Download application of the variational principle to quantum

APPLICATION OF THE VARIATIONAL PRINCIPLE TO QUANTUM MECHANICS
Suvrat R Rao, Student, Dept. of Physics, IIT Kharagpur
Guide: Prof. Kumar Rao, Asst. Professor, Dept. of Physics, IIT Bombay
Source: Introduction to Quantum Mechanics, David J Griffiths, Second Edition
Abstract: This project deals with the application of the Variational Principle to the field of
Quantum
Mechanics
in
what
is
known
as
the
‘Variational
Method’.
In this paper, the Variational Principle has been applied to several scenarios, with the aim being
to confirm its authenticity (see section three – Applications of the Variational Principle)
TABLE OF CONTENTS
The paper consists of three sections and each section consists of one or more chapters:
1) Preface:
i) The Infinite Square Well
ii) The Dirac Delta Function Well
iii) A Very Brief Introduction to Quantum Mechanics in Three Dimensions
iv) Multiple-Particle Systems
2) The Variational Principle
3) Applications of the Variational Principle:
i) Ground State of a Linear Potential Using a Gaussian Trial Function
ii) Ground State of Dirac’s Delta Function Well Using a Gaussian Trial Function
iii) Ground State of the Infinite Square Well Using a Triangular Trial Function
iv)A Crude Approximation of the Hydrogen Atom Ground State Using a Gaussian Trial Function
v)The Exact Ground State of the Hydrogen Atom Using an Exponential Trial Function
4) Conclusion
PREFACE
Chapter i): The Infinite Square Well
We shall be dealing with a system which consists of a single particle constrained to move in one
dimension only (here, the x axis).
Along the axis, there exists a time-independent potential field, V(x), whose magnitude depends
only on the position from the origin along the axis. The Infinite Square Well is a scenario where
we have a potential field defined as follows:
𝑉(𝑥) = {
0,
∞,
0≤𝑥≤𝐿
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Thus, the particle is trapped inside the ‘potential well’  a space spanning a width ‘L’, with two
barriers of infinite force preventing it from escaping.
Thus we can say with certainty that the wavefunction,  ( x ) , has to be zero outside the well.
So, we solve the time-independent Schrodinger equation for 0 ≤ 𝑥 ≤ 𝐿, where V(x) = 0 :
ћ  2

 0  E
2m x 2
This can be written as,
 2
2

k
  ,
x 2
where k 
2mE
ћ
This is the differential equation for the classical Simple Harmonic Oscillator and has the general
solution :
( x )  Asin(kx)  Bcos(kx)
Applying the boundary conditions for  ( x ) :
1)  ( x ) is always continuous.
2) First derivative of  ( x ) is continuous except at the points where the potential is infinite.
Since the second condition is obviously not applicable to the Infinite Square Well, we apply the
first condition only:
(0)  (L)  0
Which gives us, respectively,
B = 0, hence
( x )  Asin(kx)
And,
Asin(kL)  0
Therefore
kL  n
And inputting the value of k 
(since A = 0 is an absurd solution)
k
2mE
ћ
,
We get,
n2 2 ћ2
En 
2mL2
These are the stationary state energies of the particle in the infinite square well. The ground
state energy of the system corresponds to n = 1.
Eg 
 2ћ2
2mL2
We can find the arbitrary constant ‘A’ by normalizing  ( x ) over the x axis:



L
( x ) dx  | A |
2
2
2
sin
 (kx)dx 1
0
Solving, we get,
A
2
L
significance)
(we choose the positive real root because the phase of A has no physical
Therefore, overall, we have,
n ( x) 
2
n
sin( x )
L
L
Therefore, the solution(s) to the time-dependent Schrodinger equation i.e. the stationary
states, are:
n ( x, t ) 
2 2
2
2
n
sin( x )ei ( n  ћ/2 mL ) t
L
L
And the general solution to the time-dependent Schrodinger equation is a linear combination of
the stationary states:

