Perturbation Theory & the Ground State in a 1-D Potential Box

[michael2k100]

A particle is in the ground state in a one-dimensional box given by the potential

v(x)= 0 for 0<x<a
v(x)= inifinity other wise

A small perturbation V = V(0)x/a is now introduced. Show, correct to first order in perturbation theory, hat the energy change in the ground state is V(0)/2

the normalised wavefuctions of v(x) are PHi(n)=(sqrt(2/a))sin(n(pi)x/a)i have looked up what i can on perturbation theory, and i can't get to doing this question,

if anyone could give me a few pointers on what to do and where to start, that would be great.

 

[jbsteele]

To solve this problem using perturbation theory, you need to first calculate the matrix elements of the perturbation V = V(0)x/a between the ground state and all other states. This can be done using the following formula:Vnm = <φn|V|φm> = ∫a 0 V(x)φn(x)φm(x)dxFrom this, you can calculate the energy shift of the ground state due to the perturbation V = V(0)x/a. The energy shift is given by:ΔE0 = (V00)2/(E1-E0)Where V00 is the matrix element of V between the ground state and itself, and E1 and E0 are the energies of the first excited state and the ground state, respectively.For your problem, you can calculate V00 = V(0)2/2a, and the energies of the ground and first excited states are E0=0 and E1=(nπ)2/2ma2, respectively.Therefore, the energy shift in the ground state due to the perturbation V = V(0)x/a is ΔE0 = V(0)2/(2n2π2/2ma2) = V(0)/2, correct to first order in perturbation theory.