# Section 5.8 The Harmonic Oscillator by smapdi60

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```									Section 5.8: The Harmonic Oscillator

The Schrodinger wave equation (SWE) describes the time evolution of the wave
function. The Hamiltonian for the harmonic oscillator describes a particle of mass m in a
quadratic potential. Displacing the mass from equilibrium produces a linear restoring
force. We focus on the 1-D oscillator since a 3-D oscillator can be decomposed into
three 1-D oscillators. Any coupling between the three 1-D oscillators can be included in
the Hamiltonian later if desired.
The harmonic oscillator has important applications. Many systems have
nonlinear potential functions. Expanding these nonlinear potentials in a Taylor series
often produces a quadratic term as the lowest order approximation. As an example, the
periodic motion of atoms about their equilibrium position can be modeled with the
quadratic potential. We know this motion must be related to phonons moving through
the material. We will see that the zero point motion of the atom can be described by the
quantum mechanical vacuum state.
Quantum optics provides a somewhat surprising application for the quadratic
potential. The electromagnetic fields can be modeled by quadratic kinetic and potential
terms. Of course, these do not refer to an electron in an electrostatic potential. Nor do
they refer to the position or momemtum of photons. Instead, they refer to the form
assumed by the fields in the Hamiltonian. The quantized form of the electromagnetic
fields can be immediately written by comparison with the wavefunctions for the electron

Topic 5.8.1: Introduction to the Classical and Quantum Harmonic Oscillators

For a harmonic oscillator, the quadratic potential
produces a linear restoring force
1 2
 kx 1 − D
V = 2
1
 kr 2 3 − D
2
where the equilibrium position occurs at the origin x=0, the
“spring constant” must be positive k>0 and it describes the Figure 5.8.1: The
curvature of the potential (i.e., magnitude of the force).   quadratic potential
The classical Hamiltonian has the form
p2 1 2
Hc =      + kx                          (5.8.1)
2m 2
where we consider the dynamic variables x, p to be independent of one another.
Newton’s second law can be demonstrated using Hamilton’s canonical equation (refer to
Section 5.2).
∂H
p = − c = −kx = F
∂x
The Lagrangian shows that the momentum p must be related to the velocity by
p = mv = mx .

5.41
We want to compare and contrast solutions x(t)
to the classical and quantum harmonic oscillators. The
classical Hamiltonian (the total energy) can be rewritten
using Equation 5.8.1 and p = mx
2
m  d x(t)        1
 + k (x ( t ) ) = E
2
                                (5.8.2)
2  dt            2
where E represents the total energy of the oscillator and
x(t) represents the position of the electron parameterized
by the time t. The solution has the form
x ( t ) = A sin(ωo t )       (5.8.3a)
The formula ωo = k / m relates the angular frequency of
2

oscillation ωο to the “spring constant” k. Substituting
Equation 5.8.3a into Equation 5.8.2 provides
2E       2E
A=            =                   (5.8.3b) Figure 5.8.2: Motion of a harmonic
k      mωo 2
oscillator. The probability density ρ
The amplitude A represents the points on the shows the most likely position of
potential plot V(x) where the kinetic energy becomes finding the mass m is at the turning
zero (see Figure 5.8.2)                                           points     where      the    oscillator
momentumarily comes to rest.
1                         2E
E = kx 2             → A=
2       x =A               k
Classically, the particle can only be found in the region
x ∈ [−A, A] and never outside that region. The
probability density ρ for finding the particle at a point x
appears similar to a delta function near the endpoints of
the motion; this behavior occurs because the particle
slows down near those points and spends more time
there.
Several differences exist between the classical
and quantum mechanical harmonic oscillators. Figure
5.8.3 shows the quantum mechanical solution to
Schrodinger’s equation with the quadratic potential.
Unlike the classical particle, the quantum particle can
be found in the classically forbidden region. The figure
shows how the wavefunction exponentially decays in
these classically forbidden regions. Classically, the
particle doesn’t have enough energy to enter the
forbidden region. The basis functions have the form             Figure 5.8.3: The first two quantum
1
mechanical solutions to the
      α    2                  α x 
2 2
φn ( x ) =  1/ 2    n 
H n ( αx ) Exp  −      (5.8.4)    harmonic        oscillator.      The
 π n! 2                         2               probability density ρ for finding the
particle at point x does not
where α 4 = ( mωo /     )
2
. The exponential part of the     resemble the classical one.
solution ensures the wave function decreases in the

