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Simple question of matter waves

  1. Jul 22, 2006 #1
    Just get started with quantum so I have a simple question.

    Does a matter wave describe the probability of the particle to be in a certain place, or is it the wave squared?

    In other words, is
    y(r,t) = Probability(finding particle at r and time t)


    y(r,t)^2 dV = Probabilty(finding particle at r and time t)

    I know it is the second, since that is what the book lists. But then what does the wave function y(r,t) represent?
  2. jcsd
  3. Jul 22, 2006 #2
    I'll try to venture a guess myself. The comparison with electromagnetic waves is an example in the book I'm using, though I don't know if I follow it correctly.

    The eletromagnetic wave e(r,t) is a measure of the eletric field at that point.
    The electromagnetic wave's energy is proportional to e(r,t)^2. But we know electromagnetic energy is quantized as photons, there e(r,t)^2 must be a measure of the number of photons at r at time t, or, if normalized, is proportional to the probability of the number of photons in that area.

    Likewise, y(r,t)^2 is proportional to the "intensity" of the particles's <what?> and therefor is the proportional to probability that the particle is in that location. Therefor, simple y(r,t) is a measure of the wave's "strength", which is undefined by is somehow anagolous to an eletric field's strength?
  4. Jul 22, 2006 #3


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    The wave function has a complex number value (or it could be an object like a vector or a spinor with complex number components). So it can't represent a probability itself; probabilities are real numbers between zero and one.

    You can take any complex number a + bi and multiply it by its conjugate a - bi and get a real number a2 + b2, and you can arrange things so that the complex values you deal with have their "complex squares" like this between zero and one, so the square of your wave function value (which is called an amplitude) can be regarded as a probability.

    As to what the amplitude "represents", quantum theory doesn't say, and the various interpretations fill many of the discussion threads on this forum.
  5. Jul 22, 2006 #4
    thank you, that is very helpful.

    I have another question. When physicists say that the uncertainy of a particles position is not simply a matter of ignorance but a statement that the particle does not have a defined position, does that refer to the whole space that the wave equation describes itself? In other words, if I have a wave packet describing the particle, the particle is actually not in a more defined position that the entire packet? Or as an example, if an electron is bound to a nuclues, described by a certain standing wave, the particle itself is actually in all of those locations (in some sense I guess) that is described by that wave (with different proportions or "probability") I guess.

    Why is it, btw, that physicists believe the position is not defined, and that it is not simply a matter of are ignorance? Is it because a bound electron does not lose energy and therefor does not radiate and therefore must not be an accerlerating or decelerating charge and therefor must be a constant charge distrabution and therefor the electron must actually in some sense be in all of those location? (sorry for the long sentence:bugeye: )
  6. Jul 22, 2006 #5


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    Bear with me here, I have to do some explaining before I can answer you.

    Since the wave function for position is complex it doesn't by itself determine position in our spacetime, which again can be described exactly by real numbers. What is needed is something to turn the complex amplitude for position into an actual value of a spacetime position. That something is an observation (or I believe, more generally, an interaction). Corresponding mathematically to the observation is an object called an operator, and the operator acts on the wave function (the model here is the way a matrix acts on a vector, if you are familiar with that formalism). If you have just the right operators they will act on the amplitudes to produce a string of real numbers. These are called eigenvalues of the operator, and in quantum mechanics you often see the term eigenstates. Each of the eigenstates is one possible outcome of your measurement of position, and you will see one of them. The operator-amplitude action will also produce those probabilities, one for each eigenstate, giving the probability that your measurement will see the corresponding eigenstate.

    All of this explanation is necessary to answer your question. A quantum system does not have a real position unless this interaction or measurment happens. And because the result of the measurement is probabilistic, the position you will see is uncertain.

    I hope the above explanation will help to answer this question too. There's a lot more to describe, but why don't you take this much and go back to your studies and see if they are clearer?
    Last edited: Jul 22, 2006
  7. Jul 22, 2006 #6
    Oh boy! I am going to need to spend some time with your explanation. It seems you have jumped ahead to more general quantum mechanics including the full treatment of eigenvalues and such. I am still working through a sophmore year Modern Physics text which doesn't include such detail. Can I ask one final question to try to map the more general case into the simplified version presented in this book. If the operator-amplitute action gives both the possible positions and probability for each position, how does this relate to squaring the wave function in order to get the probability of being in a certain volume?

    Or in other words, how does this qoute from Modern Physics by Taylor and Zafiratos relate to your fuller explanation, "Associated with each individual quantum particle there is a wave function y(r,t) whose intensity at any position r determines the probability P of finding the particle at r (at time t)".
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