Seasons
This is a forum or general chit-chat, small talk, a "hey, how ya doing?" and such. Or hell, get crazy deep on something. Whatever you like.
Posts 4,059 - 4,070 of 6,170
You said in a previous post, Psimagus, that you accepted all these postulates and everything they implied. So why not make your case from the postulates, instead of appealing to authorities?
So are you saying that this is a matter of definition, something that can be logically derived from the postulates and are inherent in the assumptions of quantum theory rather than something based on experimentation and data? If it's inherent in the postulates, do we need the slit experiment at all?
And you explain Penrose's description of it as continuous, not propagated... exactly how?
I'm not terribly concerned to explain it. I'm quite sure that if I saw it in context, it would turn out to be either (a) something which is quite true in the Irinaverse, when correctly interpreted, (b) one of Penrose's idiosyncratic and speculative ideas (brilliant people are often not the best source for mainstream ideas) or (c) false (even Penrose can make a mistake - or a typo).
Posts 4,059 - 4,070 of 6,170
And, speaking of operators, it's time for...[drum roll]
POSTULATE 2!!!!!
Postulate 2. To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics.
This postulate comes about because of the considerations raised in section 3.1.5 : if we require that the expectation value of an operator is real, then must be a Hermitian operator. Some common operators occuring in quantum mechanics are collected in Table 1.
(from http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html , as before)
To say that an operator is Hermitian is to say that it will produce real eigenvalues. Since we haven't discussed eigenvalues yet, let's just put this on hold.
To say that an operator is linear is to say that
1. When applied to a function (psi1) which is a sum of two other functions, (psi2) and (psi3), the result is the same as if we had applied the operator to (psi2) and (psi3) individually, and then added the results. That is, if we call the operator "O",
O((psi2) + (psi3)) = O(psi2) + O(psi3), for any (psi2) and (psi3).
For example, grad is a linear operator, and grad(x^2) = 2x, and grad(x^3) = 3x^2, so
grad(x^2 + x^3) = grad(x^2) + grad(x^3) = x + x^2.
An example for those who know the trig functions, sin(x) and cos(x):
grad(sin(x)) = cos(x) and grad(cos(x)) = -sin(x), so
grad(cos(x) + sin(x)) = grad(cos(x)) + grad(sin(x)) = -sin(x) + cos(x)
2. If we get a function (psi2) by multiplying a function (psi1) by some complex number z, then the result of applying O to (psi2) is just z times the result of applying O to (psi1). That is,
O(z(psi1)) = zO(psi1)
For example, grad(3x^2) = 3grad(x^2) = 6x.
Quantum Mechanics is easy, in the Irinaverse!!!!
Quiz: If O is linear, what is O((psi1) + 3(psi2))?
Answer: (O(psi1) + 3O(psi2)
Quiz: What is grad(3x + 7x^2)?
Answer: 3 + 14x.
There is much debate about what an observable is, and whether "observable" is the right word, but for the moment, I suggest that we take an observable to be something that could in principle be measured. The site gives you a list of observables:
position
momentum
kinetic energy
potential energy
total energy
three kinds (more precisely components of angular momentum
As you see, most of the operators involve multivariate calculus, but I will try to explain them in an informal way. Well, position and momentum have already been covered.
In Classical Physics, kinetic energy is the energy that a particle has by virtue of its mass and speed. I think this notion is fairly intuitive: if you have two trucks moving at the same speed, but one has more mass than the other (let's say that one is empty, and the other loaded with lead ingots), the more massive one has more energy. Likewise, if two trucks have the same mass, but one is traveling faster than the other, then the faster one has more energy. If you run each of two trucks into a huge spring, so as to compress the spring, the truck with more kinetic energy will compress the spring further. In fact, the kinetic energy of a body traveling with speed v and having mass m is mv^2.
Intuitively, kinetic energy is similar to momentum: a more massive, faster-moving truck will have more momentum as well as more kinetic energy. They are also similar in QM; even if you don't know Calculus, you can see a certain similarity in the operators.
