- #1
BOAS
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Homework Statement
A free particle moving in one dimension is initially bound by a very narrow potential well at the origin. At time ##t = 0## the potential is switched off and the particle is released; its wave function is:
##\psi (x,0) = N e^{-\frac{|x|}{\lambda}}##
where λ is a positive real constant giving the decay length of the initial wave function away from the origin.
(a) Calculate N.
(b) The wave function ##\psi(x, 0)## can be written as a superposition of stationary states, which for a free particle are planes waves,
##\psi (x,0) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{\infty} dk c(k) e^{ikx}##
Calculate c(k) by applying the inverse Fourier transformation.
(c) How does ##c(k)## behave (i) if the initial spread of the wave function is very small (that is: if ##\lambda## is very small), and (ii) if ##\lambda## is very large? Bearing in mind that for a free particle ##k## is ##\frac{p}{\hbar}##, can you see a connection between the dependence of ##c(k)## on ##λ##, the width of the initial distribution, and the Heisenberg uncertainty principle?
Homework Equations
The Attempt at a Solution
I am stuck with part c of this question.
I find that ##N = \frac{1}{\sqrt{\lambda}}##, so that ##\psi (x,0) = \frac{1}{\sqrt{\lambda}} e^{-\frac{|x|}{\lambda}}##
##c(k) = \frac{1}{\sqrt{2 \pi} \sqrt{\lambda}} (\frac{2 \lambda}{1 + \lambda^2 k^2})##
I think that for very small ##\lambda##, ##(1 + \lambda^2 k^2) \approx 1##, so ##c(k) \approx \frac{1}{\sqrt{2 \pi} \sqrt{\lambda}} 2 \lambda##. So for small ##\lambda##, ##c(k)## is also small.
For large ##\lambda##, ##(1 + \lambda^2 k^2) \approx \lambda^2 k^2## whick seems to suggest that ##c(k)## is also small for large ##\lambda##.
I don't think I'm seeing the physics of what's going on here, and would really appreciate a conversation about the situation.
Thank you!