( x, t )   Cn
n 1
2 2
2
2
n
sin( x )ei ( n  ћ /2 mL )t
L
L
The arbitrary constants,
Cn , can be calculated by the following method, which takes
advantage of the orthonormality of the wavefunction :

( x,0)   Cn
n 1
Multiplying on both sides by

where
m

2
n
sin( x )  Cn  n ( x )
L
L
n 1
 m ( x )* and integrating over the x axis, we get,


n 1
n 1
( x ) * ( x,0)dx   Cn   m ( x ) * n ( x )dx   Cn mn  Cm
mn
is the Kronecker delta defined in the usual way.
Hence, replacing the dummy index ‘m’ by ‘n’ for consistency,
Cn    n ( x ) *( x,0)dx
Or in Dirac’s bra-ket notation, this can be written as:
Cn  n ( x) ( x,0)
Chapter ii): The Dirac Delta Function Well
The Dirac Delta function is defined as
∞,
 (𝑡) = {
0,
𝑡=0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
such that,

 t )dt  1

The delta function has the following properties:
1) f (t ) t  a )  f(a)t  a ) …. (this follows because the delta function is zero
everywhere except at t=a. Therefore at all other points, it is the same as multiplying f(t) by zero)
2) Integrating both sides of the above equation, we get,




f (t )t  a )dt  f(a)  t  a )dt  f ( a )

The Delta Potential Well is a scenario wherein (again, we consider a system consisting of a
single particle constrained to move in one dimension only) there exists a time-independent
potential V(x) such that
V(x)   a  x )
We will consider only the bound states i.e. where the total energy of the particle E < 0,
Thus, the particle is bound/trapped inside a potential well that is like an infinitely deep, narrow
spike.
Therefore, the time-independent Schrodinger equation reads,
ћ  2

 a  x )  E 
2
2m x
In the region x < 0, V(x) =0, so
 2
2m E


   2
2
2
x
ћ
, where  
2m E
ћ
Since E is negative,  is real and positive.
Therefore the general solution is,
( x )  Aekx  Be kx
Since the second term blows up to infinity as x   , it cannot be a realizable solution.
Hence, we neglect it. Thus for x < 0,
( x )  Aekx
In the region x > 0 also V(x) =0, and so the time-independent Schrodinger equation and its
solution take the same form as above. So,
( x )  Cekx  De kx
Here, the first term blows up to infinity and therefore is neglected. So for x > 0,
( x )  De kx
Now, we apply the boundary conditions:
1)
 ( x ) is always continuous.
Therefore,
(0 )  (0 )
That gives us A  D . But this does not tell us anything about the stationary state energies.
2) The first derivative of  ( x ) is continuous except at the points where the potential is infinite.
The second condition does not apply here since the potential blows to negative infinity at x=0.
Hence we use a different technique to derive the stationary state energies.
We take the time independent Schrodinger equation and integrate it from -  to  , (where 
is a positive constant) and then set a limit for  to tend to zero.



ћ2  2 x )

dx   V ( x )( x )dx  E   (x )dx
2

2m  x


As  0 ,

The first term becomes
since
ћ 2  2  x )
ћ2  d 
d   ћ 2 A

 x 2 dx   2m  dx |0  dx |0   m ,
2m 
( x )  Ae kx for x > 0, and ( x )  Aekx for x < 0, as proved above.

E  ( x )dx

becomes zero, because  x ) is finite everywhere.

 V ( x )( x )dx would have been zero also, if V(x) had been finite, but since that is not so,

we get






 V ( x )( x )dx  a  ( x)( x )dx  a)  ( x)  aA
Thus the equation becomes,
ћ2 A
 aA  0
m
So we get,

Inputting the value,
ma
ћ2
2m E
, we end up with,
ћ

ma 2
E
2ћ 2
Hence we see that the particle can only exist in a single stationary state inside the Dirac Delta
Potential Well.
We can find the constant ‘A’ by normalizing  x ) :

 ( x)


