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

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


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

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?


2 Answers 2


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.


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

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

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


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


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