[SOLVED] Amath 481/581 homework 2 – quantum harmonic oscillator/boundary value problems
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the partial differential equation:
i~ψt +
~
2
2m
ψxx − V (x)ψ = 0, (1)
where ψ is the probability density and V (x) = kx2/2 is the harmonic confining potential. A typical
solution technique for this problem is to assume a solution of the form
ψ(x, t) = X
N
n=1
anφn(x) exp
−i
En
2~
t
, (2)
and is called an eigenfunction expansion solution (φn=eigenfunction, En > 0=eigenvalue). Pluggingin this ansatz into Eq. (1) gives the boundary value problem
d
2φn
dx2
− [Kx2 − εn]φn = 0 (3)
where we expect the solution φn(x) → 0 as x → ± ∞ and εn > 0 is the quantum energy. Note
that K = km/~
2 and εn = Enm/~
2
. In what follows, take K = 1 and always normalize so that
R ∞
−∞ |φn|
2dx = 1.Calculate the first five normalized eigenfunctions (φn) and eigenvalues (εn) (up to tolerance
of 10−4
) in increasing order such that the first eigenvalue is the lowest one using a shooting
scheme. For this calculation, use x ∈ [−L, L] with L = 4 and choose xspan = −L : 0.1 : L. Save
the absolute value of the eigenfunctions in column vectors (vector 1 is φ1, vector 2 is φ2 and so
on) and the eigenvalues in a separate 5×1 vector.Hint: Derive the boundary conditions at ±L as if these are the infinite boundaries, i.e. replacing
x = ±∞ with x = ±L and performing the derivation that we did in class. Start with initial guess
for the solution at x = −L as y(−L) = 1.ANSWERS: Should be saved as A1–A5 for the eigenfunctions and A6 for the eigenvalues.(2) Calculate the first five normalized eigenfunctions (φn) and eigenvalues (εn) in increasing
order such that the first eigenvalue is the lowest one using the direct method. For this
calculation, use x ∈ [−L, L] with L = 4 and choose xspan = −L : 0.1 : L. Save the absolute
value of the eigenfunctions in column vectors (vector 1 is φ1, vector 2 is φ2 and so on) and the
eigenvalues in a separate 5×1 vector.Hint 1: Formulate the harmonic oscillator problem as a differential e. value problem, i.e.,
−
d
2
dx2
+ Kx2
φn = εnφn (4)
1
and discretize it using 2nd order central difference for interior points (without first and last points)
to receive an e. value problem Aφ~
n = εnφ~
n where φ~
n = [φn(x2), …, φ(xN−1)]. Such problems can
be solved in MATLAB using the eig command.Hint 2: Use a bootstrap approach to determine the boundary equations: To construct the matrix A use the derived boundary conditions (from question 1) and approximate the first and last
points using 2nd order forward or backward difference and assume that ∆x is small such that
∆x
√
KL2 − εn ≈ 0. After you found the values of φn in the interior do not forget to compute the
first and last points (φn(x1) and φn(xN )) using full forward or backward-difference approximation
(without assuming ∆x
√
KL2 − εn ≈ 0).Be sure to save the eigenvectors including the first and
last points, i.e., φ~
n = [φn(x1) φn(x2) , …, φ(xN−1) φn(xN )].ANSWERS: Should be saved as A7–A11 for the eigenfunctions and A12 for the eigenvalues.
(3) There has been suggestions that in some cases, nonlinearity plays a role such that
d
2φn
dx2
− [γ|φn|
2 + Kx2 − εn]φn = 0 (5)Depending upon the sign of γ, the probability density is focused or defocused. Find the first two
normalized modes for γ = ±0.05 using shooting. For this calculation, use x ∈ [−L, L] with L = 2
and choose xspan = −L : 0.1 : L. Save the absolute value of the eigenfunctions in column vectors
(vector 1 is φ1, vector 2 is φ2) and the eigenvalues in a separate 2×1 vector.ANSWERS: For γ = 0.05, should be written out as A13-A14 (eigenfunctions) and A15 (eigenvalues). For γ = -0.05, should be written out as A16-A17 (eigenfunctions) and A18 (eigenvalues).Notes: Use 10−4
for the tolerance in shooting methods.