0
2
dx  | A |
2
e
2 kx
 | A|
2

0
Thus,
A  
e
ma
ћ
2 kx
| A |2

1

So, the one and only stationary bound state of the delta function well is:
( x ) 
ma 
e
ћ
ma 2
E
2ћ 2
ma| x|
ћ2
,
And the wavefunction for this single stationary state is:
( x,t) 
ma
e
ћ

ma
| x|
ћ2
.e
ima 2
t
2 ћ3
Chapter iii): A Very Brief Introduction to Quantum Mechanics in Three Dimensions
The time dependent Schrodinger equation is
iћ
 ˆ
 H
t
The Hamiltonian operator is obtained from the classical energy by replacing the positions and
momenta in the three mutually independent directions by their corresponding operators.
H
px 
1
( px 2  px 2  px 2 )  V
2m
ћ 
ћ 
ћ 
, py 
, pz 
i z
i x
i y
or
p
ћ

i
Therefore ,
2
2
2
2
2
ћ



ћ
Hˆ  
( 2  2  2 ) V  
2  V
2m x y z
2m
Where
2
2
2
  2 2 2
x y z
2
is the Laplacian in Cartesian coordinates.
Now, the potential V and the wavefunction  are functions of r = (x,y,z) and time ‘t’
The probability of finding a particle in a volume
d 3r  dxdydz
(r, t) d 3r
2
is
Note: In spherical coordinates we can write d 3r  dxdydz  r 2 dr.sin  d .d  r 2dr.d (cos ).d
. The change of variables is carried out using the Jacobian | J |  d 3r  dxdydz | J | drd d .
The transformation from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z), is
given by the function F: ℝ+ × [0, π] × [0, 2 π) → ℝ3 with components:
The Jacobian matrix for this coordinate change is
The determinant is r2 sin θ. Thus, as dV = dx dy dz this determinant implies that the differential
volume element dV = r2 sin θ dr dθ dφ. Nevertheless this determinant varies with coordinates.
Integrals are to be performed one by one with respect to each variable, with  going from 0 to
2 ,  going from 0 to  and r going from 0 to  .
The condition for normalization then becomes,

With the integral taken over all space.
(r, t) d 3r  1
2
If the potential is independent of time, we can use the variables-seperable method to solve the
time dependent Schrodinger equation where we get a complete set of stationary states. Thus,
we have,

n (r, t)   n (r)e
iEnt
ћ
Where the spatial wave function  n (r) satisfies the time-independent Schrodinger equation,
ћ2 2

   V  
2m
And the general solution to the time-dependent Schrodinger equation is a linear combination of
the particular solutions i.e. the stationary states

n (r,t)   Cn n (r)e

iEnt
ћ
n 1
Where the constants Cn are determined in the same way there were in Chapter 1 - The Infinite
Square Well:
Cn   n (r) (r,0)    n (r)*(r,0)d 3r
If the potential admits continuum states, then the sum in the summation in the general solution
of the Schrodinger equation becomes an integral and in that case, the constants Cn can be
determined using the Fourier transform of
 (r,0)
Note: Typically, the potential is a function only of the distance from the origin and hence is
centro-symmetric. In that case it is better to adopt spherical coordinates ( r, , ). The Laplacian
in terms of the spherical coordinates can be obtained using transformation of variables
(explained above) :
 2 
1   2  
1
 
 
1
  2 r
 2
 sin 
 2 2  2 
r r  r  r sin   
  r sin    
2
The Hamiltonian becomes,
2

Hˆ 
2m
1   2  
  2 
1
 
 
1
 2 r
 2
 sin 
  2 2  2  V
r

r

r
r
sin








 r sin     

And the time-independent Schrodinger equation reads:
 2
2m
 1   2 
 2 r
 r r  r
1
 



