How to Quantize Simple Harmonic Oscillator: A Tutorial
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How to Quantize Simple Harmonic Oscillator: A Tutorial

Quantum harmonic oscillator has a great importance in physics because of its application in a wide variety of problems. In this article I have discussed how to quantize a harmonic oscillator and find its eigenstates and eigenenergies by the using the so called ladder operator prescription. Then the result is generalized to the case of multidimensional harmonic oscillator.

Understanding the physics of a classical harmonic oscillator is of great importance because it appears in a variety of problems. The classical harmonic oscillators completely obey the Newton’s law of motions and we can determine all relevant physical quantities by applying Newton’s laws. But the nature of quantum harmonic oscillators is quite different from the classical one and nature in fact follows the quantum mechanical description of harmonic oscillators at the microscopic level. In this article we are going to learn how to quantize a simple harmonic oscillator and find out some bizarre results.

 

THE HAMILTONIAN

The Hamiltonian of a particle of mass m and angular frequency w showing harmonic oscillations is given by

H = p2/2m + mw2x2/2                  (1)

Here, the first term represents kinetic energy and the second term represents potential energy. p is the momentum of the particle. Classically p and x are simple variables and commute with each other, i.e. px = xp.

 

QUANTIZATION

In the quantum mechanical treatment the Hamiltonian, momentum and position will become operators and p and x will no longer commute with each other.

[ x , p ] = xp – px = i h~               (2)

Here, i is the square root of (-1) and h~ is the reduced Planck’s constant; h~ = h/2(Pi).

This is the very first step where quantum mechanics differs from the classical description. Now to quantize, we will factorize the Hamiltonian (1).

H = [ p2 + (mwx)2 ] /2m = ( mwx + ip) (mwx – ip) /2m                     (3)

 

THE LADDER OPERATORS

Let us define two operators as the following.

a = c ( mwx + ip)

a+ = c ( mwx - ip)                        (4)

Here c is a (real) constant determined by demanding that the commutation relation between a and a+ is unity. The reason of calling them ladder operators will be clarified little later. Let us first find the constant c. We have,

aa+ = c2 ( mwx + ip) ( mwx - ip)  = c2 { (mwx)2 + p2 – imw [ x, p ] }

or,

aa+  = c2 ( 2mH + mwh~ )                (5)

Similarly,

a+a = c2 ( 2mH - mwh~ )                  (6)

Subtracting (6) from (5) and demanding

[ a , a+ ] = 1                                    (7)

we get c = ( 2mwh~ )-1/2

Thus we obtain the expressions

a =  ( mwx + ip) / ( 2mwh~ )1/2

a+ = ( mwx - ip) / ( 2mwh~ )1/2         (8)

 

COMMUATION RELATIONS WITH THE HAMILTONIAN

Now, adding (5) and (6) we can easily obtain

H = ( aa+ + a+a ) h~w/2  =  h~w ( a+a + ½ ) =  h~w ( aa+ - ½ )               (9)

where the last two expressions are obtained using (7). Then,

[ a , H ] = h~w ( [ a , a+a ] + [ a , ½ ] ) = h~w ( [ a , a+ ] a + a+ [ a , a ] ) + 0  ( any operator commutes with a constant)

or,   [ a , H ] = h~wa           (10)

Taking Hermitian conjugate of (10) we get

[ a+ , H ] = - h~wa+             (11)

 

ACTION OF LADDER OPERATORS

Since the eigenvalue of Hamiltonian is the energy of the eigenstate, we represent our energy state with eigenenergy E by the ket |E >. Then,

H|E > = E |E >                    (12)

Then,

H ( a+|E > ) = ( H a+ )|E >  =  ( a+H + h~wa+ )|E >

  = a+ ( H|E > ) + h~wa+|E >  =  E a+|E > + h~w a+|E >

Therefore, 

H ( a+|E > ) = (E + h~w) a+|E >                      (13)

Similarly,

H ( a|E > ) = (E - h~w) a|E >                          (14)

From (13) and (14) we can see that the operator a+ increases the energy of a state by an amount h~w and the operator a decreases energy by the same amount. Thus, a+ acts as a creation operator and a acts as an annihilation operator. This is why they are called them ladder operators.

