Seasons

Dear Bev (4049) :

    I think that there is a great deal of overlap between the words "probability" and "chance," but they are not quite the same. When we say that the probability of a coin's coming up heads is 1/2, we mean that in the long run, it will come up heads about half the time. If we say that it has "half a chance" of coming up heads, we are saying the same thing. On the other hand, we frequently say that something "happened by chance," whereas, to my ear, it sounds rather odd to say, "It happened by probability."
To say that it happened by chance is to say that it cannot be fully explained, and could not have been predicted. Sometimes the particles go through, sometimes they do not. Suppose we run the experiment and they go through. We then run another experiment, and we take great care to prepare it in exactly the same way, but it may happen that they do not go through. So far, we have just not been able to find any variable or factor that allows us to determine in advance whether they will go through or not. There may be such a factor, but if so, we have not discovered it yet; hence the term 'hidden variable.'

Some people have very strong intuitions that ultimately, there is no chance in the universe, and that if there appears to be, this is only because of our ignorance. Einstein was such a person. At the present time, however, we can only say that we have not yet discovered such a hidden variable.
    On the other hand, we can say with certainty that if one particle goes through, the other one will also. So that is not a matter of chance. If we create the two particles in the correct way (and we know how to do this), then either they will both go through or they will both fail to go through. So I would not say that it is a coincidence that they do the same thing. To say that something happens 'by coincidence' is, to my ear, about the same as saying that it happens 'by chance.' You cannot rely on it happening again, quite the contrary! But in fact, you can rely on the particles' either both going through or both not going through. If you find this odd, you are in good company; Einstein, for example.
    When I said, in my previous post, "the rest is explained by chance," I meant that whether or not they both go through is a matter of chance. QM predicts with certainty that they will both do the same thing (go through or not); that is not chance, not a coincidence, according to QM. But whether they will both go through, or both fail to go through, appears at present to be a matter of pure chance.
    In general, the way Quantum Mechanics works is this: in a given situation, it says what the possible outcomes are, and it assigns a probability to each of those outcomes. In this case, it says there are two possible outcomes: (A) both particles go through and (B) neither particle goes through. QM says that each of these outcomes receives probability 1/2. QM is not willing to say more. Many people are inclined to say that if both of the particles are bound to do the same thing, it must be because either (1) the one particle going through (or not) causes the other to do the same, or (2) there must be something that makes it determined in advance whether they go through, although this something is hidden to us. Einstein could not accept (1) because it would involve causation at greater than light speeds, and so he (and others) thought that QM must be leaving something out, because QM merely assigns a probability of 1/2 to each of the two possible outcomes. Perhaps QM is wrong, but this is all that it says. What QM says and whether it is right are two different issues. I am here concerned primarily with the former.

Walk in Beauty, Irina

Psimagus:

Yes, you are correct: the phrase "psi" does appear in many scientific papers. For example, it is a common abbreviation for "pounds per square inch." My mistake. I was wrong.

OK, here is my attempt to explain, without using higher math, what Postulate 1 means:

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.
That is, given a point of space (r) and a point of time (t), the function (Psi) assigns a value to this number. Or, to put it in a manner similar to Psimagus’ checkerboard metaphor, for any given time t, (Psi) assigns to every point r of space a given value. The value assigned to a given point of space may vary from one time to another, however; this is what I mean by saying that the wave function propagates: as time goes by, it assigns different values to various space points at different times.

Alas, I must disagree with what I understand Psimagus to be saying, namely that the values of (Psi) are restricted to positive real numbers. In fact, the values of (Psi) are in general complex numbers, that is, numbers which can be put in the form, (a + bi), where a and b are real numbers, and i is the square root of -1. [A Wikipedia article on real numbers may be found here: http://en.wikipedia.org/wiki/Real_number , and a Wikipedia article on complex numbers may be found here: http://en.wikipedia.org/wiki/Complex_numbers ] I gather (perhaps incorrectly) that the reason Psimagus wants to restrict Psi to real numbers is because he is under the impression that the value of (Psi) expresses a probability, which has to have a real value between 0 and 1. It is true that (Psi) is sometimes called a probability wave, but as Penrose has pointed out, this is a misnomer. As we shall see in a moment, (Psi) itself does not express the probability of anything, although (Psi) may be used as a starting-point from which to calculate various probabilities.

This function, called the wave function or state function, has the important property that
(Psi)*(r,t) (Psi) (r,t)is the probability that the particle lies in the volume element d(tau) located at r at time t.

The notation (Psi)* refers to the complex conjugate of Psi. The complex conjugate of a complex number (a + bi) is the complex number (a – bi). The fact that the authors use this notation shows that they agree with me that the value of (Psi) is, in general, complex. When we multiply a complex number times its complex conjugate, however, the result is always a real number. [Discussion of this may be found in the abovementioned Wikipedia article, although their notation is different; they indicate the complex conjugate of a number by drawing a line above it.] So
(Psi)*(r,t) (Psi) (r,t) is a real number, different from (Psi).

Now,to grasp the exact meaning of the expression d(tau) requires a knowledge of Calculus; however, you will not be far wrong if you think of d(tau) as a very, very small region of space surrounding the point r. S what the authors are saying is, that (Psi)*(r,t) (Psi) (r,t) is the probability that the particle lies in this small region (“volume element’). In other words: If you want to know the probability that the particle would be found in a tiny region around r, at time t, were a suitable detector to be placed there, find the value of (Psi) at r and time t, and multiply that value by its complex conjugate. This gives you a real number which is the desired probability.

