# Negative Probabilities

I found this great article from 2013 posts of Azimuth blog. I didn’t understand it completely but I should say that is amazing! I really enjoy it. I must read some other times.

## Negative Probabilities

The physicists Dirac and Feynman, both bold when it came to new mathematical ideas, both said we should think about negative probabilities.

These days, Kolmogorov’s axioms for probabilities are used to justify formulating probability theory in terms of measure theory. Mathematically, the theory of measures that take negative or even complex values is well-developed. So, to the extent that probability theory is just measure theory, you can say a lot is known about negative probabilities.

But probability theory is not just measure theory; it adds its own distinctive ideas. To get these into the picture, we really need to ask some basic questions, like: what could it mean to say something had a negative chance of happening?

I really have no idea.

In this paper:

• Paul Dirac, The physical interpretation of quantum mechanics, Proc. Roy. Soc. London A 180 (1942), 1–39.

Dirac wrote:

Negative energies and probabilities should not be considered as nonsense. They are well-defined concepts mathematically, like a negative of money.

In fact, I think negative money could have been the origin of negative numbers. Venetian bankers started writing numbers in red to symbolize debts—hence the phrase ‘in the red’ for being in debt. So, you could say negative numbers were invented to formalize the idea of debt and make accounting easier. Bankers couldn’t really get rich if negative money didn’t exist.

A negative dollar is a dollar you owe someone. But how can you owe someone a probability? I haven’t figured this out.

Unsurprisingly, the clearest writing about negative probabilities that I’ve found is by Feynman:

• Richard P. Feynman, Negative probability, in Quantum Implications: Essays in Honour of David Bohm, eds. F. David Peat and Basil Hiley, Routledge & Kegan Paul Ltd, London, 1987, pp. 235–248.

He emphasizes that even if the final answer of a calculation must be positive, negative numbers are often allowed to appear in intermediate steps… and that this can happen with probabilities.

Let me quote some:

Some twenty years ago one problem we theoretical physicists had was that if we combined the principles of quantum mechanics and those of relativity plus certain tacit assumptions, we seemed only able to produce theories (the quantum field theories) which gave infinity for the answer to certain questions. These infinities are kept in abeyance (and now possibly eliminated altogether) by the awkward process of renormalization. In an attempt to understand all this better, and perhaps to make a theory which would give only finite answers from the start, I looked into the” tacit assumptions” to see if they could be altered.

One of the assumptions was that the probability for an event must always be a positive number. Trying to think of negative probabilities gave me cultural shock at first, but when I finally got easy with the concept I wrote myself a note so I wouldn’t forget my thoughts. I think that Prof. Bohm has just the combination of imagination and boldness to find them interesting and amusing. I am delighted to have this opportunity to publish them in such an appropriate place. I have taken the opportunity to add some further, more recent, thoughts about applications to two state systems.

Unfortunately I never did find out how to use the freedom of allowing probabilities to be negative to solve the original problem of infinities in quantum field theory!

It is usual to suppose that, since the probabilities of events must be positive, a theory which gives negative numbers for such quantities must be absurd. I should show here how negative probabilities might be interpreted. A negative number, say of apples, seems like an absurdity. A man starting a day with five apples who gives away ten and is given eight during the day has three left. I can calculate this in two steps: 5 -10 = -5 and -5 + 8 + 3. The final answer is satisfactorily positive and correct although in the intermediate steps of calculation negative numbers appear. In the real situation there must be special limitations of the time in which the various apples are received and given since he never really has a negative number, yet the use of negative numbers as an abstract calculation permits us freedom to do our mathematical calculations in any order simplifying the analysis enormously, and permitting us to disregard inessential details. The idea of negative numbers is an exceedingly fruitful mathematical invention. Today a person who balks at making a calculation in this way is considered backward or ignorant, or to have some kind of a mental block. It is the purpose of this paper to point out that we have a similar strong block against negative probabilities. By discussing a number of examples, I hope to show that they are entirely rational of course, and that their use simplifies calculation and thought in a number of applications in physics.

First let us consider a simple probability problem, and how we usually calculate things and then see what would happen if we allowed some of our normal probabilities in the calculations to be negative. Let us imagine a roulette wheel with, for simplicity, just three numbers: 1, 2, 3. Suppose however, the operator by control of a switch under the table can put the wheel into one of two conditions A, B in each of which the probability of 1, 2, 3 are different. If the wheel is in condition A, the probability of 1 is p1A = 0.3 say, of 2 is p2A = 0.6, of 3 is p3A =0.1. But if the wheel is in condition B, these probabilities are

p1B = 0.1, p2B = 0.4, p3B = 0.5

say as in the table.