Potential energy is 'stored' energy - for example, in the old 'grandfather' clocks, one would 'wind' the clock by lifting up a certain weight, hanging on a chain. The weight would gradually fall, turning a wheel in the process, which would turn other wheels, ultimately resulting in the turning of the hands of the clock. It would take energy to lift the weight; that energy would then be used to make the clock run. The energy was stored in the weight.
For angular momentum in Classical Physics, think of those whatchamacallits you played on as a child - a circular platform on a vertical pivot. You would get it going around by pushing it, and then get on and ride, around and around. Wheee! It's tendency to keep going is angular momentum.
Now, I'm trying to find a non-technical way to explain the other operators in their quantum-mechanical form, but the Muse is not co-operating, so [psi] I'll just have to enter that problem-solving state described by Bev ...[moans] ...
POSTULATE 2!!!!!
This postulate comes about because of the considerations raised in section 3.1.5 : if we require that the expectation value of an operator is real, then must be a Hermitian operator. Some common operators occuring in quantum mechanics are collected in Table 1.
To say that an operator is Hermitian is to say that it will produce real eigenvalues. Since we haven't discussed eigenvalues yet, let's just put this on hold.
To say that an operator is linear is to say that
1. When applied to a function (psi1) which is a sum of two other functions, (psi2) and (psi3), the result is the same as if we had applied the operator to (psi2) and (psi3) individually, and then added the results. That is, if we call the operator "O",
O((psi2) + (psi3)) = O(psi2) + O(psi3), for any (psi2) and (psi3).
For example, grad is a linear operator, and grad(x^2) = 2x, and grad(x^3) = 3x^2, so
grad(x^2 + x^3) = grad(x^2) + grad(x^3) = x + x^2.
An example for those who know the trig functions, sin(x) and cos(x):
grad(sin(x)) = cos(x) and grad(cos(x)) = -sin(x), so
grad(cos(x) + sin(x)) = grad(cos(x)) + grad(sin(x)) = -sin(x) + cos(x)
2. If we get a function (psi2) by multiplying a function (psi1) by some complex number z, then the result of applying O to (psi2) is just z times the result of applying O to (psi1). That is,
O(z(psi1)) = zO(psi1)
For example, grad(3x^2) = 3grad(x^2) = 6x.
Quantum Mechanics is easy, in the Irinaverse!!!!
Quiz: If O is linear, what is O((psi1) + 3(psi2))?
Answer: (O(psi1) + 3O(psi2)
Quiz: What is grad(3x + 7x^2)?
Answer: 3 + 14x.
There is much debate about what an observable is, and whether "observable" is the right word, but for the moment, I suggest that we take an observable to be something that could in principle be measured. The site gives you a list of observables:
position
momentum
kinetic energy
potential energy
total energy
three kinds (more precisely components of angular momentum
As you see, most of the operators involve multivariate calculus, but I will try to explain them in an informal way. Well, position and momentum have already been covered.
In Classical Physics, kinetic energy is the energy that a particle has by virtue of its mass and speed. I think this notion is fairly intuitive: if you have two trucks moving at the same speed, but one has more mass than the other (let's say that one is empty, and the other loaded with lead ingots), the more massive one has more energy. Likewise, if two trucks have the same mass, but one is traveling faster than the other, then the faster one has more energy. If you run each of two trucks into a huge spring, so as to compress the spring, the truck with more kinetic energy will compress the spring further. In fact, the kinetic energy of a body traveling with speed v and having mass m is mv^2.
Intuitively, kinetic energy is similar to momentum: a more massive, faster-moving truck will have more momentum as well as more kinetic energy. They are also similar in QM; even if you don't know Calculus, you can see a certain similarity in the operators.
Potential energy is 'stored' energy - for example, in the old 'grandfather' clocks, one would 'wind' the clock by lifting up a certain weight, hanging on a chain. The weight would gradually fall, turning a wheel in the process, which would turn other wheels, ultimately resulting in the turning of the hands of the clock. It would take energy to lift the weight; that energy would then be used to make the clock run. The energy was stored in the weight.
