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Its been some years since I did the infinite square well. I am doing an econimics problems with probability distributions and I vaguly remember there being a name for either the wave functions in the inifinte square well, or the probability distributions.

What is the equation if the well is between [0,1]? The equation of all possible probablity distributions, including the harmonics.

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  • $\begingroup$ Is this what you're looking for? physics.stackexchange.com/q/328370 $\endgroup$ Feb 8 at 21:30
  • $\begingroup$ If you are just after the name, rather than the formulae, were you perhaps thinking of eigenfunctions, or eigenstates? $\endgroup$
    – Tony
    Feb 8 at 21:54
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TL;DR

The energy eigenstates for the infinite square well between $0$ and $a$ are $$E(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$ with energies $$E_n=\frac{n^2\pi^2\hbar^2}{2ma}$$ The general solution for any initial wavefuntion $\psi(x,0)$ is $$\psi(x,t)=\frac{2}{a}\sum_{n=1}^\infty \left[\exp\left(-i\frac{n^2\pi^2}{2ma}t\right)\sin\left(\frac{n\pi x}{a}\right)\int^\infty_{-\infty}dx\sin\left(\frac{n\pi x}{a}\right)\psi(x,0)\right]$$ Simply set $a=1$ for your desired equations.


Derivation

For any time-independent system, after calculating the eigenkets of $H$, we have $$\begin{align} i\hbar\partial_t\langle E | \psi\rangle&=E\langle E | \psi\rangle\\ \langle E | \psi(t)\rangle&=e^{-iEt/\hbar}\langle E | \psi(0)\rangle\end{align}$$ Thus, assuming discrete values of $E$, we have $$\begin{align}\langle x | \psi\rangle &=\sum_{n=0}^\infty\langle x | E_n\rangle\langle E_n | \psi\rangle\\ &=\sum_{n=0}^\infty\langle x | E_n\rangle e^{-iE_nt/\hbar}\langle E_n | \psi(0)\rangle \\ &=\sum_{n=0}^\infty e^{-iE_nt/\hbar}\langle x | E_n\rangle\int_{-\infty}^\infty dx'\langle E_n | x'\rangle\langle x'| \psi(0)\rangle\end{align}$$ Converting to function notation, we have $$\psi(x,t)=\sum_{n=0}^\infty e^{-iE_nt/\hbar} E_n(x)\int_{-\infty}^\infty dx E^\star_n(x) \psi(x,0)$$ Now, for the infinite square well, the eigenvalue equation is $$\frac{1}{2m}P^2|E\rangle=(E-V_0)| E\rangle$$ Outside the well, the probability of the particle being found there is zero (which I will not prove). Inside, $V_0=0$. Converting to the position basis, we have $$\begin{align} -\frac{\hbar^2}{2m}d_x E(x)&=E\cdot E(x) \\ E(x)&=A\sin\left(\frac{\sqrt{2mE}}{\hbar}x\right)+B\cos\left(\frac{\sqrt{2mE}}{\hbar}x\right)\end{align}$$ However, we must have $E(0)=E(a)=0$; thus, $B=0$ and either $A=0$ or $\sin(a\sqrt{2mE}/\hbar)=0$. Since $\sin(x)=0$ iff $x=n\pi,n\in\mathbb{Z}$ for $x\in\mathbb{R}$, we have $$E_n=\frac{n^2\pi^2\hbar^2}{2ma}$$ and $$E(x)=A\sin\left(\frac{n\pi x}{a}\right)$$ After normalization, $$E(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$ Plugging these into the general formula derived above returns the answer given above. This derivation is rather terse; refer to a standard undergraduate text in the field for more clarity (such as Shankar or Griffiths).

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