# A Thanksgiving Question

Discussion in 'Physics' started by studiot, Nov 27, 2008.

1. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Since there has been a resurgence of interest amongst members about the mechanics of waves, relativity and simultaneity I would like to pose a question.

We are generally familiar with Fourier series and perhaps know that to represent a step function we require the series to reach all the way to infinity.

However there is one more requirement, understood but not normally stated.

All the components must be present at the same time (simultaneously), for the series to sum to the required function.

Now consider an energy density function, such a solution to Schroedingers wave equation for a step function that occurs when an elementary particle appears (is created)

This has components all the way to infinity, which must suddenly come into being and exist simultaneously. (sound familiar?)

Any changes to the state of the particle must change all these components simultaneously.

So
1) How does this happen if light is the limiting speed?
2) How does the infinite series fit into our allegedly finite universe?

2. ### triggernum5 Active Member

May 4, 2008
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Are you pointing out the difficulty of expressing the energy density as a fourier series.. Experiements have confirmed the existence of traveling wave packets for the energy density while simultaneously studying the time evolution of energy within a finite cavity..
You may find this pdf interesting..
http://www.sbfisica.org.br/bjp/files/v35_211.pdf

3. ### blazedaces Active Member

Jul 24, 2008
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I'm not extremely familiar with schrodinger's wave equation for a step function, but I know what you're talking about.

But as for the fourier series, I would like to point out that this is simply an approximation for periodic functions where the approximation becomes closer and closer to the real function as the series reaches infinity (where it literally is the function when it reaches infinity).

So talking about our universe in such a manner is not really relevant. We only use these series to help us understand and manipulate these concepts.

-blazed

4. ### triggernum5 Active Member

May 4, 2008
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Yea, my point is kind of along what blazed said (I think).. The universe signed no agreement that ensures its properties can be expressed as a fourier series in any domain..

5. ### BillO Well-Known Member

Nov 24, 2008
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Schrodinger's equation , allows you a means to calculate the systems wave function , which should not be interpreted as the equation of motion as in classical mechanics, although it looks similar.

Rather than representing the physical specifics (position, momentum, existence) of a particle in a system, it represents the probability distribution that a particular state of the system will occur at (x,t). Squaring it gives the probability that such a particle will be in such a state at such a location and at such time. Therefore it is valid over all time.

So, it does nothing to tell you the specific step function, but it will tell you how likely the step in that function is to occur at position x at time t (x, by the way, represents the fully qualified position is 3 dimensional space).

In my experience, this is the single most difficult concept to understand in QM, so don't feel bad if you find it confusing.

6. ### triggernum5 Active Member

May 4, 2008
216
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Im pretty sure studiot understands that, but waves are waves..

7. ### BillO Well-Known Member

Nov 24, 2008
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Then, evidently, you don't understand.

The wave equation regarding this system, as given by Schrodinger's equation, would in no way look like a step function and it never changes shape, even after the event occurs.

8. ### blazedaces Active Member

Jul 24, 2008
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You say "shape". What do you mean by that? Isn't schrodinger's wave equation not something that can be visualized?

I'm asking out of curiosity. I want to know more about it, but my understanding of it is quite lacking...

-blazed

9. ### triggernum5 Active Member

May 4, 2008
216
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I realize the requirements for continuity, the pdf I posted outlined some of the work ppl like Dirac did to reconcile those needs with these situations..
And blazed, perhaps the physical signifigance of a probability wave cannot be visualized, but obviously a graphical representation of it can be..

10. ### BillO Well-Known Member

Nov 24, 2008
985
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Well, every continuous and well-behaved function definitely has a shape within the space it is defined, but in this case, the probability density wave function's shape does not represent the 'shape' of the equation of motion of the system in question.

For instance, we have all likely seen the shapes of the electron orbital of the first few s, p and d, states of the hydrogen atom. These spheroid and lobed shapes are found by defining a constant probability surface within the probability density wave function for the specified state that contains (usually) 67% of the total probability. In other words, for hydrogen in a given state, the electron should be found 67% of the time within that shape. However, there is absolutely no information given about how the electron moves within that space nor what its flight path would look like, or even if it could be described as a flight.

11. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Well thank you all for carrying on the discussion while I was away.

Unfortunately you have not really addressed the point I was trying to make; obviously I didnt explain very well so I will try again.

Firstly though to dispel some inaccuracies that have been stated.
I am still struggling to get math onto an AAC page so please bear with me. I originally typed the following in Word.

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12. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Now to repeat my original point, slightly differently.

