Consider the infinite square well situation where the potential is infinite at positions $|x| > a$ and $0$ otherwise.

When solving the Time independent Schrodinger Equation (TISE) we can come to the conclusion that $u(x) = A e^{ikx} + B e^{-ikx}$.

However now we need to fulfill the condition that $u(a) = 0$.

In order to do this we can use some complex numbers manipulation and then comparing Reals and imaginary as follows:

$$A e^{ika} + B e^{-ika} = A \cos(ka) + A i \sin(ka) + B \cos(ka) -i B \sin(ka) = 0 + 0 i$$ $\Rightarrow$ $$ (A+B)\cos(ka) + i(A-B)\sin(ka) = 0 + 0i$$

This Boils down to:

(1) $\cos(ka) = 0$
(2) $\sin(ka) = 0$.

From (1): $ka = \frac{n\pi}{2} \Rightarrow k = \frac{n\pi}{2a}$

From (2): $ ka = n\pi \Rightarrow k = \frac{n\pi}{a}$

The above seems to agree with online sources.

However I was under the impression that both (1) and (2) have to be satisfied in order to solve the TISE and determine k, and this is not possible. Could anyone clarify this idea for me.

  • 2
    $\begingroup$ Please note that you can use MathJax to typeset formulae here. $\endgroup$ – ACuriousMind Oct 23 '17 at 11:38

Obviously the solution $u(x)=Ae^{ikx}+Be^{-ikx}$ is equivalent to $u(x)=A\cos(kx)+B\sin(kx)$. If you enforce $u(a)=u(-a)=0$ you will get $$ u(x)=\left\{\begin{array}{c} A\cos(ka)+ B\sin(ka)=0\\ A\cos(ka)- B\sin(ka)=0\, .\end{array}\right. $$ Summing yields the condition $2A\cos(ka)=0$ hence $ka=(2n+1)\frac{\pi}{2}$ or $k\to k_{2n+1}=(2n+1)\frac{\pi}{2a}$. Subtracting yields $2B\sin(ka)=0$ hence $ka=n\pi$ or $k\to k_{2n}=\frac{n\pi}{a}$.

Since these conditions are mutually independent in the sense that $A$ does not determine $B$ or vice versa, the solution $u_{2n}$ is obtained by choosing $B=0$ and the solution for $u_{2n+1}$ is obtained by choosing $A=0$: $$ u_q(x)=\left\{\begin{array}{cc} A\cos\left(\frac{(2n+1)\pi x}{2a}\right)\, ,&q=2n+1 \quad\hbox{is odd}\\ B\sin\left(\frac{n\pi x}{a}\right), ,&q=2n \quad\hbox{is even} \end{array}\right. $$


Well 1 and 2 can't both be true, because you arrive at a contradiction. Instead look at each part individually:

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

Now, we know from algebra that if $xy=0$ then $x=0$ or $y=0$. So let's say we take $\cos(ka)=0$ and we get $k=(2n+1)\pi/2a$ for integer $n$. Well then in our second equation, we are left to choose $A-B=0$ because we can't have $\sin(ka)=0$ when $\cos(ka)=0$

Therefore, $A=B$ and we get $$u(x)=2A\cos\left(\frac{(2n+1)\pi}{2a}x\right)$$

Of course, we could have chosen $\sin(ka)=0$ and $A+B=0$ instead. And we would get to $$u(x)=2iA\sin\left(\frac{n\pi}{a}x\right)$$

And these are the two families of valid solutions. (I personally like to define the box to be from $x=0$ to $x=L$ so that you only get $\sin$ functions this way)


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