Why amplitude squared gives probability / Schrodinger Equation

[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)

 

[ArjenDijksman]

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.

Probability is amplitude squared because we project the total state-space (N*N separate basis states |A,A>, |A,B> ,|B,A>, |B,B>) on the detectable subspace (N separate events |A,A>, |B,B>).

 

[vitruvianman]

An extra info:

Let's say in the famous double-slit exp., a photon (when only one of the slits is opened) has a probability A(s,t) which is the complex "probability" number z, to come the the point t from the photon source s. And the probability A(t,p) which is the complex number w. These are the possibilities of coming to the point t from the source s, and after that, to the point p (on the screen) lastly.

Multiplying is surely so:
|zw|^2=|z|^2|w|^2=|A(s,t)|^2|A(t,p)|^2=|A(s,t)A(t,p)|^2

But if the photon has 2 different options we should add the probabilities and here apperas the characteristics of quantum mechanics. We should square the amplitude of w+z, and here comes a correction term, \cos\theta. The sum will be:
|z+w|^2=|w|^2+|z|^2+2|w||z|\cos\theta
\theta is the angle between w and z points on the Argand plane. After a quick (and very basic) geometric calculation you'll find out why the correction term is 2|w||z|\cos\theta , because of Pythagoras theorem :)

 

[jambaugh]

Probability is amplitude squared because we project the total state-space (N*N separate basis states |A,A>, |A,B> ,|B,A>, |B,B>) on the detectable subspace (N separate events |A,A>, |B,B>).

No. That won't do it. The projection is linear and one linearly renormalizes when projecting. "You caint get thare frum here!"

It is clear that the squaring comes from going from mode vectors to the operators
|A> ----> |A><A|
which is the proper domain to speak of boolean observables and their expectation values (probabilities are expectation values of logical bits which are in turn observables of a given physical system.)

More properly said the square-rooting comes from representing mode operators (density operators) for sharp modes by just giving the eigen-vectors. The Hilbert space is "the square root" of the operator algebra. Its the operator algebra which is the proper mathematical object in which to represent the physics.

 

[ArjenDijksman]

It is clear that the squaring comes from going from mode vectors to the operators
|A> ----> |A><A|
which is the proper domain to speak of boolean observables and their expectation values (probabilities are expectation values of logical bits which are in turn observables of a given physical system.)

Well, what you say, is just what I mean but stated otherwise. Let me restate it applied to https://www.physicsforums.com/showpost.php?p=2482669&postcount=40": the squaring of the amplitudes (factors of the Hilbert state vectors representing Alice's possible locations) comes from the measurement process which yields probabilities (Bob detecting Alice at a particular location = operating on Alice's state with an operator).

As a non-trivial example: Alice can be at any two places M or N, with respective weights w(M)=3/5 and w(N)=4/5 (choosing these different weights avoids LaTeX script for square roots). For Bob the same. The respective weights for the composite system (Alice,Bob) are thus w(M,M)=9/25, w(M,N)=12/25, w(N,M)=12/25, w(N,N)=16/25.

We can therefore write the Hilbert space state vector |Alice>= 3/5 |M> + 4/5 |N>. But the proper domain to speak of expectation values is when Bob observes the state of Alice → Bob meets Alice at the same place → Bob operates with an operator on |Alice>. That yields prob(M,M)=9/25 and prob(N,N)=16/25 which are the squares of the Hilbert state vector amplitudes.

Greetings,
Arjen

 

[jambaugh]

Well, what you say, is just what I mean but stated otherwise. Let me restate it applied to https://www.physicsforums.com/showpost.php?p=2482669&postcount=40": the squaring of the amplitudes (factors of the Hilbert state vectors representing Alice's possible locations) comes from the measurement process which yields probabilities (Bob detecting Alice at a particular location = operating on Alice's state with an operator).

As a non-trivial example: Alice can be at any two places M or N, with respective weights w(M)=3/5 and w(N)=4/5 (choosing these different weights avoids LaTeX script for square roots). For Bob the same. The respective weights for the composite system (Alice,Bob) are thus w(M,M)=9/25, w(M,N)=12/25, w(N,M)=12/25, w(N,N)=16/25.

We can therefore write the Hilbert space state vector |Alice>= 3/5 |M> + 4/5 |N>. But the proper domain to speak of expectation values is when Bob observes the state of Alice → Bob meets Alice at the same place → Bob operates with an operator on |Alice>. That yields prob(M,M)=9/25 and prob(N,N)=16/25 which are the squares of the Hilbert state vector amplitudes.

Greetings,
Arjen

No. We are not describing the same thing though the math is paralleling. Yes those weights work out to square to probabilities. But you now cannot interpret those weights as rescaled probabilities themselves. Your exposition provides no enlightenment as to why e.g. weights of 3/5|M> + 4i/5|N> express a distinct case from yours above.