5.42
classically forbidden region. The Hermite polynomials H n primarily control the
behavior in the classically allowed region near the center. They can be conveniently
generated by differentiating an exponential according to
H n (ξ ) = (− 1) exp(ξ 2 ) n exp(− ξ 2 )
n         dn
(5.8.5a)
dξ
where ξ=αx. The first three Hermite polynomials are
H o (ξ ) = 1, H1 (ξ ) = 2ξ, H 2 (ξ ) = 4ξ 2 − 2       (5.8.5b)
Continuing with Figure 5.8.3, perhaps most striking of all, the probability density
function for the quantum particle decays to zero near the endpoints of motion and reaches
its peak value (or values) near the center of the classical region [-A,A]. However, the
classical probability of finding the classical particle assumes its minimum value near the
origin.
Here’s another difference between the classical and harmonic oscillator solutions.
The classical oscillator energy can be increased by applying a driving force and
increasing the oscillation amplitude E = A 2 mωo / 2 . The angular oscillation frequency
2

ωo = k / m remains constant for a fixed spring constant k. The energy of the quantum
oscillator increases by also absorbing energy
    1
E n = ωn = ωo  n +  n = 0,1, 2,...              (5.8.6)
    2
The integer “n” can be interpreted as either the “basis function number” or as the number
of quanta stored in the motion. Contrary to the classical case, the angular frequency ωn
of the quantum oscillator changes even though the value ωo remains fixed. The angular
frequency does not refer to the rate at which the quantum particle bounces from side to
side. We view the quantum particle as a stationary wave function. Larger numbers of
quanta “n” result in larger “displacements” from equilibrium meaning the probability
density has more peaks that move closer to the classically forbidden region.
We find similar plots for quantized EM waves. The energy of an EM oscillator
(the EM waves) can be changed by changing the angular frequency (or wavelength) or by
changing the amplitude (i.e., the number of quanta in the mode). We will see that the
“position x” and “momentum p” become the “in-phase” and “out-of-phase” electric
fields. Therefore, the wavefunctions in the EM case describe the probability of finding a
particular value of the electric field.

Topic 5.8.2: The Hamiltonian for the Quantum Harmonic Oscillator

The quantum mechanical Hamiltonians come from the classical ones by replacing
the dynamical variables x,p with the corresponding operators x , p in          ˆ ˆ
H c = p 2 / 2m + kx 2 / 2 to find

ˆ          ∂                   p2 1 2 
ˆ                    ∂
H Ψ (t) = i    Ψ (t)              + kx  Ψ ( t ) = i
ˆ                  Ψ(t )         (5.8.7)
∂t                  2m 2                ∂t
        

5.43
∂
Operatin with the “coordinate” projection operator x produces x → x and p →
ˆ         ˆ
i ∂x
(refer to Appendix 6) to obtain the Schrodinger equation
ˆ            ∂Ψ (x, t)          − 2 ∂2 1 2                     ∂
H Ψ (x, t) = i              or               + kx  Ψ ( x, t ) = i    Ψ ( x, t ) (5.8.8)
∂t             2m ∂x 2 2                      ∂t
                   
The boundary conditions for the Schrodinger wave equation for the harmonic oscillator
require the wave function to approach zero as "x" goes to infinity
Ψ (x → ±∞, t ) → 0                             (5.8.9)
There are two methods for solving the Schrodinger equation for the harmonic
oscillator. The first method uses a power series solution, which becomes very
algebraically involved. The solution starts by separating variables in Equation 5.8.8 and
using a power series to find the solutions to the Sturm-Liouville problem (the eigenvector
problem). The second method uses the linear algebra of raising and lowering operators.
We present the method of raising and lowering operators (commonly referred to as the
algebraic approach). We will find the stationary solutions given in Equation 5.8.4 and
the energy eigenvalues in Equation 5.8.6.