 Cond. A Cond. B 1 0.3 0.1 2 0.6 0.4 3 0.1 0.5

We, of course, use the table in this way: suppose the operator puts the wheel into condition A 7/10 of the time and into B the other 3/10 of the time at random. (That is the probability of condition A, pA = 0.7, and of B, pB = 0.3.) Then the probability of getting 1 is

Prob. 1 = 0.7 (0.3) + 0.3 (0.1) = 0.24,

etc.

[…]

Now, however, suppose that some of the conditional probabilities are negative, suppose the table reads so that, as we shall say, if the system is in condition B the probability of getting 1 is -0.4. This sounds absurd but we must say it this way if we wish that our way of thought and language be precisely the same whether the actual quantities pi α in our calculations are positive or negative. That is the essence of the mathematical use of negative numbers—to permit an efficiency in reasoning so that various cases can be considered together by the same line of reasoning, being assured that intermediary steps which are not readily interpreted (like -5 apples) will not lead to absurd results. Let us see what p1B = -0.4 “means” by seeing how we calculate with it.

He gives an example showing how meaningful end results can sometimes arise even if the conditional probabilities like p1B are negative or greater than 1.

It is not my intention here to contend that the final probability of a verifiable physical event can be negative. On the other hand, conditional probabilities and probabilities of imagined intermediary states may be negative in a calculation of probabilities of physical events or states. If a physical theory for calculating probabilities yields a negative probability for a given situation under certain assumed conditions, we need not conclude the theory is incorrect. Two other possibilities of interpretation exist. One is that the conditions (for example, initial conditions) may not be capable of being realized in the physical world. The other possibility is that the situation for which the probability appears to be negative is not one that can be verified directly. A combination of these two, limitation of verifiability and freedom in initial conditions, may also be a solution to the apparent difficulty.

The rest of this paper illustrates these points with a number of examples drawn from physics which are less artificial than our roulette wheel. Since the result must ultimately have a positive probability, the question may be asked, why not rearrange the calculation so that the probabilities are positive in all the intermediate states? The same question might be asked of an accountant who subtracts the total disbursements before adding the total receipts. He stands a chance of going through an intermediary negative sum. Why not rearrange the calculation? Why bother? There is nothing mathematically wrong with this method of calculating and it frees the mind to think clearly and simply in a situation otherwise quite complicated. An analysis in terms of various states or conditions may simplify a calculation at the expense of requiring negative probabilities for these states. It is not really much expense.

Our first physical example is one in which one· usually uses negative probabilities without noticing it. It is not a very profound example and is practically the same in content as our previous example. A particle diffusing in one dimension in a rod has a probability of being at $x$ at time $t$ of $P(x,t)$satisfying

$\partial P(x,t)/\partial t = -\partial^2 P(x,t)/\partial x^2$

Suppose at $x =0$ and $x =\pi$ the rod has absorbers at both ends so that $P(x,t) = 0$ there. Let the probability of being at $x$ at $t = 0$ be given as $P(x,0) =f(x).$ What is $P(x,t)$ thereafter? It is

$\displaystyle{ P(x,t) = \sum_{n=1}^\infty P_n \; \sin x \;\exp(-n^2 t) }$

where $P_n$ is given by

$x \displaystyle{ f(x) = \sum_{n = 1}^\infty P_n \; \sin n x }$

or

$x \displaystyle{ P_n = \frac{2}{\pi} \int_0^\pi f(x) \sin nx \; dx }$

The easiest way of analyzing this (and the way used if $P(x,t)$ is a temperature, for example) is to say that there are certain distributions that behave in an especially simple way. If $f(x)$ starts as $\sin nx$ it will remain that shape, simply decreasing with time, as $e^{-n^2 t}$ Any distribution $f(x)$can be thought of as a superposition of such sine waves. But $f(x)$ cannot be $\sin nx$ if $f(x)$ is a probability and probabilities must always be positive. Yet the analysis is so simple this way that no one has really objected for long.

He also gives examples from quantum mechanics, but the interesting thing about the examples above is that they’re purely classical—and the second one, at least, is something physicists are quite used to.

Sometimes it’s good to temporarily put aside making sense of ideas and just see if you can develop rules to consistently work with them. For example: the square root of -1. People had to get good at using it before they understood what it really was: a rotation by a quarter turn in the plane.

Along those, lines, here’s an interesting attempt to work with negative probabilities:

• Gábor J. Székely, Half of a coin: negative probabilitiesWilmott Magazine (July 2005), 66–68.

