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.

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