Quantum Waves

de Broglie’s Postulates

de Broglie's Postulates

de Broglie postulated that particles of matter have wave nature: $\begin{array}{rl}E=\hslash \omega & =h\nu ⟹\nu =\frac{E}{h}=\frac{p^2}{2mh}\\ p=\hslash k& =\frac{h}{\lambda }⟹\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}\end{array}$

Proposition

The dispersion relation for non-relativistic particle is $\omega =\frac{p^2}{2m\hslash }=\frac{\hslash k^2}{2m}$Thus the group velocity can be derived:$v_g=\frac{\partial \omega }{\partial k}=\frac{\hslash k}{m}$

Free Particle de Broglie Waves

From the standpoint of our quantum mechanics, there is no quantity which causally fixes the effect of a collision in an individual event.

Comment

In Max Born’s statistical interpretation, the state of a particle is represented by a complex function $\Psi (r,t)$ such that $|\Psi (r,t){|}^2\text{dd}V$ is the probability of finding that particle at the time $t$ in the volume element $\text{dd}V$ at the point $r$.

e.g. In one dimensional case, we have the probability of finding the particle between $a$ and $b$ at $t$ is given by${\int }_a^b|\Psi (x,t){|}^2\text{dd}x$

Proposition

The mean position is given by $⟨x⟩={\int }_a^{b}x|\Psi (x,t){|}^2\text{dd}x$

Momentum Space Wave Function

Given the wave function $\Psi (k)$, we can use Fourier transform to obtain the wave function in the momentum space:$\Phi (p,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\Psi (x)e^{-ikx}\text{dd}x$