Let us now, for the brevity of notations, denote a particular state with energy E by |n >. Then considering h~w a single quanta of energy, we can write,

a+|n > = C+ |n + 1 >

a|n > = C |n - 1 >                              (15)

Here, the constants C and C+ are to be determined by the condition of orthogonality of the newly formed states. Taking Hermitian conjugate of first equation of (15) and multiplying with it we get,

< n|aa+|n >  =  |C+|2 < n + 1|n + 1 >  =  < n| H/h~w + ½ |n > 

or,

|C+|2 = E/h~w + ½

C+ = ( E/h~w + ½ )1/2                         (16)

Similarly,

|C|2 = E/h~w - ½

C = ( E/h~w - ½ )1/2                           (17)

Therefore, now we need to determine value E of a particular state |n >.

 

THE EXISTENCE OF GROUND STATE

We have already found that the annihilation operator decreases the energy of a state by one quantum, but can it do that infinitely? The answer is no, because you can easily see from (17) that for energy E less than h~w/2 the state no longer remains normalizable. A better argument is as follows.

< n|H|n > = h~w < n|a+a|n > + (h~w/2) < n|n > = h~w |C|2 < n – 1|n – 1 > + (1/2) h~w < n|n >

or,

< n|H|n > = h~w (|C|2 + ½ )                        (18)

From (18) we can conclude that the Hamiltonian is positive definite, so that energy of the system is always greater than or equal to zero. Thus, there must be a minimum energy state – the ground state, which we denote by the ket |0 >. Therefore,

a|0 > = 0                              (19)

THE GROUND STATE ENERGY

Let us apply the Hamiltonian on |0 >.

H|0 > = h~w (a+a + ½ )|0 > = (1/2) h~w|0 >                            (20).

H|0 ~w (a+a + ½~w|0 >

The first term vanishes because of result (19). Hence, we see that ground state energy of a harmonic oscillator is E0 = (1/2)h~w, which is in contradiction with classical oscillator, which can have minimum energy of zero.

 

CREATING AN ARBITRARY EXCITED STATE

Now, we can use (15) and (16) repeatedly to obtain

a+|0 > = ( ½ + ½ )1/2|0+1 > = ( 1 )1/2 |1 >

E1 = E0 + h~w = (3/2)h~w                               (21)

( a+ )2|0 > = ( 1 )1/2 a+| 1 > = ( 2 X 1 )1/2 |2 >

E2 = E1 + h~w = E0 + 2h~w = (5/2)h~w            (22)

Thus, energy of the nth excited state is

En = ( n + ½ )h~w                         (23)

Then we have from (16) and (17),

a+|n > =  ( n + 1 )1/2|n + 1 >           (24)

a|n > = ( n )1/2 |n - 1 >                   (25)

Clearly,

( a+ )n|0 > = ( n X n-1 X … 2 X 1 )1/2 |n >

or,

|n > = ( n! )-1/2 ( a+ )n|0 >                (26)

This is the expression for the nth excited state of a quantum harmonic oscillator.

THE NUMBER OPERATOR

Let us try to find the effect of the operator a+a on any state |n >

a+a|n > = ( n )1/2 a+|n – 1 > = ( n )1/2 ( n – 1 + 1 )1/2|n >

a+a|n > = n |n >                                (27)

Thus we see that the eigenvalue is the level of occupied excited state. This is the reason why the operator a+a is called the number operator.

MULTIDIMENSIONAL HARMONIC OSCILLATOR

The above prescription can be easily generalized to the case of multidimensional harmonic oscillator. In the occupation number representation a system of p-dimensional oscillators can be labeled as |n1 , n2 , … , np > and its energy will be given by

E(n1,n2,…,np) = h~w1 ( n1 + ½ ) + h~w2 ( n2 + ½ ) + … + h~wp ( np + ½ )                   (28)

Here w1, w2 etc. are frequencies of the harmonic oscillators.

Thus, we have learnt how to quantize a quantum harmonic oscillator.

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