A particle starts out in the ground state of the infinite square well (on the interval 0 ≤...

A particle starts out in the ground state of the infinite square well (on the interval 0 ≤ x ≤ a). Now a wall is slowly erected, slightly off-center:

Where f(t) rises gradually from 0 to ∞. Accoroding to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.

(a) Find (and sketch) the ground state at t → ∞. Hint: This should be the ground state of the infinite square well with an impenetrable barrier at a/2 + ∑. Note that the particle is confined to the (slightly) larger left “half” of the well.

(b) find the (transcendental) equation for the ground state of the Hamiltonian at time t. Answer:

Where z ≡ ka, T ≡ maf(t)/h2, δ ≡ 2∑/a, and k ≡ .

(c) Setting δ = 0, solve grahically for z, and show that the smallest z goes from π to 2 π as T goes from 0 to ∞. Explain this result.

(d) Now set δ = 0.01, solve numerically foe z, using T = 0,1,5,20,100. And 1000.

(e) Find the prbalility pr that the particle is in the right “half” of the well, as a function of z and δ. Answer : Pr = 1/[1+(I+/I-)], where I± ≡ [1±δ-(1/z) sin (z(1±δ))] sin2[z(1  δ)/2]. Evalute this expression numerically for the T’s in part (d). Comment on your results.

(f) Plot the ground state wave function for those same values of T and δ. Note how it gets squeezed into the left half of the well, as the barrier grows.