The wavefunction must satisfy certain mathematical conditions because of this probabilistic interpretation. For the case of a single particle, the probability of finding it somewhere is 1, so that we have the normalization condition


(110)    [Integrate from negative infinity to infinity](Psi)*(r,t)(Psi)(r,t)d(tau)(Psi) (r,t)(Psi)(r,t)d(tau)

To understand “integrate” precisely requires a knowledge of Calculus. However, you will not be far off if you think of it as “adding up.” We saw above how to find the probability that the particle would be found in a very tiny region; to find what the probability is that the particle would be found in some larger region, just divide that larger region up into tiny ones, and add up the probabilities for the tiny ones!

To integrate from negative infinity to infinity means, in this context, to integrate over all space [actually, IMHO, the authors are a little imprecise here, but this is surely what they mean]. So

(111)    [Integrate from negative infinity to infinity](Psi)*(r,t)(Psi)(r,t)d(tau)(Psi) (r,t)(Psi)(r,t)d(tau)

Means, “Divide all space into tiny regions, and add up all the probabilities for the regions, and you get the probability that the particle is somewhere in all space.” But, given that the particle exists at all, it’s gotta be somewhere, so this probability is equal to 1.

It is customary to also normalize many-particle wavefunctions to 1.

The wavefunction must also be single-valued, continuous, and finite.

“Single-valued” means that (Psi) can only assign only one value to a given point of space and a given time. That is, at each point in time, (Psi) has a single value for each point in space.
“Continuous” means that Psi is define everywhere, at all times, and that its value does not make sudden jumps; it changes smoothly from one point to another, and from one time to another.
“Finite” means that (Psi) never takes on infinite values; that is, it only takes on values
(a + bi), where a and b are both finite.

OK, that is my humble attempt to explain what this postulate means. Please let me know if you have any questions!


Walk in Beauty, Irina

Oh, there is one more point I would like to make:

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 function, called the wave function or state function, has the important property that
(Psi)*(r,t) (Psi) (r,t)is the probability that the particle lies in the volume element d(tau) located at r at time t.

It should be evident from this passage that "(Psi)" and "the wave function" are, to the authors, just two different names for the same thing.

I did know what Ø was in terms of it being a Greek letter

I fear some of these characters are getting a bit mangled. We probably have different code pages installed - my default settings seems to show Ψ (Greek psi) perfectly, and by the sounds of it you recognized it, but it's got changed to an "O" with a slash through it

Irina,

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 function, called the wave function or state function, has the important property that (Psi)*(r,t) (Psi) (r,t)is the probability that the particle lies in the volume element d(tau) located at r at time t.

It should be evident from this passage that "(Psi)" and "the wave function" are, to the authors, just two different names for the same thing.

Yes. And it should also be obvious that this is nonetheless not remotely the same as the de Broglie electromagnetic wave that diffracts and interferes. Its purpose (in the words of the authors) is precisely to define "the probability that the particle lies in the volume element d(tau) located at r at time t." NOT to perform the the diffracting and interfering that the photon exhibits.
It tells us how likely the chess piece is to be on any given square - it is not a PART of the chess piece.

I did know what Ø was in terms of it being a Greek letter

I fear some of these characters are getting a bit mangled. We probably have different code pages installed - my default settings seems to show Ψ (Greek psi) perfectly, and by the sounds of it you recognized it, but it's got changed to an "O" with a slash through it

So, you reject my hypothesis of a trickster quantum bunny messing with the posts? I suppose I can grudgingly admit the possibility of code page issues, but your theory, while better suited to known facts, is not half as much fun.

The weird thing is that when I see my post, I see the Greek letters I intended (psi, delta and pi). When I read Irina's post where she pasted in quotes from mine, I see the circle with the line through (I wanted it to be an omicron or something, but I am not sure what the angled slash means). I constantly have issues with question marks for apostrophes and quotes too. *shrugs*

Dear Psimagus (4056):

Yes.

Why Psimagus, I believe you are actually agreeing with me! What am I going to do with you? [hugs]

[Addresses crowd] Yes, folks, many are the wonders of the Irinaverse! 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! Just apply Postulate 1! But that is only the beginning! Using the wonder of OPERATORS, you can also find other properties of the particle, for example, its
momentum!!! All you have to do is apply the momentum operator to the wave function!!! And you can get the momentum operator, ABSOLUTELY FREE, right here:

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

Just scroll down to where it says "Momentum in quantum mechanics." In fact, when the wave function takes on the particularly simple (one-dimensional) form, Ae^i(kx-wt), you can just read off the momentum - it is k!!! And as a bonus, you can read off the amplitude of the wave, namely A!!!

Quiz question: What are the amplitude and momentum associated with the wave function 7e^(5x - 17t)?
Answer: The amplitude is 7, and the momentum is 5.

That triangle you see, in the momentum operator as shown on the Wikipedia site, is, as they say, the gradient operator. It is often written as "grad", and that's the way I will do it. The h with a little slash through it, the reduced Planck's constant, is just h/(2pi), where h is Planck's constant itself, and pi is the ratio of a circle's circumference to its diagonal! So the momentum operator can be written as

(h/2pi)grad

which means that we can write it in chat boxes here on the Forge!!!
But what is the grad operator? Well, imagine placing a marble on the side of a hill. Which way would it roll? It would roll in the direction wherein the downward slope was steepest! And that's what grad does, folks - it points in the direction of greatest change! It also tells us how great that change is!

If you place a marble on a flat surface, it doesn't go anywhere - so grad is zero in this case! For example, grad(3)=0, because 3 doesn't change!
Now consider the line y=x. It has a slope of 1, and
grad(x) is indeed 1. The line y=2x has a slope of 2, and grad(2x) is 2! And so on!

Quantum Mechanics is easy, in the Irinaverse!

So, you reject my hypothesis of a trickster quantum bunny messing with the posts?

I think it was a chocolate psi - it melted :O

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] ...

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.

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.