If a wave packet is given by:

enter image description here

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 1


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.

  • $\begingroup$ 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 $x_0$ $\endgroup$
    – DrEntropy
    Jul 26, 2014 at 14:22
  • $\begingroup$ @DrEntropy: indeed. $\endgroup$ Jul 26, 2014 at 16:39

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