Topic 5.8.3: Introduction to the Operator Solution of the Harmonic-Oscillator

The operator approach (i.e., algebraic approach) to solving Schrodinger's equation
for the harmonic oscillator is simpler than the power series approach. In addition, it
provides a great deal of insight into the mathematical structure of the quantum theory.
The algebraic approach uses “raising a + and lowering a operators” (i.e., ladder
ˆ                    ˆ
operators, or sometimes called promotion and demotion operators). Later chapters
demonstrate the similarity between the ladder operators and the
“creation/annihilation” operators most commonly found in
We will rewrite the Hamiltonian in terms of the raising
and lowering operators in the form of the number operator
ˆ ˆ ˆ
N = a + a . The raising and lowering operators map one basis
vector into another one according to
a+ n = n +1 n +1
ˆ                              a n = n n −1
ˆ                      (5.8.10)
as suggested by Figure 5.8.4. The lowering operator produces           Figure 5.8.4: Raising
and lowering operators
zero when operating on the vacuum state a 0 = 0 . The number
ˆ                              move the harmonic
operator has two interpretations for the harmonic oscillator.          oscillator from one state
First we will show the energy eigenvectors are also eigenvectors       to another.
ˆ
for the number operator according to N n = n n . The number operator therefore tells
us the number of the eigenstate occupied by a particle.
The number operator also tells us the number of energy quanta in the system as its
second interpretation. We can say that a particle occupying one of the energy basis states
n ∈ BV = { 0 = E 0 , 1 = E1 , …} has n quanta of energy according to

5.44
E n = ωo ( n + 1/ 2 ) . Therefore the vacuum state 0    corresponds to a particle state
without any quanta of energy n=0. Interestingly, there exists energy in the vacuum state
E 0 = ωo / 2 . The vacuum energy corresponds to zero point motion of atoms in a solid,
for example. The atoms continue to move even though all of the extractable energy has
been removed (i.e., n=0). Absolute zero can never be achieved since it is a classical
concept corresponding to stationary atoms. Studies in quantum optics indicate that the
electric field also experiences vacuum fluctuations; these fluctuations produce
spontaneous emission from an ensemble of excited atoms.
In the next few topics, we wish to find the energy eigenvectors
B V = { 0 = E 0 , 1 = E1 ,…} and eigenvalues for the harmonic oscillator. We assume
non-degenerate eigenvalues En, which means that for each energy En, there corresponds
ˆ
exactly one eigenstate φ n satisfying H φ n = E n φ n . We further assume an order for
the energy levels E 0 < E1 < E 2 < . The operator approach must reproduce the results
found with the power series approach.
We first show how the Hamiltonian incorporates the raising-lowering operators.
We briefly discuss the mathematical description of the ladder operators and demonstrate
the origin of their normalization constant. We then easily solve for the energy
eigenvalues and eigenvectors.

Topic 5.8.4: Ladder Operators in the Hamiltonian

The Hamiltonian for the harmonic oscillator is
Hˆ = p + mωo x
ˆ2        ˆ2
(5.8.11)
2m       2
We define the lowering a and the raising a + operators in terms
ˆ                   ˆ
of the position xˆ and momentum operators p .      ˆ
mωo           i pˆ
a=
ˆ               x+
ˆ                    (5.8.12a)
2m ωo         2m ωo
mωo          i p ˆ
a+ =
ˆ                x−
ˆ                  (5.8.12b)
2m ωo        2m ωo
The raising operator in Equation 5.8.12b comes from taking the
adjoint of the lowering operator in Equation 5.8.12a and using
ˆ ˆ
the fact that both x , p must be Hermitian since they correspond
Figure 5.8.5: Physical
to observables. Notice that the raising and lowering operators examples showing the
are not Hermitian a ≠ a + . These two equations for the lowering effect of a raising
ˆ ˆ
and raising operators can be solved for the position and operators defined for an
atom (top) and square
momentum operators to find                                         well (bottom) rather
mωo                     than for the harmonic
x=
ˆ
2mωo
( a + a + ) p = −i 2 ( a − a + ) (5.8.13) oscillator.
ˆ ˆ           ˆ                 ˆ ˆ

We need the Hamiltonian written in terms of the ladder operators. We must first
determine the commutation relations. We can demonstrate that the raising operator

5.45
commutes with itself as does the lowering operator while the raising operator does not
commute with the lowering operator
[a, a ] = 0 = a + , a + 
ˆ ˆ           ˆ ˆ              a, a +  = 1
ˆ ˆ                (5.8.14)
These last two relations can be proven using the commutation relations between
the position and momentum operators
[x, x ] = 0 = [p, p] [x, p] = i
ˆ ˆ           ˆ ˆ   ˆ ˆ                        (5.8.15)
We prove  a, a +  = 1 by first substituting Equations 5.8.12.
ˆ ˆ 
 mωo             ˆ
i p       mωo         i p 
ˆ
 a, a +  = 
 ˆ ˆ                 x+
ˆ            ,          x−
ˆ        
 2m ωo
              2m ωo     2m ωo       2m ωo 

