# The probability of something happening

1. May 31, 2014

### Duhoc

is equal to the amplitude squared in quantum mechanics.

Why?

2. May 31, 2014

### Simon Bridge

... not quite, but a fair summary.

Because thems the rules.
That's the way Nature works.

Last edited by a moderator: Sep 25, 2014
3. May 31, 2014

### Staff: Mentor

Because we developed quantum mechanics to match the way the world works. The probability is what it is; we can determine what it is by doing repeated measurements; and it turns out that it is always the square of an amplitude that obeys particular rules. We've chosen, for historical reasons, to call these rules "quantum mechanics".

4. May 31, 2014

### Staff: Mentor

"But why does Nature work that way?"

We don't know. What, you're surprised that physicists don't know everything?

5. Jun 1, 2014

### Staff: Mentor

Gleason's Theorem:
http://kof.physto.se/cond_mat_page/theses/helena-master.pdf [Broken]

The bottom line reason - non contextuality.

The paper above gives two versions of the theorem.

The first version is Gleason's original version which is quite difficult but based on resolutions of the identity.

The second version is much simpler but based on the stronger assumption of POVM's.

For simplicity I will use the second.

A POVM is a set of positive operators Ei ∑ Ei =1.

The fundamental axiom of QM is a generalized measurement is described by a POVM such that outcome i is determined by Ei, and only by Ei.

Here only by Ei means regardless of what POVM the Ei belongs to the probability is the same. This is the assumption of non contextuality and the rock bottom essence of the probability rule of QM, known as Born's rule.

You can run through the proof in the link above. Its proof of continuity is a bit harder than it needs to be so I will give a simpler one. If E1 and E2 are positive operators define E2 < E1 as a positive operator E exists E1 = E2 + E. This means f(E2) <= f(E1). Let r1n be an increasing sequence of rational's whose limit is the irrational number c. Let r2n be a decreasing sequence of rational's whose limit is also c. If E is any positive operator r1nE < cE < r2nE. So r1n f(E) <= f(cE) <= r2n f(E). Thus by the pinching theorem f(cE) = cf(E).

Hence a positive operator P of unit trace exists such that probability Ei = Trace (PEi).

This is called the Born rule and by definition P is the state of the system.

You probably haven't seen it in that form. To put it in a more recognisable form by definition a Von Neumann measurement is described by a resolution of the identity which is a POVM where the Ei are disjoint. Associate yi with each outcome to give O = ∑ yi Ei. O is a Hermitian operator and via the spectral theorem you can recover uniquely the yi and Ei. By definition O is called the observable associated with the measurement. The expected value of O E(O) = ∑ yi probability outcome i = ∑ yi Trace (PEi) = Trace (PO).

A state of the form |u><u| is called pure. A state that is the convex sum of pure states is called mixed. It can be shown (it's not hard) all states are either mixed or pure. For a pure state E(O) = trace (|u><u|O) = <u|O|u> which is the most common form of the Born rule.

JTBell is correct - I haven't really proven the Born Rule in the sense of doing it without some other assumption. The assumption I made is non-contextuality. Its just it seems less like being pulled out of nowhere than Born's rule. Also as a by-product it shows what states are and why they must exist - sort of two things for the price of one. And its good to know as well from discussions you may see about Kochen-Specker etc. Its a simple corollary of Gleason.

Thanks
Bill

Last edited by a moderator: May 6, 2017
6. Jun 1, 2014

### tom.stoer

Bill, I don't think that Gleason's theorem answers this question completely.

Gleason's theorem explains that if there's a probabilistic theory to be formulated on Hilbert spaces, then the probability is given by Born's rule. But the theorem doesn't answer the if.

Last edited: Jun 1, 2014
7. Jun 1, 2014

### Simon Bridge

I agree - it just changes the words used to express the question.
One could get a slightly different kind of answer though if one asked: "How is it that a probabilistic theory based on Hilbert spaces is so useful we have a whole field of science to use it?"

This kind of thing leads to philosophy though.
I don't have that much whiskey right now.

@Duhoc: how do you like the replies so far?

8. Jun 1, 2014

### Staff: Mentor

It follows from the assumption that goes into it which I will state explicitly - I wasn't as careful as I should have been.

An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

Whether such actually resolves anything is a matter of opinion. I believe it does because its no longer pulled out of the air so to speak - you can see what goes into it - the most important being non-contextuality - and from the outset see exactly what you have to do violate it ie it must be contextual- and things like BM are explicitly contextual.

Thanks
Bill

Last edited: Jun 1, 2014
9. Jun 1, 2014

### tom.stoer

agreed

10. Jun 3, 2014

### naima

QM theory does not give a probability value to events.
We only have amplitude for couples <f|g>
So it is a theory of conditional probabilities. Who knows Luders Rule?

11. Jun 3, 2014

### Zarqon

I think the best answer to this question is: Because using squares as probabilities is the only way to create a universe with a probability rule set which is more interesting than the classical one (with the amplitudes themselves as probability).