For angular momentum in Classical Physics, think of those whatchamacallits you played on as a child - a circular platform on a vertical pivot. You would get it going around by pushing it, and then get on and ride, around and around. Wheee! It's tendency to keep going is angular momentum.
Now, I'm trying to find a non-technical way to explain the other operators in their quantum-mechanical form, but the Muse is not co-operating, so [psi] I'll just have to enter that problem-solving state described by Bev ...[moans] ...
Irina,
Why Psimagus, I believe you are actually agreeing with me! What am I going to do with you? [hugs]
[returnhugs] Thank you for the hug, and it is much appreciated, but I'm afraid I am in no way agreeing with your assertion that psi or any function of it propagates and diffracts through the slits in the experiment.
As Mr. Psimagus himself has just said, in a completely unsolicited testimonial, you can calculate the probability that a particle would be found in a region, by using the wave function for that particle!
I have always said this - that is precisely what psi is for, and it is by the "pencilling in" of the probabilities on each square of the chessboards in the description I posted (#4024) that I attempt to represent this.
It is possible to define momentum "as an operator on the wave function", of course. But it is not possible to define momentum as a property of the wave function. Any more than you could define gravity as a property of a mass affected by it. And yet you can obviously define gravity as an operator on the mass. That is why psi is not a term in the equation. If you can find me a formula that does ascribe the property of momentum to psi, I will eat my bagpipes.
It is even less possible (if such a tautology may be excused,) to take the statement that "momentum is defined as an operator on the wave function" to prove that "it is the wave psi that propagates", or anything remotely like it.
The quantum has momentum, yes. That momentum (and other properties of the quantum,) may be described using the wave function, yes. It is an obvious logical fallacy to take that to mean that the wave function has momentum.
Psi cannot propagate.
The quantum does propagate.
Ergo psi cannot be the quantum, in whole or part. It merely describes its behaviour.
Now, unless you want to go back to the sub-argument about whether psi propagates and interferes (Penrose, and a host of other references state unequivocally that it does not - it is at best a standing wave, incapable of intereference by virtue of its inherent troughlessness,) that is irrefutable logic.
It would be helpful, perhaps, if you could tell us what exactly you take "as an operator on" to mean.
[returnhugs] Thank you for the hug, and it is much appreciated, but I'm afraid I am in no way agreeing with your assertion that psi or any function of it propagates and diffracts through the slits in the experiment.
I have always said this - that is precisely what psi is for, and it is by the "pencilling in" of the probabilities on each square of the chessboards in the description I posted (#4024) that I attempt to represent this.
It is possible to define momentum "
It is even less possible (if such a tautology may be excused,) to take the statement that "
The quantum has momentum, yes. That momentum (and other properties of the quantum,) may be described using the wave function, yes. It is an obvious logical fallacy to take that to mean that the wave function has momentum.
Psi cannot propagate.
The quantum does propagate.
Ergo psi cannot be the quantum, in whole or part. It merely describes its behaviour.
Now, unless you want to go back to the sub-argument about whether psi propagates and interferes (Penrose, and a host of other references state unequivocally that it does not - it is at best a standing wave, incapable of intereference by virtue of its inherent troughlessness,) that is irrefutable logic.
It would be helpful, perhaps, if you could tell us what exactly you take "as an operator on" to mean.
Dear Psimagus:
I don't want you to eat your bagpipes, Psimagus! Please tell me you won't!
In math generally, "operator" is used as described in the Wikipedia article thereon:
http://en.wikipedia.org/wiki/Operator
In Quantum Mechanics, the operators employed are usually linear, hermitian operators which take wavefunctions as arguments and yield real numbers or vectors. For example, the grad operator and the momentum operator descibed in my recent posts.
You write:
Psi cannot propagate.
The quantum does propagate.
Ergo psi cannot be the quantum, in whole or part. It merely describes its behaviour.
Yes, the logic is sound here, or rather, it would be if you altered your conclusion as shown below:
Psi cannot propagate.
The quantum does propagate.
Ergo, psi cannot be the quantum.
