# Infinite square well – phase difference

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.6.

As another example of an explicit case of a particle in the infinite square well, suppose we modify the example in the last post so that the second stationary state has a constant phase relative to the first. That is:

$\displaystyle \Psi(x,0)=A[\psi_{1}(x)+e^{i\phi}\psi_{2}(x)] \ \ \ \ \ (1)$

we can first normalize ${\Psi}$ by noting that ${A}$ is the same as in the last example since ${\psi_{1}}$ and ${\psi_{2}}$ are orthogonal. So

$\displaystyle A=\frac{\sqrt{2}}{2} \ \ \ \ \ (2)$

The time-dependent form is then:

$\displaystyle \Psi(x,t)=\frac{\sqrt{2}}{2}\left(\psi_{1}(x)e^{-i\omega t}+\psi_{2}(x)e^{i(\phi-4\omega t)}\right) \ \ \ \ \ (3)$

where ${\psi_{n}(x)}$ is given by the infinite square well formula.

The modulus is calculated in the usual way

 $\displaystyle |\Psi(x,t)|^{2}$ $\displaystyle =$ $\displaystyle \Psi^*\Psi\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}[\psi_{1}^{2}(x)+\psi_{2}^{2}(x)+2\psi_{1}(x)\psi_{2}(x)\cos(\phi-3\omega t)] \ \ \ \ \ (5)$

The mean position ${\langle x\rangle}$ is calculated by integrating this expression multiplied by ${x}$:

 $\displaystyle \langle x\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int_{0}^{a}x[\psi_{1}^{2}(x)+\psi_{2}^{2}(x)+2\psi_{1}(x)\psi_{2}(x)\cos(\phi-3\omega t)]dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{2}-\frac{16a}{9\pi^{2}}\cos(\phi-3\omega t) \ \ \ \ \ (7)$

where we have used the infinite square well formula to substitute for the ${\psi_{n}}$ functions, and done the integral with Maple. This reduces to the answer from before when ${\phi=0}$.

For ${\phi=\pi/2}$ we have

$\displaystyle \langle x\rangle=\frac{a}{2}-\frac{16a}{9\pi^{2}}\sin(3\omega t) \ \ \ \ \ (8)$

and for ${\phi=\pi}$we have

$\displaystyle \langle x\rangle=\frac{a}{2}+\frac{16a}{9\pi^{2}}\cos(3\omega t) \ \ \ \ \ (9)$

Thus the phase ${\phi}$ has no effect on the amplitude of the oscillation; but it shifts the oscillation in time.