Download Energy Levels Of Hydrogen Atom Using Ladder Operators

Energy Levels Of Hydrogen Atom Using Ladder
Operators
Ava Khamseh
Supervisor: Dr. Brian Pendleton
The University of Edinburgh
August 2011
1
Abstract
The aim of this paper is to first use the Schrödinger wavefunction methods
and then the ladder operator methods in order to study the energy levels of
both the quantum Simple Harmonic Oscillator (SHO) and the Hydrogen
atom.
Introduction
Starting from the Simple Harmonic Oscillator[1] we first work through the solution of the bound-state wavefunctions for the harmonic oscillator in order to
obtain the energy eigenvalues, E = (n + 12 )!ω, energy eigenfunctions and the
Hermite polynomials by solving differential equations. Then, we will use algebraic ladder operator methods in order to find the energy eigenvalues which
should result in the same answer.
In the second section the aim is to find the the energy levels of the Hydrogen
atom[1] using the two methods[2] explored in the previous section. However,
this difference between the SHO and the Hydrogen atom is that in the case of
the latter we need to consider “three dimensional” space. Finally the results of
both methods will be compared in the end.
1
The Harmonic Oscillator
1.1
Differential Equations Method
For the SHO the Hamiltonian is of the form
p̂2
1
Ĥ =
+ kx2
2m 2
So the time independent Schrödinger equation 1.1 becomes
!2 d2 ψ(x) 1 2
+ kx ψ(x) = Eψ(x)
2m dx2
2
Now, introducing the frequency of the oscillator
!
k
ω=
m
writing
2E
ε=
!ω
−
(1.1)
(1.2)
(1.3)
(1.4)
and substituting for k and E into equation 1.2 we get
−
!2 d2 ψ(x) 1 2 2
ε!ω
d2 ψ(x) ωm "
ωm 2 #
+
ω
mx
ψ(x)
=
ψ
(x)
⇒
+
ε
−
x ψ (x) = 0
2m dx2
2
2
dx2
!
!
(1.5)
3
Changing variables to
y=
equation 1.5 can then be written as
!
mω
!
(1.6)
#
d2 ψ "
+ ε − y2 ψ = 0
dy 2
(1.7)
As y 2 → ∞, ε becomes negligible in comparison with y 2 and we can rewrite
equation 1.7 as
d2 ψ0 (y)
− y 2 ψ0 (y) = 0
(1.8)
dy 2
0
Now if we multiply the equation by 2 dψ
dy and use the chain rule we get
&$
'
$
%2
%2
"
#
d dψ0
d
d
dψ
(x)
0
− y2
ψ2 = 0 ⇒
− y 2 ψ02 = −2yψ02
dy dy
dy 0
dy
dy
(1.9)
In order to make this equation simpler, we assume that the term on the right
side of the equation can be neglected. However, we will check later that this
assumption is correct. So we have
&$
'
%2
d
dψ0 (x)
− y 2 ψ02 = 0
(1.10)
dy
dy
Integrating both side of the above equation with respect to (y) and rearranging,
dψ0 (x)
= (C + y 2 ψ02 )1/2
dy
Both ψ0 and
dψ0
dy
(1.11)
must vanish at infinity, so C = 0 so
dψ0 (x)
= ±yψ0
dy
(1.12)
and the acceptable solution, which vanishes at infinity is
ψ0 (y) = e−y
If we now observe the term
d
2 2
dy (y ψ0 )
2
/2
(1.13)
which appears in equation 1.9,
2
2
d 2 2
d 2 −y2
(y ψ0 ) =
(y e ) % −2y 3 e−y + 2ye−y
dy
dy
2
(1.14)
2
it can be said that 2yψ02 = 2ye−y is indeed negligible compared with −2y 3 e−y
for large y so the assumption made above is correct.
Hence, we can make an ansatz that the general solution for ψ(y) takes the form
ψ(y) = h(y)e−y
2
/2
(1.15)
NB Note that this is a general solution for ψ(y) not ψ0 (y) which is only defined
for large y.
