# The Infinite square well

I am studying Quantum Mechanics by David J. Griffiths. I am kind of stuck on the part of "The infinite square well" where griffith says

"negative solutions give nothing new, since $\sin(-\omega) = - \sin \omega$ and we can absorb the minus sign into $A$."

To be more specific, he uses boundary conditions on the solution $y(x) = A \sin(kx) + B \cos(kx)$, to show that $B$ is $0$ and $K$ can only take values $\frac{n\pi}{a}$ where $n$ can take values $1,2,3,4$...

My question is: Since $A$ is $\sqrt{\frac{2}{a}}$, how can you "absorb" minus sign into it.

Ps: i couldn't find symbols for psi and theta, so i have used $y$ and $\omega$ in their stead.

• For formatting math, use MathJax Apr 14 '18 at 16:29
• It means that for negative solutions you have a different value of $A$. But that "new" solution is identical to the original solution. Apr 14 '18 at 16:31
• Thanks garyp. I think i understand it now. Since, A^2 = 2/a, it will give 2 values for A, -√(2/a) and +√(2/a). Apr 14 '18 at 17:03
• not sure what you mean by boundary conditions on the solution $y(x)=A\sin(kx)+B\sin(kx)$. Is this not just $(A+B)\sin(kx)$? Apr 14 '18 at 17:13
• Ow i am so sorry. It should be Bcos(kx). I will edit it right away. Thanks for pointing it out. Apr 14 '18 at 17:22

Griffiths finds that the nontrivial solutions of the stationary Schroedinger equation in the quantum well with infinite barriers at $x=0$ and $x=a$ (where $\psi =0$) are $$\psi(x)=A\sin(k_nx) \tag 1$$ with $$k_n=±n\frac {\pi}{a} \tag 2$$ where $n=1,2,...$ However only the solutions with positive signs for $k_n$ in (2) have to be considered because the solutions with negative $k_n$ are equivalent to the ones with positive sign as the negative sign for $k_n$ in eq. (1) can be taken out of the argument of the sinus and considered to be part of a new arbitrary constant $A'$ so that the solution is the same as the solution with positive $k_n$.
The variable $k,$ which later in the book will be interpreted as momentum, can be either positive or negative. However, $A\sin(\pm kx)$ is the same as $\pm A\sin(kx)$, which is the same as just $A\sin(kx)$ as long as we don't make any restrictions on $A$'s sign.