Probability of finding particle in shifted ground state

[AegisAndAtrophy]

Homework Statement

At t<0 a particle is in the ground state of the potential V(x)= \frac{1}{2} mw^2x^2. At t=0 the potential is suddenly displaced by an amount x0 to V(x)= \frac{1}{2} mw^2(x-x_0)^2 .

a) What is the probability of the particle being in the ground state; the first excited state?
b) At t=\frac{2pi}{w} write the wave function

Homework Equations

The Attempt at a Solution

I think that the wave function should be Ψ_0(x,0)= (\frac{mw}{πh})^\frac{1}{4} e^ \frac{-mw(x-x_0)^2}{2h}
I'm not sure what I should do though in order to find the probability in the ground state.
For the first excited energy I know I need to use the raising operator in order to get the wave function, which would give me (\frac{mw}{h})^\frac{1}{2} times Ψ_0 but again I'm unsure how to find the probability.

 

[Simon Bridge]

Express the original state in terms of the new ones.

 

[AegisAndAtrophy]

I'm not exactly sure what you mean in order to go about doing that.

 

[Simon Bridge]

The particle still has the same wavefunction after the potential has been changed.
However, that no longer corresponds to an energy eigenstate.
Expand the original wavefunction as a linear sum of the new energy eigenstates.
What do the coefficients of the expansion represent?

 

[AegisAndAtrophy]

So I should expand Ψ =∑CnΨn where |Cn|^2 is the probability of the new energy eigenstate?

 

[AegisAndAtrophy]

Does the change in time from t<0 to t=0 not affect it in any way?

 

[Simon Bridge]

The effect is in the possible outcome of future measurements.

 

[vela]

Homework Statement

At t<0 a particle is in the ground state of the potential V(x)= \frac{1}{2} mw^2x^2. At t=0 the potential is suddenly displaced by an amount x0 to V(x)= \frac{1}{2} mw^2(x-x_0)^2 .

a) What is the probability of the particle being in the ground state; the first excited state?
b) At t=\frac{2pi}{w} write the wave function

Homework Equations

The Attempt at a Solution

I think that the wave function should be Ψ_0(x,0)= (\frac{mw}{πh})^\frac{1}{4} e^{-\frac{mw(x-x_0)^2}{2h}}

This is the ground state of the system after the potential is displaced. It's not the state of the particle at $t=0$, which is still in the state
$$\Psi(x,0) = \left(\frac{m\omega}{πh}\right)^\frac{1}{4} e^{-\frac{m\omega x^2}{2h}}.$$ By the way, the frequency is denoted by the Greek letter $\omega$, not $w$.

I'm not sure what I should do though in order to find the probability in the ground state.
For the first excited energy I know I need to use the raising operator in order to get the wave function, which would give me (\frac{mw}{h})^\frac{1}{2} times Ψ_0 but again I'm unsure how to find the probability.

Look up the definition of the probability amplitude. This is a basic definition you need to know.