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IS every quantum mechanical wave function a wave packet, i.e an infinite superposition of differently wave numbered sinusoidal waves, irrespective of whether the potential admits bound states or scattering states?

Then for the free particle while considering the stationary state of the form (or at least containing the term) $x ± vt =$ constant, thereby implying a constant phase velocity of $v $. What includes in the computation of the wave function the taking into account of all the sinusoidal wave functions?

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Not every wave function is a wave packet. Wave functions are frequently considered to include functions that are not properly normalizable, like $\operatorname{e}^{ikx}$, so we describe them as "normalized to the delta function." That is, we demand $$\delta(k - k') = \int_{-\infty}^\infty \operatorname{d} x \psi^\star(x, k') \psi(x, k),$$ leading to things like $\psi(x,k) = \operatorname{e}^{ikx} / \sqrt{2\pi}$.

With a wave packet we make the more strict requirement that it be normalizable and localized in both $x$ and $p$. The canonical example of a wave packet is:$$\psi(x) \propto \exp \left(-\frac{1}{4} \frac{(x - x_0)^2}{\sigma^2} + i \frac{p_0 x}{\hbar}\right),$$ which has $\langle x\rangle = x_0$ and $\langle p \rangle = p_0$, and is normalizable with finite width in both $x$ and $p$ space.

Another example of a wave packet is the $\operatorname{sinc}$ function: $$\psi(x) \propto \frac{\sin\left(\frac{x-x_0}{2\hbar}\Delta p\right)}{(x-x_0)} \operatorname{e}^{ip_0 x / \hbar}.$$ It is localized in $x$, in the sense that it is normalizable and has definite quantiles, but it has divergent variance. In $p$ it is much better behaved because it is a boxcar function that stretches from $p_0 - \Delta p / 2$ to $p_0 + \Delta p / 2$.

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  • $\begingroup$ Are you saying that we have for the wave packet, the non normalisable exponential should be localised ? And that implies, being solutions to the time independent Schrödinger equation for those wave functions carrying any sort of exponential like the canonical example you just gave, therefore being a wave packet. $\endgroup$
    – TESLAGEN
    Oct 10 '16 at 3:37
  • $\begingroup$ Sorry, @TESLAGEN, I don't understand your question. $\endgroup$ Oct 10 '16 at 3:38

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