# Chapter Seven The Quantum Mechanical Simple Harmonic Oscillator

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```					                                Chapter Seven
The Quantum Mechanical Simple Harmonic Oscillator

Introduction
The potential energy function for a classical, simple harmonic oscillator is given
by Z ÐBÑ œ " 5B# where 5 is the spring constant. Such a classical oscillator has an
angular frequency = œ È5Î7 , where 7 is the mass of the oscillator. Writing the
#

potential energy function in terms of the angular frequency, rather than the spring
constant, gives
"
Z ÐBÑ œ     7=# B#                          (7.1)
#
The time-independent Schrodinger equation can then be written in the form
h# ` #          "
<ÐBÑ       7=# B# <ÐBÑ œ I <ÐBÑ                 (7.2)
#7 `B#           #
or, in operator notation,
s#
:            "
<ÐBÑ        7=# s # <ÐBÑ œ I <ÐBÑ
B                                  (7.3)
#7           #
The potential energy function for this problem is different from the ones we have
considered thus far. This potential energy function essentially spans all space. In the
diagram below, you can see that there are regions where the total energy is greater than
the potential (regions where the wave function would be sinusoidal) and regions where
the total energy is less than the potential (regions where proper wave functions will be
exponentially decreasing). Thus, a proper solution to the simple harmonic oscillator
problem will be sinusoidal in the central region, but must be exponentially decreasing as
B tends toward both positive and negative infinity.

Fig. 7.1 The Simple Harmonic Oscillator Potential
Quantum Harmonic Oscillator                                                             2

Analytical Method
We begin with the time-independent Schrodinger equation for the simple harmonic
oscillator, expressed in terms of the angular frequency
h# ` #               "
<ÐBÑ            7=# B# <ÐBÑ œ I <ÐBÑ                (7.4)
#7 `B#                #
In order to simplify this equation somewhat, we will let
B œ α0                                 (7.5)
This introduces an arbitrary constant α which can then be defined is such a way as to
create a dimensionless equation. The Schrodinger equation becomes
h# " ` #                "
< Ð0 Ñ          7=# α# 0# <Ð0Ñ œ I <Ð0Ñ             (7.6)
#7 α# ` 0#               #
Multiplying by    #7α# Îh # gives
`#              7# =# α% #                     #7α#
< Ð0 Ñ             0 < Ð0 Ñ œ                  I < Ð0 Ñ    (7.7)
` 0#              h#                            h#
Now using our freedom to define α, we choose α to be
α œ ÈhÎ7=                                  (7.8)

so that our equation simplifies to the dimensionless form
`#                                    #
< Ð0 Ñ     0 # < Ð0 Ñ œ            I < Ð0 Ñ         (7.9)
` 0#                                 h=
You can tell this equation is indeed dimensionless, since the energy has units of h=.
Letting O œ #IÎh= our equation reduces to

< Ð 0 Ñ œ ˆ0 #      O ‰< Ð 0 Ñ
`#
(7.10)
` 0#
Note that

For large 0 (i.e., large displacements from the equilibrium position where 0 p ∞) we
expect the wave function to have the approximate form
"0#
< Ð0 Ñ ¸ /                               (7.11)
since we know that the wave function must tend toward zero for both positive and
negative values of 0. Let's take this approximate form of the wave function as a trial
Quantum Harmonic Oscillator                                                                                                   3

solution to our differential equation and see how well it satisfies our differential equation.

’/ "0      #"0 “
`#       "0#       `         #
/          œ                                                                      (7.12)
` #0              `0
œ / "0 ˆ%" # 0# ‰ / "0 #"
#              #

œ %" # / "0 Œ0#       
#        "
#"

Notice that if we set " œ " ß this last equation reduces to
#

œ ˆ0#      " ‰/
`#         " #                         " #
/     #0                          #0                                     (7.13)
` #0
which is almost exactly the form of our dimensionless oscillator equation except for the
energy constant O . We expect the exact solution of our differential equation, therefore,
to be of the form
" #
<Ð0Ñ œ 2Ð0Ñ/          #0                                               (7.14)
and hope that pulling out the exponential term will help us find a simple expression for
2Ð0Ñ. (One can always hope!)
Our dimensionless diffential equation for the quantum oscillator is

