# The probability interpretation of the wavefunction?

1. Aug 21, 2010

### jeebs

Hi,
I'm about to go into my 4th and final year of my undergraduate physics degree and after all the quantum mechanics we've done so far, I still get this nagging feeling that I'm answering homework and exam problems blindly. Apologies for the length of this question by the way.

For example, take the infinite square well problem. The particle is confined in a 1-dimensional region of space in an infinite potential. It's easy enough to mess around with the Schrodinger equation and out pops a nice, simple little wavefunction, say, $$\psi (x) = Asin(n\pi x/a)$$ for a particle confined between x=0 and x=a. Now apparently this can be used to determine how likely you are to find the particle in a specific position x, if you simply square the wavefunction.

I have never understood where this came from. No text book I have read has explained why this is done or why it works. Also, in the infinite square well example that I have used, it seems to me to give a bit of a dubious result.
When you square this wavefunction, you find that at x=0 and x=a, you get $$\psi^2 = 0$$. Assuming that squaring the wavefunction really does give you the probability of the particle's position, then this result of zero probability at the walls of the well makes sense - since how could the particle possibly escape an infinite potential and cross over to this impossible-to-reach region? This much I am happy with.
However, take the n=2 (first excited) eigenstate. At x=a/2 we are going to get $$sin^2(\pi) = 0$$ which means that we have a zero probability of finding the particle right in the middle of the well. How does that make sense? This does not seem like a sensible result to me. What is wrong with being in the middle of the well when you are at this energy level? Does this not suggest that the probability interpretation of the square of the wavefunction is less than perfect?

Oh, another thing I have been wondering about is to do with Heisenburg's uncertainty principle. I have watched a few quantum documentaries (eg. with that Michio Kaku guy) on the internet and they always go on about quantum particles existing in 2 places at once, but they seem to be aimed at a less-educated audience so they just state their results with no explanation.
The understanding I have about this and the uncertainty principle is that, if you have particle sitting still in some position, you have to bounce a photon off it to detect it. When the photon leaves the particle, the particle recoils under momentum conservation, so by measuring the position of the particle you have altered its momentum - hence the position and momentum cannot both be known with absolute certainty at the same time.
My question about all of this is, are these documentaries being inaccurate/misleading when they say that the particle is in two positions at the same time?
Just because our imperfect method of measuring introduces this uncertainty in a particle's position, surely that doesn't mean the particle literally exists in 2 positions at once, right?
Shouldn't these documentaries really be trying to say that there is a probability of finding the particle exclusively in each position?
Is there really an intrinsic uncertainty in a particle's position, or is it just that human methods of collecting position data introduce this uncertainty that in reality is not there?
I'm not sure if I've explained what I mean here (hence the long, rambling question ^_^) but I appreciate any attempt at an answer.

Thanks.

2. Aug 21, 2010

### skippy1729

If you view the particle in a potential well as a physical experiment rather than a mathematical exercise the problem of finding the particle at the center of the well disappears. Any classical particle detector at the center of the well will detect particles within a certain small but finite range about the center. Over the course of many trials, you will detect the probability density for that range, never the probability at a point.

Is the particle ever actually at the center of the well? According to the CI (Copenhagen Interpretation) the particle does not have any position until it is observed. The question is meaningless according to CI. Of course, there are other interpretations, the deBroglie-Bohm pilot wave theory maintains that there is an unobservable but definite trajectory for all particles. While I am no authority on this theory it would seem that if the particle was in the left side of the well and was later found to be in the right half then at some instant of time it must have been at the center.

As you can see, in neither of these two interpretations is the particle ever at two places at once. Perhaps someone else knows what Dr. Kaku was alluding to.

You are also interested in the origin of the probability interpretation (the Born rule). The strongest justification for this is Gleason's theorem. Which is covered in "Quantum Theory: Concepts and Methods" by Asher Peres and many other advanced texts. If you are at all interested in the Copenhagen viewpoint, it is the best (non-introductory) quantum text I have read. If you have had one formal course in QM, it is readable and suitable for self study.

Cheers, Skippy

3. Aug 21, 2010

### jeebs

The Born rule eh? I'll be sure to look it up, although I'm guessing it's complicated since it's been conveniently ignored in every lecture and textbook I've seen. Thanks man.

One thing though, going back to the infinite potential well, what about for higher excited states that have zero probability positions in places other than the centre of the well? If I'm not mistaken you were talking about having some detector positioned right at the centre of it, right? well what about in the 3rd excited state (n=4) where $$\psi^2 = Asin(4\pi x/a)$$, which has a zero value at x=a/4 and x=3a/4 ?

or is this just one of the mad things physics throws at us that is so different to what we might expect in everyday life (like the falseness of galilean transformation of velocities that relativity reveals) but that we have no choice but to accept?

