Why amplitude squared gives probability / Schrodinger Equation

[phina]

I actually read QED about a year ago.
I watched that part of that lecture, where it seemed Feynman basically said "Why does it work this way? I dunno! No one does!" so... I guess I'm not finding my answer, haha.

 

[ArjenDijksman]

I actually read QED about a year ago.
I watched that part of that lecture, where it seemed Feynman basically said "Why does it work this way? I dunno! No one does!" so... I guess I'm not finding my answer, haha.

Come on, don't let yourself be impeded by such negative statements, even by Feynman. Quantum mechanics stems from a very simple fundamental principle: a quantum particle is represented by a spinning arrow. The motion of the tip of a spinning arrow is always perpendicular to the arrow itself. Draw this on paper. In complex notation, perpendicular means exp(i.pi/2) or in shorthand i. So the equation that describes the spinning of the arrow |psi> is simply:

|psi(t+dt)> - |psi(t)> = i.|psi(t)>.omega.dt

with omega the angular velocity of the tip of the arrow. This is a generalized form of the time dependent Schrödinger equation for arrow-like objects (needle, rod, twirling baton...).

 

[RUTA]

Come on, don't let yourself be impeded by such negative statements, even by Feynman. Quantum mechanics stems from a very simple fundamental principle: a quantum particle is represented by a spinning arrow.

“All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics. It has survived all tests and there is no reason to believe that there is any flaw in it. We all know how to use it and how to apply it to problems; and so we have learned to live with the fact that nobody can understand it.” Murray Gell-Mann in L. Wolpert, The Unnatural Nature of Science (Harvard University Press, Cambridge, MA, 1993), p. 144.

 

[ArjenDijksman]

“All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics. It has survived all tests and there is no reason to believe that there is any flaw in it. We all know how to use it and how to apply it to problems;..."

Yes.

"...and so we have learned to live with the fact that nobody can understand it.” Murray Gell-Mann in L. Wolpert, The Unnatural Nature of Science (Harvard University Press, Cambridge, MA, 1993), p. 144.

Too sad. I can understand that senior physicists express this disillusion but in order to ensure progress for the future, it's important to encourage junior physicists to question it.

"Murray Gell-Mann said that we all know how to calculate, how to use QM, but none of us really understands what is behind the formalism, what it is saying about nature. That has to be answered in some way." Basil Hiley in an http://www.goertzel.org/dynapsyc/1997/interview.html".

 

[ytuab]

“All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics. It has survived all tests and there is no reason to believe that there is any flaw in it.

We often see these phrases which are full of confidence in the QM textbooks.
But I think we need more explanation about QM.

For example, the probability density of the electron of the point at infinity (near the point at infinity) in the hydorogen atom is not zero.
And when an electron is released by an apparatus, if the electron's position at first is expressed by a wave packet,
(or the instant the delta function which means the electron's position becomes a wave packet,)
the probability density of the electron of the point at infinity(near the point at infinity) is not zero.

I want to know how QM textbooks explain this phenomina.
Are there any flaw in QM?

 

[Fra]

there is no reason to believe that there is any flaw in it.

This is obviously where some of us differ.

I'd prefer to hold such statements, at least until we have coherent theory of measurement that also includes gravity? Which we do not yet have.

In such a context, there are rather several good reasons IMHO to believe that QM is more like an approximation, or special case, just like SR was as special case of GR.

I think it's exactly the fact that a deeper understanding is lacking that makes us not konwing wether QM is fundamental/universal, or just a special case.

/Fredrik

 

[ArjenDijksman]

For example, the probability density of the electron of the point at infinity (near the point at infinity) in the hydorogen atom is not zero.
And when an electron is released by an apparatus, if the electron's position at first is expressed by a wave packet,
(or the instant the delta function which means the electron's position becomes a wave packet,)
the probability density of the electron of the point at infinity(near the point at infinity) is not zero.

I want to know how QM textbooks explain this phenomina.
Are there any flaw in QM?

That's a good remark. In fact, the current way of solving analytically the radial part of the Schrödinger equation for the hydrogen atom needs assumptions (on the shape of the Coulomb field) and approximations (asymptotic behaviour of power series) that are instilled by classical physics. I haven't seen any pure quantum derivation in the textbooks. So I don't consider these solutions to be purely quantum mechanics and the flaw at infinity doesn't come from the 5 (or 6) quantum principles, but from our approximate methods of solving.

