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).

      Reply

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