define the direction of the string at a point, $D(x,y)$ to be +1 if $\frac{dx}{ds} > 0$ and -1 if $\frac{dx}{ds} < 0$. Then we can define a new way of counting the string:

$N(y,x) = D(x,y) P(y,x)$

As shown in the bottom figure, the places where the string doubles back always results in some sections having a negative value for $N(y,x)$ (the sections shown in red) and some sections having a positive value (the sections shown in black). For a fixed value of $x$, the red and black sections cancel, except for one black section. So we conclude that

In terms of this quantity, it will be true that $\int_{-\infty}^{+\infty} N(y,x) dx = 1$

If you had a formula for computing $N(y,x)$, then in the simple cases (no doubling back), you can identify $N$ with $P$ and interpret it as a probability density. But in the more complex cases, it can be negative at points, and it has to be interpreted as some kind of "directed string density", which can be positive or negative.

23. Sep 24, 2015

Demystifier

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