# Harmonic oscillator – example starting state

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.41.

Suppose a particle in the harmonic oscillator potential starts out in the state

$\displaystyle \Psi(x,0)=A\left(1-2\sqrt{\frac{m\omega}{\hbar}}x\right)^{2}e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

We can find the expectation value of the energy by expressing the given wave function as a linear combination of Hermite polynomials, since these form the orthonormal basis of solutions in the harmonic oscillator potential. Using the definitions

 $\displaystyle \alpha$ $\displaystyle \equiv$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\ \ \ \ \ (2)$ $\displaystyle \xi$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)$

we have

$\displaystyle \Psi(x,0)=A(1-2\xi)^{2}e^{-\xi^{2}/2} \ \ \ \ \ (4)$

Normalizing ${\Psi(x,0)}$, by hand or using Maple, we find

 $\displaystyle \int_{-\infty}^{\infty}\left|\Psi(\xi,0)\right|^{2}d\xi$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (5)$ $\displaystyle |A|^{2}$ $\displaystyle =$ $\displaystyle \frac{\alpha^{2}}{25} \ \ \ \ \ (6)$

The stationary states of the Harmonic oscillator are

$\displaystyle \psi_{n}(\xi)=\alpha\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\xi\right)e^{-\xi^{2}/2} \ \ \ \ \ (7)$

Taking ${A}$ to be real and positive, and expanding in terms of the Hermite polynomials, we have

 $\displaystyle \Psi(x,0)$ $\displaystyle =$ $\displaystyle \frac{\alpha}{5}(4\xi^{2}-4\xi+1)e^{-\xi^{2}/2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha}{5}(H_{2}-2H_{1}+3H_{0})e^{-\xi^{2}/2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{5}(2\sqrt{2}\psi_{2}-2\sqrt{2}\psi_{1}+3\psi_{0}) \ \ \ \ \ (10)$

The expectation value for the energy can now be calculated from the energies of the lowest 3 states of the harmonic oscillator: ${E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega}$. We get:

$\displaystyle \left\langle E\right\rangle =\frac{\hbar\omega}{25}\left(8\times\frac{5}{2}+8\times\frac{3}{2}+9\times\frac{1}{2}\right)=\frac{73}{50}\hbar\omega \ \ \ \ \ (11)$

Now suppose that at some later time ${T}$ the wave function has changed to

 $\displaystyle \Psi(x,T)$ $\displaystyle =$ $\displaystyle B\left(1+2\sqrt{\frac{m\omega}{\hbar}}x\right)^{2}e^{-m\omega x^{2}/2\hbar}\ \ \ \ \ (12)$ $\displaystyle \Psi(\xi,T)$ $\displaystyle =$ $\displaystyle B\left(1+2\xi\right)^{2}e^{-\xi^{2}/2} \ \ \ \ \ (13)$

(The only changes from ${\Psi(x,0)}$ are the plus sign in the middle factor and the different normalization constant ${B}$.)

By normalizing the new wave function, we find that ${|B|^{2}=\frac{\alpha^{2}}{25}=|A|^{2}}$. This time, we can’t assume that ${B}$ is real, but it can differ from ${A}$ by a complex factor ${q}$ such that ${|q|^{2}=1}$. If we convert the given wave function into Hermite polynomials as in part (a), we get:

 $\displaystyle \Psi(\xi,T)$ $\displaystyle =$ $\displaystyle q\frac{\alpha}{5}(H_{2}+2H_{1}+3H_{0})e^{-\xi^{2}/2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{5}(2\sqrt{2}\psi_{2}+2\sqrt{2}\psi_{1}+3\psi_{0}) \ \ \ \ \ (15)$

Thus the only difference between this function and the initial function (apart from the factor ${q}$) is that the coefficient of ${\psi_{1}}$ has changed sign. Since the general time-dependent solution is given by (with the lower limit on the sum changed to 0 because the ground state for the harmonic oscillator has index 0 rather than 1):

$\displaystyle \Psi(x,t)=\sum_{n=0}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (16)$

we need to find the time ${T}$ such that ${-e^{-iE_{1}T/\hbar}=e^{-iE_{0}T/\hbar}=e^{-iE_{2}T/\hbar}}$. Substituting the energies, we find:

$\displaystyle -e^{-3i\omega T/2}=e^{-i\omega T/2}=e^{-5i\omega T/2} \ \ \ \ \ (17)$

That is

 $\displaystyle \frac{3\omega T}{2}$ $\displaystyle =$ $\displaystyle \frac{\omega T}{2}+\left(2m+1\right)\pi\ \ \ \ \ (18)$ $\displaystyle \frac{5\omega T}{2}$ $\displaystyle =$ $\displaystyle \frac{\omega T}{2}+2r\pi \ \ \ \ \ (19)$

for some integers ${m}$ and ${r}$. From the first equation, we get ${\omega T=\left(2m+1\right)\pi}$ and from the second, ${2\omega T=2r\pi}$. The smallest ${T>0}$ occurs when ${m=0}$ and ${r=1}$, giving ${T=\pi/\omega}$. Then ${e^{-3i\pi/2}=+i}$, and ${e^{-\pi i/2}=e^{-5\pi i/2}=-i}$. By setting ${q=-i}$, we get the wave function at ${t=T=\pi/\omega}$:

$\displaystyle \Psi(x,T)=-\frac{i}{5}(2\sqrt{2}\psi_{2}+2\sqrt{2}\psi_{1}+3\psi_{0}) \ \ \ \ \ (20)$

## 2 thoughts on “Harmonic oscillator – example starting state”

1. gwrowe Post author

Since (16) is the general time-dependent solution, we need to find ${c_{n}}$ and ${T}$ so that it is equal to (15). This means that ${c_{n}=0}$ if ${n\ge3}$, and the rest follows as explained following (16).