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Physics Education April June 2006 61

PHYSICS THROUGH PROBLEM SOLVING IV

Problems from ElementaryQuantum Mechanics

PRASANTAKPANIGRAHI

Physical Research Laboratory, NavrangpuraAhmedabad 380009([email protected])

QUESTION 1

Consider a one-dimensional harmonic oscilla-

tor in the presence of a constant electric fieldE

along theX-axis:

Hm x

m x eEx= + h2 2

2

2 2

2

1

2

,

where we may associate an electrical dipole

moment for the oscillator along the X-axis

given by d= ex:

a) Find the exact energy eigen values andeigen functions ofH. Interpret the energy

shift relative to the levels for the non interacting oscillator.

b) Consider the ground state ofH. Show thatit is an eigen state of the lowering operatora defined as,

am

xm x

=

2h

h

c) Find the eigen states of a2 and interpretthem physically.

d) Starting from H0 find the first and secondorder corrections to the ground state

energy. What is the correction to theground state energy from higher orders?

e) Consider the complex Hamiltonian:H

m xm x i x= + +

h2 2

2

2 2

2

1

2

.

Find the energy eigen values and eigenfunctions. What is the difference between

eigen values ofHand H?


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62 Physics Education April June 2006

f) Find the transformation property of Hand the eigen functions under the parityand time reversal operations:

P:x x and T: i i.

Assuming that is invariant under PT,draw conclusions about the energy eigenvalue.

ANSWER 1

(a) Given Hamiltonian is:

Hm x

m x eEx= + h2 2

2

2 2

2

1

2

H m x= +

h2 2

22

1

22

2

22 2

2 4m x

eE

m

e E

m

Put y xeE

m=

2which gives

Hm y

m ye E

m= +

h2 2

2

2 22 2

22

1

2 2

Thus, energy eigen values are,

E n e Em

n = + 12 2

2 2

2h

It is clear that, energy gets reduced in the

presence of a constant electric field as

compared to a non-interacting oscillator.

The eigen functions are:

n nc

m

hx

eE

m=

exp2 2

2

Hm

hx

eE

mn

2

One can clearly see that equilibrium pointhas shifted to the left.

(b) Consider the ground state ofH:

0 0 2

2

2=

cm

hx

eE

mexp

Now,

am

xm x

c

0 02

=

h

h

exp

mx

eE

m

2 2

2

h

=

=m eE

m

2 2

0 0h

Therefore, 0 is an eigen state of annihilationoperator, which makes it the well-known

Coherent State.

(c) There are two eigen states of a2 : | > and|>. We note that

a2| = 2|

and

a 2 2[| | ] [| | ] + = +

The first one is the coherent state, a shifted

Gaussian function as seen above. The second

one is a linear superposition of two displaced

Gaussians with opposite displacements. This is

the known as the Cat state invented bySchrdinger.

(d) The Hamiltonian can be written as:

H=H0 +H1


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where,

Hm x

m x i x= + +h2 2

2

2 2

2

1

2

H1 = eEx

Now, we know that

xm

a a= + +h

2 ( ) ,

also

H n E nn00| |( ) =

Here, E n hn( )0

1

2= +

denotes the unper-

turbed eigen value.

First order correction to energy is,

E n H nn( ) | |1 1=

E eEm

n x nn( ) | |1

2=

h

,

E eEm

n a a nn( ) |1

20= + =+

h

.

So, there is no change in energy in the firstorder.

Second order correction can be evaluatedas,

Ek H n

E En

n kk

( )

( ) ( )

| |2

1

2

0 0=

Now,

k H n eE m

12

= h

( )n nk n k n , , ++ +1 11 .

Thus,

E e Em

n nE E

n

k

k n k n

n k

( ) , ,

( ) ( )( )2 2 2 1 1

0 021= + +

+

h ,

=

++

+

Ee E

m

n

E E

n

E En

n n n n

( )

( ) ( ) ( ) ( )2

2 2

01

0 01

02

1h

,

= Ee E

mn( )2

2 2

22 .

Hence, the entire correction to energy

comes from the second order. All higher order

corrections (containing higher powers of eE)

are zero.

(e) Given Hamiltonian is:

Hm x

m x i x= + +h2 2

2

2 2

2

1

2

,

with > 0. It can be written as,

Hm x

m xi

m= + +

h2 2

2

2

2

2

2

1

2

+

2

22m.

Now, put y x im= + 2

which gives,

Hm y

m ym

= + +h2 2

2

2 22

22

1

2 2

.

Energy eigen values and eigen functionsare as follows:

E nm

n = +

+1

2 2

2

2h

n nc

mx

i

m= +

exp2 2

2

h

Hm

xi

mn

h

+

2

.


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64 Physics Education April June 2006

In this case, the eigen values are still realbut energy levels get shifted upwards incomparison to the first example.

(f) Under parity operation, x changes sign and

under time reversal operation, i changes sign.

Hamiltonian H is not invariant under parityand time reversal operations separately but is

invariant under joint operations. It can be seen

that the above wave functions are invariant

under the actions of parity and time reversaloperations together. This accounts for the realnature of the eigen values even when the

perturbing term is imaginary. Starting from the

eigen value equation, one can show that under

parity and time reversal operations E E*.Hence, when Hamiltonian and the wave

functions remain invariant under PT, the eigen

values will be real.

QUESTION 2

For the ground state of harmonic oscillator:

0 0

2

2( ) expx c

m x=

h

(a) Compute

W x p dai

pa( , ) exp=

1

2h h

* xa

xa

+

2 2

(b) Show that-

W x p dp x( , ) | ( )|=

0 2 .

Note: W(x, p) is called the Wigner function,

which is a quasi probability distribution

function

.

ANSWER 2:

(a) Putting the value of0(x), we have

W x p daipa

c( , ) exp | |=

1

20

2

h h

exp exp

+

mx

a mx

a 2 2 2 2

2 2

h h

=

W x pc m x

( , )

| |

exp

02 2

2

h h

daipa m a

exp exp

h h 2

4

=

W x p

c m x( , )

| |exp0

2 2

2

h h

dam x ipa

daexp

2

4h h

Finally, we get

W x pc

m

m x p

m

( , )| |

exp=

02 2 2

h h h

(b) Now, we will integrate this Wigner

function with respect to momentum as follows:


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W x p dp cm

m x( , ) | | exp=

0

22

h h

exp

p

mdp

2

h

=

W x p dpc

m( , )

| |02

h

exp

m x

m

2

hh

=

W x p dp c m x( , ) | | exp02

2

h

=| ( )| 0 2x

which is the probability density in coordinatespace.

QUESTION 3

(a)Compute the wave function in the

momentum representation,

0 01

2( ) ( ) expp x

ipxdx=

h h

Interpret your result.

(b) Show that,

W x p dx( , )

= 02

( )p

.

ANSWER 3

(a) The Fourier transform of a Gaussian exp (

x2) goes like exp(k2/4). It is worth notingthat has appeared in the denominator. Alocalized wave function in the coordinate space

is more spread out reciprocally in the

momentum space, which indeed is the originaluncertainty relation.

(b) Integrating Wigner function with respect tocoordinate gives,

W x p dxc

m

p

m( , )

| |exp=

02 2

h h

exp

m x dx2

h

=

W x p dx( , )

| |exp

c

m

p

m m0

2 2

h hh

=

W x p dx( , )

| |exp | |

c

m

p

mp

02 2

2

=

h

which is the probability density in momentumspace.

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