4
It is important to mention that the solution is exact for the ground state where
h (y) = 1 (see equation 1.17, for m = 0, h (y) = a0 = 1)
2
We insert this back into equation 1.7 and devide both sides by ey /2 to get
d2 h(y)
dh(y)
− 2y
+ (ε − 1)h(y) = 0
dy 2
dy
(1.16)
Now that we have considered the behavior at infinity, we should observe the
behavior near y = 0 by starting from the power series expansion
h(y) =
∞
!
am y m
(1.17)
m=0
which when substituted into equation 1.16, rearranged and devided by y m results
in the recursion relation
(1.18)
(m + 1)(m + 2)am+2 = (2m − ε + 1)am
For large m , ie m > N where N is a large number, we have
m2 am " 2mam ⇒ am+2 "
2
am
m
(1.19)
which implies that the general solution to h(y) is
h(y) = ( solution for m < N ) + ( solution for m ≥ N )
so h(y) = (a polynomial in y)
"
#
2
22
23
+aN y N + y N +2 +
y N +4 +
y N +6 + . . .
N
N (N + 2)
N (N + 2)(N + 4)
(1.20)
where the first term just comes from the power series expansion and last terms
is derived as follows
1. m = N ⇒ aN y N
2. According to equation 1.19 , aN +2 =
3. Applying 1.19 again, aN +4 =
2
N aN
2
N +2 aN +2
2
2
N +4
N (N +2) y
=
⇒
2
N +2
N aN y
2
2
N +2 ( N aN )
=
22
N (N +2) aN
⇒
and so on. Note that, for simplicity, we have only considered the even solution.
The second term in equation 1.20 can, for reasons that will soon become clear,
be written in the following form
$
'
1
1
& y N +2 + . . .
aN y 2 y N −2 + N y N + % N & % N
2
2
2 +1
'
%N
& $
% 2 & N2 −1
1 2 N
1
2 2 −1 !
2 N
+1
2
2
&
& (y )
= aN y % N
y
+ N (y ) + % N & % N
+ ...
2 −1 !
2
2
2 +1
5
= aN y 2
!


" % 2 & N2 −1
2 N
2 N
2
2 +1
y
N
(y
)
(y
)
& + %N & + %N
&
− 1 ! %N
2
−
1
!
!
+
1
!
2
2
2
(1.21)
We introduce k such that N = 2k so the equation above takes the form
) % &k−1
*
% 2 &k
% 2 &k+1
y2
y
y
2
y (k − 1)!
+
+
+ ...
(k − 1)!
k!
(k + 1)!
)
)
% 2 &2
% 2 &k−2 **
y
y
2
y2
2
= y (k − 1)! e − 1 + y +
+ ...
(1.22)
2!
(k − 2)!
Where we have made use of the Taylor series for the exponential function ex =
2
3
1 + x + x2! + x3! + . . .
In other words, equation 1.22 can be interpreted as (h(y) = a polynomail + a
2
constant × y 2 ey ).
When this is substituted back into equation 1.15 ,
ψ(y) = (a polynomial × e−y
2
/2
) + (a constant × y 2 ey
2
/2
)
(1.23)
which does not vanish at infinity so it is not an acceptable solution. Therefore,
the recursion relation 1.18 must terminate for some n, ie,
2n − ε + 1 = 0 ⇒ ε = 2n + 1
If we know go back to the definition of ε on page 3 , E = ε!ω
2 ,
1
E = !ω(n + ) ; n = 0, 1, 2, . . .
2
which is the energy eigenvalue for the Simple Harmonic Oscillator.
(1.24)
(1.25)
The polynomials h(y) (apart from the nomalisation constants) are known as the
Hermite polynomials Hn (y) and some of their important properties are stated
below,
d2 Hn (y)
dHn (y)
− 2y
+ 2nHn (y) = 0
(1.26)
dy
dy
and the recursion relations
Hn+1 − 2yHn + 2nHn−1 = 0
Hn+1 +
dHn
− 2yHn = 0
dy
(1.27)
(1.28)
Both of the above relations can be proved by induction and it is left as an
exercise for the reader.
The first few polynomials are listed below:
H0 (y) = 1
H1 (y) = 2y
6
H2 (y) = 4y 2 − 2
H3 (y) = 8y 3 − 12y
1.2
Operator Methods
The Hamiltonian for the SHO is of the form
p̂2
1
Ĥ =
+ mω 2 x̂2
2m 2
(1.29)
∂
where the operators for momentum and position are p̂ = −i! ∂x
and x̂ = x
respectively.
