Variational calculations for Hydrogen and Helium
Recall the variational principle. See Chapter 16 of the textbook. The variational
theorem states that for a Hermitian operator H with the smallest eigenvalue E0 , any
normalized |ψi satisfies
E0 ≤ hψ|H|ψi .
Please prove this now without opening the text. Given a Hamiltonian the method consists
in starting with a clever and tractable guess for the so-called trial wave function with
one (or more) free (variational) parameters. Normalize the wave function and evaluate
hψ|H|ψi. This yields a function of the variational parameters. Find the value of the
parameters that minimizes this function and this yields the variational estimate for the
ground state energy.
Gaussian trial wave function for the hydrogen atom:
Try a Gaussian wave function since it is used often in quantum chemistry. Start from
the normalized Gaussian:
µ ¶3/4
2
2
α3/4 e−αr .
ψ(r) =
π
The kinetic energy can be evaluated to be
Z ∞
h̄2
h̄2 ~
~
4π
∇ψ(r) · ∇ψ(r) =
3α .
r dr
2m
2m
0
Observe that since the dimensions of α is L−2 we can guess the dependence on dimensional
considerations only. The potential energy is given by
2
s
Z ∞
e2
2 2 1/2
1
2
2
−
4π
q α .
r dr ψ (r) = −2
4π²0
r
π
0
Minimize the energy with respect to α.
8m2 q 4
.
9πh̄4
Substituting we find that variational estimate for the energy is
α∗ =
4 me q 4
.
3π h̄2
The exact energy has a factor of 1/2 in front and the error is 15.1%.
EG = −
2
2
If one uses the sum of two Gaussians Ae−αx + Be−βx one can do better but this is
tedious to implement.
The Hamiltonian is given by
H = −
h̄2 2
Zq 2
Zq 2
q2
h̄2 2
∇1 −
∇2 −
−
+
2m
2m
|~r1 |
|~r2 |
|~r1 − ~r2 |
1
where the first two terms correspond to the kinetic energies of the two electrons and the
last term is the Coulombic repulsion between the two electrons. For Helium the nuclear
charge Z = 2.
As a variational trial wave function we place the two electrons in a 1s-like state with
the spatial part being symmetric (both electrons are in the same state) and the spin part
in the antisymmetric spin singlet. So we write the spatial part of the wave function as
φ(~r1 ) φ(~r2 ). 1 . We try a form inspired by the 1s state of the hydrogen atom which is
exponentially decaying:
1
φ(r) = √ 2α3/2 e−αr .
4π
This is normalized and has one variational parameter α which controls the spatial extent
of the wave function. The angular part is a constant, i.e., ` = 0. We need to evaluate
hΨ|H|Ψi. The first term we need is
Z
Z
3
d r1
Ã
h̄2 2
∇
d r2 φ (~r1 ) φ (~r2 ) −
2m 1
3
∗
!
∗
φ(~r1 ) φ(~r2 ) .
Since the operator only involves the first coordinate ~r1 the integral over d3 r2 yields 1 since
φ(~r2 ) is normalized. We are left with
Z
h̄2 ~
d r1
[∇φ(r)]2
2m
3
and this is the same calculation as the kinetic energy of the electron in the hydrogen atom
and we find
h̄2 α2
2m
; this must be multiplied by a factor of 2 for the kinetic energy of the
second electron which involves identical integrals.2 Now for the potential energies. For
the attractive energy of the first electron and the nucleus we have
Z
Z
3
d r1
Ã
Zq 2
d r2 φ (~r1 ) φ (~r2 ) −
r1
3
∗
∗
!
φ(~r1 ) φ(~r2 ) .
Again the d3 r2 integral yields 1 and we are left with
Z
Ã
Zq 2
d r1 φ (~r1 ) −
r1
3
∗
!
φ(~r1 )
which can be done since this is spherically symmetric and yields −αZq 2 ; this should
also be multiplied by 2 for the contribution from the other electron.
1
Note that the actual wave function will not be of this form and can depend upon |~r1 − ~r2 | which
cannot be written as a product of a function of ~r1 and ~r2 separately. However, we choose the product
form for calculational convenience. If we want a better approximation to the ground state energy we can
use other forms of the wave function.
2
Observe that when the Hamiltonian involves a single-particle term, i.e., depends on the coordinates of
one of the particles only, the expectation value involves a three-dimensional integral over the coordinates
of that particle only.