< Ð 0 Ñ œ ˆ0 #         O ‰< Ð 0 Ñ
`#
#
(7.15)
`0
If we assum that the correct solution for this equation has the form of equation 7.14, we
need the evaluate the second partial of this function. The first partial is given by

Š2Ð0Ñ/            ‹œŒ     2Ð0Ñ/ # 0  2Ð0ÑŠ                                      ‹
`              " #        `         " #                                       " #
#0                                                      0/     #0                   (7.16)
`0                        `0

œŒ         02Ð0Ñ/ # 0
`2Ð0Ñ            " #

`0
Taking the partial again, we obtain

Š                     ‹           Š                      ‹
` # < Ð0 Ñ   ` 2 2Ð0Ñ            " #       `2Ð0Ñ                 " #         `2Ð0Ñ                  " #
œ          /          #0                        0/    #0                            0/   #0        (7.17)
` 0#         ` 0#                         `0                                `0
2Ð0ÑŠ                       0Š               ‹‹
" #                    " #
/    #0                0/   #0

œ                                      2Ð0Ñˆ0#        "‰
` # < Ð0 Ñ         " #       ` 2 2Ð0Ñ             `2Ð0Ñ
œ/      #0                       #0
` 0#                         ` 0#                `0
Our dimensionless differential equation for the quantum oscillator becomes

œ                                      2Ð0Ñˆ0#        "‰ œ ˆ0#          O ‰2Ð0Ñ/
" #     ` 2 2Ð0Ñ            `2Ð0Ñ                                                                " #
/   #0                    #0                                                                         #0       (7.18)
` 0#               `0
Quantum Harmonic Oscillator                                                                                                          4

This leads to a differential equation for the function 2Ð0Ñß since the exponential drops
out, given by
` 2 2Ð0Ñ             `2Ð0Ñ
#0                         O          " 2Ð0Ñ œ !                             (7.19)
` 0#                `0
This equation we now solve using the assumption that the solution can be expressed in
terms of a power series in 0 of the form
∞
#
2Ð0Ñ œ +!                +" 0        +# 0          âœ              +4 0 4                      (7.20)
4œ!

Differentiating the series term by term, onceß gives
∞
`2Ð0Ñ                                              #
œ +"                 #+# 0           \$+\$ 0          âœ             4+4 04         "
(7.21)
`0                                                                4œ!

Differentiating again gives
` 2 2Ð0Ñ
œ " † # +#                           # † \$ +\$ 0           \$ † % +% 0 #           â              (7.22)
` 0#
∞
œ          4        " 4             # +4 # 0 4
4œ!

Substituting this into the differential equation for 2Ð0Ñ gives
∞                                                       ∞                                       ∞
4                            4 "
4       " 4          # +4 # 0            #0             4+4 0             O        "            +4 0 4 œ !   (7.23)
4œ!                                                     4œ!                                     4œ!

Collecting terms of the same power and rearranging finally gives
∞
4        " 4        # +4 # 0 4              #4+4 04          O         " +4 0 4 œ !                (7.24)
4œ!

or

c4                                                                " +4 d0 4 œ !
∞
" 4        # +4          #       #4+4            O                                      (7.25)
4œ!

Now if the differential equation must be valid for all values of the variable 0, then the
only way this last equation can be valid for any arbitrary choice of 0 is for the coefficients
to be indentically zero, or
4     " 4        # +4       #         #4+4          O        " +4 œ !                          (7.26)
Quantum Harmonic Oscillator                                                                                 5

Now this equation relates the coefficient +4                      #   to the coefficient +4 according to the
equation
#4 " O
+4   #   œ            +4                                        (7.27)
4 " 4 #
This recursion formula enables us to write any coefficient if we know just two, +! and
+" . All other coefficents can be determined based upon these two. We therefore write
2Ð0Ñ œ 2even Ð0Ñ                2odd Ð0Ñ                         (7.28)
where
2even Ð0Ñ œ +!             +# 0 #        +% 0 %    â                    (7.29)
is the even function of 0, built upon +! , and where
2odd Ð0Ñ œ +" 0            +\$ 0 \$        +& 0 &    â                    (7.30)
is the odd function of 0, built upon +" . Thus, we have the solution of the differential
equation in terms of two undetermined constants, +! and +" , which is what we would
expect from a second-order differential equation. The final form of the solution, then, is