Note that both of these operators are Hermitian, ie, they have real eigenvalues.
It is useful to prove that these operators are hermitian:
For any linear operator Ô , the hermitian conjugate Ô† is defined by
ˆ ∞
ˆ ∞
!
"∗
dxψ ∗ (x) Ôψ (x) =
dx Ô† ψ (x) ψ (x)
In other words,
−∞
(1.30)
−∞
〈ψ|O|ψ〉 = 〈Ô† ψ|ψ〉
(1.31)
Hence, for the momentum operator p̂ we have
#
$
ˆ ∞
ˆ ∞
ˆ ∞
d
dψ
dxψ ∗ (x) p̂ψ (x) =
dxψ ∗ −i!
ψ = −i!
dxψ ∗
dx
dx
−∞
−∞
−∞
Using integration by parts
%
ˆ
∞
= −i! [(ψ ∗ ψ)]−∞ −
∞
−∞
dx
#
dψ ∗
dx
$ &
ˆ
ψ = i!
∞
−∞
dx
#
dψ ∗
dx
andassuming that ψ and ψ ∗ vanish at ±∞,
#
$∗
ˆ ∞
ˆ ∞
dψ
∗
=
dx −i!
ψ=
dx (p̂ψ (x)) ψ (x)
dx
−∞
−∞
$
ψ
(1.32)
d
Therefore, p̂ = p̂† = −i! dx
as required.
The same argument holds for x.
The commutation relation between the two operators is
[p̂, x̂] = p̂x̂ − x̂p̂ = −i!
(1.33)
One might think that the Hamiltonian ( 1.29 ) can be written in the form
#'
$ #'
$
mω
p̂
mω
p̂
Ĥ = ω
x̂ − i √
x̂ + i √
(1.34)
2
2
2mω
2mω
However, since p̂ and x̂ do not commute, equation 1.34 is not correct.
7
In fact,
!"
# !"
#
mω
p̂
mω
p̂
mω 2 2 iω
iω
p̂2
ω
x̂ − i √
x̂ + i √
=
x̂ + x̂p̂ − p̂x̂ +
2
2
2
2
2
2m
2mω
2mω
=
mω 2 2
p̂2
iω
mω 2 2
p̂2
iω
x̂ +
− (p̂x̂ − x̂p̂) =
x̂ +
− (−i!)
2
2m
2
2
2m
2
1
= Ĥ − !ω
2
(1.35)
Now, we introduce the operators  and † such that
"
mω
p̂
 =
x̂ + i √
(1.36)
2!
2mω!
"
mω
p̂
†
 =
x̂ + i √
(1.37)
2!
2mω!
$
where the factor of !1 is introduced in order to make the operators dimensionless.
Note that since x̂ and p̂ are both Hermitian, † is the Hermitian conjugate of
Â.
Since
x̂
%
& commutes with itself, and so does p̂, in order to obtain the relation
A, A† we only need to consider
"
* )
*
'
( )" mω
p̂
p̂
mω
i
i
†
Â, Â =
x̂, −i √
+ i√
,
x̂ = − [x̂, p̂]+ [p̂, x̂] = 1
2!
2!
2!
2!
2mω!
2mω!
(1.38)
Equation 1.35 implies that
!
#
1
†
Ĥ = !ω Â Â +
(1.39)
2
We also have
'
(
'
(
'
(
Ĥ,  = !ω † Â,  = !ω †  − †  = −!ω Â
'
(
'
(
'
(
Ĥ, † = !ω † Â, † = !ω † † − †  = !ω †
(1.40)
(1.41)
The eigenvalue equation for the Hamiltonian is
Ĥ|E〉 = E|E〉
(1.42)
Now, let us consider the state Â|E〉.
Using the commutation relation 1.40
Ĥ Â|E〉 = ÂĤ|E〉 − !ω Â|E〉 = (E − !ω)Â|E〉
(1.43)
Therefore, it can be said that Â|E〉 is also an eigenstate of Ĥ but with the energy
lowered by !ω. Each time we apply  again, the energy will be lowered by !ω
but there is a point where this process must terminate since the expectation
8
value of Ĥ is positive. The state of lowest energy is called the ground state
and is denoted by |0〉 .
So for the ground state
Â|0〉=0
(1.44)
where the energy cannot be lowered any more. Using equation 1.39 the energy
of the ground state is
!