sin

 2


 r sin   
  2  
1

  2 2  2    V  E
 r sin     
Chapter iv): Multiple-Particle Systems
Consider a two-particle system where the subscript ‘1’ corresponds to the first particle and ‘2’
corresponds to the second particle.
The wavefunction for this system will depend, (apart from the time, ‘t’) on the coordinates of
both the particles - r1 and r2 :
(r1 ,r2 ,t)
The time dependent Schrodinger Equation will then read:
iћ
 (r1 , r2 , t) ˆ
 H  (r1 , r2 , t)
t
Where Hamiltonian for the whole system is given by:
ћ2 2 ћ2 2
ˆ
H 
1 
2  V ( r1 r2  t)
2m
2m
It is obtained, as usual, by replacing the positions and momenta by their corresponding
operators in the classical energy expression of the system. The classical energy of the system is
of course, the sum of the kinetic energies of all the particles plus the potential energy of the
whole system. The subscript in the Laplacian indicates the particle with respect to whose
coordinates the differentiation is being done.
As an example, consider the Helium atom as shown in the diagram. Two electrons (particle 1
and particle 2) are shown revolving the nucleus consisting of two protons and two neutrons. r1
stands for the coordinates of electron no. 1 and r2 , for electron no. 2. r12 represents the
position of electron 1 with respect to electron two. i.e. r12 | r1  r2 | is the separation between
the two electrons.
The classical energy of this system comprises of the kinetic energies of the two electrons, and
the potential energies due to the electron-proton attractions and the electron-electron
repulsions. Therefore the Hamiltonian for this system reads,
ћ2 2 ћ2 2
e2 2 2 1
ˆ
H 
1 
2 
   
2m
2m
4 0 r1 r2 r12
Moving on,
The statistical interpretation for a two particle system is:
(r1 ,r2 ,t) d 3 r1 d 3r2
2
is the probability of finding particle 1 in volume d 3 r1 and particle 2 in d 3r2 .
And the normalization process reads:
3
3

(r
,r
,t
)
d
r
d
r2  1
1
2
1

2
If the potential is independent of time, we can use separation of variables to achieve
(r1 ,r2 ,t)  (r1 ,r2 )e

iEnt
ћ
Where the spatial wave function (r1 ,r2 ) satisfies the time independent Schrodinger equation:
ћ2 2
ћ2 2

1  
2   V   
2m
2m
The Variational Principle
The variational principle is applied to quantum mechanics in what is known as the ‘variational
method’ It is a useful tool for situations where you are unable to solve the Schrodinger equation
because of its mathematical complexity.
A variational principle is a scientific principle used within the calculus of variations, which
develops general methods for finding functions which minimize or maximize the value of
quantities that depend upon those functions.
In quantum mechanics, the variational method is one way of finding approximations to the
lowest energy eigenstate or ground state. This allows calculating approximate wavefunctions for
systems such as molecular orbitals.
The method consists in choosing a "trial wavefunction" depending on one or more parameters,
and finding the values of these parameters for which the expectation value of the energy is the
lowest possible. The wavefunction obtained by fixing the parameters to such values is then an
approximation to the ground state wavefunction, and the expectation value of the energy in that
state is an upper bound to the ground state energy.
If this upper bound is exactly equal to the actual ground state energy of the system under
observation, then the trial wavefunction is the actual wavefunction of the ground state of the
system. Otherwise, it is just an approximation of the ground state.
In simple language, in the variational method, we choose a ‘trial wavefunction’,  , and
assume (maybe incorrectly) that the system under consideration is in this state. The Variational
Principle states that the expectation value of the Hamiltonian,
assumed state
H , of the system, in the
 ,will always be greater than or equal to the actual ground state energy, Eg ,
of the system. Mathematically,
Eg    H |   H
Proof: The general solution to any differential equation is a linear combination of its particular
solutions. For the Schrodinger equation, this translates to the fact that (as we have seen in the
previous sections),
 can be expressed as a linear combination of its stationary states:

   Cn  n
n 1
Where each presumably normalized
n
satisfies the time independent Schrodinger equation ,
H n  En n
 gives us,
The normalization of
 