Distributing the terms provides
2
 mωo                     [ p, p] + i p, x − i x, p
ˆ ˆ
 a, a +  = 
 ˆ ˆ                     [ x, x ] +
ˆ ˆ                   [ˆ ˆ]    [ˆ ˆ]
 2m ω                   2m ωo 2            2
          o 

Substituting the commutation relations from Equation 5.8.15, we find the desired results
i           i
 a, a  = 0 + 0 + 2 ( −i ) − 2 ( i ) = 1
ˆ ˆ+ 
In the case of an ensemble of independent harmonic oscillators, each one has its own
ˆ ˆ
degrees of freedom x i , p i that obey their own commutation relations.
ˆ ˆ                ˆ ˆ 
 x i , x j  = 0 =  pi , p j      ˆ ˆ 
 x i , p j  = i δij
As a result, there will be raising and lowering operators for each oscillator
 a i , a j  = 0 =  a i + , a j+ 
ˆ ˆ               
ˆ ˆ

 a i , a j+  = δij

ˆ ˆ

Using the definitions of the position and momentum operators, the Hamiltonian
for the single harmonic oscillator can be rewritten by substituting relations 5.8.27.
2                               2
p2 1
ˆ                1        mωo                                          
H =ˆ
2m 2
+ mωo x 2 =
2
ˆ
2m 
 −i
2
( a − a + ) + 1 mωo2  2mω ( a + a + ) (5.8.16a)
ˆ ˆ                            ˆ ˆ
 2               o        
Squaring the constants provides
ω                   ω
H = − o (a − a+ ) + o (a + a+ )
2                2
ˆ             ˆ ˆ               ˆ ˆ
4                   4
Squaring the operators and taking care not to commute them gives us
ˆ     ω
4
{             ˆ ˆ ˆ
2
H = o −a 2 + aa + + a + a − a + + a 2 + aa + + a + a + a +
ˆ ˆˆ                       ˆ ˆˆ       ˆ ˆ ˆ
2
}
Combining the squared terms
ω
H = o { aa + + a + a}
ˆ             ˆˆ      ˆ ˆ                             (5.8.16b)
2
We must always use commutation relation to change the order of operators. Finally, by
using the commutation relation  a, a +  = 1 → aa + = 1 + a + a , the Hamiltonian becomes
ˆ ˆ             ˆˆ      ˆ ˆ
ω                 ω
H = o {aa + + a + a} = o {2a + a + 1}
ˆ          ˆˆ    ˆ ˆ          ˆ ˆ
2                 2
As a result, the Hamiltonian for the single harmonic oscillator can be written as

5.46
ˆ      ˆ ˆ 1
H = ωo  a + a +                                                     (5.8.17a)
      2
ˆ ˆ ˆ+
We can define the number operator N = a a and rewrite Equation 5.8.17a as
ˆ        ˆ 1
H = ωo  N +                                                         (5.8.17b)
     2

Topic 5.8.5: Properties of the Raising and Lowering Operators

Next, we demonstrate the relations
a+ n = n +1 n +1
ˆ                                          a n = n
ˆ            n −1
by first showing n − 1 ~ a n and n + 1 ~ a + n are eigenvectors of the number
ˆ                    ˆ
ˆ
operator N corresponding to the eigenvalues n-1 and n+1, respectively. We next find the
constants of proportionality. We will need two commutation relations. Using
ˆˆ ˆ       ˆ ˆ ˆ         ˆ ˆ ˆ          ˆ ˆ         ˆ ˆ
 AB, C  = A  B, C  +  A, C  B and  A, B = −  B, A  and Equation 5.8.14, we find
                                                  
ˆ ˆ
 N, a  = a + a, a  = a + , a  a = −a
ˆ ˆ ˆ ˆ ˆ ˆ               ˆ       ˆ ˆ
 N, a +  =  a + a, a +  = a + a, a +  = a + (5.8.18)
ˆ ˆ ˆ  ˆ ˆ ˆ  ˆ
                                                     