2
The interaction energy is much harder to calculate (Please see the end of this note if
you are mathematically curious) and yields 5q 2 α/8. Together we have
µ
h̄2 α2
5
E(α) =
− q 2 α 2Z −
m
8
¶
.
Minimizing with respect to α we obtain3
mq 2
α = 2
h̄
µ
5
Z−
16
∗
¶
.
The minimum energy is given by E(α = α∗ ) and is
Eg = −
mq 4
h̄2
µ
Z−
5
16
¶2
.
Recall that the ground state of the hydrogen atom is given by
mq 4
.
2h̄2
E0 = −
Thus we write our estimate of the ground state energy is
µ
5
Z−
16
Eg = −2E0
¶2
.
If we had just naively used the 1s hydrogen atom energy except with nuclear charge Z,
each electron would have an energy −E0 × Z 2 and so for two electrons we would have
−2E0 Z 2 excluding the repulsive energy. Instead the nuclear charge is reduced from Z = 2
to a lower value thanks to the shielding effect of the other electron. Evaluating we find an
energy of −2.8477Ry per electron while the experimental value is −2.904Ry per electron
for a 2% error which is rather good. The variational estimate for the ground state energy
is −77.5eV in contrast to the experimental value of −79eV .
S. Chandrasekhar and G, Herzberg used 10 variational parameters and then 18. Later
T. Kinoshita used 39 and later 80 parameters.
If we had set Z = 1 this calculation corresponds to the variational estimate for H − :
note that our calculation for Z = 1 yields −121/128 Ry which is higher than that of the
hydrogen atom and so this makes the prediction that H − is unstable with respect to the
hydrogen atom and the electron far away. However, experimentally, H − is stable. Bethe
was one of the first to ask this question. Superior trial wave functions have been tried. For
3
µ
¶¯
2h̄2 α
5 ¯¯
2
− q 2Z −
= 0
m
8 ¯α=α∗
3
example, multiply by (1+W r12 ) where r12 = |~r1 −~r2 |.4 Using two variational parameters
we obtain an energy that yields a stable ion with an energy of −1.01756 Ry.
Calculation of the interaction energy for the Helium atom with the simplest
variational ansatz
The trial wave function is given by
1
u(~r) = √ 2α3/2 e−αr .
4π
We need to evaluate the six-dimensional inetgral
Z
Z
3
I ≡
d r1
d3 r2 |u(~r1 )|2
q2
|u(~r2 )|2 .
|~r1 − ~r2 |
There are many ways of doing this integral. Here is one: We use the fact that the
(three-dimensional) Fourier transform of the Coulomb potential is 4π/k 2 , a very important
result5 :
Z
1
d3 k 4π i~k·(~r1 −~r2 )
=
e
.
|~r1 − ~r2 |
(2π)3 k 2
Substituting this in the integral and changing the orders of integration
Z
I =
Z
d3 k 4πq 2 Z 3
~
2 i~k·~
r1
d
r
|u(~
r
)|
e
d3 r2 |u(~r2 )|2 e−ik·~r2 .
1
1
(2π)3 k 2
Clearly, we need we need the Fourier transform of |u(~r)|2 which occurs twice:6
Z
~
d3 r1 eik·~r1
Z
Using
~
dΩ eik·~r1 ≡
Z π
0
α3 −2αr1
e
.
π
dθ sin θ
Z 2π
0
~
dφ eik·~r1 = 4π
sin(kr1 )
kr1
we have
Z
3
i~k·~
r1
d r1 e
Z ∞
α3 −2αr1
sin(kr1 ) −2αr1
16α4
e
=
dr r2 4α3
e
=
.
π
kr1
(4α2 + k 2 )2
0
4
This is a non-trivial calculation done most simply by using spherical coordinates for ~r1 and cylindrical
coordinates for ~r2 with the “z” axis aligned along ~r1 and ρ = r⊥ . Even with this one has to work hard
to evaluate the different terms.
5
One way of convincing yourself of this is to Fourier transform the equation
µ ¶
1
2
= −4πδ(~r) .
∇
r
6
Since we can change ~r2 → −~r2 the sign of the argument in the exponential does not matter.
You have done (essentially) this integral in an earlier homework when you considered the momentum
distribution of the hydrogen atom ground state.
4
So we finally need the integral7
Z
7
d3 k 4πq 2
(2π)3 k 2
Ã
16α4
(4α2 + k 2 )2
!2
=
5 2
q α.
8
This is not difficult to do or you can be use Maple or Mathematica.
5