"
1
1
†
Ĥ|0〉 = !ω Â Â +
|0〉 = !ω|0〉
(1.45)
2
2
Let us now consider the energy for the state † |0〉,
!
"
#
$
1
Ĥ † |0〉 = † Ĥ + !ω † |0〉 = !ω
+ 1 † |0〉
2
(1.46)
This time the energy is !ω more than that of the ground state. When † is
applied again, the energy will increase by a further !ω, ie two units of energy
above the ground state, and so on.
In this context, † and  are known as the ladder operators. The figure
below [1] indicates the raising and lowering operators where ε = !ω.
Hence it can be concluded that
!
"
1
E = n+
!ω
2
(1.47)
which is exactly the same as the enegy eigenvalue (equation 1.25 ) derived in
the previous section using the differential equations method. So we have derived
the energy levels for the SHO using two different methods.
2
2.1
The Hydrogen atom
Differential equations method
We start from the 3-D Schrödinger equation
! 2
"
p̂
Ĥψ(r) =
+ V (r) ψ (r) = Eψ (r)
2µ
9
(2.1)
!
"
∂
∂
∂
where p̂2 = p̂2x + p̂2y + p̂2z and p̂ = (p̂x , p̂y , p̂z ) = −i! ∂x
, −i! ∂y
, −i! ∂z
and
µ is the reduced mass1 and the Schrödinger equation is written as a partial
differential equation
−
!2 2
∇ ψ (x, y, z) + V (x, y, z, ) ψ (x, y, z) = Eψ (x, y, z)
2µ
2
2
(2.2)
2
∂
∂
∂
With ∇2 = ∂x
2 + ∂y 2 + ∂z 2
In spherical polar coordinates[3] this becomes2
# 2
$
∂2
2 ∂
1
∂
∂
1 ∂2
∇2 = 2 +
+ 2
+
cot
θ
+
∂r
r ∂r r
∂θ2
∂θ sin2 θ ∂ϕ2
(2.3)
which can be written in the form3
∂2
2 ∂
L2
+
−
(2.4)
∂r2
r ∂r !2 r2
where L is the orbital angular momentum.
We will only consider the case of a central potential so that the potential
V = V (r) only depends on the distance from the origin. So the Schrödinger
equation becomes
#
$
!2 ∂ 2
2 ∂
L2
−
+
ψ (r) +
ψ (r) + V (r) ψ (r) = Eψ (r)
(2.5)
2
2µ ∂r
r ∂r
2µr2
∇2 =
The wavefunction ψ (r) can be split into radial and angular components
ψ (r) = Rnl (r)Ylm (θ, ϕ)
which is then substituted back into equation 2.5 to get
#
$
!2 ∂ 2
2 ∂
l (l + 1)
−
+
−
Rnl (r) + V (r) Rnl (r) = Enl Rnl (r)
2µ ∂r2
r ∂r
r2
(2.6)
(2.7)
Note: we have used the fact that Ylm (θ, ϕ) are eigenfunctions of L2 with eigenvalue !2 l (l + 1). 4
For the case of Hydrogen atom we have the attractive Coulomb potential in SI
units
V (r) = −
Ze2
4πε0 r
(2.8)
so the Schrödinger equation 2.7 becomes
1 For
M
a two-body system we need to consider the reduced mass µ = mme+M
where me is
e
the mass of the electron and M is the mass of the proton.
2 These are all fully derived in “Mathematics for Physics 4:
Fields” lecture
notes. Alternatively refer to Suppliment 7-B of the “Quantum Physics” (Ref [1]) on
www.wiley.com/college/gasiorowicz
3 Refere to equation (7B-9) of the Suppliment.
4 Refer to Chapter 7 of “Quantum Physics” by S.Gasiorowicz.
10
!
d2
2 d
2µ
+
+ 2
dr2
r dr
!