C  C 
m
m
m
n
n
  Cm*Cn  m  n   Cm*Cn mn   | Cn |2  1
n
m
n
m
n
n
Now,
H   H | 
C 
m
m
Since
H 
n
m
H  Cn n 
n
satisfies the time independent Schrodinger equation,
 C   C (E
m
m
m
n
n
n
C  C (H
m
m
m
n
n
)
n
H n  Enn
. Thus,
 n )    E n Cm*Cn  m  n   E n Cm*Cn mn   E n | Cn |2
m
n
m
n
n
We know that the ground state energy,
Eg , corresponds to n=1 in En
state. All other higher energy excited states have a greater value of
and is the lowest energy
E.
Therefore,
Eg  En
Multiplying both sides by | Cn |2 and carrying out a summation, we get,
E
g
| Cn |2  Eg | Cn |2   En | Cn |2
n
n
And we just proved above that
| C
n
n
|2  1 and
n
E
n
| Cn |2  H . Hence the inequation
n
reduces to the statement of the Variational Principle:
Eg    H |   H
Applications of the Variational Principle
Chapter i) Ground State of a Linear Potential Using a Gaussian Trial Function
Consider a one particle system constrained to move in one dimension under a potential given
by
V (x)  a | x |
Although it is possible to directly solve the Schrodinger equation for this potential without
much trouble, that is not the aim of this project. The aim is to apply the variational principle to
this potential using a Gaussian trial function and get an approximate value of the ground state
energy and the wavefunction.
Hence, we assume that the ground state wave function is of the form:
 (x)  Aebx
2
Where ‘b’ is the parameter which can be tweaked as per our requirement and ‘A’ is determined by
normalizing  (x) :

A
2
2 bx
 e dx  A
2

2

2b
1
That gives us,
A
4
2b

Now,

H | A |
2
e

2
ћ2  2 ebx
(
 a | x | ebx )dx | A |2
2
2m x
2
 bx 2

 ћ2 

2
2 2 2 bx 2
2 bx 2
(4b x e
 2b e
)dx  a  | x | e 2bx dx 


 2m 



For the last term, we can write
 | x|e

 2 xe2bx
2 bx2

2
. Substituting the term in the exponent
0
by a different variable, say, ‘u’ will simplify this integral. Upon solving, this term reduces to
1
4b
For the middle term, we have the standard Gaussian integral:

2 bx
e
dx 

2


2b
For solving the first term, we can use a very elegant technique called ‘Feynman’s Trick’. This
involves taking the standard Gaussian integral (the above equation) and differentiating both
sides with respect to the parameter ‘b’. Applying that to the first term gives us:


x 2 e2bx dx 
2

1  3/2
b
4 2
Thus, the equation reduces to:
ћ 2b
a
H 

2m
8 b
Now, according to the variational principle,
ћ 2b
a
Eg  H 

2m
8 b
So if we take the tightest upper bound by selecting ‘b’ such that the RHS is minimum, we can
approximate this minimum upper bound to be equal to the ground state energy. Therefore, if
we differentiate the RHS and equate it to zero we will get ‘b’ for which the RHS is minimum.
Solving and resubstituting this value of ‘b’ into the inequality gives us:
1/ 3
bmin
 m2 a 2 

4 
 8 ћ 
1/ 3
H
min
3  ћ2 a 2 
 

2  8 m 
Hence,
1/3
Eg
3  ћ2 a 2 


2  8 m 
Chapter ii) Ground State of Dirac’s Delta Function Well Using a Gaussian Trial Function
From Preface - Chapter ii), we know that the Hamiltonian for the Delta Function Well is
ћ2  2
ˆ
H 
 a ( x )
2m x 2
ma 2
And we also know the exact ground state energy to compare the results with : Eg 
2ћ 2
And since we will be using a Gaussian Trial Function,  (x)  Ae
bx2
the normalization constant : A 
4
we have also worked out
2b

