a linear harmonic oscillator is described by the wave function

FAQs & their solutions for Module 3:

Linear Harmonic Oscillator

Question1: At t 0, a linear harmonic oscillator is described by the wave function

) 3 (

) 2 ( 3 ) 1 0 ,

(x ₀ x i ₂ x (1)

where _n( )x are the normalized eigenfunctions of the Schrodinger equation for the linear harmonic oscillator problem.

(a) Is the wave function normalized???

(b) Write the wave function describing the time evolution of x, 0 .

(c) If we make a measurement of energy, what will be the probabilities of finding it in the ground state (with ¹

E 2  ), first excited state (with ³

E 2  ) and the second excited state (with ⁵

E 2  ),?

(d) Calculate the average energy value.

Solution1:

(a) The eigen functions satisfy the following orthonormality condition

,...

2 , 1 , 0 ,

; )

( )

*(

n m dx

x

x _{n mn}

m

Using the above orthonormality condition, one can easily show that

1 )

0 ,

(x ² dx

Thus the wavefunction is normalized.

(b) Since ¹

n 2

E n  , we will have exp exp ¹

2 iE tn

i n t

 . Thus

the time evolution of the wave function would be given by

2 / 5 2 2

/

0 ( )

3 ) 2

3 ( ) 1 ,

(x t x e ^{i t} i x e ^{i t}

(c) If we make a measurement of energy, there will be

1 rd

3 probability of finding it in the ground state (with ¹

E 2  ), zero probability of finding it in the second excited state (with ³

E 2  ) and

2 rd

3 probability of finding it in the second excited state (with ⁵

E 2  ).

(d) The average energy will be given by





 6 11

2 5 3 2 2

1 3 E 1

Question2: For the linear harmonic oscillator problem, write ( , )x t in the following form

x d x t x x K t

x, ) ( , , ) ( ,0)

( (2)

whereK(x,x,t) is known as the propagator. Derive an expression for K x x t( , , ). Solution 2: The most general solution of the Schrodinger equation for the linear harmonic oscillator problem is given by

0,1,...

( , ) _{n n} exp ⁿ

n

x t c x iE t

h ⁽³⁾

or

1 2 0,1,...

, ( )

i n t

n n

n

x t c x e (4)

If we know x, 0 , we can determine ( , )x t as follows:

0,1,...

, 0 _{n n}

n

x c x

we multiply the above equation by m

*(x) and integrate to obtain

* , 0

n n

c x x dx

where we have used the orthonormality condition. If we substitute the above expression for c_n in

1 2 0,1,...

, ( )

i n t

n n

n

x t c x e

we may write it in the form:

x d x t x x K t

x, ) ( , , ) ( ,0) (

where

* 0,1,...

( , ', ) ( ) ( ) exp 1

n n 2

n

K x x t x x i n t

^{2 2 2}

0,1,...

1 1

exp 2 2

n n n

n

N H H i n t

The above summation can be carried out and the final result is

2 2 2

( , ', ) 1/ 2 exp ( ' ) cos 2 '

(2 sin ) 2 sin

K x x t x x t xx

i t i t

Question3: Show that the function

) 2

(x Ae ^bx

satisfies the Schrodinger equation

0 ) 2 (

1

2 _{2 2}

2 2 2

x x

E dx

d



for a particular value of b; determine the value of b and the corresponding value of the energy eigenvalue.

Solution 3:

2 2 2

2

2 2

2 ^bx 2 ^bx 4 2 ^bx

d d

e bx e b x b e

dx dx

Substituting in the Schrodinger equation

0 ) 2 (

1

2 _{2 2}

2 2 2

x x

E dx

d



we get

2 2 2 2 2

2 1

4 2 0

b x b E 2 x



For the above equation to be valid for all values of x, we must have

 2

b 1 and

2 1

2 E b

 .

Question4: Show that the function e ^cx4 can never be an eigenfunction of the Schrodinger equation.

Solution4:

4 4 4

2

3 2 6 2

2 ^cx 4 ^cx 16 12 ^cx

d d

e cx e c x cx e

dx dx

Substituting in the Schrodinger equation

0 ) 2 (

1

2 _{2 2}

2 2 2

x x

E dx

d



we get

2 6 2 2 2

2

2 1

16 12 0

c x cx E 2 x



which can never be satisfied. Thus e ^cx4 can never be an eigenfunction of the Schrodinger equation.

