Potential Well

Now we take a look at a particle in a potential well with the length $L$ and infinitely high walls which can be written as $$E_\mathrm{pot}(x) = \left\{\begin{array}{ll}0&\text{ for } 0 \leq x \leq L\\ \infty&\text{ otherwise}\end{array}\right.$$ Inside the well, the particle is free. We can therefore write the Schrödinger equation as $$-\frac{\hbar^2}{2m}\psi''(x) = E\psi(x)$$ because the potential vanishes. This equation can be rearranged according to $$\psi''(x) + k^2\psi(x) = 0$$ where $k^2$ is defined as $2mE/\hbar^2$ The general solution is given as $$\psi(x) = Ae^{ikx} + Be^{-ikx}$$ The two parts with $A$ and $B$ describe waves traveling in both directions. Using the boundary condition $$\psi(x\leq 0) = \psi(x\geq L) = 0$$ leads to $$A + B = 0$$ and $$Ae^{ikL}+ Be^{-ikL} = 0$$ From the first condition it follows $$\psi(x) = A(e^{ikx} - e^{-ikx}) = 2iA\sin kx$$ The second condition leads to $$2iA\sin kL = 0$$ which means the product $ka$ has to be equal to $n\pi$ with $n$ being an integer from 1 upwards: $$kL = n\pi$$ The result can therefore be written as $$\psi_n = 2iA\sin \left(n\frac{\pi}{L}x\right)$$ Now we still have to normalize our wave function to get the value for the ampliude $A$. The probability density function is then given as the squared amplitude $\psi_n^2(x)$. The integral over this probability density over the full space has to be equal to 1: $$\int_{-\infty}^{\infty} \psi^2_n(x)\,\mathrm{d}x = \int_{-\infty}^{\infty} \psi^*(x)\psi(x)\,\mathrm{d}x$$ Inserting the function $\psi(x)$ leads to the following integral: $$A^2\int_0^L \sin^2\left(n\frac{\pi}{L}x\right)\,\mathrm{d}x$$ The integral of $\sin^2(x)$ can be looked up in integration tables or solved with the help of a computer algebra system. The result can be written as follows: $$A^2\left(\frac{L}{2}-\frac{L}{4n\pi}\sin(2n\pi)\right) = 1$$ Since the sine is always equal to 0, the right term vanishes and only the following equation remains $$A^2\frac{L}{2} = 1$$ which leads to $$A = \frac{2}{L}$$ This then leads to the following wave function as a function of $x$: $$\boxed{\psi(x) = \frac{2}{L}\sin\left(n\frac{\pi}{L}x\right)}$$ The result are standing waves with the following wavelengths: $$\boxed{\lambda_n = n\frac{\pi}{L}}$$ The results of the first 3 wave functions are shown in figure with a vertical offset to differentiate between them easier. One can easily see the nodes at the walls of the potential well. For the energy levels, the formula $$E_n = \frac{p^2}{2m} = \frac{\hbar^2 k_n^2}{2m}$$ is used. Inserting $k_n$ then leads to $$\boxed{E_n = \frac{\hbar^2 \pi^2 n^2}{2mL^2}}$$ The minim energy of the trapped particle can therefore be also not 0, but instead $$E_1 = \frac{\hbar^2 \pi^2}{2mL^2}$$