ˆ ˆ ˆ
We now show N = a + a and H =         ˆ    ω N + 1/ 2 have eigenvectors n − 1 ~ a n
o
ˆ(       )                                    ˆ
+
and n + 1 ~ a n . Suppose n represents one eigenvector then
ˆ

    
ˆˆ       {
N a n  = Na n =  N, a  + aN n = −a + aN n = {−a + a n} n = (n − 1)  a n 
ˆ ˆ

ˆ ˆ

ˆˆ }          {
ˆ ˆˆ         }ˆ ˆ              ˆ 
Therefore a n must be an eigenvector of N with eigenvalue ( n − 1) . We can similarly
ˆ                                    ˆ

show that N a + n  = ( n + 1) a + n  (see the chapter review exercises). Therefore, we
ˆ ˆ
                ˆ     
+
conclude a n = Cn n + 1 and a n = D n n − 1 since the eigenvalues are not
ˆ                           ˆ
degenerate where Cn and Dn denote constants of proportionality.
ˆ ˆ ˆ            ˆ        ˆ 1
The eigenvalues of N = a + a and H = ωo  N +  must be real because
     2
N = a a is Hermitian according to N = ( a a ) = a a = N . Further the eigenvalues n
+
ˆ ˆ ˆ
+                              ˆ +    +
ˆ ˆ     ˆ ˆ ˆ
+

must be greater than or equal to zero since the length of a vector must always be positive
2
ˆ
n = n N n = n a + a n = a n ≥ 0 . We can also show that only integers represent the
ˆ ˆ       ˆ
eigenvalues n.
Next, we find the normalization constants Cn and Dn occurring in the relations.
a + n = Cn n + 1
ˆ                   a n = Dn n − 1
ˆ
Let’s work with the lowering operator. To find Dn, consider the string of equalities
+                         +
 ˆ  ˆ 
D* D n n − 1 n − 1 =  D n n − 1   Dn n − 1  =  a n   a n 
n                               
ˆ
= n a +a n = n N n = n n n = n n n
Now use the fact that all eigenvectors are normalized to one so that
n −1 n −1 = 1 = n n

5.47
Therefore, the coefficient Dn must be
2
Dn = n           → Dn = n
where a phase factor has been ignored. Similarly, an expression for Cn can be developed
+                         +
C* C n n + 1 n + 1 =  Cn n + 1   Cn n + 1  =  a + n  a + n 
n                                         ˆ         ˆ     
= n a a n = n a a +1 n = n N + 1 n = n n + 1 n = ( n + 1) n n
+             +                  ˆ
where a commutator has been used in the fifth term. Once again using the eigenvector
normalization conditions and comparing both sides of the last equation
2
Cn = n + 1 → Cn = n + 1
as expected. We therefore have the required relations.
a+ n = n +1 n +1
ˆ                             a n = n
ˆ                     n −1   (5.8.19)
The set of eigenvectors
0 , 1 ,...
can be obtained by repeatedly using the relation a + n = n + 1 n + 1 as
ˆ

( a + ) 0 , …, n = ( a + ) 0 ,….
2                       n
a+
ˆ           a+
ˆ         ˆ                  ˆ
1 =     0 , 2 =     1 =                                           (5.8.20)
1           2       2 1                  n!

Some Commutation Relations

H , a  = ωo a + a, a  = ωo a + [ a, a ] + ωo  a + , a  a = − ωo a
ˆ
(1)
            ˆ ˆ ˆ          ˆ ˆ ˆ             ˆ ˆ ˆ              ˆ

(2)      ˆ ˆ
H , a +  = ωo a +  a, a +  = ωo a +
ˆ ˆ ˆ             ˆ
        

(3)      ˆ ˆ       ˆ ˆ ˆ
 N, a  = −a  N, a +  = a +
ˆ
                    

Topic 5.8.6: The Energy Eigenvalues

The Hamiltonian for the harmonic oscillator can be written in terms of the ladder
operators as given in Topic 5.8.4, Equation 5.8.17b.
ˆ         ˆ 1
H = ωo  N +                                     (5.8.21)
    2
ˆ
We already know the eigenvalues of the number operator to be N n = n n . The
allowed energy values can be found as follows
ˆ           ˆ 1         1
H n = ωo  N +  n = ωo  n +  n
     2       2
Therefore the energy values must be

