This is a final exam problem. Here is what I can remember: We know that if an electron's wavefunction starts out as a narrow wavepacket, and moving in a region of constant potential, then the wavepacket will disperse very quickly. A wider wavepacket will disperse slower that a narrow one. Find the smallest initial width of the wavepacket, $\Delta x_{min}$, so that the wavepacket's dispersion at a later time $T$ is minimized. Ignore any constant multipliers.

Here is my approach:

Width of wavepacket at time $T$ = initial width.

$\Delta x_T = \Delta x_{min}$

Uncertainty principle for the initial wavepacket (ignoring multipliers).

$\Delta x_{min} \Delta p = \hbar$

Width of wavepacket at time $T$ is time * velocity.

$\Delta x_T = \frac {T \Delta p} {m_e}$

Substitute in initial uncertainty of momentum ($\Delta p$ doesn't change over time).

$\Delta x_T = \frac {\hbar T} {m_e \Delta x_{min}}$

Since $\Delta x_T = \Delta x_{min}$:

$ \Delta x_{min}^2 = \frac {\hbar T} {m_e}$

$ \Delta x_{min} = \sqrt {\frac {\hbar T} {m_e} }$

Am I doing it correctly?

This problem is very difficult. I asked what other students got after the exam, and almost everyone said they left it blank, but one person said he got a square root in his answer.

I believe this question requires knowledge of the second order expansion of $E = \frac{p^2}{2m}$, which is $\frac{\operatorname d^2 E}{{\operatorname d p}^2} = \frac 1 m$ in the derivation of group velocity, which was a bonus assignment I didn't do.

  • $\begingroup$ It seems correct. Just I wonder whether $\Delta x_T$ shouldn't be equal to the initial value + the subsequent expansion, i.e. $\Delta x_{min} + \frac {T \Delta p}{m_e}$ $\endgroup$ – Sofia Dec 19 '14 at 3:45

I think your solution is correct.

The width $\Delta x(T)$ of the wave packet at any time $T$ can be obtained from $$\Delta x(T)=\Delta x(T=0)+\Delta v_g T\text{ ,}$$ where $\Delta v_g$ is the uncertainty in the group velocity $v_g$, which causes the widening.

You have implicitly used $$v_g=\frac{p}{m_e}\Longrightarrow \Delta v_g=\frac{\Delta p}{m_e}=\frac{\hbar}{\Delta x(T=0)m_e}$$ which gives you \begin{equation} \Delta x(T)=\Delta x_{min}+\frac{\hbar}{\Delta x_{min}m_e} T \end{equation} with $\Delta x_{min}:=\Delta x(T=0)$. Taking the derivative with respect to $\Delta x_{min}$ yields your result $$\Delta x_{min}=\sqrt{\frac{\hbar T}{m_e}}\text{ .}$$

You passed :-)


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