In an actual measurement process the quantum (if you want to call her Alice) is observed each trial in either |M> or |N> modes. Those cases where "Bob and Alice are not at the same slot" are not simply discarded trials that didn't yield valid results. Some Bob always sees Alice. In thinking of Bob as the measuring device, there are in fact an array of "Bob"'s at each location announcing "Hey! Alice just bumped into Me!" (For example a cloud chamber is a room of Bobs who so state by condensing a droplet when ionizing Alice passes by.)

I think your exposition provides "false enlightenment". The better interpretation would be to take the Bra's and Ket's to be say arrays of Alice's and Bob's observing Elvis. An Alice announces "Elvis has entered the building!" and a Bob announces "Elvis has left the building!" We then know Elvis was in building 1 when we get a A1,B1 announcement pair represented by |1><1|. We must calibrate our Alices and Bobs by repeating enough Elvis concerts to see which Alice and Bob pairs correlate exactly. (Mathematically this defines the metric on our Hilbert space) That provides the logic of Elvis sightings. Then to express probabilities we weight in-out correlations:

rho = = w1 P1 + w2 P2 + ... w1 |1><1| + w2|2><2| ...
(P = projector = logical predicate about Elvis.)

We start with the projectors and then we may for convenience use eigen-vectors which requires we square-root the weighting system (not just individual weights) as we construct the mathematical space on which the projectors act to define these eigen-vectors. The actual bras and kets are one level of abstraction further removed from the physical system and not more directly representative of it.

It is in the algebra of these predicates where we get the odd quantum behavior because such predicates=projectors don't all commute. PQ != QP. We can mathematically view the details by observing that an eigen-vector of P must be expressed as a linear combination of eigen-vectors of Q with coefficients which are not directly interpretable because we're doing all this in an abstract mathematical structure farther removed from the physics. It is the operators which are most closely connected to the physics and that's our logical starting point. We shouldn't (and as it turns out can't) deconstruct them further physically. Mathematical deconstruction, though necessary for mathematical reasons, is moving farther from the physical representation of what's happening in the lab.

A point here is that we never directly represent the quantum Elvis. The Bra's and Ket's express channels of Elvis behavior not Elvis states. Alice and Bob as symbols refer to parts of our measuring process not directly to the entity being measured.

Now on a tangential note. We can trivially quantify the system (speak of varying the number of instances) to be one or zero. In that extension we can treat the Hilbert space vectors and dual vectors (kets and bras) as respectively system creation actions and system annihilation actions. Then system transitions are expressed via pairs of annihilation-creation actions. This as I see it is the only way to "put a little more meat into them" i.e. give a bit more metaphysical meaning to the bra's and kets. But in standard quantum mechanics these actions are virtual, not physical. You can think of these acts of creation and annihilation rather as boundary crossings and an |a><b| transition channel as our painting a little cut in the system domain where the system instantaneously leaves and re-enters the domain. But here again they do not refer to the system but act as labels for modes by which the quantum leaves or enters this abstract domain of "being".

 

[ArjenDijksman]

No. We are not describing the same thing though the math is paralleling. Yes those weights work out to square to probabilities. But you now cannot interpret those weights as rescaled probabilities themselves. Your exposition provides no enlightenment as to why e.g. weights of 3/5|M> + 4i/5|N> express a distinct case from yours above.

Well, I was discussing Carl's classical analogy, which I don't see as a complete exposition of quantum probabilities. It merely gives insight in the fact that quantum probabilities arise in case of measurement, from weighted coefficients pertaining to the eigenstates of the system. That analogy provides by no means insight in the complex-valued nature of quantum amplitudes and you've raised the relevant objections. On that point, I have further ideas that I mentioned earlier in this thread.

In an actual measurement process the quantum (if you want to call her Alice) is observed each trial in either |M> or |N> modes. Those cases where "Bob and Alice are not at the same slot" are not simply discarded trials that didn't yield valid results. Some Bob always sees Alice. In thinking of Bob as the measuring device, there are in fact an array of "Bob"'s at each location announcing "Hey! Alice just bumped into Me!" (For example a cloud chamber is a room of Bobs who so state by condensing a droplet when ionizing Alice passes by.)

I think your exposition provides "false enlightenment". The better interpretation would be to take the Bra's and Ket's to be say arrays of Alice's and Bob's observing Elvis. An Alice announces "Elvis has entered the building!" and a Bob announces "Elvis has left the building!" We then know Elvis was in building 1 when we get a A1,B1 announcement pair represented by |1><1|. We must calibrate our Alices and Bobs by repeating enough Elvis concerts to see which Alice and Bob pairs correlate exactly. (Mathematically this defines the metric on our Hilbert space) That provides the logic of Elvis sightings. Then to express probabilities we weight in-out correlations:

rho = = w1 P1 + w2 P2 + ... w1 |1><1| + w2|2><2| ...
(P = projector = logical predicate about Elvis.)