5.48
    1
E n = ωo  n +                                        (5.8.22)
    2

Topic 5.8.7: The Energy Eigenfunctions

We know the energy eigenvectors can be listed in the sequence
( a + ) 0 , …, n = ( a + ) 0 ,….
2                   n
a+
ˆ             a+
ˆ          ˆ                  ˆ
1 =      0 , 2 =       1 =                                    (5.8.23)
1             2        2 1                  n!
from Equation 5.8.20. However, we would like to know the functional form of these
functions. There exists a simple method for finding the energy eigenfunctions for the
harmonic oscillator using the ladder operators. Starting with
0=a 0 ˆ
operate on both sides using the bra operator x and insert the definition for the lowering
operator
mωo xˆ          ˆ
ip              mωo xˆ                   ˆ
ip
0= x a 0 = x
ˆ                    +            0 = x             0 + x                    0 (5.8.24)
2 mωo        2 mωo             2 mωo                 2 mωo
Factor out the constants from the brackets and use the relations
∂              ∂
x x 0 = x x 0 = x φ0 ( x )
ˆ                             x p0 =
ˆ             x 0 =         φ0 ( x )
i ∂x           i ∂x
where x 0 = φ 0 (x ) is the first energy eigenfunction in the set of eigenfunctions given by
{ φ 0 (x ), φ1 (x ),… }
Equation 5.8.24 now provides
mωo x                         ∂
0= xa 0 =                 φ o (x ) +            φ o (x )
2 mωo                 2 mωo ∂x
which is a simple first-order differential equation
dφ 0 mωo
+        xφ o = 0
dx
We can easily find the solution
 mωo 2 
φ 0 (x ) = φ 0 (0) exp −      x 
 2        
which represents the first energy eigenfunction. The normalization constant φ 0 (0) is
found by requiring the wave function to have unit length 1 = φ 0 (x ) φ 0 (x ) which gives
1/ 4
 mωo      mωo 2 
φ 0 (x ) =       exp −    x 
 π        2       
Now the other eigenfunctions can be found from φ1 (x ) using the raising operator
a+        mωo x
ˆ         ˆ
ip
φ1 (x ) = x 1 = x         0 = x        −            0
1        2 mωo     2 mωo

5.49
where the constants can be factored out and the coordinate representation can be
substituted for the operators to get
mωo x                      ∂             mωo x                    ∂
φ1 ( x ) =            x 0 −                x 0 =             φ0 ( x ) −          φ0 ( x )
2 mωo             2 mωo ∂x                2 mωo            2 mωo ∂x
Notice that we do not need to solve a differential equation to find the eigenfunctions
φ1 , φ2 ,... in the basis set. Differentiating φ 0 (x ) provides
1/ 4
∂φ o   ∂  mωo          mωo 2         mωo x
=            exp −    x =−            φo (x)
∂x ∂x  π               2       
Consequently the n=1 energy eigenfunction becomes
mωo x                   mωo x            2 mωo
φ1 (x ) =           φ 0 (x ) +             φ 0 (x ) =        x φ 0 (x )
2 mωo             2 mωo                    2
1/ 4
2 mωo  mωo             mωo x 2 
=               x exp −             

2  π                    2      
The n=2 energy eigenfunctions can be found repeating the procedure using
a+
ˆ
φ2 (x) =      φ1 ( x )
2
Notice that the above procedure only requires the relation between the ladder
operators and the momentum/position operators. At this point, the energy eigenvalues
can be found using the time independent Schrodinger equation.

Special Integrals

The raising and lowering operators can be used to show the following integrals.

∞                  d                      n +1            n        d  i
(1)   ∫       dx φn ( x )      φm ( x ) = α δm,n +1      − δ m,n −1   since    = pˆ
−∞                    dx                        2             2       dx

∞
(2)   ∫       dx φn ( x ) x φm ( x ) =        δ m,n +1 n + 1 + δ m,n −1 n           where Problem 5.24 was
−∞                                 2mωo                             
combined with integral (1) above and with E n +1 − E n = ωo

∞                                      n +1              ( n +1)( n + 2 )
(3)   ∫        dx φn ( x ) x 2 φm ( x ) = δm,n        + δm,n ± 2                      The closure relation can
−∞                                     2α 2                  2α 2
be used to prove this last one.

5.50

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