A particle moves non-relativistically
in a three-dimensional harmonic oscillator potential. The potential energy
function is U(r) = αr2. Use
(a) Use the trial function ψ = A exp(-br2) in the variational method to find the ground-state energy and the normalized wave function.
(b) Comment on the quality of the following trial functions. Do not do the calculations, just comment on the functions' good and bad points.
ψ = A exp(-br)
ψ = A sech(-br) [sech(x) = 1/cosh(x) = 2/(ex + e-x)]
ψ = A (1 + b2r2)-1
ψ = A sin(br)/(br)
Note: [∫0∞ r4exp(-r2)dr = 3π½/8. ∫0∞ r2exp(-r2)dr = π½/4.]
Assume that you are asked to use the variational method to estimate the
ground state energy of a particle with mass m, given a one-dimensional potential
energy function U(x). Assume U(x) = -Bx, x < 0, U(x) = Fx, x > 0,
where B and F are positive constants with the appropriate units.
(a) Assume B ≠ F. Choose a trial wave function ψ(x) with two adjustable parameters and explain why it is an acceptable wave function. (You do not have to find the ground-state enery.)
(b) Assume B = F. Choose a trial wave function ψ(x) with one adjustable parameter and explain why it is an acceptable wave function. (You do not have to find the ground-state enery.)
(c) Let B --> ∞. Choose a trial wave function ψ(x) with one adjustable parameter and estimate the ground state energy
Assume the potential for the deuterium is given by
U(r) = -U0exp(-r2/a2), with U0 = 37 MeV and a = 2.2*10-15 m.
Estimate the ground state energy of the deuterium.
Details of the calculation:
The relative motion of the proton and neutron is described in the same way as the motion of a fictitious particle of reduced mass m in a central potential. The ground state has zero angular momentum. Therefore we have
Φk00(r) = [uk0(r)/r]Y00(θ,φ).
((-ħ2/2μ)(∂2/∂r2) - U0exp(-r2/a2))uk0(r) = Ek0uk0(r).
Choose uk0(r) = A r exp(-r2/b2). Here b is an adjustable parameter.
uk0(r) must be zero at r = 0.
A2∫0∞ r2 exp(-2r2/b2)dr = 1, A2 = 8*2½/[π½b3].
<H> = A2∫0∞ r exp(-r2/b2)((-ħ2/2μ)(∂2/∂r2) - U0exp(-r2/a2)) r exp(-r2/b2)dr
= A2(-ħ2/2μ)∫0∞ [-6r2/b2 + 4r4/b4]exp(-2r2/b2)dr - A2U0∫0∞ r2exp(-r2(2/b2 + 1/a2))dr
= A2[(3ħ2/μ)(b/23/2)∫0∞ r2exp(-r2)dr - A2[(ħ2/2μ)(b/2½)∫0∞ r4exp(-r2)dr
+ A2U0[a2b2/(2a2-b2)]3/2∫0∞ r2exp(-r2)dr
= A2[(3ħ2/μ)(b/23/2)(π½/4) - A2[(ħ2/2μ)(b/2½)(3π½/8)
[∫0∞ r4exp(-r2)dr = 3π½/8. ∫0∞ r2exp(-r2)dr = π½/4.]
<H> = 3ħ2/(2μb2) - U0(2a2/(2a2+b2))3/2.
Let x = 2a2/b2.
<H> = 3ħ2x/(4μa2) - U0/(1 + 1/x)3/2.
d<H>/dx = 0 --> 3ħ2/(4μa2) - (3U0/2x2)(1 + 1/x)-5/2 = 0.
ħ = 6.58*10-22 MeV-s, μ = 938/2 MeV/c2, a = 2.2*10-15 m
ħ2/(μa2) = 17.2 MeV,
0.348 = (3/2)(1/x2)(1 + 1/x)-5/2. 0.232 = (1/x2)(1 + 1/x)-5/2,
x = 0.64.
b2 = 3.125 a2.
<H>/U0 = 0.223 - 0.244 = -0.0213
E(upper limit) = -0.786 MeV
The system is barely bound.
Consider the half space linear potential in one dimension
U(x) = ∞, x < 0, U(x) = Fx, x > 0.
Use the variational method to estimate the ground state energy. Choose your own variational wave function.
Details of the calculation:
H = p2/(2m) + Fx, x > 0.
The eigenfunctions of H must be zero at x = 0 and x = ∞. The derivative dψ/dx must be continuous at all x except x = 0. Any function that satisfies these boundary conditions can be written as a linear combination of eigenfunctions of H and is an acceptable trial function giving <H> > E0.
We may choose as the trial function for x > 0 the eigenfunction of the first excited state of the harmonic oscillator. It satisfies the boundary conditions. We choose
ψ(x) = (√a/π¼) 2ax exp(-½a2x2), with a2 = mω/ħ, for x >0,
ψ(x) = 0 for x < 0.