Thus, the expectation value of the Hamiltonian is

H | A |
2
e



2
ћ 2  2 e bx
ћ2  2 e bx
 bx 2
2
 bx 2
(

a
e

(
x
)
)
dx

|
A
|
(
e

dx  a  e 2bx  ( x )dx )
2
2

2 m x
2m x


2
 bx 2
Solving, we get,
ћ 2b
2b
H 
a
2m

Minimizing RHS gives us
d H
ћ2 a 2


0
db
2m 2  b
Therefore ,
bmin
H
2m2 a 2

 ћ4
min
ma 2

 ћ2
2
Comparing
H
min
with the actual ground state energy, Eg 
that the variational principle holds true since
Eg  H
min
ma 2
, we can conclude
2ћ 2
Chapter iii) Ground State of the Infinite Square Well Using a Triangular Trial Function
We already know the exact value of the ground state energy of the Infinte Square Well:
Eg 
 2ћ2
2mL2
For the Infinite Square Well, we shall use a triangular wave function as a trial wave function
since it vaguely resembles the sinusoidal wavefunction of the ground state of the infinite
square well:
 (x) =
{
𝐴𝑥,
𝐴(𝐿 − 𝑥),
0
0 ≤ 𝑥 < 𝐿/2
𝐿/2 ≤ 𝑥 < 𝐿
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
We can determine ‘A’ by normalizing  (x) :
L/2
| A| ( 
2
0
L
L3
x dx   (L x ) dx )  | A |
1
12
L/2
2
2
A  2 3L3/2
2
Now, we will require the second derivative of  (x) when calculating the Hamiltonian. But
 (x) is a piecewise defined function and its first derivative is discontinuous. Hence, we
employ an ingenious technique to represent the second derivative:
𝐴,
0 ≤ 𝑥 < 𝐿/2
𝐿/2 ≤ 𝑥 < 𝐿
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
d (x)
= −𝐴,
dx
0
{
d (x)
d 2 (x)
(which is nothing but
), becomes infinite at x = 0, a/2
dx
dx 2
d 2 (x)
and a. but is zero everywhere else. So, we can ingeniously write
as a sum of Dirac Delta
dx 2
functions in the following way:
We see that the slope of
d 2 (x)
 A ( x )  2 A ( x  L / 2)  A ( x  L)
dx 2
and hence,
ћ2 d 2 (x)
ћ2 A
H   (x).( 
)dx  
(  [ ( x )  2 ( x  L / 2)   ( x  L)]. (x)dx )
2
2
m
dx
2
m
0
0
L
L
ћ2 A
ћ2 A2 L 12ћ2

[ (0)  2 (L/ 2)  (L)] 

2m
2m
2mL2
12ћ 2
We can see that the expectation value of the Hamiltonian, H 
is greater than the
2mL2
actual ground state energy of the Infinite Square Well, Eg 
Principle holds true.
 2ћ2
2mL2
. Thus the Variational
Chapter iv) A Crude Approximation of the Hydrogen Atom Ground State Using a Gaussian Trial
Function
We take the trial wave function  ( r ) 
Aebr
2
Normalization over all space gives us the constant ‘A’:
 | (r ) | d r | A |  e
2
3
2
r dr.d (cos ).d  4 | A |
2 br 2 2
2

r e
2  br 2
0
 
dr | A |  
 2b 
3/2
2
The last integral can be calculated using ‘Feynman’s Trick’ (see Chapter i) Ground State of a Linear
Potential Using a Gaussian Trial Function)
 2b 
A 
 
3/4
Now, the Hamiltonian for the Hydrogen Atom is:
Hˆ  
2
e2 1
 
2m
4 0 r
2
1
Thus,
Hˆ   A( 
2
2m
e2 e br

)
4 0 r
2
 2e br
2
Where the Laplacian in terms of spherical coordinates is:
2
1  2 e br
 2 (r
)  (4b 2 r 2  6b )e br
r r
r
2
 2e br
2
(the angular differential terms in
the Laplacian become zero since  ( r ) is independent of  and  and hence they have been
ignored.)
Hence,
 | H |  | A |  {
2
2
2m
2
e2 e2br 3
e2 e2br 2
2
2 2 2 br 2

}d r | A |  { (6b  4b r )e

}r dr.d (cos  ).d
4 0 r
2m
4 0 r
2
(6b  4b r )e
2 2
2 br 2
 2
e2 e2br
2 2 2 br 2
 4 | A |   (6b  4b r )e 
4 0 r
 2m
2
2
2
3/2



