Question

Evaluate Eigen function and energy Eigen values for a particle moving in one dimension inside an infinite potential well of width L. Also present a scheme of energy levels and probability densities of the particle inside the well.

Answer 1

Answer:

Most of the math for this problem should have been covered in lectures (or text) for the single δδ -function well. Otherwise the question looks a lot like a homework exercise that’s sort of cute. I’ve outlined the pieces that follow from the single well treatment, extending them to the double-well for the problem set-up, leaving a lot of bits and pieces to be picked up.

Given a Hamiltonian H=−ℏ22md2dx2+γδ(x−L)+γδ(x+L)H=−ℏ22md2dx2+γδ(x−L)+γδ(x+L) a parity operator Pψ(x)=ψ(−x)Pψ(x)=ψ(−x) will commute [H,P]=0[H,P]=0 . So eigenvector solutions will be eignevectors of both operators simultaneously. Given Pψ=pψPψ=pψ and that P2=1P2=1 , it follows p2=1p2=1 so p=±1p=±1 . Then the two cases are even functions corresponding to p=1p=1 , and odd functions corresponding to p=−1p=−1 .

In the region where x≠Lx≠L , −ℏ22md2ψdx2=Eψ−ℏ22md2ψdx2=Eψ . So for bound states, where E<0E<0 , the solutions are exponential rather than oscillatory.

Even solutions are ψ(x)=Acoshkxψ(x)=Acosh⁡kx for x<Lx<L , and odd solutions are ψ(x)=Asinhkxψ(x)=Asinh⁡kx for x<Lx<L , with ψ(x)=Bek(L−x)ψ(x)=Bek(L−x) for x>Lx>L . (why?)

The function has to be continuous at x=Lx=L so that ψ(L−)=ψ(L+)ψ(L−)=ψ(L+) . Taking an integral from x=L−ϵx=L−ϵ to x=L+ϵx=L+ϵ gives −ℏ22m(ψ′(L+)−ψ(L−))−γψ(L)=0−ℏ22m(ψ′(L+)−ψ(L−))−γψ(L)=0 as ϵ→0+ϵ→0+ .

The odd functions will satisfy (do the math) k=2mγℏ2tanhkLk=2mγℏ2tanh⁡kL , and the even ones will satisfy k=2mγℏ2cothkLk=2mγℏ2coth⁡kL . Note that tanhx<1tanh⁡x<1 and cothx>1coth⁡x>1 where x>0x>0 . Also, in −ℏ22md2ψdx2=Eψ−ℏ22md2ψdx2=Eψ , don’t forget the minus sign. Note also, the eigenvectors require transcendental solutions (without simple algebraic solutions) - but you have enough info to answer the question without a complete solution.