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)

    Reply

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