Why amplitude squared gives probability / Schrodinger Equation

[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.

 

[jambaugh]

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

No that won't wash. If they need to be normalized then normalize them before using them in a joint probability. All the P's in P(ab) = P(a)P(b), are normalized probabilities. Now you can reweigh and counter-weigh the factors but that's not the case for amplitudes. Both scale (by real factors) the same not reciprocally.

It just doesn't work that way.

Amplitudes are (phased) square roots of probabilities. The "why" is in the fact that we (unnecessarily but historically) work with the left and right ideals (spaces on which the operators act) of an operator algebra rather than at the level of the algebra (where the physics is expressed).

If you stick to Hermitian operators for observables and density operators for modes then there is no puzzle about squaring amplitudes. The density operator is the quantum analogue of the classical probability distribution. No squaring, no square-rooting.

Ask first why we take the square root of the operator algebra to get the Hilbert space (ket space) and its dual (bra space). The main answer so far as I can tell is that the math is easier at that level.

Note we can still formulate sharp system modes ("states") via density operators.
We can still formulate the eigen-value principle:
$$X\rho = x\rho$$
We can still formulate equations of motion.
We don't even need to invoke the modes when expressing HUP.

Though the historical formulation starts with the Hilbert space the operational sequence (how closely the mathematical objects link to physical actions) starts with the operator algebra.

 

[ArjenDijksman]

No that won't wash. If they need to be normalized then normalize them before using them in a joint probability. All the P's in P(ab) = P(a)P(b), are normalized probabilities. Now you can reweigh and counter-weigh the factors but that's not the case for amplitudes. Both scale (by real factors) the same not reciprocally.

It just doesn't work that way.

I fear we have a different understanding of Carl's analogy. With 2 possible outcomes, the way I understood his analogy is:

Let particle 1 be the particle to be detected. Let particle 2 be the particle that detects particle 1 at a specific location A or B. The state of the joint system is then:
|particle 1, particle 2> = 1/2|A,A> + 1/2|A,B> + 1/2|B,A> + 1/2|B,B>

The |A,B> and |B,A> basis states are undetectable because detecting particle 2 doesn't share the same location as particle 1, so you can drop them out for the measurement process. Renormalize and you get |particle 1, particle 2> = 0.707|A,A> + 0.707|B,B> but the initial amplitudes just add to 1.

 

[carllooper]

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 "A" and "B" in my analogy are not two components of a superposition (of a single particle). Rather, Alice and Bob are two particles that meet at some given point in space, ie. you wouldn't add their amplitudes. You would (if anything) multiply them.

Another way of posing the problem is: Given that the probability of two particles meeting at a given/specified/a priori point in space, is proportional (inversely or otherwise) to the square of the number of positions at which each could meet (so to speak), in what way might one otherwise describe this square in relation to the wave function, if not as a proportion of the wave function squared?

Carl

 

[jambaugh]

I fear we have a different understanding of Carl's analogy. With 2 possible outcomes, the way I understood his analogy is:

Let particle 1 be the particle to be detected. Let particle 2 be the particle that detects particle 1 at a specific location A or B. The state of the joint system is then:
|particle 1, particle 2> = 1/2|A,A> + 1/2|A,B> + 1/2|B,A> + 1/2|B,B>

The |A,B> and |B,A> basis states are undetectable because detecting particle 2 doesn't share the same location as particle 1, so you can drop them out for the measurement process. Renormalize and you get |particle 1, particle 2> = 0.707|A,A> + 0.707|B,B> but the initial amplitudes just add to 1.

Yes I didn't parse it this way. As you just explained it though its just composite quantum probabilities (projected onto a subspace and renormalized) and has nothing to say about why square amplitudes.

The "A" and "B" in my analogy are not two components of a superposition (of a single particle). Rather, Alice and Bob are two particles that meet at some given point in space, ie. you wouldn't add their amplitudes. You would (if anything) multiply them.

OK I was using the A and B index values to represent "positions".

Another way of posing the problem is: Given that the probability of two particles meeting at a given/specified/a priori point in space, is proportional (inversely or otherwise) to the square of the number of positions at which each could meet (so to speak), in what way might one otherwise describe this square in relation to the wave function, if not as a proportion of the wave function squared?

