**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 . If nothing changes in this system, the oscillator will remain in the ground state. However, if we quadruple the spring constant , this will cause a change in the frequency, since . Thus will double to a new value . In terms of this new frequency, we have

After the change in the spring constant, the new stationary states will be the functions which have the same form as the original functions except is replaced by . Thus the new ground state is

The energy of this state is . Since this is the new ground state, the probability of finding the energy at the old value of 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 , we expand the original wave function in terms of the new stationary states and find the coefficient of the first term in the expansion. That is

Since the are orthogonal functions, we can use the usual method of calculating :

The probability of being in the ground state is thus

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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)

gwrowePost authorThe latter. I’ve added a …