That was to express it very shortly, but a good description of that thought can be found at http://www.scottaaronson.com/democritus/lec9.html

12. Jun 3, 2014

### tom.stoer

Of course. Given a quantum system prepared in state |f> we can calculate the probability to find it in a state |g>.

13. Jun 3, 2014

### naima

Classically, conditional probability p(B knowing A) is p(A and B) / p(B).
But in QM if the operators of A and B do not commute, there is a problem.
The Luders Rule gives a mathematical definition for this situation with Hilbert spaces. Click on appendix B read the paper.

14. Jun 3, 2014

### Duhoc

I originally asked the question because one time I heard a physics lecture and the lecturer was talking about some parameter in an equation and he said, "well we can make it positive by simply squaring it." And this bothered me because I began to think that nature could behave in a way that is determined by mathematical logic. In any case, thinking about quantum mechanics might be the most interesting topic of our day. I know very well that I could have looked up what I wanted to know on wikepedia, which by the way, I have done. But, yes, I enjoyed the discussion and all of the replies. If quantum mechanics is an information system which determines physics at the "very small level" I am interested in knowing about that system and what it implies. Maybe the ability to calculate the probability of an electron being in any one place means that it has to be someplace, perhaps even in another universe. But anyway, here is the most germaine answer I found on wikepedia, and it makes some sense to someone who doesn't know much about math.

Probability amplitudes have special significance because they act in quantum mechanics as the equivalent of conventional probabilities, with many analogous laws, as described above. For example, in the classic double-slit experiment, electrons are fired randomly at two slits, and the probability distribution of detecting electrons at all parts on a large screen placed behind the slits, is questioned. An intuitive answer is that P(through either slit) = P(through first slit) + P(through second slit), where P(event) is the probability of that event. This is obvious if one assumes that an electron passes through either slit. When nature does not have a way to distinguish which slit the electron has gone though (a much more stringent condition than simply "it is not observed"), the observed probability distribution on the screen reflects the interference pattern that is common with light waves. If one assumes the above law to be true, then this pattern cannot be explained. The particles cannot be said to go through either slit and the simple explanation does not work. The correct explanation is, however, by the association of probability amplitudes to each event. This is an example of the case A as described in the previous article. The complex amplitudes which represent the electron passing each slit (ψfirst and ψsecond) follow the law of precisely the form expected: ψtotal = ψfirst + ψsecond. This is the principle of quantum superposition. The probability, which is the modulus squared of the probability amplitude, then, follows the interference pattern under the requirement that amplitudes are complex: P = |\psi_{\rm{first}} + \psi_{\rm{second}}|^2 = |\psi_{\rm{first}}|^2 + |\psi_{\rm{second}}|^2 + 2 |\psi_{\rm{first}}| |\psi_{\rm{second}}| \cos (\varphi_1-\varphi_2). Here, \varphi_1 and \varphi_2 are the arguments of ψfirst and ψsecond respectively. A purely real formulation has too few dimensions to describe the system's state when superposition is taken into account. That is, without the arguments of the amplitudes, we cannot describe the phase-dependent interference. The crucial term 2 |\psi_{\rm{first}}| |\psi_{\rm{second}}| \cos (\varphi_1-\varphi_2) is called the "interference term", and this would be missing if we had added the probabilities. However, one may choose to devise an experiment in which he observes which slit each electron goes through. Then, case B of the above article applies, and the interference pattern is not observed on the screen. One may go further in devising an experiment in which he gets rid of this "which-path information" by a "quantum eraser". Then the case A applies again and the interference pattern is restored.

15. Jun 3, 2014

### Jilang

Naima, I hadn't heard of Luders Rule so thanks for that. I have often wondered if joint probabilities came into the equation but have never found an answer for it. For the Born rule what is coupling to what? I.e. It seems to be the probability of a particle being In a certain state, given that the particle is in a certain state.....

16. Jun 3, 2014

### Duhoc

I believe this relationship was developed by mathematical calculation and most likely predicted its resuls. I think the individual won the Nobel Prize for it.

17. Jun 4, 2014

### naima

If we know that the state of a system is given by a unit trace density matrix $\rho$ for everiy measurement G the mean value of the outputs is $Tr(\rho G)$
If G is a projector (G2 = G) the eigenvalues are 0 or 1 so $Tr(\rho G)$ = <G> is the probability that G is true (G = 1)

I found how to use Ludres Rule to generalize this to conditional probability "eprints.ucm.es/10636/1/T31869.pdf" [Broken]
In the author's notation $\rho$ is W

Last edited by a moderator: May 6, 2017
18. Jun 4, 2014

### naima

We see that when knowing Q, W -> QWQ / Tr (QWQ)
It is close to another formula with POVM:
If a POVM gives a "i" output the density matrix W is replaced by $Q_i WQ_i^\dagger / Tr (Q_i W Q_i^\dagger)$