We start with the projectors and then we may for convenience use eigen-vectors which requires we square-root the weighting system (not just individual weights) as we construct the mathematical space on which the projectors act to define these eigen-vectors. The actual bras and kets are one level of abstraction further removed from the physical system and not more directly representative of it.

I consider Carl's analogy as a "partial" enlightenment, rather than a "false" enlightenment: it's better to have an analogy than no analogy, in as much as one is aware of its limits. Your "bra/ket announcement pair" interpretation is interesting as well. Is there any literature (or posts) developing that insight?

Cheers,
Arjen

 

[Dunedain]

To be frank you are talking about an equation that works after the fact, and isn't derived from anything other than quantum mechanics, so it's always going to agree with what is predicted in that wave function. It works experimentally that's all you can ask from it, it's not derivable from anything but itself. We can talk interpretation but frankly that's not science that's philosophy, we are stuck with relating equations to experiment, not predicting experiments with equations, at least for now.

It's kind of like asking why c? Why indeed it just is it seems...

 

[carllooper]

To be frank you are talking about an equation that works after the fact, and isn't derived from anything other than quantum mechanics, so it's always going to agree with what is predicted in that wave function. It works experimentally that's all you can ask from it, it's not derivable from anything but itself. We can talk interpretation but frankly that's not science that's philosophy, we are stuck with relating equations to experiment, not predicting experiments with equations, at least for now.

It's kind of like asking why c? Why indeed it just is it seems...

Science and philosophy are not mutually exclusive terms. The equations of quantum mechanics were not developed "after the fact" (as if a function of the facts) but have their roots in creative assumptions regarding the facts. For example, Plank 'discovered' the quantum through speculative assumptions he made regarding experimental results. His equations represented those assumptions (as much as the results). Indeed the assumptions he made were so speculative he could not (initially) see how they could ever fit into the "science" of the time. Plank rejected his own equations because of 'science'. But his equations show that he had (in both fact and philosophy) discovered the quantum. If you look at how Einstein, Heisenberg, Bohr, Shrodinger (etc) approached the development of their equations you will see they also engaged in philosophy when doing so.

I can't see how it would be possible to do otherwise.

Carl

 

[jambaugh]

...
I consider Carl's analogy as a "partial" enlightenment, rather than a "false" enlightenment: it's better to have an analogy than no analogy, in as much as one is aware of its limits.

Fair enough.

Your "bra/ket announcement pair" interpretation is interesting as well. Is there any literature (or posts) developing that insight?

Cheers,
Arjen

Hmmm... I kind of came up with it on the spot based on my understanding of QM. (The announcement part). I've been thinking of the vectors and dual vectors as "kind of like boundary operators" for some time. I see a strong analogy between the duality of operators (as actions) and density operators used with trace to form dual cooperators and the duality of chain and differential form in calculus.

I'm of course greatly influenced by my thesis advisor and mentor David Finkelstein. He has a book "Quantum Relativity" which you might find interesting.

 

[ArjenDijksman]

I'm of course greatly influenced by my thesis advisor and mentor David Finkelstein. He has a book "Quantum Relativity" which you might find interesting.

David Finkelstein, Georgia Tech. I better understand why you emphasize the role of operators. I created http://en.wikiquote.org/wiki/David_Finkelstein" [/I]" and acknowledge that we speak a different quantum dialect: I'm more inclined to put forward analogies, whether mathematical or mechanical, but always experimentally testable. Nevertheless, there are interesting thoughts on your website.

Greetings,
Arjen

 

[carllooper]

... In an actual measurement process the quantum (if you want to call her Alice) is observed each trial in either |M> or |N> modes. Those cases where "Bob and Alice are not at the same slot" are not simply discarded trials that didn't yield valid results. Some Bob always sees Alice. In thinking of Bob as the measuring device, there are in fact an array of "Bob"'s at each location announcing "Hey! Alice just bumped into Me!" (For example a cloud chamber is a room of Bobs who so state by condensing a droplet when ionizing Alice passes by.) ...

Lets suppose we have a Bob array with only two elements (Bob0 and Bob1) and one Alice. And that there are two locations Alice and Bob can occupy (X and Y). For simplicity when Bob is occupying location X, assume Bob0 occupys location X and his spatially correlated partner Bob1 occupys location Y - and vice versa (ie. one oreintation of Bob is upside down with respect to the other orientation)

Holding X,Y stationary (so to speak) there are four possible outcomes:

X: Bob0 Y: Alice, Bob1
X: Bob0,Alice Y: Bob1
X: Bob1 Y: Alice, Bob0
X: Bob1,Alice Y: Bob0

So the probability of anyone of these outcomes remains the joint probability: 1/4.

Now while we can say, for example, that Alice meets, say Bob0, 2 out of 4 times. Or "some Bob always sees Alice" 4 out of 4 times, such statements are conflating otherwise different (ie. inconsistent) worlds.

In other words, the analogy has a little more scope than one might assume.

Carl