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Newbie questions: wave and particle viewpoints

  1. Nov 30, 2011 #1
    I'm still in high school. I'm reading about physics as a hobby.

    In this document the limitations of the particle and wave viewpoints are described.

    I've read that the probability of an event is the squared absolute value of the probability amplitude. Why would you do that? Wouldn't just taking the square work as well?

    On the first page of that document there is this probability amplitude: e[itex]^{i(ωt-k \bullet r)}[/itex]
    Why is kr substracted from ωt? Why aren't the phases added to eachother?
    It also says that the squared absolute value of this amplitude is constant but aren't t and r variables? Why would it be constant?
  2. jcsd
  3. Nov 30, 2011 #2


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    Staff: Mentor

    The probability amplitude is a complex number, in general. Squaring a complex number gives another complex number. Probability has to be a real number. The squared absolute value is always real.

    It gives you a wave that travels in the +r direction.

    They can be added, in which case the wave travels in the -r direction.

    To get the squared absolute value of a complex number, we multiply the number by its complex conjugate (replace i with -i and vice versa).

    [tex]e^{i\theta}e^{-i\theta} = e^{i\theta-i\theta} = e^0 = 1[/tex]
    Last edited: Nov 30, 2011
  4. Nov 30, 2011 #3
    Why? Does a negative phase mean that the wave is travelling in a positive direction?
  5. Nov 30, 2011 #4
    I think what they explain is true. there are implicit physics characteristics of wave that represent by complex number.
    for more understanding purpose, you can read Marry L.Boas, Mathematical Method for Physics. You will understand with what Chaucy approach about complex number
  6. Nov 30, 2011 #5


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    Staff: Mentor

    To convince yourself of this, I think it's best to make some graphs. Pick numbers for [itex]\omega[/itex] and k, set t = 0, and draw a graph of y versus x for [itex]y = \cos (\omega t - kx) = \cos (-kx)[/itex]. Then set t = 0.1, say, and draw another graph. Then set t = 0.2 and do another one.

    Repeat for [itex]y = \cos (\omega t + kx)[/itex].

    You need to use the cosine (or sine) instead of [itex]e^{i(\omega t - kx)}[/itex] because it's kind of hard to draw graphs of complex numbers.
  7. Dec 11, 2011 #6
    Hi, I've just started a Masters in Acoustics after a ten year gap from education, so I've spent the last few months getting my head around this subject too.

    One really important thing I realised recently is that ωt = kx. Here's the proof,

    c=fλ, f=c/λ, λ=c/f
    c=x/t, x=ct, t=x/c
    ω = 2πf
    → ω=2πc/λ → ω=ck → ω=kx/t → ωt = kx

    So what's the point then? Shouldn't the whole thing just be cos(0)=1?

    Engineering Noise Control: Theory and Practice is a good book for this. It gives this expression for the wave equation, φ=Acos(k(ct±x)+β

    Here ± is used because a wave can go in either direction, but more importantly, k has been taken out, clearly showing the two most important things within the bracket - ct & x. What this does is allow us to consider the wave when time is zero but at a distance x, or when time is 20 ms and the distance is zero - or at some combination of the two.

    k=2π/λ and this 2π makes the cos bracket give a solution that will always be between -1 and 1.

    β is the phase - and is easiest considered as a value that shifts the entire wave left or right by a given amount. The phase was also a part of something that confused me for a long time, which was the addition of waves. Adding two incoherent sources together (two violinists as opposed to one) increases the SPL by 3 dB, but adding two coherent sources (two speakers as opposed to one - putting out exactly the same signal which is in phase) increases the SPL by 6 dB. When two signals are in phase with each other you get an extra 3 dB increase.

    There isn't actually a standard convention for using positive or negative in this respect (Cox, 2009, s. 1.4.1).
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