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Eigenfunction expansion, time evolution, Hamiltonian

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Eigenfunction expansion, time evolution, Hamiltonian.


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You should have it in your textbook(s) or lecture notes that the normalized stationary states for operator

-(hbar^2/2m)*(d^2/dx^2) on L^2([0,b]) with boundary conditions f(0)=f(b)=0 are

psi_n(x,t) = sqrt(2/b)*e^{-iw_n*t}*sin(k_n*x), n = 1, 2, .... (1)
k_n = pi*n/b (2)
is the wave vector,
w_n = E_n/hbar = hbar*pi^2*n^2/(2m*b^2) (3)

is the angular frequency and E_n is the energy.
Therefore, the lowest energy (n=1) state for b = a has energy

E_1 = hbar^2*pi^2/(2m*a^2) (4)
and wave function

psi_1(x,t) = ...

Solution Summary

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