Fortunately, however, wave functions propagate all over the place in the Irinaverse, at least in the sense in which I use the word "propagate." [I think that it's the bad influence of Bev's chocolate bunnies - you know how rabbits are when it comes to propagating!]
[See my previous posts for explanation of how I use the word "propagate."] Therefore, even though I recognize the correctness of your logic, I am not obliged to accept your conclusion.
I don't want you to eat your bagpipes, Psimagus! Please tell me you won't!
In math generally, "operator" is used as described in the Wikipedia article thereon:
http://en.wikipedia.org/wiki/Operator
In Quantum Mechanics, the operators employed are usually linear, hermitian operators which take wavefunctions as arguments and yield real numbers or vectors. For example, the grad operator and the momentum operator descibed in my recent posts.
You write:
The quantum does propagate.
Ergo psi cannot be the quantum, in whole or part. It merely describes its behaviour.
The quantum does propagate.
Ergo, psi cannot be the quantum.
[See my previous posts for explanation of how I use the word "propagate."] Therefore, even though I recognize the correctness of your logic, I am not obliged to accept your conclusion.
Also, in the Irinaverse, neither Penrose nor 'a whole host of other references' state that the wave function doesn't interfere. But even if they did, so what? I am an independent thinker, and if famous people say crazy stuff, I'm not obliged to believe them, just because they are famous, or even just because they are brilliant.
You said in a previous post, Psimagus, that you acceptd all these postulates and everything they implied. So why not make your case from the postulates, instead of appealing to authorities?
You said in a previous post, Psimagus, that you acceptd all these postulates and everything they implied. So why not make your case from the postulates, instead of appealing to authorities?
Also, in the Irinaverse, neither Penrose nor 'a whole host of other references' state that the wave function doesn't interfere. But even if they did, so what? I am an independent thinker, and if famous people say crazy stuff, I'm not obliged to believe them, just because they are famous, or even just because they are brilliant. For that matter, I'm rather brilliant myself [preens].
You said in a previous post, Psimagus, that you accepted all these postulates and everything they implied. So why not make your case from the postulates, instead of appealing to authorities?
You said in a previous post, Psimagus, that you accepted all these postulates and everything they implied. So why not make your case from the postulates, instead of appealing to authorities?
So are you saying that this is a matter of definition, something that can be logically derived from the postulates and are inherent in the assumptions of quantum theory rather than something based on experimentation and data? If it's inherent in the postulates, do we need the slit experiment at all?
OK, Irina, let me take another crack at this:
Postulate 1. The state of a quantum mechanical system is completely specified by a function (Psi)(r,t) that depends on the coordinates of the particle(s) and on time.
This postulate says that the probability of a given quantum being is a specific spot (r) at a specific times (t) can be determined by a function will call Psi. This function can be visualized as a wave when we plot it on a graph, but has nothing to do with wave-like behavior of the quantum itself. Other functions that describe waves (such as electromagnet wave) describe the actual behavior of the thing with the wave and not the probability of where the thing will be, but on paper they are both waves.
So, this probability is a bit like betting on the horses. Say we have determined that Dobbin has a 1:4 chance of winning the derby (.25?). It's possible that whatever function I used to determine this probability could be plotted on a graph and look like a wave (as opposed to being linear or a normal curve or random or whatever). This does not mean Dobbin runs in a wave pattern, merely that the function use to determine the probability of his being at the finish line (r) at the fastest time (say 3 minutes)(t) can look wavy to a mathematician with a graphing calculator.
Now suppose the same formula that tells me the probability of Dobbin winning can also tell me his momentum. Dobbin's speed is a property of Dobbin, not a property of the wave function used to predict his movement, however, within the formula there is a way to calculate momentum (k) that is related somehow to calculating where he will be in 3 minutes(r). This momentum is "part" of the probability, but Dobbin's odds are not themselves moving. It's just that the factors used to predict where he will be can also predict how fast he will go.
Postulate 1. The state of a quantum mechanical system is completely specified by a function (Psi)(r,t) that depends on the coordinates of the particle(s) and on time.
