If a wave packet is given by:
My question is basically how do we choose the write $A(k)$ to fit the particle we are looking at, or does it not matter (as my matter as my textbook seems to imply) which seems counterintuitive?
1 Answer
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If you solve the Schrodinger equation for a free particle the solutions are plane waves, and any sum of plane waves is also a solution. Since any wave packet profile can be constructed by summing plane waves then your equation with any $A(k)$ is also a solution of the Schrodinger equation.
The $A(k)$ is not determined by the Schrodinger equation but rather it's a boundary condition. You choose the $A(k)$ that matches the system you're trying to describe. For example if your particle is completely delocalised the $A(k)$ is a delta function, which physically means there is a precise momentum but the position is completely unknown. The other extreme would be if you pinpoint the particle's position precisely, in which case $A(k) = 1$ so the momentum is completely uncertain.
In practice we'd probably usually choose an $A(k)$ that is Gaussian, because Gaussians are easy to work with. In that case the width of the Gaussian would correspond to the uncertainty in momentum.
answered Jul 26, 2014 at 10:24
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$\begingroup$ A(k) is 1 (well really $\frac{1}{\sqrt{2 \pi}}$ to get the normalization right) in the case of localizing the particle at x=0. More generally it would be $\frac{e^{-ikx_0}}{\sqrt{2 \pi}}$ for a particle at $x_0$ $\endgroup$ Commented Jul 26, 2014 at 14:22
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