# Set of probabilities distributions inside a infinite square well [duplicate]

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.

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

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.
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).