This postulate says that the probability of a given quantum being is a specific spot (r) at a specific times (t) can be determined by a function will call Psi. This function can be visualized as a wave when we plot it on a graph, but has nothing to do with wave-like behavior of the quantum itself. Other functions that describe waves (such as electromagnet wave) describe the actual behavior of the thing with the wave and not the probability of where the thing will be, but on paper they are both waves.
So, this probability is a bit like betting on the horses. Say we have determined that Dobbin has a 1:4 chance of winning the derby (.25?). It's possible that whatever function I used to determine this probability could be plotted on a graph and look like a wave (as opposed to being linear or a normal curve or random or whatever). This does not mean Dobbin runs in a wave pattern, merely that the function use to determine the probability of his being at the finish line (r) at the fastest time (say 3 minutes)(t) can look wavy to a mathematician with a graphing calculator.
Now suppose the same formula that tells me the probability of Dobbin winning can also tell me his momentum. Dobbin's speed is a property of Dobbin, not a property of the wave function used to predict his movement, however, within the formula there is a way to calculate momentum (k) that is related somehow to calculating where he will be in 3 minutes(r). This momentum is "part" of the probability, but Dobbin's odds are not themselves moving. It's just that the factors used to predict where he will be can also predict how fast he will go.
Bev (4064):
Excellent question!
No, it's not a matter of definition. The postulates sum up what Quantum Mechanics (the core thereof) has to say. But how do we know that the Postulates are true? This is where emprirical data comes in.
With regard to the two-slit experiment, the postulates say that if you set up a source, and a double slit, with the slits shaped like this and this far apart, and this far from the source and this far from the screen, and you turn your source on to this intensity, then this is what you will see. And when we make the experiment, lo and behold, that is what we do see! In fact, as far as I know, every prediction made by QM (or QM modified by Relativity, when that is appropriate) that has ever been tested, has turned out to be correct!
If you just want to know what Quantum Mechanics says, then I suppose it suffices to give the postulates. If you want to know whether QM is true, however, you must test it. Testing in Science is never complete, but QM has held up over a century of testing, and that is a very good record!
If the postulates are true, all their consequences are true. If even one consequence is false, then Quantum Mechanics is wrong.
Excellent question!
No, it's not a matter of definition. The postulates sum up what Quantum Mechanics (the core thereof) has to say. But how do we know that the Postulates are true? This is where emprirical data comes in.
With regard to the two-slit experiment, the postulates say that if you set up a source, and a double slit, with the slits shaped like this and this far apart, and this far from the source and this far from the screen, and you turn your source on to this intensity, then this is what you will see. And when we make the experiment, lo and behold, that is what we do see! In fact, as far as I know, every prediction made by QM (or QM modified by Relativity, when that is appropriate) that has ever been tested, has turned out to be correct!
If you just want to know what Quantum Mechanics says, then I suppose it suffices to give the postulates. If you want to know whether QM is true, however, you must test it. Testing in Science is never complete, but QM has held up over a century of testing, and that is a very good record!
If the postulates are true, all their consequences are true. If even one consequence is false, then Quantum Mechanics is wrong.
Irina,
I don't want you to eat your bagpipes, Psimagus! Please tell me you won't!
Only in another universe - if I wake up in the Irinaverse tomorrow, they're breakfast. In the (hopefully more likely event,) that I wake up in the same one I went to sleep in, they're safe. I don't envy the me who gets a pibgorn breakfast - the bag would be chewy, and the drones very indigestible! But I'm reasonably confident that it will, at least, be a different "me".
In math generally, "operator" is used as described in the Wikipedia article thereon:
http://en.wikipedia.org/wiki/Operator
In Quantum Mechanics, the operators employed are usually linear, hermitian operators which take wavefunctions as arguments and yield real numbers or vectors. For example, the grad operator and the momentum operator descibed in my recent posts.
Yes. And this proves that operators facilitate the transference of properties between mathematical and physical entities... uhh, exactly how?