Carl

Let me try again. The Alice particle represents the system to be measured? And the Bob particle is doing the measuring? If so then let's parse the two "position" case (labeled positions + and - as in above or below a box divider).

To avoid confusion I will not use ket notation except when explicitly jumping back into standard QM.

The Alice particle has probability of being in positions + or - with 1/2 probability each:
P_A(+) = P_A(-)=1/2.
Similarly for the Bob particle.

Thence you get the equally 1/4th probabilities for:
P_{AB}(++), (+-), (-+), and (--) cases.
That's fine and that's standard classical independent joint probabilities. We indeed see that for(assumed necessarily?) symmetric dual A and B halves of a measurement the joint probabilities end up as squares of factor probabilities.

Similarly we can (in a totally different setting) refer to quantum amplitudes:

$$[\sqrt{2}/2(|A+\rangle + |A-\rangle)]\otimes[\sqrt{2}/2(|B+\rangle + |B-\rangle)]=$$

$$=1/2|++\rangle +1/2|+-\rangle +1/2|-+\rangle +1/2|--\rangle$$

Squaring amplitudes in the quantum case agrees with the classical case (as it will when we consider a commuting subset of observables. We always get a classical correspondence in this case.)

But again I don't see how the multiplication in the forming of a joint (classical) pdf which has the quantum correspondent of multiplication of amplitudes in a tensor product for composite systems in any way relates to squaring amplitudes to get probabilities.

You must explain how a root two over two = 0.707... amplitude represent the normalized probability of some (unobserved) state of affairs. Otherwise you are comparing apples and oranges while keeping your eyes closed to not notice the inconsistency.

I suggest you try again with unequal probability cases (say a 1/3 vs 2/3) so that accidental correspondences of values do not get confused with intended real and general correspondences you are trying to assert.

EDIT: For example I can observer 2+2 = 4 and 2x2 = 4 but err in asserting this means + = x . Maybe you're not doing this but it is easy to do in the hairier world of QM. I don't yet see that you aren't making such a mistake.

 

[carllooper]

...
I suggest you try again with unequal probability cases (say a 1/3 vs 2/3) so that accidental correspondences of values do not get confused with intended real and general correspondences you are trying to assert.

EDIT: For example I can observer 2+2 = 4 and 2x2 = 4 but err in asserting this means + = x . Maybe you're not doing this but it is easy to do in the hairier world of QM. I don't yet see that you aren't making such a mistake.

Yes, for example, one can consider a detecting particle (an electron) with a probability confined to a region of space much smaller than the particle to be detected (eg. powered by an independant power supply) uncorrelated (other than spatially) to any other detector cell. And in general the chances of two particles having the same probability over the same region of space would be very small anyway, ie. even without orchestrating bad odds.

And I'd accept that as a very reasonable argument except that the wave function (and associated probability function) is normally defined in relation to a classically defined detector (ie. with a probability == 1).

In the tentative model I'm proposing (and it remains to be formalised) both the particle to be detected, and the detector system as a whole (eg. a solid state array of detectors) - are each in a state of uncorrelated superposition - but with respect to what?

I'm just treating space (or spacetime) itself as the frame of reference.

In other words, the role otherwise played by a classical detector system is now played by an 'empty' spatial frame of reference with a probability == 1 (ie. non-selectable).

Of course, if we allow spacetime itself to be selectable (ie. assume a point of view on multiple universes!) then the probability would be the wave function cubed (or higher) - but we'll leave that for the gods to ponder.

Carl

 

[jambaugh]

[...]
In the tentative model I'm proposing (and it remains to be formalised) both the particle to be detected, and the detector system as a whole (eg. a solid state array of detectors) - are each in a state of uncorrelated superposition - but with respect to what?

I'm playing (orthodoxy) devil's advocate but I am quite interested. One word of warning about being "in a state of superposition". In the orthodox view this isn't a property of the particle but rather of our choice of resolution of that particle.

In short given the mode vector psi is a linear combination of several basis vectors in one basis and thus a quantum in that mode is "in a superposition of states" with respect to the (relative) classical logic defined by that basis, it is none-the-less "not in a superposition" in a basis in which psi itself is an element.