                            +#4 " 0#4 " Ÿ
∞                  ∞
" #                  " #
< Ð0 Ñ œ /   #0    2Ð0Ñ œ /       #0              +#4 0#4                            (7.31)
4œ!                4œ!

where we use the recursion relation to determine the +'s from +! and +" .
Before we celebrate too much, however, we must determine if this wave function
converges as 0 p ∞; otherwise the solution we have obtained is unacceptable. If there is
a problem with this function as 0 gets larger it would obviously be due to the larger
powers of 0, i.e., larger values of 4. Let's see what the recursion formula gives us as 4
gets large. The recursion formula can be written
#4 " O
+4    #   œ          +4                                               (7.32)
4 " 4 #
4 # "Î4 OÎ4
œ #            +4
4 " "Î4 " #Î4
or, in the limit of large 4
#
lim +4        #   œ     +4                             (7.33)
4Älarge                4
or
+4 #   #
lim        œ                                          (7.34)
4Älarge +4    4

If we let N be the value of 4 where this limit begins to be valid, then we can divide the
summation up into two separate pieces; the region below N where the recursion formula
Quantum Harmonic Oscillator                                                                                                                                 6

must be used, and the region above N where the approximation to the recursion formula
for large 4 can be used. The summation over the even terms, then, becomes
∞                          N                         ∞
2even Ð0Ñ œ                  +#4 0#4 œ                  +#4 0#4                  +#4 0#4                             (7.35)
4œ!                          4œ!                  4œN "

or
∞                           N
#4
+#4 0 œ                   +#4 0#4          +#N    #0
#N #
+#N       %0
#N %
+#N        '0
#N '
â   (7.36)
4œ!                     4œ!

œ"                                                         â
N
+#N       % #           +#N       ' %
œ           +#4 0#4          +#N    #0
#N #
0                       0
4œ!
+#N       #             +#N       #

But for large 4 we have
+#N       %     + #N #          #                  #
œ                       ¸                                                                                                (7.37)
+#N       #      + #N #                  #N #
+#N       '     + #N %          #       + #N % # + #N                          %                   #                       #
œ                       œ         †                                   ¸                         †
+#N       #      + #N #                  + #N %   + #N                         #              #N        %             #N       #

giving

œ"                                                                        â
∞                    N
#4                                                                       0#                             0%
+#4 0 œ                +#4 0#4       +#N     #0
#N #
(7.38)
4œ!               4œ!
N         "         N             # N         "

N xœ                                                                                  â
∞                    N
#4                                                              "                0#                                      0%
+#4 0 œ                +#4 0#4       +#N     #0
#N #

4œ!               4œ!
Nx             N " Nx                       N          # N          " Nx

N xœ                                                                      â
∞                    N
#4                                                              "                   0#                       0%
+#4 0 œ                +#4 0#4       +#N     #0
#N #

4œ!               4œ!
Nx             N         "x             N         #x

N xœ                                                                     â
∞                    N
0#N              0#N #                    0#N %
+#4 0#4 œ              +#4 0#4       +#N     #0
#

4œ!               4œ!
Nx               N "x                     N #x

# 0 N x                                                                           âŸ
∞                    N
0#N              0# N "                   0# N #
+#4 0#4 œ              +#4 0#4       +#N        #

4œ!               4œ!
Nx               N "x                     N #x

# 0 N x                                                                               âŸ
∞                    N                                                     N                   ÐN "Ñ                      ÐN #Ñ
0#                 0#                       0#
+#4 0#4 œ              +#4 0#4       +#N        #

4œ!               4œ!
Nx                 N         "x             N         #x

N xš / 0 ›
∞                    N
#
+#4 0#4 œ              +#4 0#4       +#N     #0
#

4œ!               4œ!
Quantum Harmonic Oscillator                                                                                            7

The even solution to the quantum harmonic oscillator problem, therefore, has the
form

                                        N xš/0 ›Ÿ
N
" #                         " #                                                 #
<even Ð0Ñ œ /   #0    2even Ð0Ñ œ /         #0               +#4 0#4     +#N           #
#0                    (7.39)
4œ!
N
" #                         " #                                " #
<even Ð0Ñ œ /   #0    2even Ð0Ñ œ /         #0          +#4 0#4        G 0# / # 0
4œ!