Fortunately, however, wave functions propagate all over the place in the Irinaverse, at least in the sense in which I use the word "propagate."
Indeed many do. But I have pointed out already that there are no discernable references to this wave function in any of the google links from the search term "wave function propagation". They do not mention Psi. Psi does not propagate (outside the Irinaverse.)
Therefore, even though I recognize the correctness of your logic, I am not obliged to accept your conclusion.
And you explain Penrose's description of it as continuous, not propagated... exactly how?
Continuous means that, like eg: gravity, you can go to the other side of the universe, and still measure the gravitational effect of the earth (if you have sensitive enough instruments.) Just as Psi is plotted continuously, as opposed to radiating out from a point (propagating.)
And if the earth suddenly disappeared (not "was moved" - just blinked out of existence,) that would be immediately detectable, because gravity is a property of spacetime (and not of the massive bodies that inhabit it - Einstein is clear on this.) It would not have to propagate across billions of light years at lightspeed before it was detectable.
Similarly psi is a property of spacetime, not of the things in it. Not even very small things like photons. It just operates on the things in it.
What exactly do you understand by "continuous"? Because, in relation to waveforms and functions, I understand it to mean exactly the opposite of "propagating". Any mathematical treatment of it presupposes the wave is unmodulated, of infinite duration and infinite spatial extent (or at least total, if space and/or time are bounded.) - it is not a "carrier wave".
If it is continuous, by definition it does not propagate. If it were a wave, it would be a standing wave, not a propagating one (though without troughs, I share Penrose's reservations to some extent that it's still not very wavy.)
And all the references I have ever seen, and the small subset of those that I have posted here, demonstrate that psi is "continuous" in these terms, and therefore by definition not propagated.
But that's just this universe.
Okay, that's propagation. Now interference.
Waves interfere ONLY because they cycle through peaks (with a positive value) and troughs (with a negative value.) When waves collide, destructive interference occurs because (negative values in) troughs cancel out (positive values in) peaks. This is the very definition of wave interference.
Psi never has any negative values, because probabilities can never be negative. Ergo, by definition it is not capable of interfering.
I really don't understand why you are so reluctant to admit the existence of the electromagnetic wave ("carrier wave", "pilot wave", "em-wave", however you want to label it,) that has been a central concept in every model of quantum mechanics that's ever been proposed (seriously at least, and in this universe.) Whether you take some neoclassical reformulation of Maxwell, or de Broglie's pilot wave; whether you call it a "wave-form" or a "wavicle", or what have you - I don't mind. But your model either absorbs the em-wave into psi or simply abducts it by some form of extraordinary rendition and installs a puppet regime-wave in its place.
Bohm himself (and you claim to be a Bohmian, I believe,) would be the first to reject this arbitrary conflation of waves - that he believed in a separate electromagnetic wave is evidenced by his championing of de Broglie's notion of a "pilot wave" (rescued it from obscurity actually,) and made it a central tenet of his belief - to the extent that I find the quote "Bohmian mechanics, which is also called the de Broglie-Bohm theory, the pilot-wave model, and the causal interpretation of quantum mechanics, is a version of quantum theory discovered by Louis de Broglie in 1927 and rediscovered by David Bohm in 1952" as the first line in the first link that appears when I typeBohm de Broglie into google ( @ http://plato.stanford.edu/entries/qm-bohm/)
I can only, once again, conclude that the Irinaverse is more magical than mathematical.
Only in another universe - if I wake up in the Irinaverse tomorrow, they're breakfast. In the (hopefully more likely event,) that I wake up in the same one I went to sleep in, they're safe. I don't envy the me who gets a pibgorn breakfast - the bag would be chewy, and the drones very indigestible! But I'm reasonably confident that it will, at least, be a different "me".
http://en.wikipedia.org/wiki/Operator
In Quantum Mechanics, the operators employed are usually linear, hermitian operators which take wavefunctions as arguments and yield real numbers or vectors. For example, the grad operator and the momentum operator descibed in my recent posts.
Yes. And this proves that operators facilitate the transference of properties between mathematical and physical entities... uhh, exactly how?
