HW4 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. 1. T&L 6.P.03 In a region of space, a particle has a wave function given by

! (x) = Ae" x2/(2L

2) and energy

E =!2

2mL2.

(a) Find the potential energy V as a function of x. Sketch V(x) vs. x. (b) What is the classical potential that has this dependence? Solution: a) To find the potential energy, we use the time-independent

Schrodinger equation:

!!2

2m

d2" (x)

dx2

+V (x)" (x) = E" (x) . For the given

wave function, we have d! (x)dx

= A "1

2L2

#$%

&'(2xe

" x2 /(2L2 )= "

Ax

L2e" x2 /(2L2 )

= "x

L2! (x) , and thus

d2! (x)

dx2

= "1

L2! (x) +

x2

L2! (x) . Therefore, the Schrodinger equation

becomes

!!2

2m!1

L2+x2

L4

"#$

%&'( (x) +V (x)( (x) =

!2

2mL2( (x), resulting in

!!2x2

2mL4" (x) +V (x)" (x) = 0 . Hence,

V (x) =!2x2

2mL4. This is a parabola

centered at x=0. b) The classical potential with this dependence is the simple harmonic oscillator potential. 2. T&L 6.P.22 Show that the wave functions of a particle in a one-dimensional infinite square well are orthogonal: i.e., for n ! m , !

n(x)!

m(x)" dx = 0 .

Solution: As the wave functions of the infinite square well are given by

!n(x) =

2

Lsin

n" x

Linside the well and zero outside, the integral of

interest takes the following form: !n(x)!

m(x)" dx =

2

Lsin

n# xLsin

m# xL

dx

0

L

" .

Using the identity sinasinb =1

2cos(a ! b) ! cos(a + b)( ) , the integral

becomes 1

Lcos

(n ! m)" xL

! cos(n + m)" x

L

#$%

&'(dx =

1

" (n ! m)0

L

) sin(n ! m)" x

L 0

L

!1

" (n + m)sin(n + m)" x

L 0

L

= 0

for n ! m . 3. T&L 6.P.33 Compute < x >0= !

0

*(x)x" !

0(x)dx and < x2 >0= !

0

*(x)x

2" !0(x)dx for the

ground state of a harmonic oscillator:

!0(x) =

m"#!

$%&

'()1/4

e*m" x2 /(2!) .

Solution: For the ground state of the harmonic oscillator, the expectation value of the position operator x is given by

< x >0= !

0

*(x)x" !

0(x)dx =

m#$!

xe%m# x2 /!

%&

&

" dx = 0 . The fact that this

expression vanishes can be seen either by brute force calculation, or by symmetry (the integrand is odd and the limits of integration are symmetric). The expectation value of x2, however, doesn’t vanish:

< x2>0=

m!

"!x2e#m! x2 /!

#$

$

% dx .

Integrals of this type are given in Appendix B1 of the textbook. However, here is a useful trick: knowing the fundamental integral

e!cx2

!"

"

# dx =$

c, we can take a derivative with respect to c to obtain

x2e!cx2

!"

"

# dx = !d

dcx2e!cx2

!"

"

# dx = !d

dc

$

c=

$

2c3/2

.

Here

c =m!

!, and thus

< x2>0=

m!

"!x2e#m! x2 /!

#$

$

% dx =m!

"!

1

2

"!3

m3! 3

=!

2m!.

4. T&L 6.P.33

a) Using

!0(x) =

m"#!

$%&

'()1/4

e*m" x2 /(2!) , write down the total wave function

!0(x,t) for the ground state of a harmonic oscillator.

b) Compute the expectation value of the momentum operator

< p >0= !

0

*(x)("i!)

#!0(x)

#x$ dx for this state.

Solution:

a) The ground state energy is

E0=!!

2, and hence

!0(x,t) ="

0(x)e

# iE0 t /! =m$%!

&'(

)*+1/4

e#m$ x2 /(2!)

e# i$ t /2 .

b) The expectation value of the momentum is given by

< p >0= !

0

*(x)("i!)

#!0(x)

#x$ dx . As

d!0(x)

dx=

m"#!

$%&

'()1/4 *m"x

!

$%&

'()e*m" x2 /(2!)

=*m"x!

$%&

'()!0(x) ,

< p >0=

m!"!

#$%

&'(1/2m!!

(i!)) e*m! x2 /!

xdx = 0 , again by symmetry.

The expectation value of p2 doesn’t vanish. We have

d2!

0(x)

dx2

="m#

!!0(x) +

m2# 2

x2

!2

!0(x). Therefore,

< p2>0= !!2

m"#!

$%&

'()1/2

m"!

e!m" x2 /! !1+

m"x2

!

$%&

'()dx

!*

*

+ . The first term is

!m!m!"!

#$%

&'(1/2

e)m! x2 /!

dx

)*

*

+ = !m!m!"!

#$%

&'(1/2 "!

m!#$%

&'(1/2

= !m! , and the second

term is

!(m" )2m"#!

$%&

'()1/2

x2e!m" x2 /!

dx

!*

*

+ = !(m" )2

2

m"#!

$%&

'()1/2 #!3

m3" 3

$%&

'()

1/2

= !!m"2

.

Hence, the final answer is

< p2>0=!m!

2.

5. T&L 6.P.54 A particle of mass m is in an infinite square potential given by V = !, x < ! L 2, V = 0, ! L 2 < x < L 2, V = !, x > L 2. Since the potential is symmetric about the origin, the probability density |! (x) |2 must also be symmetric.

a) Show that this implies either ! ("x) =! (x) or ! ("x) = "! (x) . b) Show that the proper solutions of the time-independent

Schrodinger equation can be written

! (x) =2

Lcos

n" x

L, n = 1,3,5,7,… and ! (x) =

2

Lsin

n" x

L, n = 2,4,6,8,… .

c) Show that the allowed energies are

En=n2!2!2

2mL2.

Solution

a) Since ! 2(x) =! 2

("x) =! ("x)! ("x) , we must have ! ("x) = ±! (x) .

b) The analysis of the time-independent Schrodinger equation is similar to that of the infinite well located between x=0 and x=L, as done in class.

The main differences are that the wave function is nonvanishing only for ! L

2< x < L

2, and the boundary conditions on the wave

functions are ! " L2( ) =! L

2( ) = 0 . More precisely, the wave function is given by ! x( ) = Acoskx + Bsin kx , where A and B are constants to be determined by the boundary conditions and normalization.

The potential is symmetric, and hence the solutions can be classified as even, ! ("x) =! (x) , or odd, ! ("x) = "! (x) . For the even solutions, B=0, and thus the wave function takes the form

! (x) = Acoskx . The boundary conditions ! (x = ± L 2) = AcoskL

2= 0 then

imply that k =n!

L, with

n = 1,3,5,… . The wave function for the even

solutions is therefore given by ! (x) = Acosn" x

L,

n = 1,3,5,… .

For the odd solutions, A=0, and the boundary conditions

! (x = ± L 2) = ±BsinkL

2= 0 imply that k =

n!

L, with

n = 2,4,6,… . The

wave function for the odd solutions is ! (x) = Bsinn" x

L,

n = 2,4,6,… .

c) The energy is given by

E =!2k2

2m=!2

2m

n!L

"#$

%&'2

=n2! 2!2

2mL2

.

.