E.g. with spins a particle with specific spin x state is in a superposition of spin z states and vise versa. "being in superposition" isn't a property of the quantum but rather a relationship between the quantum and of our choice of basis modes. Have this in mind when trying to formulate an explanation of quantum superposition.

 

[carllooper]

I'm playing (orthodoxy) devil's advocate but I am quite interested. One word of warning about being "in a state of superposition". In the orthodox view this isn't a property of the particle but rather of our choice of resolution of that particle.

Ah yes, the "super-position" of a particle is how you'd describe (with due difficulty) a particle within an otherwise classical context. I'm doing much the same thing, but reducing the classical context to it's bare minimum - an empty (Newtonian/Kantian or "Einsteinian") frame of reference - possibly altering the meaning of "super-position" somewhat but hopefully clarifying it in due course.

cheers
Carl

 

[jambaugh]

The relativity concept is a powerful tool for understanding quantum theory.

In SR we have relativity of time. One observer's time-step is another's increment through both space and time. Each observer resolves motion into a "superposition" of spatial displacements and temporal durations. We can think of a moving object relative to a given observer as having been physically transformed away from an object co-moving with the observer and thus having proper time and observer time no longer in 1-1 correspondence.

In QM we have relativity of objective logic (extended to classical probabilities). One (complete) observable's boolean logic of observed values is another's superposition which must be reconciled probabilistically. The observer frame here is the eigen-basis of a given complete observable.

We can think of a quantum in superposition w.r.t. a given frame (in objective state w.r.t. another frame) as having been physically transformed away from one in an objective state with respect to the given frame (via general linear transformation). The specific transformation is the one which diagionalizes the quantum's density matrix.

In a sense both cases involve time in a certain way in that one must resolve transitional behavior in a language of static state of reality. In special relativity we express an object's state of motion via its 4-velocity. Objects which are not stationary have non-temporal components of motion. In quantum theory we express a quantum's dynamic mode via its density operator. If the quantum is not "objectively stationary" with respect to an observer frame (complete observable=observable with complete non-degenerate spectrum) it has off diagonal components expressing its being in a mode of transition between the objective states defined by the given observable's logical frame.

The parallel in application of a relativity principle is tight. However the level of application is very different. QM relativizes at a higher level of abstraction. We can't represent transitional modes in terms of time parametrized objective states because it is the objective states which we are relativizing. Just as in SR we cannot represent all clocks as simply running slower or faster in an absolute time frame since it is time itself we are relativizing.

With that in mind look at the time component of an objects 4-velocity as the resolution of the component of the object's proper time in the observer's time. Then consider the diagional components of a quantum's density operator as the resolution of the quantum's objective logic in the probability extended logic of the observer frame defined by the choice of basis. In the SR case we must invoke time dilation i.e. scaling of relative times. In QM we must invoke probability i.e. scaling of the logical certainty of objective outcomes of observations. In both cases how these mesh is dictated by the observables transform under the group of frame transformations. (Lorentz/Poincare in SR and U(N) in QM).

 

[vitruvianman]

First of all the probability MUST be between 0 and 1, so be positive but sometimes when we solve the Schrödinger equ. for all its 4 dimensions, we may find a complex result which may be negative too (for example all the results for pozitrons).

actually squaring the amplitude, i mean for a z=x+iy it means $$|z|^{2}=x^{2}+y^{2}$$ and if we add an extra probability like w=u+iv, we'll find $$|zw|^2=|z|^2 |w|^2$$ because of $$|w|^2+|z|^2=1$$, is normalizing the wave function, so making it stand between 0 and 1 (or one of them), as a "real" probability.

 

[vitruvianman]

But.. argh. It's so frustrating. This seems like such a simple question, but there doesn't seem to be any simple answer.

What is amplitude, then?

Anyone know of any books that'll talk about this with some depth, but for a beginner? All books I've read have just sort of... mentioned it briefly and moved on. I want further explanations and discussions, but wouldn't be able to handle advanced jargon and stuff.

you should read Roger Penrose's The Emperor's New Mind serie's "Minds, and The Laws of Physics"
In 6. part, you'll find many answers to your questions (as I've seen yet)