The variational method

Problem:

A particle moves non-relativistically in a three-dimensional harmonic oscillator potential.  The potential energy function is U(r) = αr².  Use spherical coordinates.
(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/(e^x + e^-x)]
ψ = A (1 + b²r²)^-1
ψ = A sin(br)/(br)

Note:  [∫₀^∞ r⁴exp(-r²)dr = 3π^½/8.  ∫₀^∞  r²exp(-r²)dr = π^½/4.]

Solution:

• Concepts:
  The variational method
• Reasoning:
  The variational method often yields a very good estimate for the ground state energy of a system.
• Details of the calculation:
  (a)  The ground state has zero angular momentum.  Therefore we have
  ψ(r)  = [u(r)/r]Y₀₀(θ,φ).
  ((-ħ²/2m)(∂²/∂r²) + ar²))u_k0(r) = E_k0u_k0(r).
  Choose u_k0(r) = B r exp(-br²).  Here b is an adjustable parameter.
  u(r) must be zero at r = 0 and at r = infinity and should not have any nodes in the region 0 < r < infinity.
  B²∫₀^∞ r² exp(-2br²)dr = 1,  B² = 8*2^½b^3/2/π^½.
  Then
  <H> = B²∫₀^∞ r exp(-br²)((-ħ²/2 m)(∂²/∂r²) + αr²) r exp(-br²)dr
  = B²(-ħ²/2m)∫₀^∞ [-6br² + 4r⁴b²]exp(-2br²)dr + B²α∫₀^∞ r⁴exp(-2br²)dr
  = B²[(3ħ²/m)(b^-½2^-3/2)∫₀^∞ r²exp(-r²)dr - B²[(ħ²/2m)(b^-½2^-½)∫₀^∞ r⁴exp(-r²)dr
  + B²α(b^-5/22^-5/2)∫₀^∞ r⁴exp(-r²)dr
  = (ħ²/2m)3b + 3α/(4b).
  [∫₀^∞ r⁴exp(-r²)dr = 3π^½/8.  ∫₀^∞  r²exp(-r²)dr = π^½/4.]
  d<H>/db = 0, (3ħ²/2m) = 3α/(4b²) , b² = αm/(2ħ²).
  <H> = 3ħα^½/(2m)^½.  E(upper limit) = 3ħα^½/(2m)^½. 
  Setting α = ½mω², we see that E(upper limit) = (3/2)(ħω), the exact result.
  If ψ = A exp(-br²), then the normalization constant A = B/(4π)^½.
  (b) The wave function must satisfy the boundary conditions and be normalizable. 
  rψ (r) must be zero at r = 0 and at r = infinity.  In the region 0 < r < infinity rψ and its derivative with respect to r must be continuous.  In the region 0 < r < infinity rψ for the ground state has no nodes.
  
  ψ = A exp(-br) is an acceptable trial function
  ψ = A sech(-br)  is an acceptable trial function
  ψ = A (1 + b²r²)^-1 is an acceptable trial function.
  ψ = A sin(br)/(br) is not an acceptable trial function.  It has nodes and is not normalizable.

Problem:

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

Solution:

• Concepts:
  The variational method
• Reasoning:
  The trial wave functions ψ(x) must satisfy the boundary conditions at infinity, ψ(±∞) = 0, or ψ(x) = 0 in regions where U(x) = ∞, the derivative dψ/dx must be continuous at all x except where U(x) --> ∞, and the ground state wave function has no nodes except at x = ±∞ or where U(x) --> ∞.  Any function that satisfies these conditions can be written as a linear combination of eigenfunctions of H and is an acceptable trial function giving <H> > E₀.
• Details of the calculation:
  (a)  The trial wave function must be asymmetric.  For example, we can use a linear combination of exp(-c²x²) and x*exp(-c²x²).
  ψ(x) = N[exp(-cx²) + b*x*exp(-cx²)].  This wave function satisfies the conditions stated above.  Here b and c are the adjustable parameters.  We minimize <H> with respect to b and c to find an upper limit for the ground state energy.
  (b)  The trial wave function must be symmetric.  For example, ψ(x) = N exp(-c²x²).  This wave function satisfies the conditions stated above.  Here c is the trial parameter.
  (c)  Choose the trial function ψ(x) = N x exp(-cx) to make the integrals easy to evaluate.  This wave function satisfies the conditions stated above.  Here c is the adjustable parameter.
  Normalize the wave function:   N²∫₀^∞ x² exp(-2cx) dx = N²/(4c³) = 1,  N² = 4c³.
  <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).

Problem:

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.

Solution:

• Concepts:
  The variational method
• Reasoning:
  The variational method often yields a very good estimate for the ground state energy of a system.