2
 2

e2
 2b  
2 2 br 2
2
4 2 br 2
2 br 2
re
dr
 r dr  4 .    [6b  r e dr  4b  r e dr] 

4 0 0
    2m 0

0

The integrals can be solved using Feynman’s Trick, or integration by substitution, or integration by
parts, or using the properties of gamma functions. Solving, we get,
3 2b
e2
H 

2m 2 0
2b

Minimizing the RHS gives us,
d H
3 2
e2
2
m2 e 4


 0  bmin 
db
2m 4 0  b
18 3 4 02
Plugging ‘b’ back into the equation gives us:
Eg
H
min
me4
me3

J
eV  11.5eV
12 3 2 02
12 3 2 02
We see that the Variational Principle still holds true, since the actual ground state energy of the Hydrogen
atom is known to be -13.6 eV, whereas we have overestimated it in our approximation of -11.5eV, just like
the Variational Principle states.
Chapter v) The Exact Ground State of the Hydrogen Atom Using an Exponential Trial Function
If, instead of a Gaussian (as in the previous case), we assume the trial wavefunction to be of a linear
exponential form:
 ( r )  Aebr
Normalization of  ( r ) over all space gives us the constant ‘A’:
 | (r) | d r | A |
2
3
2
e
r dr.d (cos ).d  4 | A |
2 br 2
2

2  br
2
r
e
dr

|
A
|

0
A
b3

Now, the Hamiltonian for the Hydrogen Atom is:
Hˆ  
2
e2 1
 
2m
4 0 r
2

b3
1
Hence,
2

e2 ebr 
2  br
ˆ
H  A  
e 

2
m
4


r

0

Where, the Laplacian is given by:
2  br
e
1  2 ebr
2b
 2 (r
)  (b2  )ebr
r r
r
r
( the angular differential
terms in the Laplacian become zero since  ( r ) is independent of  and  and hence they have
been ignored.)
Therefore, the expectation value of the Hamiltonian is given by,
2
 2 2b 2 2br
e2 e2br  3
2b 2 2br
e2 e2br  2
2 
 | H |  | A |   (  b )e 
 d r | A |   (  b )e 
r dr.d (cos ).d
4 0 r 
4 0 r 
 2m r
 2m r
2



 2

 2 2b 2 2br
e2 e2br  2
e2
3
2 br
2
2 2 br
 4 | A |  (  b )e 
re2 br dr 
 r dr  4b  [2b  re dr  b  r e dr] 

4 0 r 
4 0 0
 2m r
 2m 0
0

2
The integrals can be solved using integration by parts, or using the properties of gamma functions.
Solving, we get,
2 2
b
e2b
H 

2m 4 0
Minimizing the RHS gives us,
2
d H
b
e2
me2
1


 0  bmin 

db
m 4 0
4 2 0 a
Where, ‘a’ is the Bohr radius, a  0.529 10
10
m
And therefore, plugging this value of ‘b’ back in the equation for the Hamiltonian gives us,
2
H
min
This time, we have H
 m  e2 
 2
  E1  13.6eV
2  4 0 
min
exactly equal to the ground state energy of the Hydrogen atom,
which is 13.6 eV. This means that our assumed trail function  ( r ) 
ground state wavefunction of the Hydrogen atom.
Aebr
Eg ,
is the exact, actual
Substituting the values of ‘A’ and ‘b’ in  ( r ) gives us the exact ground state of the Hydrogen atom:
 g (r) 
1
 a3
e r / a
CONCLUSION
The Variational Principle is a powerful tool in Quantum Mechanics, and at the same time, it is very
simple to use. It is especially useful for situations wherein one is unable to directly solve the
Schrodinger equation due to its mathematical complexity corresponding to some proportionally
complicated quantum system. The only drawback(s) of this technique are that we never know how
close we are to the actual ground state energy of the system – we only have an upper bound.
Moreover, the technique applies only to the ground state, and sometimes, to certain excited
states.