which clearly diverges. This same type behavior can be demonstrated for <odd Ð0Ñ as
well. So what do we do?
Obviously, our solution is not the infinite series we have derived, because this
infinite series diverges. However, if the series were to terminate for some value of 4,
then the function would remain finite. Let's look again at the recursion formula
#4 " O
+4   #   œ           +4                                                    (7.40)
4 " 4 #
If we demand that the term +4      #   œ ! for 4 œ - (where +4 Á !Ñ, we obtain
#- " O
œ!                                                            (7.41)
- " - #
or
#-         "       Oœ!                                                (7.42)
from which we obtain an equation which fixes the energy of the oscillator
O- œ #IÎh = œ #-                       "                                   (7.43)
or

I- œ Œ-             h =
"
where - œ !ß "ß #ß \$ß á                         (7.44)
#
This gives us a denumerable energy spectrum and implies that the correct solution
for our problem is not an infinite series solution for 2Ð0Ñ, but a polynomial solution for
2Ð0Ñ, expressed as
-
2- Ð0Ñ œ +!              +" 0         +# 0 #       âœ             +4 0 4                   (7.45)
4œ!

with different order polynomials being associated with different energies. Now the
coefficients satisfy the recursion formula
Quantum Harmonic Oscillator                                                              8

#4 " O
+4   #   œ           +4                            (7.46)
4 " 4 #
where O œ #- ", so the recursion relation depends upon - as well as 4. We express
this recursion formula with this substitution to obtain

-             #- 4
+4   #   œ          +-                             (7.47)
4 " 4 # 4

For - œ !, Ðcorresponding to O! œ ", which gives I! œ " h =), and we see that the
#
recursion formular

!             #! 4
+4   #   œ          +!                             (7.48)
4 " 4 # 4
!
gives zero for all even +'s other than +! . But what about the odd values of +? We require
that +" be set equal to zeroà otherwise, we would have an infinite series which diverges.
Likewise, +! is set equal to zero for the odd polynomials. The polynomials, then can be
expressed as:
!
2! Ð0Ñ œ +!                                         (7.49)
"
2" Ð0Ñ œ +" 0
#   #
2# Ð0Ñ œ +! +# 0#
\$     \$
2\$ Ð0Ñ œ +" 0 +\$ 0\$
%   %     %
2% Ð0Ñ œ +! +# 0# +% 0%
ã
or, in terms only of the first coefficient in each of the even or odd sequences (note that
!     #     %
+ ! Á + ! Á + ! Ñ:
!
2! Ð0Ñ œ +!                                          (7.50)
"

2# Ð0Ñ œ +! ˜"    #0# ™
2" Ð0Ñ œ +" 0
#

2\$ Ð0Ñ œ +" œ0      0 
\$       # \$
\$

2% Ð0Ñ œ +! œ"              0 
%               % %
%0#
\$
ã

Remember that multiplying the wavefunction solution to Schrödinger's equation
by some arbitrary constant does not change the differential equation. This is the reason
we are always at liberty to normalize the wave function. The leading coefficient, then,
(the one outside the braces) in each of these polynomials will be determined by
normalization of the wave function. This means that we are at liberty to multiply or
divide the coefficients in each polynomial by some common factor without changing the
Quantum Harmonic Oscillator                                                                         9

equation in any significant way (we just incorporate that arbitrary change into the
arbitrary coefficient). It is customary to choose the constants in the polynomials so that
the coefficient of the highest order term of 0 is # 8 . Doing this removes the fractions in
the polynomial equations. In addition, we can mulitiply by a negative so that the sign of
the highest order term is always positive. With these changes, the polynomials can be
written as
!
2! Ð0Ñ œ +!                                                     (7.51)
"

2# Ð0Ñ œ +! ˜%0#                #™
2" Ð0Ñ œ +" #0
#

2\$ Ð0Ñ œ +" ˜)0\$
\$
"#0™
2% Ð0Ñ œ +! ˜"'0%
%
%)0#          "#™
ã
The terms inside the braces turn out to be the so-called Hermite polynomials which arise
from the Taylor series expansion
∞
D # #D 0                 D8
/              œ                L8 Ð 0 Ñ                     (7.52)
8œ!
8x

and can be generated using the Rodrigues formula

" 8 /0 Œ            /
8
#     .             0#
L8 Ð 0 Ñ œ                                                      (7.53)
.0
_______________________________________________________________________
Problem 7.1
Show that you can obtain the terms in the braces from the Taylor series expansion. Also,
prove that the Rodrigues formula works for the first few Hermite polynomials.
_______________________________________________________________________

