The Quantum Harmonic Oscillator

Rachel Dudik

May 5, 2004

Harmonic motion is one of the most important examples of motion in all of physics. Any vibration with a restoring force equal to Hooke’s law is generally caused by a simple harmonic oscillator. The potential for the harmonic ocillator is the natural solution every potential with small oscillations at the minimum. Almost all potentials in nature have small oscillations at the minimum, including many systems studied in quantum mechanics. Here, harmonic motion plays a fundamental role as a stepping stone in more rigorous applications.

The Harmonic Oscillator is characterized by the its Schrödinger Equation. This equation is presented in section 1.1 of this manual. The harmonic oscillator has only discrete energy states as is true of the one-dimensional particle in a box problem. The equation for these states is derived in section 1.2. An exact solution to the harmonic oscillator problem is not only possible, but also relatively easy to compute given the proper tools. It is one of the first applications of quantum mechanics taught at an introductory quantum level. Systems with nearly unsolvable equations are often broken down into smaller systems. The solution to this simple system can then be used on them. A firm understanding of the principles governing the harmonic oscillator is prerequisite to any substantial study of quantum mechanics.


1 Solution of the Schrodinger Equation
 1.1 The Schrodinger Equation for the Harmonic Oscillator
 1.2 The Power Series Method
2 Math Moves and Helpful Hints
3 Solved Harmonic Oscillator Problems

1 Solution of the Schrodinger Equation

1.1 The Schrodinger Equation for the Harmonic Oscillator

The classical potential for a harmonic oscillator is derivable from Hooke’s law. It is conventionally written:

       1  2
V (x) = 2kx

Where w is the natural frequency, k is the spring constant, and m is the mass of the body.

     V~ --
w =   m-

For convenience in this calculation, the potential for the harmonic oscillator is written

       1   2 2
V(x) = 2mw  x

Placing this potential in the one dimensional, time-independent Schrödinger equation, it yeilds:

  -h2d2Y-   1   2 2
- 2m  dx2 + 2mw  xY = EY

which equals:

 2    (         2 2  )
d-Y2-+  2mE2--- m--w2-x2  Y = 0
dx      h       h

The Equation for the Quantum Harmonic Oscillator is a second order differential equation that can be solved using a power series. In following section, 2.2, the power series method is used to derive the wave function and the eigenenergies for the quantum harmonic oscillator.

1.2 The Power Series Method

The first step in the power series method is to perform a change of variables by introducing the dimensionless variable, y1:

    V~  mw
y =  -h- x

Substituting this new variable into Equation (5) above yeilds:

dY(y2)-+ (2E-- y2)Y(y) = 0
 dy      hw

For very large values of y, the term 2E
hw is negligible in comparison to the y2 term. With this fact we can guess that Y(y) will be as e 2
y2. The general solution to the differential equation is:

Y(y) = u(y)e-y22

We must now calculate the derivatives of Y that will be substitued into the Schrödinger equation.

                2             2
dY(y) = du(y)e-y2- + (- y)u(y)e-y2-
  dy      dy


d2Y(y) = d2u(y)e -y22-+(- y)du(y)e-2y2 +(- y)du(y)e-y22+ u(y)(-1+ y2)e-y22-
 dy2      dy2             dy             dy

Putting the values from (8) and (10) in equation (7):

 2       2               2                 2                 2
d-u(y)e-y2-+ (-2y)du(y)e-y2-+ u(y)(- 1+ y2)e-y2-+ (2E-- y2)u(y)e-y2- = 0
 dy2              dy                           hw

Canceling terms:

d2u(y)--y22      du(y) -y22       -y22  2E-     -y22-
 dy2 e    -(2y) dy  e   - u(y)e   + hw u(y)e   = 0

