**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

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

we have

Normalizing , by hand or using Maple, we find

The stationary states of the Harmonic oscillator are

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

The expectation value for the energy can now be calculated from the energies of the lowest 3 states of the harmonic oscillator: . We get:

Now suppose that at some later time the wave function has changed to

(The only changes from are the plus sign in the middle factor and the different normalization constant .)

By normalizing the new wave function, we find that . This time, we can’t assume that is real, but it can differ from by a complex factor such that . If we convert the given wave function into Hermite polynomials as in part (a), we get:

Thus the only difference between this function and the initial function (apart from the factor ) is that the coefficient of 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):

we need to find the time such that . Substituting the energies, we find:

That is

for some integers and . From the first equation, we get and from the second, . The smallest occurs when and , giving . Then , and . By setting , we get the wave function at :

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RobertHow did you go from (16) to (17)?

gwrowePost authorSince (16) is the general time-dependent solution, we need to find and so that it is equal to (15). This means that if , and the rest follows as explained following (16).