So, apart from the normalization constant, the solutions to the harmonic oscillator
problem are just the Hermite polynomials. We next need to determine the normalization
constants. The wave functions associated with the various energy states have the form
" #
<8 Ð0Ñ œ 28 Ð0Ñ/              #0                         (7.54)
The normalization condition for the ground state wave function is

(                          œÊ    (
∞                                              ∞
*                     h                        *
<! ÐBÑ<! ÐBÑ .B                                <! Ð0Ñ<! Ð0Ñ . 0 œ "   (7.55)
∞                         7=                   ∞
Quantum Harmonic Oscillator                                                              10

where we recall that B œ α0, with α œ ÈhÎ7=. Substituting in for the wave function,
we have

Ê   (      ˆ+! ‰# / 0 . 0 œ "
∞
h             !      #
(7.56)
7= ∞

Ê     ˆ+! ‰# #(
∞
h     !               #
/ 0 .0 œ "
7=            !

Ê       ˆ+! ‰# #œ 1 #  œ "
h     !        " "
7=              #
using the Gaussian integral tables. Solving for the coefficient, we have

œ’    “
!      7= " %
+!                                          (7.57)
1h
Normalization of the next wave function gives

Ê (
∞
h
<* Ð0Ñ<# Ð0Ñ . 0 œ "                     (7.58)
7= ∞ #

Ê      (      ˆ+" #0‰# / 0 . 0 œ "
∞
h            "          #

7= ∞

Ê     ˆ+" ‰# #(
∞
h    "                    #
%0 # / 0 . 0 œ "
7=            !

Ê       ˆ+" ‰# )œ 1 #  œ "
h      "        " "
7=               %
or, solving for the coefficient,

’   “ œ       ’   “
" 7= "            7= "
È # 1h     È#" "x 1h
"             %    "         %
+"   œ                                                     (7.59)

Continuing this process, we find that the normalization constant can be written

’   “
7= "
È#8 8x 1h
"      %
E8 œ                                           (7.60)

Thus, the solution to the quantum harmonic oscillator problem can be written

’    “ L8 0 /
7= "
È#
"     %                 0# Î#
<8 Ð0Ñ œ                                                    (7.61)
8 8x   1h

or since B œ α0, with α œ ÈhÎ7=, we write 0 œ È7=Îh B so that the equation, in
terms of B can be written

’    “ L8 ˆÈ7=Îh B‰/
7= "
È#
"     %                                 7=B# Î#h
<8 ÐBÑ œ                                                            (7.62)
8 8x   1h
Quantum Harmonic Oscillator                                                                            11

Note 1: Normalization of the second order wave function gives

Ê      (
∞
h              *
<# Ð0Ñ<# Ð0Ñ . 0 œ "
7=     ∞

Ê    (        +! ˜%0#         #™‘ /
∞
h            #                      #   0#
.0 œ "
7=     ∞

Ê    ˆ+! ‰# #(       ˜"'0% "'0# %™/ 0 . 0 œ "
∞
h    #                                #

7=            !

Ê    ˆ+! ‰# #œ"' † † 1 # "' † † 1 # % † † 1 #  œ "
h   #           \$      "        "   "     "    "

7=               )               %         #

Ê     ˆ+ ‰ #e' % #f œ "
1h # #
7= !

Ê    ˆ+ ‰ ) œ "
1h # #
7= !
or, solving for the coefficient,

’   “           ’   “
" 7= "              7= "
È ) 1h       È## #x 1h
#               %     "         %
+! œ               œ                                                 (7.63)

which follows the form stated above.

Note 2:         Two additional relationships for the Hermite polynomials
which often prove useful in calculations are
L8   "   0 œ # 0 L8 0        #8L8       "   0                     (7.64)
where one polynomial can be related to the sum of two previously
calculated polynomials, and
`L8 0
œ #8L8        "   0                                (7.65)
`0
where the derivative of a polynomial is related to the previous polynomial.

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