 

[jambaugh]

As far as determining square amplitude probabilities from observed wave-like phenomena. That is a good point as far as logical inference goes. But it can also be turned around in that the "why" of the wave phenomena is explainable in terms of the fact that probabilities are squares of the amplitudes.

Upon more though I think the following best explains the fundamental "why" of Born probability formula.

Recall we can dispense with the wave function per se and work with the density operator in a more general setting. It is the density operator which is the quantum correspondent to a classical probability distribution. More to the point the density operator (along with the trace operation) is used as a "co-operator" i.e. a dual to the elements of the space of observables.

System modes are operationally defined by the expectation values they give for all observables.

That duality matches up perfectly with the duality of passive vs. active transformations on the system-measuring device relationship. The dynamics follows necessarily from this duality relationship and how we posit the observables themselves transform.

Once we decide to represent observables with operators, and thence system modes with "co-operators", we can then "take the square root" of this representation and look at the (ideals) Hilbert spaces which these entities are mathematically defined to act upon. When we stick to the more general density (co)operator language there is no mystery as one is indeed working with probabilities (and correlations) and not their "square roots".

To give meaning to the Hilbert space elements we can view their representation of (sharp) system modes as a special type of quantification (extension to variable particle number with either 0 or 1 "particle" number) wherein we describe creation and annihilation of the system in a specific mode via Bra's and Ket's treated as as operators. (I call this "singular statistics".) It better matches what we actually do experimentally in that we not only consider non-destructive measurements (such as birefringent crystal polarization measurement) but also semi-destructive measurements (such as polaroid filters). We should represent these differently as their effect on systems is empirically distinguishable.

In short, the squaring of amplitudes to define probabilities is a manifestation of the less than ideal square-rooting of the general system mode representations, the density (co)operators.

 

[atyy]

Wave interference phenomena, ie. the relative phase between waves, is important in quantum mechanics. When we take the absolute value, the absolute phase is not important, but the relative phase is.

Here theta is the absolute phase, and the result is independent of theta:
|exp(i.theta).psi|.

Here theta is the relative phase, and the result is not independent of theta:
|exp(i.theta).psi + phi|.

 

[carllooper]

What I really don't understand is why you square the amplitude to get the probability. I understand that the amplitude is a complex number, and squaring it would solve that (I believe... I've never formally learned about complex numbers), but I'm really confused as to why you would square it.

I haven't found any published explanation for squaring the wave function - other than the usual - ie. that it makes the values conform to what is observed - but I can venture a better reason.

The wave function (not squared) describes the probability of a particle occupying a particular location in space - but this needs to be multiplied by the probability of a particle being detected at that same location in space.

By way of analogy, if Alice and Bob can be at one of four places, then the probability of Alice (or Bob) being at anyone place is 1:4 (the wave function) and the probability of Alice meeting Bob is also 1:4, BUT the probability of Alice meeting Bob at a specificied location is 1:16, ie. 4 squared.

That's how I interpret the wave function squared.
Carl Looper
8 December 2009

 

[bcrowell]

I haven't found any published explanation for squaring the wave function - other than the usual - ie. that it makes the values conform to what is observed

It's required by the correspondence principle. For example, the energy density of the electromagnetic fields is proportional to the square of the fields, so it would violate the correspondence principle if the probability of detecting a photon in a certain location was not proportional to the square of the fields. For a longer discussion: http://www.lightandmatter.com/html_books/6mr/ch03/ch03.html

- but I can venture a better reason.

The wave function (not squared) describes the probability of a particle occupying a particular location in space - but this needs to be multiplied by the probability of a particle being detected at that same location in space.

No, this is incorrect. probabilities are scalar, real, and positive. An electron's wavefunction is complex. A photon's wavefunction is a vector.

 

[meopemuk]

Before answering the question of why the wave function must be squared you should ask yourself more general questions: why Hilbert spaces? why states are represented by wave functions? why observables are represented by Hermitian operators? I found the most satisfying answers in the discipline called "quantum logic". This approach derives the entire quantum mechanical formalism from a set of simple and natural postulates. In this approach the QM formalism is just a generalization of the classical probability theory. The probability interpretation of the squared wave function comes out rather naturally.

Eugene.