"
#$
Ze2
!2 l (l + 1)
E+
−
R (r) = 0
4πε0 r
2µr2
(2.9)
We only consider the solutions with E < 0 , ie the bound states. To make the
above equation simpler, let us define the dimensionless parameters ρ and λ such
that
%
8µ | E |
ρ=
r
(2.10)
!2
and
&
%
Ze2
µ
µc2
λ=
= Zα
(2.11)
4πε0 ! 2 | E |
2|E|
with α =
e2
4πε0 !c
= 1/137. Now, equation 2.9 takes the form
"
#
d2 R (ρ) 2 dR (ρ) l (l + 1)
λ 1
+
−
R
(ρ)
+
−
R (ρ) = 0
dρ2
ρ dρ
ρ2
ρ 4
(2.12)
In order to solve this we will first consider the behavior for large ρ , ie
d2 R 1
− R=0
dρ2
4
(2.13)
the acceptable solution that behaves properly at infinity, ie goes to zero, is
R ≈ e−ρ/2
(2.14)
R (ρ) = e−ρ/2 G (ρ)
(2.15)
therefore, for all ρ , we make an ansatz for the general solution to be of the form
for some function G (ρ) . Substituting back into equation 2.12 we get
"
#
!
$
d2 G
2 dG
λ − 1 l (l + 1)
− 1−
+
−
G=0
dρ2
ρ dρ
ρ
ρ2
(2.16)
In order to be able so solve this, first consider very small ρ , so the equation
becomes
d2 G 2 dG l (l + 1) ∼
+
−
G=0
dρ2
ρ dρ
ρ2
(2.17)
By observing the equation one can conclude that G (ρ) ∝ ρl . So the general
solution will be of the form
G (ρ) = ρl H (ρ)
therfore
d2 H
+
dρ2
"
2l + 2
−1
ρ
#
dH
λ−l−1
+
H=0
dρ
ρ
(2.18)
(2.19)
which is implied by substituting equation 2.18 into 2.16 .
In order to solve this differential equation, we use the same method used in part
1 of section1, ie the power series expansion
11
H (ρ) =
∞
!
ak ρk
(2.20)
k=0
which can be substituted into 2.19 and rearranged to yield
"
#
$
%
∞
!
2l + 2
ak k (k − 1) ρk−2 + k
− 1 ρk−1 + (λ − l − 1) ρk−1 = 0
ρ
k=0
⇒
∞
!
k=0
ρk−1 [(k + 1) (k + 2l + 2) ak+1 + (λ − l − 1) ak ] = 0
(2.21)
So the coefficient of ρk−1 must be zero which implies the recursion relation below
ak+1
k+l+1−λ
=
ak
(k + 2l + 2) (k + 1)
(2.22)
Again, the general solution of H (ρ)= (solution for k < N ) + (solution for
k > N ) where N is a large number.
For small k, ie k < N , we have a polynomial in ρ . However for large k > N
the above relation can be written as
ak+1
k
1
→ 2 =
(2.23)
ak
k
k
It can easily be shown that for very large k
1
ak+1 =
(2.24)
k!
Hence H (ρ) = ( a polynomial in ρ ) + eρ which is not acceptable since R (ρ)
will have the term ρl e−ρ/2 × eρ which means that R (ρ) will grow as eρ/2 which
does not vanish at infinity. So the recursion relation 2.22 must terminate, ie
k+l+1−λ=0
It can be said that for a given l there exists an integer k , denoted by nr , such
that
(2.25)
λ = nr + l + 1
In fact, the principal quantum number n is defined as
(2.26)
n = nr + l + 1
From λ = n and equation 2.11,
2
1
(Zα)
E = − µc2
(2.27)
2
n2
Also, since nr ≥ 0 , one can conclude that n is an integer and n ≥ l + 1.
The energy can also be written in the following form using 2.11 and substituing
for α and λ,
#
$2
Z
µe4
E=−
× 2 2
(2.28)
4πε0
2n !
For the case of Hydrogen, Z = 1.
12
2.2
Operator Method
Finally, we use operator methods to derive the Bohr levels of the hydrogen
atom. This is an extension and generalisation of the techniques used in part 2
of section 1.
The Hamiltonian can be written as
Ĥ =
p2r
L2
e2
+
−
2µ 2µr2
r
where
pr = −i!
!
∂
1
+
∂r
r
"
(2.29)
(2.30)
and the potential is in cgs units.
It can be checked that this is the same as the Hamiltonian used in section 2.1 (
compare with equations 2.4 and 2.5 ) , in fact
#
$
!
"#
$
∂ψ ψ
∂
1
∂ψ ψ
2
2
pr [ψ (x)] = pr [pr (ψ)] = −i!pr
+
= (−i!)
