Assume we have the following potential :

$$ V\left(x\right)=\begin{cases} 0 & -\frac{L}{2}\leq x\leq\frac{L}{2}\\ \infty & else \end{cases} $$

The wave function for a particle in this well potential, is given by :

$ \psi_{even}=\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right),\thinspace\thinspace\psi_{odd}=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) $

Where $n$ is even for the sin function, and $n $ is odd for the cos function, as we can see the calculation here

(the last comment)

But I tried to find the odd and even wavefunctions by shifting the wavefunction of a particle in an infinite box (which is not symmetric), and got different result. I'll be glad if someone can explain the difference. Here's my calculation:

We know that for a particle in a potential $$ V\left(x\right)=\begin{cases} 0 & 0\leq x\leq L\\ \infty & else \end{cases} $$

The wave function is given by

$ \psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x\right) $

For a symmetric potential we'll shift the function to the left by $ x\to x+\frac{L}{2} $ (we can see that for $x=-L/2 $ and for $x=L$ the wave function is $0$ so it preserved).

Now $ \psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}\left(x+\frac{L}{2}\right)\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x+\frac{\pi n}{2}\right) $

If we'll write for odd $n$ , $ n=2m+1 $ and for even $n$, $n=2m$, we'll get the solution:

$ \psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}\left(x+\frac{L}{2}\right)\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x+\frac{\pi n}{2}\right)=\begin{cases} \psi_{n=2m}=\left(-1\right)^{m}\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x\right)\\ \psi_{n=2m+1}=\left(-1\right)^{m}\sqrt{\frac{2}{L}}\cos\left(\frac{\pi n}{L}x\right) \end{cases} $

Which is different from the previous result (by a factor of $-1 $ in some of the functions)

What went wrong?

Thanks in advance.


1 Answer 1


Nothing went wrong, the difference between both results is just a global constant phase, which has no meaning. If $\psi_n(x)$ is an eigenfunction of the Hamiltonian, so is $e^{i \alpha}\psi_n(x)$. In your case, $e^{i\alpha}=(-1)^m$.

  • $\begingroup$ So should we multiply by $ (-1)^m $ in order to get the normalize condition? $\endgroup$
    – FreeZe
    Commented Mar 4, 2021 at 10:25
  • 2
    $\begingroup$ It is not necessary, as $|(-1)^m\psi_n(x)|^2=|\psi_n(x)|^2$ $\endgroup$
    – AFG
    Commented Mar 4, 2021 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.