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I've been asked to derive the eigenfunctions for the infinite square well, but by using boundaries of $L/2$ and $-L/2$ instead of $0$ and $L$

I'm happy with the general solution found using the TISE

$\psi(x)= A\sin(kx)+B\cos(kx)$

and that the wavefunction must be zero at the boundaries:

$\psi(-L/2)=\psi(L/2)=0$

however everywhere I've looked the next step involves using a boundary at $x=0$ to show that

$\psi(0)=B=0$

in order to remove the $\cos$ term, which I can't so simply do here since neither of my boundaries are at $x=0$.

It seems without doing that I'm left with two equations and 3 unknowns.

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  • $\begingroup$ The third unknown is usually determined by requiring $\psi(x)$ be normalized, i.e., $\int |\psi(x)|^2 dx = 1$. $\endgroup$
    – Michael Seifert
    Commented Mar 15, 2017 at 13:39
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    $\begingroup$ If your boundaries are at $x=\pm L/2$, why do you use a condition like $\psi(0)=0$, which is clearly NOT correct? Have you considered working backwards: starting from the solution between $x=0$ and $x=L$ and making the substitution $x\to x+L/2$ so as to shift the potential between $-L/2$ and $L/2$? $\endgroup$
    – ZeroTheHero
    Commented Mar 15, 2017 at 13:39
  • $\begingroup$ Your "general solution" is not. You don't allow for higher wave vectors. Also, if your solution is instead a "particular" solution for a particular eigenvalue, you should be able to find that eigenvalue from Schrodinger's eqn. Try to find the eigenvalue for your solution and see what you get. $\endgroup$
    – garyp
    Commented Mar 15, 2017 at 13:51
  • $\begingroup$ @garyp the higher wave vectors would come out "naturally" by enforcing the conditions at the boundaries, which could have multiple solutions. $\endgroup$
    – ZeroTheHero
    Commented Mar 15, 2017 at 13:57
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    $\begingroup$ @ZeroTheHero Yes. Typing faster than the speed of thought. $\endgroup$
    – garyp
    Commented Mar 15, 2017 at 14:00

1 Answer 1

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The simplest approach is to write $$ \psi(x)=A \sin\left(k(x+L/2)\right) $$ since this form satisfies the boundary condition at $x=-L/2$. You can expand using trig identities to get this as a sum of sine and cosine, like you initially suggested, but this is not essential.

You can finish the job by choosing $k$ so that $$ \psi(L/2)=A\sin(kL)=0\, , $$ as you normally do.

Note of course that the energy eigenvalues, which are obtained from the restriction of $k$ to enforce your second boundary condition, do not depend on the location of your wall, but only on the width of the well. This makes sense intuitively as the position of origin of the coordinate is arbitrary: locating the origin at different places does not affect the physics of the problem and thus cannot change the possible energies.

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