Here a is the adjustable parameter.
Note the change in the normalization constant, since we are only looking at x > 0.
<H> = <p2/(2m)> + F<x>, x > 0.
<p2/(2m)> = ½<T>harm. osc = ¼<E>harm. osc = (3/8)ħω = (3/8)ħ2a2/m,
since <T> = <U>.
<x> = (a/√π)∫0∞ 4a2x3exp(-a2x2) dx = (4/(a√π))∫0∞ x3exp(-x2) dx
= (2/(a√π))∫0∞ y exp(-y) dy = 2/(a√π).
<H> = (3/8)ħ2a2/m + 2F/(a√π).
d<H>/da = 0 --> a3 = (8/3)Fm/(ħ2√π).
<H> = (1/a)3F/√π = (F(2/3)ħ(2/3)/m(1/3))[(π(1/3)/√π)(3/8)(1/3)3]
= 1.86 (F(2/3)ħ(2/3)/m(1/3)).
E = 1.85 F(2/3)ħ(2/3)/m(1/3).
Let us choose a different trial function.
ψ(x) = N x exp(-cx), N2∫0∞ x2 exp(-2cx) dx = N2/(4c3) = 1, N2 = 4c3.
Here c is the adjustable parameter.
<H> = N2∫0∞ x exp(-cx)[(-ħ2/2m)(∂2/∂x2) + Fx]x exp(-cx) dx
= N2(-ħ2/2m)∫0∞ (-2cx + c2x2) exp(-2cx) dx + N2F∫0∞ x3 exp(-2cx) dx
= N2(ħ2/4mc - ħ2/8mc + 6F/(2c)4).
d<H>/dc = 0 --> c3 = 3Fm/(2ħ2).
Insert c back into the equation for <H>.
E = 1.96 F(2/3)ħ(2/3)/m(1/3).
A particle is bound to a site at the origin by a
linear "string" potential, U(r) = br, where b is the string tension.
In 3D the Hamiltonian is H = a∇2 + br, where
a = -ħ2/2m.
(a) The energy eigenvalues are all proportional to aqbp. Use dimensional analysis to determine both p and q.
(b) Assume a trial wave function of the form ψ(r) = Bexp(-d2r2/2).
(1) Determine the normalization constant B.
(2) Calculate the expected energy and find the optimum variational parameter d.
Details of the calculation:
(a) The eigenvalue equation for H is Hψ(r) = (a∇2 + br)ψ(r) = Caqbpψ(r).
Here C is a dimensionless constant.
The units of aqbp are the units of energy, in SI units aqbp has units of J.
a = -ħ2/2m has units of Jm2 and b must have units of J/m.
For the units we therefore have (Jm2)q(J/m)p =Jq+pm2q-p = J.
q + p = 1, 2q - p = 0, q = (1/3), p = (2/3).
(b) Let ψ(r) = Bexp(-d2r2/2).
Normalization: 4πB2∫0∞ r2exp(-d2r2/)dr = 1, B2 = (d/π½)3.
Using the given wave function we find <H> = <ψ|H|ψ>.
<H> = 4πaB2∫0∞ exp(-d2r2/2)(1/r2)(d/dr)[(r2(d/dr)exp(-d2r2/2)]r2dr
+ 4πbB2∫0∞ exp(-d2r2)r3dr
= (4πaB2/d)∫0∞(x4-3x2)exp(-x2)dx + (4πbB2/d4)∫0∞x3exp(-x2)dx
= (4πaB2/d)[3π½/8 - 3π½/4] + (2πbB2/d4)∫0∞yexp(-y)dy
= (4πaB2/d)[3π½/8 - 3π½/4] + (2πbB2/d4)
= -(4d2)[3a/8 - (b/(2d3π½)].
We now minimize <H> with respect to d.
d<H>/d(d) = 0. -8d[3a/8 - (b/(2d3π½)] - (4d2)(3b/(2d4π½) = 0.
d3 = -2b/(3aπ½) = 2b/(3|a|π½).
d = [2b/(3|a|π½)](1/3) is the optimum variational parameter.
<H> = ([2b/(3|a|π½)](2/3))[9|a|/2] is an upper limit for the ground state energy.
It is known that the stable H- exists (two electrons bound to a proton). Estimate the ground state energy of H- using the variational method. Assume that each electron moves in a 1s orbit. Neglect spin.
ψ1s(r) = (1/πa03)½exp(-r/a0)
< ψ1s |(1/r)| ψ1s > = 1/a0
< ψ1s |∇2| ψ1s > = -1/a02
∫∫d3r d3r'|ψ1s(r)|2|ψ1s(r')|2(1/|r - r'|) = 5/(8a0)