+
+
∂r
r
∂r
r
∂r
r
! 2
! "
"
! 2
"
∂ψ
∂ ψ
1 ∂ψ
ψ
∂ψ
2 ∂ψ
= −!2
+
+
+ 2 = −!2
+
(2.31)
2
2
∂r
∂r r
r ∂r
r
∂r
r ∂r
2
and the potential V = − er . The only difference is that the potential in the
Ze2
previouse section was taken to be V = − 4πε
which is different by a factor of
0r
1
4πε0 (for hydrogen Z=1). This is merely due to use of different units.
As it can be seen, the Hamiltonian is only written in a different way.
The operators, Ĥ , L2 and Lz ( the component of angular momentum in the
z-direciton ) satisfy the eigenvalue equations
H|E, #, m〉 = E|E, #, m〉
(2.32)
L2 |E, #, m〉 = !2 # (# + 1) |E, #, m〉
(2.33)
Lz |E, #, m〉 = !m|E, #, m〉
(2.34)
H|E, #, m〉 = H# |E, #, m〉 = E|E, #, m〉
(2.35)
where # takes non-negative integer values and m = −#, −# + 1, . . . , 0, 1, . . . , #.
The states |E, #, m〉 are in the domain of definition of operators Ĥ, L2 and Lz .
It can be observed that these operators commute.
Equations 2.32 and 2.33 imply that
where
H# =
p2r
!2 # (# + 1) e2
+
−
2µ
2µr2
r
Now, we define the operator D# such that
13
(2.36)
D! = pr − i
!
(! + 1) !
µe2
−
r
(! + 1) !
"
(2.37)
Then we define the adjoint of D! , D!† , extending equations 1.30 and 1.31 to 3-D
space. In other words, we integrate over d3 x by considering arbitrary functions
f (x), g (x) such that rf (x) and rg (x) vanish for both r → ∞ and r → 0+ . So
the aim is to find D!† such that
〈g|D! |f 〉 = 〈D!† g|f 〉
The right hand side is
ˆ
(2.38)
g ∗ D! f dx3
For simplicity, we assume that f and g are functions of r only, ie, they do not
depend on θ and φ. It is left to the reader to prove the results for the general
case, f (x) and g (x).
Using polar coordinates
ˆ π ˆ 2π ˆ ∞
=
g ∗ (r) D! f (r) r2 sin θ dr dθ dφ
= 4π
ˆ
0
∞
0
0
#
0
g ∗ pr − i
!
(! + 1) !
µe2
−
r
(! + 1) !
this can be broken into two steps:
"$
f (r) r2 dr
1. We find p†r ,
!
"$
#ˆ ∞
$
ˆ ∞ #
ˆ ∞
∂
1
∂f
〈g|pr |f 〉 = 4π
g ∗ −i!
+
f r2 dr = 4π (−i!)
g ∗ r2 dr +
g ∗ f r dr
∂r
r
∂r
0
0
0
then we integrate by parts to get
%
(
ˆ ∞
ˆ ∞
ˆ ∞
&
'∞
∂g 2
= (−i!) 4π g ∗ r2 f 0 −
r f dr − 2
rg ∗ f +
g ∗ f r dr
∂r
0
0
0
the first term goes is zero by applying boundary conditions, so
#ˆ ∞ ∗
$ ˆ π ˆ 2π ˆ ∞ !
"
ˆ ∞
∂g 2
∂g ∗
g∗
∗
= (i!) 4π
r f dr +
g f r dr =
i!
+ i!
f r2 sin θ dr dθ dφ
∂r
∂r
r
0
0
0
0
0
!
" $∗
ˆ π ˆ 2π ˆ ∞ #
∂
1
=
−i!
+
g f r2 sin θ dr dθ dφ = 〈pr g|f 〉
∂r
r
0
0
0
Hence, pr is hermitian
〈g|pr |f〉=〈pr g|f 〉 =⇒ pr = p†r
2. Using the same method it can be shown that
1
r
is also hermitian.
Therefore, (1) and (2) imply that !
"
(! + 1) !
µe2
D† = pr + i
−
r
(! + 1) !
Before we proceed, it would be useful to prove the following relations:
14
(2.39)
(2.40)
• p2r = −!2
!