 

[carllooper]

"you should ask yourself more general questions"

The following question is far more general than "why observables are represented by Hermitian operators" etc., and as a result, far more useful:

Given 4 pubs where Alice knows Bob must be, what are the odds of Alice finding Bob at anyone of those pubs?

 

[carllooper]

If the distribution of particle detections (the probability wave) is equal to the amplitude of the wave function squared, then the relationship between wave function and probability wave is quasi-equivalent since it is a simple mathematical formality, without any other variables, to transform the latter into the former (although admittedly the reverse is not true - due non-commutativity).

An alternative (and semi-equivalent) question might be why the wave function is the square root of the probability distribution?

Before approaching that question one should probably have an idea of what probability means.

 

[bcrowell]

If the distribution of particle detections (the probability wave) is equal to the amplitude of the wave function squared, then the relationship between wave function and probability wave is quasi-equivalent since it is a simple mathematical formality, without any other variables, to transform the latter into the former (although admittedly the reverse is not true - due non-commutativity).

An alternative (and semi-equivalent) question might be why the wave function is the square root of the probability distribution?

We have quantity #1 that obeys the principle of superposition (http://en.wikipedia.org/wiki/Superposition_principle), like any wave. We have quantity #2 that is proportional to probability. The question is why quantity #2 is proportional to the square of quantity #1. The answer is the correspondence principle.

 

[carllooper]

By that logic the answer to the following question might also be the correspondence principle:

Why does 8 equal to k 2 squared (where k=2)

But the answer here is that it is so by definition. Now the same could very well be said of the relationship between probalility wave and wave function, except that there is no definition. The wave function is so by definition, and the probability wave is so by observation - and the relationship between the two is by a simple mathematical transform.

There is no definition of why one is the square of the other - other than it just happens to be the case.

 

[bcrowell]

By that logic the answer to the following question might also be the correspondence principle:

Why does 8 equal to k 2 squared (where k=2)

No, please read my post #41 above where I explained that the squaring follows from the correspondence principle.

 

[meopemuk]

There is no definition of why one is the square of the other - other than it just happens to be the case.

Within "quantum logic" the connection between wave function and probability density is not an arbitrary assumption. It follows from well-defined and natural postulates. Here is the recommended reading:

G. Birkhoff and J. von Neumann, "The logic of quantum mechanics", Ann. Math., 37 (1936), 823.

G. W. Mackey, "The mathematical foundations of quantum mechanics", (W. A. Benjamin, New York, 1963), see esp. Section 2-2.

C. Piron, "Foundations of Quantum Physics", (W. A. Benjamin, Reading, 1976).

Eugene.

 

[carllooper]

The correspondence principle is not the answer.

The probability distribution is exactly the wave function squared (in the limit) - ie. not approximately.

What I put forward is not the answer either - but an analogy - an indication of the direction in which one could go.

 

[bcrowell]

The correspondence principle is not the answer.

Please explain why you think it's not.

The probability distribution is exactly the wave function squared (in the limit) - ie. not approximately.

You seem to be confusing "approximately" with "proportional to." If you look back at my post #41, you'll see the reason that I state it as a proportionality rather than an equality is that I was referring to the photon and the electron on the same footing. The photon's wavefunction is simply the field, and the square of the field differs from the probability distribution by a constant of proportionality.

 

[carllooper]

I'm not confusing proportionality with approximation.

By "approximation" I mean something altogether different - in the sense that Newtonian Physics is an approximation of Relativistic Physics - in that, at certain scales, they (appear to) correspond, ie. the correspondence principle.

But there is no suggestion that this sort of thing occurs between the wave function and the probability wave. It is not as if the probability wave is approximately equal to k times the wave function squared. It is exactly equal to such (for some constant k).

Perhaps I don't know what you mean by "correspondence principle".

Carl

 

[bcrowell]

Perhaps I don't know what you mean by "correspondence principle".

http://en.wikipedia.org/wiki/Correspondence_principle

 

[carllooper]

Well - in that case I do know what you're talking about - and my argument holds.

 

[ArjenDijksman]

So the probability of detection is proportional to the cross section of the detected particle times the cross section of the detecting particle, both projected on a fixed axis.

The wave function (not squared) describes the probability of a particle occupying a particular location in space - but this needs to be multiplied by the probability of a particle being detected at that same location in space.

