- #1
Hamiltonian
- 296
- 190
The final wave function solutions for a particle trapped in an infinite square well is written as:
$$\Psi(x,t) = \Sigma_{n=1}^{\infty} C_n\sqrt{\frac{2}{L_x}}sin(\frac{n\pi}{L_x}x)e^{-\frac{in^2{\pi}^2\hbar t}{2m{L_x}^2}}$$
The square of the coefficient ##C_n## i.e. ##{|C_n|}^2## is proportional to the probability of the system being in that state on measurement.
The wave function solution of a particle in a 2-D box
$$\Psi_{n,m}(x,y,t) = \frac{2}{\sqrt{L_x L_y}} sin(\frac{n\pi}{L_x}x)sin(\frac{m\pi}{L_y}y)e^{-\frac{i\pi^2 \hbar^2 t}{2m}[\frac{n^2}{{L_x}^2}+\frac{m^2}{{L_y}^2}]}$$
is it correct to write the final solution as:
$$\Psi (x,y,t) = \Sigma_{n=1}^{\infty} \Sigma_{m=1}^{\infty}C_n C_m\frac{2}{\sqrt{L_x L_y}} sin(\frac{n\pi}{L_x}x)sin(\frac{m\pi}{L_y}y)e^{-\frac{i\pi^2 \hbar^2 t}{2m}[\frac{n^2}{{L_x}^2}+\frac{m^2}{{L_y}^2}]}$$
will the probability of the system being in a particular state be proportional to ##|C_n C_m|^2##?
$$\Psi(x,t) = \Sigma_{n=1}^{\infty} C_n\sqrt{\frac{2}{L_x}}sin(\frac{n\pi}{L_x}x)e^{-\frac{in^2{\pi}^2\hbar t}{2m{L_x}^2}}$$
The square of the coefficient ##C_n## i.e. ##{|C_n|}^2## is proportional to the probability of the system being in that state on measurement.
The wave function solution of a particle in a 2-D box
$$\Psi_{n,m}(x,y,t) = \frac{2}{\sqrt{L_x L_y}} sin(\frac{n\pi}{L_x}x)sin(\frac{m\pi}{L_y}y)e^{-\frac{i\pi^2 \hbar^2 t}{2m}[\frac{n^2}{{L_x}^2}+\frac{m^2}{{L_y}^2}]}$$
is it correct to write the final solution as:
$$\Psi (x,y,t) = \Sigma_{n=1}^{\infty} \Sigma_{m=1}^{\infty}C_n C_m\frac{2}{\sqrt{L_x L_y}} sin(\frac{n\pi}{L_x}x)sin(\frac{m\pi}{L_y}y)e^{-\frac{i\pi^2 \hbar^2 t}{2m}[\frac{n^2}{{L_x}^2}+\frac{m^2}{{L_y}^2}]}$$
will the probability of the system being in a particular state be proportional to ##|C_n C_m|^2##?