0
$\begingroup$

Working on problem 2.40 from Griffiths but can't seem to understand the first boundary condition.

We are given the potential

$V(x) = \left\{\begin{matrix} \infty & x < 0\\ \frac{-32\hbar^{2}}{ma^{2}}& 0 \leq x \leq a\\ 0 &x>a \end{matrix}\right.$

And we want to find the bound states. Since our minimum potential is $\frac{-32\hbar^{2}}{ma^{2}}$ we know that our $E$ has to be between this potential value and 0.

The problem I'm having at the moment right now with the middle region and the boundary condition at $x=0$.

In region $1$ we have that $E - V(x) > 0$ so then we having the following form of the TISE.

$\frac{d^{2}\psi}{dx^{2}} = \frac{-2mE}{\hbar^{2}}\psi$

Letting $k$ = $\frac{\sqrt{2mE}}{\hbar^{2}}$

Then we have solutions of the form

$\psi(x) = Ae^{ikx}+Be^{-ikx}$

Or

$\psi(x) = Asin(kx) + Bcos(kx)$

If you apply the boundary condition then $\psi(0) = 0$

The thing that I'm confused by is the first equation seems to suggest that $A = -B$

And the second equation suggests $B$ is $0$ because

$A\sin(0) + B\cos(0) = 0$

I know from looking this up earlier that I'm supposed to get that A is nonzero while B is zero, but I'm not sure how the two different equations match up. I should be able to arrive at the same conclusion whether or not I use complex exponentials or sines and cosines right?

$\endgroup$
2
$\begingroup$

The choice of the constants $A$ and $B$ depends on the form of the solution. You could have denote one pair of constanst by $A$ and $B$, and the other by $C$ and $D$.

A possible solution is $\psi (x)=A\sin(kx)$. In complex form, the sine is: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

If you've never proved this formula, try it using $e^{ix}=\cos x+i\sin x$.

This indicates that if you express the solution as $\psi(x)=Ae^{ikx}+Be^{-ikx}$, the constants must obey the relation$A=-B$.

After all, both solutions are the same.

$\endgroup$
0
$\begingroup$

No they are not the same, to make them equal we have to make the constants proportionate

$ Ae^{ikx} - Ae^{-ikx}$

$A(\cos(kx) + i\sin(kx)) + A(\cos(-kx) + i\sin(-kx))$

$A(2i\sin(kx))$

Now this only proves that

that the solution

$B \sin(kx) = A(2i\sin(kx))$

$B = 2Ai$

But we cannot prove that both of them are equal without imposing these rules.Their derivative is not the same so they do not even differ by a constant so by even using Euler's formula you cannot 'prove' them to be equal

$\endgroup$

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.