And dividing through by e-y22, we obtain:

d2u(y)      du(y)-   2E-
 dy2  - (2y) dy  + (hw - 1)u(y) = 0

The next step is to solve the second order differential equation (13) above for u(y) so that we can find an exact solution for Y(y). We begin by using a power series of y as the general solution to equation (13). u(y) then takes the form below:

       sum  oo 
u(y) =   anyn

In order to substitute this solution into equation (13) we have to solve the first and second derivatives of u(y):

         oo  sum 
du(y) =   (n)anyn-1
  dy     0


d2u(y)   sum              n-2
 dy2  =  0 (n- 1)(n)any

Replacing the u(y) values in equation (13) we have:

 oo  sum             n-2    oo  sum           n-1   2E-     sum  oo   n
  (n - 1)(n)any    -   (2y)(n)any    + (hw - 1)   any  = 0
 0                   0                        0

Simplifying the second term:

 sum  oo            n-2   2E-          oo  sum     n
   (n- 1)(n)any   + (hw - 1- 2n)    any = 0
 0                                0

Replacing n with (n+2) in the first term. See section 4, Math Moves and Helpful Hints, for a discussion of this substitution.

 oo  sum                                  2E           sum  oo 
  ((n +2) -1)(n+ 2)a(n+2)y(n+2)- 2 + (hw-- 1- 2n))  anyn = 0
 0                                              0

and simplifying we have:

 sum  oo                      2E-            2n
   [(n + 2)(n +1))a(n+2) + (hw - 1- 2n)an)]y  = 0

we can then solve for an+2, because the coefficient for each power of y must equal zero (RHS of equation (20)):

        2n+ 1 - 2E
an+2 = ---------hw-an
       (n+ 2)(n+ 1)

Equation (21) is a series representation of all the expansion coefficients in terms of a0 for the power series solution to equation (13). For large values of y, n is also very large. The ratio of an+1 and an (from formula (21) for the coefficients of the power series expansion above) is very close to -2
n. Here we have a problem, because in the limit, 2
n grows faster than the exponential term in Y(y). The series must terminate in order for our solution to have any physical meaning. The best way to terminate the series is to equate the numerator in equation (21) with zero. We then have:

2n+ 1-  hw-= 0

We can now solve for energy, E:

E =  hw(2n+ 1) = (n + 1)hw
     2               2

We have found an infinite number of energies for each energy level, n. The formula for the wave function is incomplete, however, because the power series solution is incomplete. The individual eigenstates of the Hamiltonian must be made orthogonal. Hermite polynomials need be incorporated into the final solution in order to do this. What results is a function of the form:

                    [        ]
       (mw-)14 V~ -1-- (mw--)12    -m2hwx2
Y(x) =   ph     2nn!   h    x  e

The following is a table for the first 5 eigenenergies and eigenstates for the harmonic oscillator:2:

n = 0E = hw
2 Y = a0e-y22
n = 1E = 3hw-
 2 Y = a0(2y)e-y2
n = 2E = 5hw2- Y = a0(4y2 - 2)e-y2
n = 3E = 7hw-
 2 Y = a0(8y3 + 12y)e-y22
n = 4E = 9hw-
 2 Y = a0(16y4 - 48y4 + 12)e-y22
n = 5E = 11h2w-Y = a0(32y5 - 160y3 + 120y)e-y2
... ... ...

2 Math Moves and Helpful Hints

The Summation Substitution: Why is replacing n with (n+2) in the power series derivation mathematically legal?

Most textbooks do not expand on the rational for this substitution. The substitution makes perfect sense, however when each term of the summation is expanded and the derivatives for each term are taken. The steps below might help with the logic behind this part of the derivation of the solution to the harmonic oscillator equation: We are given:

u(y) =  sum  a yn
       0  n

Rewriting this summation in terms of its expansion:

u(y) = a0y0 + a1y1 + a2y2 + a3y3...+ anyn

Then taking the first and second derivatives of the expanded terms, we have:

du(y)-          1-1        2-1           n-1
 dy  = [0 + (1a1y   )+ (2a2y   )...+ (nany    )]

d2u(y)                     2- 2                n-2
 dy2  = [0+ 0+ (2- 1)(2)a2y    ...+ (n- 1)(n)any   ]

This expansion shows that the first two terms in the second derivative (equation 27 above) are zero because the coefficients are zero. The summation for the second derivative actually begins with n+2. Hence the substitution in the power series derivation above.