Question5: The probability distribution for a coherent state is given by

2 2

| ( , ) |x t exp 0 cos t

Calculate x , x² and x x² x ² . Solution5: The probability distribution is given by

2 2

| ( , ) |x t exp 0 cos t (5)

Notice that

1

| ) , (

| x t ² dx (6)

for all values of t, as it indeed should be. Now

* 2

( , ) ( , ) ( , )

x x t x x t dx x x t dx

Simple integrations will give us

0 cos

x x t (7) where

0 0

x 1

Further

2 2 2 2 2

0 2

( , ) cos 1

x x x t dx x t 2 (8)

2

2

2 

x x

x (9) Question6: The Schrodinger equation for the linear harmonic oscillator can be written as

2

2

2 0

d d

where x and

 . Make the transformation ² and look for the solution as and obtain solutions in terms of confluent hypergeometric functions.

Solution6: We will obtain the solution of the following equation

2

2

2 0

d

d (10) We introduce the variable

² (11) Thus

^{d d d d 2}

d d d d (12) and

2 2

2 4 2 2

d d d

d d d (13) Substituting in Eq. (8), we obtain

2 2

1 1

( ) 0

2 4 4

d d

d d (14)

In order to determine the asymptotic form, we let so that the above equation takes the form

2 2

1 ( ) 0 4

d d

the solution of which would be

1

e 2 . This suggests that we try out the following solution

1

y e 2 (15) Thus

1

1 2

2

d d y

d d y e (16) and

2 2 1

2

2 2

1 ( ) 4

d d y d y

y e

d d d (17)

Substituting the above equations in Eq. (12) we get

2 2

1 1

( ) 0

2 4

d y d y

d d y (18)

Now the confluent hypergeometric equation is given by

2

2 ( ) 0

d y d y

x c x a y x

d x d x (19)

where a and c are constants. For c 0, 1, 2, 3, 4,.... the two independent solutions of the above equation are

y x₁( ) ₁F a c x₁ , , (20)

and

y x₂( ) x^{1 c}₁F a c₁ 1, 2 c x, (21) where ₁F a c x₁ , , is known as the confluent hypergeometric function and is defined by the following equation

2 3

1 1

( 1) ( 1)( 2)

, , 1 ...

1! ( 1) 2! ( 1)( 2) 3!

a x a a x a a a x

F a c x

c c c c c c (22)

Obviously, for a c we will have

2 3

1 1 , , 1 ...

1! 2! 3!

x x x x

F a a x e (23)

Thus although the infinite series in Eq. (20) is convergent for all values of x, it would blow up at infinity. Indeed the asymptotic form of ₁F a c x₁ , , is given by _{1 1} , , ^{( )}

( )

a c x x

F a c x c x e

a (24)

The confluent hypergeometric series₁F a c x₁ , , is very easy to remember and its asymptotic form is easy to understand. Returning to Eq. (16), we find that

( )

y satisfies the confluent hypergeometric equation with ¹ and ¹

4 2

a c (25) Thus the two independent solutions of Eq.(8) are

1 2

1 1 1

1 1

( ) , ,

4 2

F e (26) and

1 2

2 1 1

3 3

( ) , ,

4 2

F e (27) We must remember that ². Using the asymptotic form of the confluent hypergeometric function, one can readily see that if the series does not become a polynomial then, as , ( )will blow up as

1

e2 . In order to avoid this, the series must become a polynomial. Now ₁( ) becomes a polynomial for = 1,5, 9, 13,… and ₂( ) becomes a polynomial for = 3, 7, 11, 15. Thus only when

= 1, 3, 5, 7, 9,…. (28)

we will have a well behaved solution of Eq. (8) – these are the eigenvalues of the Schrodinger equation. The corresponding wavefunctions are the Hermite Gauss functions:

( ) ( ) exp ^{1 2} ; 0,1, 2,3,....

m 2

N H m (29)

Indeed

( ) ( 1) ^{/ 2 !} _{1 1} , ¹ , ² for 0, 2, 4,....

2 2

2 !

n n

n n

H F n

n (30)

and

( ) ( 1)^{( 1) / 2 !} 2 _{1 1} ¹, ³ , ² for 1, 3,5,....

1 2 2

2 !

n n

n n

H F n

n (31)