# Infinite square well eigen-energies

## Homework Statement

Usually when we solve the problem of the infinite square well we place one wall at the origin and the other one at, say 2L (please notice the 2).
We get the eigen-energies
$$E_n = {{n^2 \pi^2 \hbar^2}\over{8ma^2}}$$
and the eigen-functions
$$\psi_n = \sqrt{1\over a}\sin{\left({n\pi\over{2a}}x\right)}$$.
My question is this: If instead we placed the well centered about the origin such that the walls are at -L and +L, do we get the same spectrum, and the wave functions shifted left by L, i.e.
$$\psi_n = \sqrt{1\over a}\sin{\left({n\pi\over{2a}}(x+L)\right)}$$.
???

## The Attempt at a Solution

Thing is, when I do it from scratch, I don't get these wavefunctions. I get a combination of sin and cos.

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I take it that $$a$$ should really be $$L$$. Your wave function is correct, since it satisfies the proper differential equation, namely $$-\frac{\hbar^2}{2m}\psi''(x) = E \psi(x)$$ and initial conditions $$\psi(-L) = \psi(L) = 0$$. Note that in general a sum of sin and cos can be written as a sin (or cos) with a phase shift. If you combined your combination of sin and cos you should get the desired wave function.

My question is this: If instead we placed the well centered about the origin such that the walls are at -L and +L, do we get the same spectrum, and the wave functions shifted left by L, i.e.
$$\psi_n = \sqrt{1\over a}\sin{\left({n\pi\over{2a}}(x+L)\right)}$$.
???

## The Attempt at a Solution

Thing is, when I do it from scratch, I don't get these wavefunctions. I get a combination of sin and cos.
Set up the Schrodinger equation and apply the new boundary conditions and see. You should get the same General solution since you are not changing the length of the well. If you are increasing the length "L" you will see a change in the spectrum. Also, the functions are combinations of sine and cosines (alternately even and odd wrt the center). Hint: Either the Sine term or cosine term disappears in each state...Can you solve it now?

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Ok. It makes sense now. Thank you for your replies.