By way of analogy, if Alice and Bob can be at one of four places, then the probability of Alice (or Bob) being at anyone place is 1:4 (the wave function) and the probability of Alice meeting Bob is also 1:4, BUT the probability of Alice meeting Bob at a specificied location is 1:16, ie. 4 squared.

That's how I interpret the wave function squared.

Yes, that's common sense. We have the same interpretation. Detection probabilities are made up of two contributions: 1) from the detected object and 2) the detecting system. This is a universal principle valid in quantum mechanics and for classical interactions: the gravitational force between two objects is proportional to 1) the mass of the first object and 2) the mass of the second object; the electric force between two objects is proportional to 1) the charge of the first object and 2) the charge of the second object.

 

[carllooper]

Yes, that's common sense.

I'm not sure it's "common" sense as such. But it is correct. When I've posed the Alice/Bob question the most common response for the probability has been (almost consistently) 1:4. It is only when the question has been reconsidered in terms of all the constraints (not just Alice meeting Bob, but meeting at a particular place) does the 1:16 solution become the conclusion (or "common sense").

cheers
Carl

 

[ArjenDijksman]

When I've posed the Alice/Bob question the most common response for the probability has been (almost consistently) 1:4. It is only when the question has been reconsidered in terms of all the constraints (not just Alice meeting Bob, but meeting at a particular place) does the 1:16 solution become the conclusion (or "common sense").

Your approach on quantum probabilities is not exactly the same but it is similar to mine. There are beables (described by amplitudes: Alice is at a location, Bob is at a location) and detectables (described by squared amplitudes: Bob being at a location, detecting Alice at a location). In your view, how does it come that Bob is described by the same amplitude as Bob? Generally the detecting system is thought of as macroscopic, and its amplitude is the same at all places of detection (e.g. a screen detecting a photon).



Arjen

 

[carllooper]

I'm not exactly sure how to map the Alice/Bob analogy to the situation in quantum mechanical experiments. One could ask Alice to play the role of a particle, and Bob to play the role of a detector - but there is a potential domain conflict in such a mapping - because the detector is typically described in classical terms whereas the particle is typically not. We could ask of Bob that he be unspecified in terms of which detector (in an array of such - a screen) that he was playing - but it starts to look more like a fix-it job than something fundamental. Or maybe not. I don't know.

I position the Alice/Bob analogy as a potentially useful starting point for how a quantum mechanical system might be re-described or re-interpreted in terms of generic probabilitys (applicable to both classical and quantum systems).

Carl

 

[ArjenDijksman]

the detector is typically described in classical terms whereas the particle is typically not.

We should see the detector in terms of quantum particles, typically electrons. The detection/measurement can then be seen as an elementary interaction between two quantum particles.

I position the Alice/Bob analogy as a potentially useful starting point for how a quantum mechanical system might be re-described or re-interpreted in terms of generic probabilitys (applicable to both classical and quantum systems).

The fact that you consider the detection probability to be the product of A's amplitude times B's amplitude is an excellent starting point. I'm sure it will bring you further.

 

[jambaugh]

The wave function (not squared) describes the probability of a particle occupying a particular location in space - but this needs to be multiplied by the probability of a particle being detected at that same location in space.
...
That's how I interpret the wave function squared.
Carl Looper
8 December 2009

That's an interesting way of looking at it but it has a serious flaw. The two probability distributions (amplitudes) for each half process do not add up correctly.

Take the case of a two outcome (say "A" and "B") observation. Let the initial mode be an equal superposition of "A" and "B". Then if the particle has equal probability of occupying either of A and B positions you must have 1/2 probability each. That's not what the superposition of amplitudes will be. Rather they'll be 0.707 and 0.707 which add to 1.414 > 1. They can't be probabilities in that sense.

 

[ArjenDijksman]

That's an interesting way of looking at it but it has a serious flaw. The two probability distributions (amplitudes) for each half process do not add up correctly.

Take the case of a two outcome (say "A" and "B") observation. Let the initial mode be an equal superposition of "A" and "B". Then if the particle has equal probability of occupying either of A and B positions you must have 1/2 probability each. That's not what the superposition of amplitudes will be. Rather they'll be 0.707 and 0.707 which add to 1.414 > 1. They can't be probabilities in that sense.

The amplitudes should be seen in this way as relative weights. You need of course to normalize them to retrieve a probability distribution.