## anonymous 3 years ago What's wrong with this infinite square well?

1. anonymous

|dw:1360436634776:dw|$φ''=-\frac{2mE}{ℏ^2} φ$ $\varphi= A \sin(\frac{\sqrt{2mE}}{ℏ}x)$ $\varphi(a)=\varphi(-a)=0$ $\pm\frac{\sqrt{2mE}}{ℏ}a=n \pi$ $\frac{\sqrt{2mE}}{ℏ}=\pm\frac{n \pi}{a}$ $\lambda= \frac{2 \pi}{\pm\frac{n \pi}{a}}=\pm\frac{2a}{n}$ Which, for n=1, is shorter than the real ground-state lavelength (4a). What have I done incorrectly?

2. JamesJ

You haven't written down the correct general solution. You can also have terms $... + B \cos(\omega x)$ with $$\omega = \sqrt{2mE}/\hbar$$.

3. JamesJ

This will give you the 4a, as cos(omega.x) = 0 => omega.x = pi/2 +- n.pi

4. anonymous

Aren't you assuming A=0 for that solution? $Asin(\pi/2 \pm n\pi) \ne 0$

5. JamesJ

The general solution of y '' + w^2.y = 0 is y(x) = A.sin(wx) + B.cos(wx) or y(x) = A.cos(wx + phi0) Either way, there are two parameters. You are applying one of the boundary conditions. You actually are going to apply another boundary condition to y'(x) in order to completely nail down the solution.

6. JamesJ

Alternatively, the two conditions you are using phi(-a) = 0, phi(a) = 0. From this, you get two families of solutions. One with sin, the other with cos. This equivalent to using the second general solution with the phi0 alternating between 0 and pi/2

7. anonymous

Ah, that's it! Thanks. What are the boundary conditions? That $\varphi'(-a)=\varphi(a)=0$ Because $\varphi(x)=0$ for $x>a, x<-a$?

8. anonymous

*$\varphi'(a)=\varphi'(-a)=0$

9. anonymous

That guess came from avoiding making the first derivative jump about at +- a

10. JamesJ

You have two conditions, which solving for the two parameters in the general solution. I recommend you start with A.cos(wx + phi0) and you'll get what you want.

11. anonymous

$A \sin( \omega a)+B \cos( \omega a)$ It is harder to make this expression = 0 in this form than yours, as both summands cannot be made = 0 or the negative of the other by eyeballing it, so yours is simpler. Thanks, I think I've got it.

12. anonymous

@JamesJ , sorry to tag you, but I have a (very) brief question: the general solution is $\varphi= A e^{\pm i \omega x}$ What is done with $\Im(\varphi)$? Is only the real part considered, as is often done in classical mechanics? Thanks again