It was known to Schrodinger, Dirac, Einstein et al that the solutions to his wave equations contained the intriguing possibility of instantaneous 'action at a distance' or remote action. This is in direct contradiction to the relativistic requirement of c being the maximum speed of a disturbance.

This contradiction has never been resolved.

When an electron (say) is promoted from one energy level to the next it 'jumps' from one x value to another. This happens instantaneously, so far as we can tell.
But the x values (measures of distance) are non zero so the electron effectively moves faster than light.

It doesn't matter what observer framework we choose to measure the difference in x, using relativistic corrections or not. The different is still non zero and non infinite.

13. ### blazedaces Active Member

Jul 24, 2008
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I've got two questions, and then a possible answer... maybe.

1: Can someone explain to me why the square of the magnitude of the wave function is the probability that a particle is at a given point x? It seems like it should be obvious... but it feels like a leap of faith to me...

2: If the function is Psi*sin(2*pi*f*t) then isn't it's double derivative in terms of time -Psi*4*pi^2*f^2*sin(2*pi*f*t)? What happened to the sine/time component when we plugged it into the original wave equation?

Now, studiot, this "contradiction", as you explained, has never been resolved. I believe though that individual particles do not move through space as we normally understand movement.

Euclidean geometry would tell us that the shortest distance between two points is a straight line, but Einstein showed us the shortest distance between two points is actually based on the curvature of spacetime through which these two points are separated.

In quantum mechanics, we can not ever know both the position and the velocity of a particle at a given time. Not only that, but these are actually randomly determined (within a certain degree of certainty) often when we measure them.

So my question to you now is this: can we truly be certain that a particle is contradictint classical physics or is it simply that we don't know what happens between the moment before it "jumps" and the moment after?

I hope my question doesn't sound ... well... stupid. Please excuse my ignorance on the subject.

-blazed

14. ### BillO Well-Known Member

Nov 24, 2008
985
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Hmm, no one said ψ was a probability function. I said it was a probability distribution function. I.E. when the amplitude of ψ is large for a given state, then the probability will be higher for that state too. BTW It's not my terminology, but it is accurate nonetheless.

However, I stand by what I said. Nothing happens to the wave function when the subject event occurs. The wave function is still just that, a wave function, with terms and parameters, calculated from the description of the system. All that does happen to ψ once the event is realized is, its relevance vanishes. It never had any physicality to begin with.

This:

Is so wrong, on so many levels.

In any case, there is a postulate similar to the one which you refer. It goes something like this (simplification):

For a simple system in which an event causes the creation of a particle and its anti-particle, the resulting wave function requires the dependent existence of both particles as conditions of its validity, for want of a better term. Or, each particle is the raison de etre of the other.

So, lets take the case where these two particles speed of in opposite directions. Even though any distance can separate them they remain a single system governed by the wave function. So if one of these particles decays out of existence, then since one of the necessary conditions of the system no longer exists, the wave function dictates that the system does not exists, which in turn means that there is no longer any validity to the existence of the other particle. So it follows the other particles must also decay at (and here is the silly part) the same time.

Experiments have been conducted to show this is the case. But does it really represent something that goes faster than the speed of light? If it does, what is going faster than the speed of light? In the same experiments attempts were made to influence the outcome. However, any attempt to force the decay of one particle altered the system such that the independent existence of the other particle could be explained. So in reality not even information could be transmitted across the gap.

Nov 9, 2007
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For Blazed

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16. ### blazedaces Active Member

Jul 24, 2008
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Ah yes, well, stupid mistake of mine. Thanks for clearing that up though.

-blazed

17. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Right now we are all agreed on the equations we can try to answer the probability question.

I rahter hoped Bill would do this since he introduced the concept.

Yes it is something of a leap of faith to associate the square of the (modulus) of the wave function with the probability of encountering a particle. Most texts just state the association as fact.

However there is a long route from the wave function to the probability. One thing that has to be done is to 'normalise' the wave function by integrating the square over the range ±∞ and setting the result equal to 1. That is the particle must be somewhere.

As a justification try this:

Consider our particle as a swinging pendulum bob. As the pendulum swings to and fro it describes simple harmonic motion, which obeys the wave equation.
The velocity of the bob varies from max at the lowest point to zero at the highest of the swing.

The time spent passing through some x coordinate x and (x+Δx) is obviously less near the point of max velocity compared to the time taken to traverse a similar (x+Δx) near the slow end.

If we were to cross the path of the bob at random our chance of encountering the bob obviously increases with the time spent passing through that particular Δx.