∂2
∂r 2
+
2 ∂
r ∂r
• [pr , r] = −i!
"
(already shown above, see equation 2.31 )
Proof:
Take some function f (r) , then
!
"
#$ ∂
%
&$ ∂
% '(
∂f
[pr , r] (f ) = −i! ∂r
+ 1r [r f ] − r ∂r
+ 1r f = −i! r ∂f
+
f
+
f
−
r
−
f
=
∂r
∂r
−i! (f )
&
'
• pr , 1r =
i!
r2
Proof:
)$
&
'
%* +
∂
pr , 1r (f ) = −i! ∂r
+ 1r fr −
i!
r2
1
r
(f )
&
• pr ,
1
r2
'
=
+,
∂f
∂r
+
f
r
*
∂f
∂r
+
= −i!
*
1 ∂f
r ∂r
−
f
r2
+
f
r2
−
1 ∂f
r ∂r
−
f
r2
+
=
2i!
r3
Proof:
)$
&
'
%* +
∂
pr , r12 (f ) = −i! ∂r
+ 1r rf2 −
2i!
r3
(f )
&
'
• p2r , 1r =
*
2!
r2
$
ipr −
!
r
1
r2
f
r
+,
= −i!
*
1 ∂f
r 2 ∂r
2f
r3
+
(f ) =
2!
r2
−
f
r3
−
1 ∂f
r 2 ∂r
+
f
r3
%
)! 2
"* +
* 2
+,
&Proof:
'
f
∂
2 ∂
1 ∂ f
2 ∂f
p2r , 1r (f ) = −!2
∂r 2 + r ∂r
r − r ∂r 2 + r ∂r
! 2
"
2
2f
2f
2 ∂f
1∂ f
2 ∂f
−!2 1r ∂∂rf2 − r22 ∂f
+
+
−
−
−
=
3
2
3
2
2
∂r
r
r ∂r
r
r ∂r
r ∂r
2!2 ∂
r 2 ∂r
Using what we have, the following four relations can be established
H"+1 D"† = D"† H"
$
ipr −
!
r
%
(f )
(2.41)
Proof: *
.
.
.
+
!2 ! (! + 1)
1
1
i (! + 1) ! 1 2
†
†
†
2
D" H" = D" , H" +H" D" =
pr , 2 −e pr ,
+
, p +H" D"†
2µ
r
r
2µ
r r
.
- .
/
0
!2 ! (! + 1) 2i!
i (! + 1) ! 2! !
2 i!
=
−e
+
× 2
− ipr + H" D"†
2µ
r3
r2
2µ
r
r
/
0 .
!2 (! + 1) i!!
iµe2
i!
2!2 (! + 1)
= H" D"† +
−
+
+
p
=
H
+
D"† = H"+1 D"†
r
"
µr2
r
(! + 1) !
r
2µr2
H" D" = D" H"+1
Proof: Using the above results, we have
15
(2.42)
+
=
!
D!† H!
"†
!
"†
†
= H!+1 D!†
=⇒ H!† D! = D! H!+1
but the Hamiltonian is hermitian, ie H = H † , so
H! D! = D! H!+1
#
D! D!†
= 2µ H! +
µe4
2
2 (! + 1) !2
$
(2.43)
Proof: Starting from the definition of D! and D!† and applying them on an
arbitrary function f (r) we have
!
"
D! D!† [f (r)]
%
&
'
&
'( )
&
'
&
'*
∂
1
(! + 1) !
µe2
∂f
f
(! + 1) !
µe2
= −i!
+
−i
−
−i!
+
+i
f−
f
∂r
r
r
(! + 1) !
∂r
r
r
(! + 1) !
& 2
'
&
'
∂ f
2 ∂f
(! + 1) ! ∂f
µe2 ∂f
µe2 f
= −!2
+
+
!
−
−
∂r2
r ∂r
r
∂r
(! + 1) ! ∂r
(! + 1) !r
&
'
&
'
&
'2
(! + 1) !
µe2
∂f
f
(! + 1) !
µe2
−!
−
+
+
−
[f ]
r
(! + 1) !
∂r
r
r
(! + 1) !
&
'2
!2 (! + 1)
(! + 1) !
µe2
†
2
=⇒ D! D! = pr −
+
−
r2
r
(! + 1) !
