# Harmonic oscillator – change in spring constant

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.14.

Suppose we have a harmonic oscillator that starts off in the ground state, with frequency ${\omega}$. If nothing changes in this system, the oscillator will remain in the ground state. However, if we quadruple the spring constant ${k}$, this will cause a change in the frequency, since ${k=m\omega^{2}}$. Thus ${\omega}$ will double to a new value ${\omega'=2\omega}$. In terms of this new frequency, we have

 $\displaystyle \Psi(x,0)$ $\displaystyle =$ $\displaystyle \psi_{0}(x)\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^{2}/2\hbar}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega'}{2\pi\hbar}\right)^{1/4}e^{-m\omega'x^{2}/4\hbar} \ \ \ \ \ (3)$

After the change in the spring constant, the new stationary states will be the functions ${\psi_{0}',\;\psi_{1}',\ldots}$ which have the same form as the original ${\psi_{n}}$ functions except ${\omega}$ is replaced by ${\omega'}$. Thus the new ground state is

$\displaystyle \psi_{0}'(x)=\left(\frac{m\omega'}{\pi\hbar}\right)^{1/4}e^{-m\omega'x^{2}/2\hbar} \ \ \ \ \ (4)$

The energy of this state is ${\hbar\omega'/2=\hbar\omega}$. Since this is the new ground state, the probability of finding the energy at the old value of ${\hbar\omega/2}$ is now zero, since the energy cannot be less than the ground state energy.

To find the probability that the energy is the ground state energy of ${\hbar\omega}$, we expand the original wave function in terms of the new stationary states and find the coefficient ${c_{0}}$ of the first term in the expansion. That is

$\displaystyle \Psi(x,0)=\sum_{n=0}^{\infty}c_{n}\psi_{n}'(x) \ \ \ \ \ (5)$

Since the ${\psi_{n}'}$ are orthogonal functions, we can use the usual method of calculating ${c_{0}}$:

 $\displaystyle c_{0}$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\Psi(x,0)\psi_{0}'dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega'}{2\pi\hbar}\right)^{1/4}\left(\frac{m\omega'}{\pi\hbar}\right)^{1/4}\int_{-\infty}^{\infty}e^{-m\omega'x^{2}/4\hbar}e^{-m\omega'x^{2}/2\hbar}dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2^{1/4}}\left(\frac{m\omega'}{\pi\hbar}\right)^{1/2}\int_{-\infty}^{\infty}e^{-3m\omega'x^{2}/4\hbar}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2^{1/4}}\left(\frac{m\omega'}{\pi\hbar}\right)^{1/2}\frac{2}{3}\left(\frac{3\pi\hbar}{m\omega'}\right)^{1/2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2^{3/4}}{3^{1/2}} \ \ \ \ \ (10)$

The probability of being in the ground state is thus

 $\displaystyle \left|c_{0}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{2^{3/2}}{3}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.9428 \ \ \ \ \ (12)$

## 2 thoughts on “Harmonic oscillator – change in spring constant”

1. Cindy

“After the change in the spring constant, the new stationary states will be the functions {\psi_{0}’,\;\psi_{1}’,}”

why can you point out there will be only two new states psi(0)’ & psi(1)’ ?
or it’s just a way to say it will turn out to be the general form psi(n)