# How to Quantize Simple Harmonic Oscillator: A Tutorial

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 = p ^{2}/2m + mw^{2}x^{2}/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 = [ p ^{2} + (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 ^{+} = c^{2} ( mwx + ip) ( mwx - ip) = c^{2} { (mwx)^{2} + p^{2 }– imw [ x, p ] }**

or,

**aa ^{+} = c^{2} ( 2mH + mwh^{~} ) (5)**

Similarly,

**a ^{+}a = c^{2} ( 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**and the operator a decreases energy by the same amount. Thus,

^{~}w**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 **n ^{th}** 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 *n ^{th}*

*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.