#
$
#
$
2
p2r
(! + 1) !2
!2 (! + 1) e2
µe4
µe4
= 2µ
+
−
−
+
= 2µ H! +
2
2
2µ
2µr2
2µr2
r
2 (! + 1) !2
2 (! + 1) !2
D!† D!
Proof: Using 2.43
#
D! D!†
(D! ) = 2µ H! +
#
= 2µ H!+1 +
µe4
2
2 (! + 1) !2
But 2.42 implies that
⇒
D! D!† D!
!
$
#
µe4
+
(D! ) = 2µ H! D! +
#
µe4
2
2 (! + 1) !2
⇒ D! D!† D! = D! 2µ H!+1 +
⇒
!
D!† D!
"
#
(2.44)
#
= 2µ D! H!+1 +
"
2
2 (! + 1) !2
$
= 2µ H!+1 +
16
µe4
D!
µe4
2
2 (! + 1) !2
$
µe4
2
2 (! + 1) !2
2
2 (! + 1) !2
$
$,
D!
$
Therefore, we have
〈E, !, m|D! D!† |E, !, m〉
!
= 2µ E +
µe4
2
2 (! + 1) !2
"
!
〈E, !, m|E, !, m〉 = 2µ E +
〈E, !, m|D! D!† |E, !, m〉 =! D!† |E, !, m〉 !2 ≥ 0
(2.46)
and the length of the vector is always positive.
In other words
µe4
(2.47)
2
2 (! + 1) !2
The minimum possible value for the energy is when! = 0, ie
µe4
(2.48)
2!2
for ! larger than this, the positivity is violated.
However, we will show that this is also an eigenvalue of the Hamiltonian Ĥ.
Using equations 2.35 and 2.41 , we have in general,
#
$
#
$
H!+1 D!† |E, !, m〉 = D!† (H! |E, !, m〉) = E D!† |E, !, m〉
(2.49)
E=−
which implies that D!† |E, !, m〉 is an eigenstate of H!+1 with the same enegry
E, so
D!† |E, !, m〉 = c†! |E, ! + 1, m〉
we can write
†
%
!
c = i 2µ E +
µe4
2
2 (! + 1) !2
"&1/2
(2.50)
(2.51)
where the factor i is chosen for convenience.
†
It can be said that D!+1
D!† |E, !, m〉 is an eigenstate of H!+2 and so on. However,
due to the positivity of D! D!† , the sequence must terminate for some ! = !max
µe4
for E ≤ E < 0. Hence, for E + 2(!+1)
2 2 to be greater than zero, there exists
!
an !max for an eigenvalue E < 0 such that D!†max |E, !max , m〉 is the zero vector.
As observed above, for the ground state energy, !max = 0 .
We can write !max = n − 1 where n = 1, 2, 3, ... since !max it is a non negative
integer. Here n = 1 corresponds to the ground state level.
Finally, since c†!max = c†n−1 must vanish, we conclude that
µe4
2!2 n2
NB This implies that the energy is independent of !.
En = −
17
2
2 (! + 1) !2
(2.45)
which is always positive since
E≥−
µe4
(2.52)
"
This is our final answer for the energy eigenvalue of Hydrogen atom using operator methods.
Referring to equations 2.43 and 2.46, the ground state energy level we must
satisfy
D!† |E, 0, m〉 = 0
(2.53)
Therefore, the ground state energy is
E=−
µe4
2!2
(2.54)
which means that the minimum possible value of energy (equation 2.48) is in
fact an eigenvalue of the Hamiltonian Ĥ.
Comparing the results with those obtained in Section 2.1 , equation 2.28 , it is
1
clear that they are the same apart from the factor 4πε
which, as mentioned
0
before, is due to use of different units for the potential.
18
References
[1] Gasiorowicz, Stephen. Quantum Physics. New Jersey: Wiley &
Sons, 2003
[2] Manoukian, Edward. B. Quantum Theory: A Wide Spectrum.
Dordrecht, The Netherlands: Springer, 2006.
[3] Pendleton, Brian. Mathematics for Physics 4: Fields lecture notes,
Edinburgh University
[4] www.wiley.com/college/gasiorowicz
[5] Griffiths, David. j. Introduction to Quantum Mechanics. Upper
Saddle River, N.J. : Pearson/Prentice Hall, 2005.
19