The free particle - TN2304 Kwantummechanica 1

V(x) = Vo —> Partícula lliure —>$|\Psi(x,t)|^2=A^2$ —> No normalitzable —> No pot representar un estat físic —> Necessitem una nova interpretació

Interpretació física: Paquets d’ona

—> Velocitat de grup i de fase

Interpretació matemàtica: Funció d’ona restringida en un pou de potencial infinit

—> $E_n = \frac{2\pi^2\hbar}{mL^2}n^2$, $E=\frac{\hbar^2k_x}{2m}$

—> Cada valor propi està degenerat 2 vegades

<aside> 🥸 In contrast with the infinite well case, for the free particle system we don't have boundary conditions on the wavefunction, and therefore the energy E can take any positive value: the only contribution to the total energy is the kinetic energy, which is positive-definite. So in the case of the free particle system the energy is not quantised and can take a continuum of values.

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<aside> 📌 The way to find physical wavefunctions is to combine solutions with different energies. In such case, Δp will take a finite value and hence Δx can be finite as well. And if a particle is localised, one will be able to construct normalisable wave functions.

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“In the case of a quantum system like the free particle, one should replace a sum by an integral.”

Most general wavefunction for a free particle —> wave packet

$$ \Psi(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} d k \phi(k) e^{i k x} $$

“Note that the wave packet itself does not have a unique energy of wavelenght.”

Using the property of ortogonalithy:

$$ \int_{-\infty}^{+\infty} d x e^{i k x} e^{-i k^{\prime} x}=2 \pi \delta\left(k-k^{\prime}\right) $$

Bla bla bla, we therefore find that the function ϕ(k) is given by:

$$ \phi(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \Psi(x) e^{-i k x} d x $$

Osigui, la transformada de Fourier

—> Aahhh aleshores allo de notació correcta és només per una free particle??

Wave packet

$$ \begin{gathered} \Psi(x, t)=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} \int_{-\infty}^{+\infty} e^{-k^2 / 4 a} e^{i\left(k x-\left(\frac{\hbar k^2}{2 m}\right) t\right)} d k=\frac{1}{\sqrt{2 \pi}} \frac{1}{(2 \pi a)^{1 / 4}} \int_{-\infty}^{+\infty} e^{-\left[\left(\frac{1}{4 a}+i \frac{h t}{2 m}\right) k^2-i x k\right]} d k \\ \Psi(x, t)=\left(\frac{2 a}{\pi}\right)^{1 / 4} \frac{e^{-a x^2 /(1+2 i \hbar a t / m)}}{\sqrt{1+2 i \hbar a t / m}} \end{gathered} $$