Now the amplitude of the wave function y(x) or y value is at its maximum when the velocity is the least, so the probability of encountering the bob is inversely proportional to the velocity, but directly proportional to the amplitude of the wave function.

Of course the probability of encountering the bob is another way of saying the probability that the bob will be in a particular place.

Note also that the speed of the bob is not the same as the wave velocity, referred to as the $\frac{1}{v^{2}}$ term in both the wave and Schrodinger's equation.

We use the square because we want the probability to be non-negative and real. There are an infinite number of solutions to any given wave or Schrodinger equation, many of them complex and/or negative functions. By squaring the modulus we avoid these difficulties.

Over to you Bill

18. ### BillO Well-Known Member

Nov 24, 2008
985
136
Sorry guys, I’ve been travelling and am in the middle of a multi-city, multi-day trip. I have a little time before I catch my flight to Montreal. Considering this let me be brief and take two shortcuts:

a. You know something about harmonics
b. You understand the difference between kinetic and potential energy
c. You know what a phonon is
d. You have a least seen the derivation of a classical wave function given by the equation of motion for a harmonic system (eerily similar to Schrodinger’s equation)

2) Gross simplification
a. One dimension
b. No imaginary components
c. No time dependence (the nature of the equation does not change over time)
The reason being, that the derivation and interpretation of both classical and quantum wave equations can take up considerable volume.

Let’s look at a bounded string. Because we can see the string, determine its materials of composition and make all manner of measurements pertaining to the parameters that effect it, we could conceivably construct a very accurate model that allow us to understand (almost) completely how it will respond and behave with respect to all possible conditions. However, let’s make the situation a little tougher in order to simplify it for this illustration. Let’s put it in a black box and only give the observer the means to measure the amplitude of potential energy along the string as it vibrates. So effectively, he can only determine what the maximum displacement is of the string at any point along its length. Further, we do not tell him it is a string – too much information. We tell him only that the box contains a phonon whose potential energy distribution is given by the equation:

Y(x,q) = asin[(x/L) pq] Where L is the length of the box.

And we ask him to determine where the phonon is within the box. Well, he can assume, since this phonon is in a box that it will be bounded and hence the equation above will only give valid solutions such that when;

x=0 and x=L, Y(x,q)=0

Which, in turn, will only be valid for integral values of q (quanta!).

He has absolutely no way of knowing the true nature of the phonon (what medium it travels in, how it travels, etc.). All he can do is use the tool you gave him to verify the equation. So, really, he cannot tell you where it is, as everything he knows about the system is given by the wave function. What he does realize is that the intensity of the wave at any given point should reflect the proximity of the phonon, just as it would in the case where getting closer to a sound source increases the intensity of the energy from that source. He knows that the square of a wave gives its intensity such that:

Y(x,q)*Y(x,q) = I(x,q)

Below is a plot ofY(x,q) vs. I(x,q) for the case where q=4

Now, since I(x,q) can viewed as a proximity distribution it should be proportional to the probability that the phonon is there. At least as close as probability theory will allow for continuous functions. So he reasons that:

Px,q = 2*I(x,q)/(L*a*a)

There are problems with this. First, there is not much in probability theory that supports it, especially in the complex (real + imaginary) wave functions of QM. Second, even within the QM community, not everyone agrees on the interpretation. Notwithstanding, it’s useful and real solutions can and have been drawn from it so the only proof of it’s validity is currently empirical.

Well, I hope this helps. Likely it doesn’t (there’s are reason I gave up teaching 25 years ago) so I’ll refer you to Feynman’s lectures and a decent discussion on the Copenhagen Interpretation (Google it).

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19. ### blazedaces Active Member

Jul 24, 2008
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Before I go delving into those lectures and other thing you suggested looking up, can I ask a few questions?

First, is intensity defined implicitly or is there is some other relationship this comes from?

What exactly is the "potential energy distribution"? Is it the potential energy at a given point x and something q?

Which brings me to my second question... what is q? Is q charge? Is it some form of energy (because you said it needs to be quantized)?

And lastly, and this is a slightly personal question, but could you elaborate on that last thing you said: why did you stop teaching?

-blazed

20. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Hard to say exactly what Bill is getting at, although the phrase

did catch my eye as I wasn't aware that potential energy had an 'amplitude'.
Apart from anything else we are back to the observation that potential energy is strictly non-negative whereas the sine function displayed has positive and negative regions.

Blazed,
How did you get on with my explanation/justification?