1
$\begingroup$

This is the problem:

A particle is restrained to move in 1D between two rigid walls localized in x=0 and x=a. For t=0, it’s described by:

$$\psi(x,0) = \left[\cos^{2}\left(\frac{\pi}{a}x\right)-\cos\left(\frac{\pi}{a}x\right)\right]\sin\left(\frac{\pi}{a}x\right)+B $$

For $t>0$, determine the probability of finding the particle between 0 and $\frac{a}{4}$.

So, using some trigonometry and the orthonormal base $\phi_{n}(x)=\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$, I can write the wave function as:

$$\psi(x,0)=\sqrt{\frac{a}{32}}\phi_{1}(x)+\sqrt{\frac{a}{8}}\phi_{2}(x)+\sqrt{\frac{a}{32}}\phi_{3}(x)+B$$

I still can’t use the evolution operator. I must find an expression to $B$, so I can put it in terms of the base.

I use: $B=\sum_{n} C_{n}\phi_{n}(x)$ where, after finding the value of $C_{n}$, and noticing that only odd values of n contributes to the wave function:

$$B\rightarrow -\frac{B}{\pi}\sqrt{8a}\sum_{0}^{\infty}\frac{1}{2n+1}\phi_{2n+1}(x)$$

So now, how could I add a constant to the wave function so it’s normalized? It’s just finding the value $B$ using $\langle\psi|\psi\rangle$? Or there is other way? Because using $\langle\psi|\psi\rangle$ I get a quadratic, and I’m not sure that is the way.

$\endgroup$
1
$\begingroup$

In a $1D$ box your wave function must vanish at $x=0,a$. In particular

$$\psi\left(x=0,t=0\right)=B=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.