Free Particle

  • Thread starter Domnu
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Problem
A free particle of mass [tex]m[/tex] moving in one dimension is known to be in the initial state

[tex]\psi(x, 0) = \sin(k_0 x)[/tex]

a) What is [tex]\psi(x, t)[/tex]?
b) What value of [tex]p[/tex] will measurement yield at the time [tex]t[/tex], and with what probabilities will these values occur?
c) Suppose that [tex]p[/tex] is measured at [tex]t=3 s[/tex] and the value [tex]\hbar k_0[/tex] is found. What is [tex]\psi(x, t)[/tex] at [tex]t > 3 s[/tex]?

Attempt at Solutions
Well one question I have is this: how is this a valid state function for a free particle if it is non-square integrable? Generally, for any free particle, doesn't the wavefunction have to be square-integrable?
 

Answers and Replies

  • #2
Dick
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Exactly the opposite. In general, free particles having wavefunctions like exp(ikx) are NOT square integrable. Since they aren't localized.
 
  • #3
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Okay, so part a) would then just be

[tex]\psi(x,t) = e^{-i \omega_0 t} \sin k_0 x[/tex]​

since [tex]\psi(x,0)[/tex] is an eigenfunction of the momentum operator. Now, when I try to evaluate part b), I get

[tex]b(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \sin k_0 x e^{-ikx} dx[/tex]

where [tex]\int_{a}^b |b(k)|^2 dk[/tex] represents the probability of a particle having a wavenumber between [tex]a[/tex] and [tex]b[/tex]. How do I actually evaluate the above integral? Also, the question asks for the probability of certain momenta... how is this allowed if momenta aren't discrete but are, instead, continuous?
 
  • #4
Dick
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You are making this much harder than it actually is. sin(k0*x) is a linear combination of exactly two momentum eigenfunctions exp(ikx). Which two? Don't think 'integral', think 'deMoivre'.
 
  • #5
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Aah, I see it now... sin(k0*x) = 1/2 exp(i*k0*x) - 1/2 exp(-i*k0*x). But we need to normalize this... so we finally have

[tex]\psi(x, t) = e^{-i \omega_0 t} \cdot \left(\frac{\sqrt{2}}{2} \phi_{k_0} - \frac{\sqrt{2}}{2} \phi_{-k_0}\right)[/tex]​

Thus, the particle has a 50% chance of having momenta [tex]\pm \hbar k_0[/tex]. This solves part b). As for part c), the wavefunction collapses onto

[tex]\psi(x, 0) = e^{-i \omega_0 t} \cdot \frac{1}{\sqrt{2\pi}} e^{i k_0 x}[/tex]​
 
  • #6
Dick
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Actually, it's sin(kx)=(exp(ikx)-exp(-ikx))/(2i). And you don't need to do any particular normalization. The only thing thats important is that the magnitude of the two coefficients are equal.
 
  • #7
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Heh, yes that's what I meant :P Thanks a bunch :smile:
 

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