• 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
  Φ_k00(r)  = [u_k0(r)/r]Y₀₀(θ,φ).
  ((-ħ²/2μ)(∂²/∂r²) - U₀exp(-r²/a²))u_k0(r) = E_k0u_k0(r).
  Choose u_k0(r) = A r exp(-r²/b²).  Here b is an adjustable parameter.
  u_k0(r) must be zero at r = 0.
  A²∫₀^∞ r² exp(-2r²/b²)dr = 1,  A² = 8*2^½/[π^½b³].
  Then
  <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/2^3/2)∫₀^∞ r²exp(-r²)dr - A²[(ħ²/2μ)(b/2^½)∫₀^∞ r⁴exp(-r²)dr
  + A²U₀[a²b²/(2a²-b²)]^3/2∫₀^∞ r²exp(-r²)dr
  = A²[(3ħ²/μ)(b/2^3/2)(π^½/4) - A²[(ħ²/2μ)(b/2^½)(3π^½/8)
  + A²U₀[a²b²/(2a²+b²)]^3/2(π^½/4).
  [∫₀^∞ r⁴exp(-r²)dr = 3π^½/8.  ∫₀^∞  r²exp(-r²)dr = π^½/4.]
  <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.

Problem:

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.

Solution:

• Concepts:
  The variational method
• Reasoning:
  We are asked to use the variational method to estimate the ground state energy.

• 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/π^¼) 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)).
  
  Exact solutions:
  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).

Problem:

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 a^qb^p.  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.

Solution:

• Concepts:
  Dimensional analysis, the variational method
• Reasoning:
  We are instructed to apply those methods

• Details of the calculation:
  (a)  The eigenvalue equation for H is Hψ(r) = (a∇² + br)ψ(r) = Ca^qb^pψ(r).
  Here C is a dimensionless constant.
  The units of a^qb^p are the units of energy, in SI units a^qb^p 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 =J^q+pm^2q-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/π^½)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)[3π^½/8 - 3π^½/4] + (2πbB²/d⁴)∫₀^∞yexp(-y)dy
  = (4πaB²/d)[3π^½/8 - 3π^½/4] + (2πbB²/d⁴)
  = -(4d²)[3a/8 - (b/(2d³π^½)].
  We now minimize <H> with respect to d.
  d<H>/d(d) = 0.  -8d[3a/8 - (b/(2d³π^½)] - (4d²)(3b/(2d⁴π^½) = 0.
  d³ = -2b/(3aπ^½) = 2b/(3|a|π^½).
  d = [2b/(3|a|π^½)]^(1/3) is the optimum variational parameter.
  <H> = ([2b/(3|a|π^½)]^(2/3))[9|a|/2] is an upper limit for the ground state energy.

Problem:

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.

Useful information:
ψ_1s(r) = (1/πa₀³)^½exp(-r/a₀)
< ψ_1s |(1/r)| ψ_1s > = 1/a₀
< ψ_1s |∇²| ψ_1s > = -1/a₀²
∫∫d³r d³r'|ψ_1s(r)|²|ψ_1s(r')|²(1/|r - r'|) = 5/(8a₀)

Solution:

• Concepts:
  The variational method, the H^- negative ion
• Reasoning:
  Neglecting spin, the Hamiltonian for the H^- negative ion is
  H = P₁²/(2m) + P₂²/(2m) - e²/r₁ - e²/r₂ + e²/|r₁ - r₂|.
  e² = q_e²/(4πε₀) in SI units.
  We are asked to find the ground state energy using the variational method.
• Details of the calculation:
  Let us use as a trial function
  ψ(r₁,r₂) = (1/πa³)exp(-(r₁ + r₂)/a)
  (We are guided by the ground state wave function of hydrogen )
  ψ(r) = (1/πa₀³)^½exp(-r/a₀)
  The adjustable parameter in our trial function is a.  The wave function is normalized.
  ψ(r₁,r₂) is a product function, ψ(r₁,r₂) = Φ₁(r₁)Φ₂(r₂).
  Φ₁(r₁) = (1/πa³)^½exp(-r₁/a),  Φ₂(r₂) = (1/πa³)^½exp(-r₂/a).
  <ψ|p²₁/(2m)|ψ> = <Φ₁|p²₁/(2m)|Φ₁> = -ħ²/(2m)< Φ₁|∇₍₁₎²|Φ₁ > = ħ²/(2ma²).
  <ψ|p²₂/(2m)|ψ> = <Φ₂|p²₂/(2m)|Φ₂> = -ħ²/(2m)< Φ₂|∇₍₂₎²|Φ₂ > = ħ²/(2ma²).
  <ψ|e²/r₁|ψ> = e²<Φ₁|1/r₁|Φ₁> = e²/a.
  <ψ|e²/r₂|ψ> = e²<Φ₂|1/r₂|Φ₂> = e²/a.
  <ψ|e²/|r₁  -r₂||ψ> = ∫d³r₁d³r₂|Φ₁(r₁)|²|Φ₂(r₂)|²(1/|r₁ - r₂|) = 5e²/(8a).
  Therefore <H> = ħ²/(ma²) - 2e²/a + 5e²/(8a) = ħ²/(ma²) - 11e²/(8a).
  d<H>/da = -2ħ²/(ma³) + 11e²/(8a²) = 0,  a = 16ħ²/(11me²)
  E_{ground state}(upper bound) = -(11/16)²me⁴/ħ² = -0.945 * 13.6 eV
  This calculation would indicate that the H^- ion is not stable, since the ground state energy is less negative than that of a free electron and a neutral hydrogen atom. 
  (H^- is a stable negative ion.)