In the context of an infinite potential well with boundaries at $(-a,a)$, where the potential is defined as follows:

\begin{equation} V(x) = \begin{cases} 0, & \text{if } -a \leq x \leq a \\ \infty, & \text{otherwise} \\ \end{cases} \end{equation}

Solving the Schrodinger equation, $$-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x)$$

With $k^2 = \frac{2mE}{\hbar^2}$, the general solution is deduced as: $$\psi(x) = A\sin(kx) + B\cos(kx)$$

The boundary conditions, $\psi(-a) = 0$ and $\psi(a) = 0$, lead to the simplified equations, $2A\sin(ka) = 0$ and $2B\cos(ka) = 0$

For the wave number we find that:

  • for $\sin(ka) =0$ ,$k =\frac{n \pi}{a} $ for n =1,2,3,4,...
  • for $\cos(ka) =0$ $k =\frac{n \pi}{2a} $ for n = 1,3,5,7,...

In most of the references which I have looked at, they try to add a factor of 1/2 in the solution for wavenumber for $\sin(ka) =0$, modifying $k$ as $$k =\frac{n \pi}{2a} \text{for} \sin(ka) =0, n =2,4,6,8...$$ As far as I know, this mathematical adaptation was to make the expression for the wavenumber consistent in both sine and cosine components. $$\psi(x) = A\sin(\frac{n \pi}{2a}x) + B\cos(\frac{n \pi}{2a}x)$$

My Question is if $sin(ka)=0$,$k=nπ/a$ can account for all the modes (n = 1,2,3,4,...),Can we write $$\psi(x) = A\sin(\frac{n \pi}{a}x)$$ as the simplified solution for the case? Wouldn`t it make the analysis easier?

Additionally, in the above mentions, boundary of an infinite potential well is at $(-a,a)$ initially, can I change the boundary to $(0,2a)$ for easier mathematical treatment?

  • 4
    $\begingroup$ Everything you have said is standard is many textbooks. And you are, of course, free to shift the interval. Just make sure to shift your variable likewise. $\endgroup$ Commented Oct 12, 2023 at 11:32
  • 1
    $\begingroup$ You can try solving for the well in a region $[0,a]$ (or $[0,2a]$), in which case all the solutions are expressed in terms of $\sin$. It might take a bit of time, but likely to answer many of your questions. $\endgroup$
    – Roger V.
    Commented Oct 12, 2023 at 11:49
  • $\begingroup$ @RogerVadim I did solve them. When I saw $(0,2a)$ giving me solution in pure sine terms and $(-a,a)$ giving me a solution with combination of sine and cosines, triggered doubts in me. $\endgroup$ Commented Oct 12, 2023 at 12:01
  • $\begingroup$ It is the other way round ($(0,2a)$ gives solutions in pure sign terms.) You can solve for $(0,2a)$ as $\psi(x)\propto \sin(kx)$ and for $(-a,a)$ as $\psi(x)\propto\sin[k(x+a)]$ and the solutions will be identical. It is a matter of translation along the $x$ axis. $\endgroup$
    – Roger V.
    Commented Oct 12, 2023 at 12:06

1 Answer 1


It depends a bit of what you want but, if you are interested in the energy levels, of course you can "shift" the potential to $[0,2a]$: surely the energies do not depend on the location of the well. In the same way, the energies of a harmonic oscillator or a hydrogen atom do not depend on the location of the origin, so there's no harm in shifting.

If you are interested in properties of the solutions, then you have to be a little more careful. By positing your well between $-a$ and $a$, your potential is now symmetric so you can break up the solutions into symmetric and antisymmetric wavefunctions, like you do for instance in a harmonic oscillator well $V=\frac12 m\omega^2 x^2$. You loose this parity property if your well does not satisfy $V(x)=V(-x)$.

Of course, one simple way to proceed is to locate the potential at a place convenient to find solutions - say between $0$ and $2a$. You then get $$ \psi_n(x)=\frac{1}{\sqrt{a}}\sin\left(\frac{n \pi x}{2a}\right)\, . \tag{1} $$ You can then shift by $a$ so to get the range between $[-a,a]$: \begin{align} \psi_n(x)&=\frac{1}{\sqrt{a}}\sin\left(\frac{n \pi (x+a)}{2a}\right)\, , \\ &= \frac{1}{\sqrt{a}}\sin\left(\frac{n \pi x}{2a} + \frac{n\pi}{2}\right)\, ,\\ &=\frac{1}{\sqrt{a}}\cos\left(\frac{n \pi x}{2a}\right)\sin\left(\frac{n\pi}{2}\right)+\cos\left(\frac{n\pi}{2}\right)\sin\left(\frac{n \pi x}{2a}\right)\, . \end{align} For $n=2,4, \ldots,$ the term in $\sin(n\pi/2)$ is $0$ and $\cos(n\pi/2)$ is just an overall phase so the solution collapses to $$ \psi_n(x)\propto \sin\left(\frac{n \pi x}{2a}\right) \qquad n \text{ even}\, , $$ whereas for $n=1,3,\ldots$, we have $\cos(n\pi/2)=0$, $\sin(n\pi/2)$ is just a phase and you now get $$ \psi_n(x)\propto \cos\left(\frac{n \pi x}{2a}\right) \qquad n \text{ odd}\, . $$ This way, the final range between $[-a,a]$ coincides with the original range in your question.


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