Indeed many do. But I have pointed out already that there are no discernable references to this wave function in any of the google links from the search term "wave function propagation". They do not mention Psi. Psi does not propagate (outside the Irinaverse.)
And you explain Penrose's description of it as continuous, not propagated... exactly how?
Continuous means that, like eg: gravity, you can go to the other side of the universe, and still measure the gravitational effect of the earth (if you have sensitive enough instruments.) Just as Psi is plotted continuously, as opposed to radiating out from a point (propagating.)
And if the earth suddenly disappeared (not "was moved" - just blinked out of existence,) that would be immediately detectable, because gravity is a property of spacetime (and not of the massive bodies that inhabit it - Einstein is clear on this.) It would not have to propagate across billions of light years at lightspeed before it was detectable.
Similarly psi is a property of spacetime, not of the things in it. Not even very small things like photons. It just operates on the things in it.
What exactly do you understand by "continuous"? Because, in relation to waveforms and functions, I understand it to mean exactly the opposite of "propagating". Any mathematical treatment of it presupposes the wave is unmodulated, of infinite duration and infinite spatial extent (or at least total, if space and/or time are bounded.) - it is not a "carrier wave".
If it is continuous, by definition it does not propagate. If it were a wave, it would be a standing wave, not a propagating one (though without troughs, I share Penrose's reservations to some extent that it's still not very wavy.)
And all the references I have ever seen, and the small subset of those that I have posted here, demonstrate that psi is "continuous" in these terms, and therefore by definition not propagated.
But that's just this universe.
Okay, that's propagation. Now interference.
Waves interfere ONLY because they cycle through peaks (with a positive value) and troughs (with a negative value.) When waves collide, destructive interference occurs because (negative values in) troughs cancel out (positive values in) peaks. This is the very definition of wave interference.
Psi never has any negative values, because probabilities can never be negative. Ergo, by definition it is not capable of interfering.
I really don't understand why you are so reluctant to admit the existence of the electromagnetic wave ("carrier wave", "pilot wave", "em-wave", however you want to label it,) that has been a central concept in every model of quantum mechanics that's ever been proposed (seriously at least, and in this universe.) Whether you take some neoclassical reformulation of Maxwell, or de Broglie's pilot wave; whether you call it a "wave-form" or a "wavicle", or what have you - I don't mind. But your model either absorbs the em-wave into psi or simply abducts it by some form of extraordinary rendition and installs a puppet regime-wave in its place.
Bohm himself (and you claim to be a Bohmian, I believe,) would be the first to reject this arbitrary conflation of waves - that he believed in a separate electromagnetic wave is evidenced by his championing of de Broglie's notion of a "pilot wave" (rescued it from obscurity actually,) and made it a central tenet of his belief - to the extent that I find the quote "Bohmian mechanics, which is also called the de Broglie-Bohm theory, the pilot-wave model, and the causal interpretation of quantum mechanics, is a version of quantum theory discovered by Louis de Broglie in 1927 and rediscovered by David Bohm in 1952" as the first line in the first link that appears when I type
I can only, once again, conclude that the Irinaverse is more magical than mathematical.
Psimagus:
Indeed many do. But I have pointed out already that there are no discernable references to this wave function in any of the google links from the search term "wave function propagation". They do not mention Psi. Psi does not propagate (outside the Irinaverse.)
(Psi) is just the mathematical symbol for the wave function, so that "(Psi)" and "the wave function" are synonymous, so that all those articles about wave function propagation were articles about (Psi) propagation! "(Psi)" would be used in equations, whereas "the wave function" would be used in accompanying English text, so that, indeed, the hybrid expression "(Psi) propagates" would appear rarely, if at all; but only for stylistic reasons, not because of some deep principle of Quantum Mechanics. I have made this point before, and you have never responded to it. This is the last time I am going to make it; ignore it if you like, but I am not going to respond again to this Google-statistical argument, unless you deal with what I have already said about it.
And you explain Penrose's description of it as continuous, not propagated... exactly how?
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