Hermite Polynomials: What are Hermite polynomials?

The Hermite polynomial is defined as the solution to Hermite’s Differential equation. This polynomial is a direct result of solving the quantum harmonic oscillator differential equation. The Hermite’s Differential equation takes the familiar form:

d-y2 - 2xdy-+ 2ny = 0
dx      dx

Where n is a real, non-negative number (n = 0, 1, 2, 3 ...)

Hermite polynomials form a complete orthogonal set on the interval - oo to + oo with respect to the function e-y2 . The orthogonality relationship can be shown as such4:

 integral   oo                    {
     e-x2Hm(x)Hn(x)dx  =      0 V~ -  m /= n
  - oo                      2nn! p   m = n

With the orthogonality condition met, piecewise continuous function such as the solution to the quantum harmonic oscillator can be expressed in terms of the equation for Hermite polynomials:

 sum  oo                f(x)     where f(x) is continuous
   CnHn(x)  =   f(x-)+f(x+)  where f(x) is discontinuous
n=0                  2


        1    integral   oo   2
Cn = 2nn! V~ p-   e-x f(x)Hn(x)dx
             - oo

The Hermite Polynomial is graphed below5:

Figure 1: The Hermite Polynomial.

3 Solved Harmonic Oscillator Problems

1. The Schrdinger equation for the one dimensional harmonic oscillator is reduced to the following equation for the polynomial u(y):

d-u(y)- (2y)du(y)+ (2E-- 1)u(y) = 0
 dy2         dy     hw

a. Determine the recursion relation separately for the even and the odd energy states.

b. Derive an equation for energy for both the even and the odd states from the recursion relations above.

c. Find the first three energy values from each of the equations obtained in (b).


Part a. We begin by considering only the even states. The general solution given in the power series derivation above may be modified so that we include only these states:

       oo  sum     2n
u(y) =    any

Taking the first and second derivative of u(y) we obtain:

du(y)    oo  sum 
----- =    (2n)any2n-1
  dy     0


 2       oo 
du(y)-=  sum  (2n - 1)(2n)any2n-2
 dy2     0

Putting these values for the derivative back into the differential equation:

 sum  oo                     oo  sum                   2E       oo  sum 
   (2n- 1)(2n)any2n- 2-   (2y)(2n)any2n- 1 + (--- 1)   any2n = 0
 0                      0                  hw      0

simplifying the second term we have:

 oo                                   oo 
 sum  (2n -1)(2n)any2n-2 + (2E-- 1- 4n) sum  any2n = 0
 0                     hw          0

Replacing n with (n+1) in the first summation, the equation can be rewritten:

 oo                                                   oo 
 sum  (2(n+ 1)- 1)2(n+ 1)a   y(2(n+1)-2 + (2E-- 1- 4n)) sum  a y2n = 0
 0                    (n+1)           hw           0   n

simplifying this:

 oo  sum  [                      2E-           ]  2n
    2(n + 1)(2n + 1)a(n+1) + (hw - 1 -4n))an y  = 0

Again, since the coefficient must equal zero, (see RHS of equation), we can solve for a(n+1) and obtain the recursion relation for the even states:

         4n+ 1 - 2E
an+1 = ----------hw--an
       2(n+ 1)(2n+ 1)

A similar method may be used to find the recursion relation for the odd states. This time we use a general solution of the form below so as to attain only odd states:

       oo  sum 
u(y) =    any2n+1

After filling this solution into the differential equation and following the same steps as shown for the even states, one obtains the recursion relation:

a    = --4n+-3---2hEw--a
 n+1   2(n+ 1)(2n+ 3) n

Part b. As in the power series derivation above, we know that the numerator must go to zero. We can therefore, solve for the energy and obtain the equation for even states:

                   (      )
E = hw-(4n + 1) = hw 2n + 1
     2                   2

and the odd states:

                   (      )
    hw-                  3
E =  2 (4n + 3) = hw 2n + 2

Part c.