Last edited: Aug 21, 2010
4. Aug 21, 2010

### alxm

There's no "probability interpretation" of quantum mechanics, there's an interpretation of what the probabilities mean. The Born rule and such is a postulate of QM and part of the formalism, not the interpretation. The difference here in interpretation is that this probability distribution is, in the deBB version a statistical distribution of quite definite states, whereas in the Copenhagen interpretation it's truly indeterminate (i.e. the probability does not represent a mere lack of knowledge about the system).

Skippy is wrong. A particle being 'in two places at once' alludes to your basic superposition of states and/or the fact that it does not have a definite location. All interpretations allow for this, and the probability distribution most certainly does have meaning even when the particle is not measured. Unless you perform a specific measurement of a specific property, the wave function does not 'collapse' (if we assume the CI interpretation for the sake of argument). You can perform measurements on a superposition state without destroying the superposition. This is your basic double-slit experiment, where determining 'which slit' will destroy the superposition, but merely observing the interference pattern does not.

5. Aug 22, 2010

Staff Emeritus
Check the index of those textbooks. You'll be surprised.

The developers of quantum mechanics knew about waves, and they knew that the intensity is the square of the amplitude. They also knew that the intensity had to integrate to particle number. Born concluded that probability is the way to interpret these facts.

6. Aug 22, 2010

### skippy1729

First off, I would like to apologize for my sloppy language. I meant to say something like "the interpretation of the quantum state as representing the probability (amplitudes) of the observer". I never meant to imply that there is some kind "Probability Interpretation of Quantum Mechanics".

Secondly, I qualified my statement about deBroglie-Bohm stating that I am not an expert on this. If they want to maintain that their particles, following definite but unknown trajectories can be in two places at the same time, it is fine with me.

Now we come to CI, where I think you have some confusion. The wave function is the summary of the observers state of knowledge of the system. It is not part of the quantum system. It exists in Hilbert space, not space-time. It does not interfere with slits. It has no ontological content. If my state is psi = alpha * |0> + beta * |1> and I perform a measurement indicating the state represented by |0> then my wave function immediatly becomes psi = |0> because my knowledge of it has changed not due to some sort of physical collapse. Perhaps you are confusing what is sometimes known as the VonNeumann-Copenhagen interpretation.

As to the double-slit example: If the experiment is performed with photons (one at a time), each photon will expose one spot on the film and the overall picture (after many photons) will show the interference pattern. No evidence here of the photon being in two places at once. Performing either the "which slit" or "slit interference" measurement destroys not only the superposition but the entire wave function. When the photon detector clicks or the film is exposed the photon ceases to exist, it has no wave function.

Cheers, Skippy

7. Aug 22, 2010

### Staff: Mentor

I'm not an expert on deBroglie-Bohm either, nevertheless I'm pretty sure that it does not maintain that a particle can be "in two places at the same time." I'll leave it to Demystifier et al. to elaborate on this if they see this thread.

8. Aug 22, 2010

### unusualname

I'm sure they taught the Born rule in first year back when I studied physics. The best strategy for an undergrad is the "shut up and calculate" one, then once you pass your exams to can contemplate the mystery of it all at leisure.

The wavefunction represents a model of the information available to us about particles, measurements give us values with a probability predicted by the amplitudes of the wave function at those values.

Why the probabilities are distributed like a wave is unknown, but it may fall out of the wavy behaviour of strings and information distributed on a holographic screen, or something even more bizarre, but certainly not mystical or unknowable.

9. Aug 22, 2010

### JesseM

Keep in mind that when you talk about position, squaring the wavefunction gives a probability density function, not a normal probability distribution (see here for a quick rundown on discrete vs. continuous probability distributions). Since there are an infinite number of distinct values of x between the walls of the well, there's no way to assign a finite probability to each one and still have all the probabilities add to one! Instead a probability density function is a function that can be integrated between two positions x1 and x2 to give the probability that the particle will be found somewhere in that range (by a measurement that only has a resolution equal to the distance from x1 to x2, for example). So although it is true that the probability density is zero at precisely x=a/2, the density is nonzero at nearby values of x, so there will be some finite probability of finding the particle in any small neighborhood of a/2.

Last edited: Aug 22, 2010
10. Aug 22, 2010

### unusualname

Yes, good point, that's true for any continuously valued observable like position. (For other observables, you can have an infinite discrete spectrum, with finite probabilities at each distinct value)

So it doesn't make sense to talk about the probability of finding the particle at a single specific point.

11. Aug 22, 2010

### alxm

They do, but their idea of a particle's 'location' is not the same as for other interpretations.

I didn't say any of that or the opposite of any of that. But in any case, the wave function exists in configuration space, not 'space-time' or Hilbert space. The states span a Hilbert space.

Performing a measurement is physical.

Yes there is - the interference pattern. And there are any number of experiments showing quantum superpositions where a particle appears to be in several places at once. Such superpositions were even known well before quantum mechanics (e.g. what chemists called 'resonance')

No, it only determines the state of the wave function, changing it from a superposition to a pure state. Depending on the circumstances, that can evolve back into a pure state. Or be 'erased' and returned to a superposition as in a quantum-eraser experiment.

Yes, a non-existent particle has no wave function. So?