He uses rigorous mathematics to study something that sounds absurd: ‘half a coin’. Suppose you make a bet with an ordinary fair coin, where you get 1 dollar if it comes up heads and 0 dollars if it comes up tails. Next, suppose you want this bet to be the same as making two bets involving two separate ‘half coins’. Then you can do it if a half coin has infinitely many sides numbered 0,1,2,3, etc., and you win $n$ dollars when side number $n$ comes up….

… and if the probability of side $n$ coming up obeys a special formula…

and if this probability can be negative sometimes!

This seems very bizarre, but the math is solid, even if the problem of interpreting it may drive you insane.

Let’s see how it works. Consider a game $G$ where the probability of winning $n = 0, 1, 2, \dots$ dollars is $g(n).$ Then we can summarize this game using a generating function:

$\displaystyle{ G(z) = \sum_{n = 0}^\infty g(n) , z^n }$

Now suppose you play two independent games like this, $G$ and another one, say $H,$ with generating function

$\displaystyle{ H(z) = \sum_{n = 0}^\infty h(n) , z^n }$

Then there’s a new game $GH$ that consists of playing both games. The reason I’m writing it as $GH$ is that its generating function is the product

$\displaystyle{ G(z) H(z) = \sum_{m,n = 0}^\infty g(m) h(n) z^{m+n} }$

See why? With probability $g(m) h(n)$ you win $m$ dollars in game $G$and $n$ dollars in game $H,$ for a total of $m + n$ dollars.

The game where you flip a fair coin and win 1 dollar if it lands heads up and 0 dollars if lands tails up has generating function

$\displaystyle{ G(z) = \frac{1}{2}(1 + z) }$

The half-coin is an imaginary game $H$ such that playing two copies of this game is the same as playing the game $G.$ If such a game really existed, we would have

$G(z) = H(z)^2$

so

$\displaystyle{ H(z) = \sqrt{\frac{1}{2}(1 + z)} }$

However, if you work out the Taylor series of this function, every even term is negative except for the zeroth term. So, this game can exist only if we allow negative probabilities.

(Experts on generating functions and combinatorics will enjoy how the coefficients of the Taylor series of $H(z)$ involves the Catalan numbers.)

By the way, it’s worth remembering that for a long time mathematicians believed that negative numbers made no sense. As late as 1758 the British mathematician Francis Maseres claimed that negative numbers

… darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.

So opinions on these things can change. And since I’ve spent a lot of time working on ‘sets with fractional cardinality’, and have made lots of progress on that idea, and other strange ideas, I like to spend a little time now and then investigating other nonsensical-sounding generalizations of familiar concepts.

This paper by Mark Burgin has a nice collection of references on negative probability:

• Mark Burgin, Interpretations of negative probability.

He valiantly tries to provide a frequentist interpretation of negative probabilities. He needs ‘negative events’ to get negative frequencies of events occurring, and he gives this example:

To better understand how negative elementary events appear and how negative probability emerges, consider the following example. Let us consider the situation when an attentive person A with the high knowledge of English writes some text T. We may ask what the probability is for the word “texxt” or “wrod” to appear in his text T. Conventional probability theory gives 0 as the answer. However, we all know that there are usually misprints. So, due to such a misprint this word may appear but then it would be corrected. In terms of extended probability, a negative value (say, -0.1) of the probability for the word “texxt” to appear in his text T means that this word may appear due to a misprint but then it’ll be corrected and will not be present in the text T.

Maybe he’s saying that the misprint occurs with probability 0.1 and then it ‘de-occurs’ with the same probability, giving a total probability of

$0.1 - 0.1 = 0$

I’m not sure.

Here’s another paper on the subject:

• Espen Gaarder Haug, Why so negative to negative probabilities?Wilmott Magazine.

It certainly gets points for a nice title! However, like Burgin’s paper, I find it a lot less clear than what Feynman wrote.

Notice that like Székely’s paper, Haug’s originally appeared in the Wilmott Magazine. I hadn’t heard of that, but it’s about finance. So it seems that the bankers, having invented negative numbers to get us into debt, are now struggling to invent negative probabilities! In fact Haug’s article tries some applications of negative probabilities to finance.

Scary.

For further discussion, with some nice remarks by the quantum physics experts Matt Leifer and Michael Nielsen, see the comments on my Google+ post on this topic. Matt Leifer casts cold water on the idea of using negative probabilities in quantum theory. On the other hand, Michael Nielsen points out some interesting features of the Wigner quasiprobability distribution, which is the best possible attempt to assign a probability density for a quantum particle to have any given position and momentum. It can be negative! But if you integrate it over all momenta, you get the probability density for the particle having any given position:

$|\psi(x)|^2$

And if you integrate it over all positions, you get the probability density for the particle having any given momentum:

$|\widehat{\psi}(p)|^2$

Reference: Azimuth Blog