The first three energy values for the even and odd states are listed below.

      Even States    Odd States

n = 0     E = hw
 2      E = 3hw

n = 1     E = 5hw
 2      E = 7hw

n = 2     E = 9hw
 2      E = 11hw-

2. Find the energy levels of a particle moving in a potential field of the shape3

V (x) =  oo ,(x < 0)

      mw2-  2
V(x) =  2 x ,(x > 0).


The first step in solving this equation is to look at the boundary conditions. As x --> 0, the wave function should fall to zero. For x > 0, the wave function satisfies the differential equation for the harmonic oscillator. Since the odd wave functions for the harmonic oscillator tend toward zero as x --> 0, we can conclude that the equation for the odd states in Problem 1 above is the solution to the problem:

                   (      )
E = hw-(4n+ 3) = hw 2n + 3       (n = 1,2,3,4...)
     2                   2

3. A Harmonic Oscillator is in the initial state: Y(x,0) = fn(x), that is, an eigenstate of ^H. What is Y(x,t)?2


From Postulate IV of Quantum Mechanics we can show that:

Y(r,t) = e   Y(r,0)

Therefore, the time-dependent wave function for the problem here is:

Y(x,t) = e-iht^H-Y(x,0) =  fn(x)e-iht^H-

4. Show that in the nth state of the harmonic oscillator:2

  2       2
<x > = <Dx >


  2       2
<p > = <Dp>


    2     2     2
<Dx > = <x >- <x>


<x2>- <Dx>2 = <x>2

It can be shown that because of orthogonality:

<x> = <n|^x|n> = 0


 2       2
<x >- <Dx > = 0


<x2> = <Dx >2

A similar case may be shown for the momentum operator

5. Using the uncertainty principles between x and p derived in Problem 3, derive the zero-point energy:

E0 = 1 hw

for a harmonic oscillator with natural frequency w0.2

Solution At E0, the kinetic energy of the system equals the potential energy of the system:

1<p->= k-<x2>
2 m    2


     V~ --
w =   k-

From the relationship developed in Problem 3,

<x2> = <Dx >2


  2       2
<p > = <Dp>


1<Dp >2   k     2
2--m-- = 2-<Dx >


<Dx> =  V~ km-


E0 = 2  1<Dp>-
         2  m


E0 = 2  -<Dx>2 = k<Dx > <Dx>


E0 =  V~ --<Dx ><Dp >


<Dx ><Dp > = <-h>

With the relationship between k and w in equation (56),

E0 =  -2--

The following list of references proved excellent resources on the topic of the quantum harmonic oscillator.


[1]   Dicke, Robert H. & Wittke, James P., Introduction to Quantum Mechanics, 1960, Addison Wesley, San Francisco, CA.

[2]   Gasiorowicz, Stephen, Quantum Physics, 1974, John Wiley & Sons, New York.

[3]   Liboff, Richard L., Introductory Quantum Mechanics, 2003, Addison Wesley, San Francisco, CA.

[4]   Saxon David S., Elementary Quantum Mechanics, 1968, Holden-Day, San Francisco, CA.

[5]   Schiff, Leonard I., Quantum Mechanics, 1955, McGraw-Hill Book Company, New York.

Books that provide practice problems concerning the harmonic oscillator

[6]   Flügge, S., Practical Quantum Mechanics, 1974, Springer-Verlag, New York.

[7]   Gol’dman, I. I. & Krivchenkov, V. D., Problems in Quantum Mechanics, 1961, Pergamon Press, Addison-Wesley Publishing Co., Reading, MA.

Websites: For information about Hermite Polynomials:



[10] polynomials

For general information about the quantum harmonic oscillator

[11] Harmonic Oscillator Lecture.pdf

[12] harmonic oscillator



Exciting Applets



[17] rubin/nacphy/CPapplets/QMWAVE/Harmos2/Harmos2.html