Chem 373- Problems related to expectation values and superposition of states (Lecture-7)

8/3/2019 Chem 373- Problems related to expectation values and superposition of states (Lecture-7)

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Problems related to expectation values and superposition of states (Lecture-7)

1. A particle is in a state described by the wavefunction

(x)= (cos )eikx

+ (sin )eikx

Please note that this is our wave function from Lecture 4 if A= cos and B= sin

where is a parameter. What is the probability that the particle will be found with a

linear momentum (a): hk, (b) : hk.

(c) What form would the wavefunction have if it were 90 per cent certain that the

particle had the linear momentum hk.

ANS:

(x)= c1eikx

+ c2eikx

= (cos )eikx

+ (sin )e ikx

The linear momentum operator is

px=

h

i

d

dx

As demonstrated in Atkins 11.5, eikx

is an eigenfunction to px with the eigenvalue hk

; likewise eikx

is an eigenfunction of px with eigenvalue. hk

Therefore, by the principle of linear superposition we have :

(a) The probability of finding the particle with momentum hk must be | c1 |2= cos2

(b) The probability of finding the particle with momentum hk must be|c2 |

2= sin

2

(c) If it is 90 percent certain that the linear momentum is hkthen


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| c1 |2= .90 = cos2 and | c2 |

2 = .10 = sin2

Thus c1 = cos = .95 ; c2 = sin = .32

The wavefunction would have the form

(x)= 0.95eikx

+ 0.32 eikx


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2. Evaluate the kinetic energy of the particle with the wavefunction (x)= (cos )e

ikx+ (sin )e

ikx

given in problem 1.

Please note that this is our wave function from Lecture 4 if A= cos and B= sin

Answer:

The kinetic energy operator , T , is obtained from the operator analog of the classical equation

EK = p2

2m

that is,

T = (p)2

2m

px =h

i

d

dx; hence p2x = h2

d2

dx2and T = -

h2

2m

d2

dx2

Then

< T > = N2 * -h

2

2m

d2

dx2

d =

* -h

2

2m

d2

dx2

d

* d N2 =

1

* d

= -h

2

2m*

d2

dx2(e ikx cos + eikx sin )d

* d

=

-h

2

2m* (k2 )(eikx cos + e ikx sin )d

* d =h

2k2

2m

* d* d =

h2k2

2m

This result should not surprise you if you look at the handout to Lecture 4 on the

free particle.


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3

.Calculate the average linear momentum of a particle described by the following wavefunctions:

(a) : eikx , ( b ) :coskx , (c) e- x2

In each case x range from - to +

px = h

i

d

dx

< px > = N2 * px dx ; N

2 = 1* dx

= * px dx

* dx=

h

i

* d

dx

dx

* dx

(a) = eikx

,d

dx= ik

Hence

< px > =

h

i ik * dx

* dx= kh

his should not surprise you since we have shown (Lecture 6)

that eikx

is an eigenfunction to p xwith eigenvalue kh


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(b) = coskx ,d

dx= ksinx

* d

dx

+ dx = -k coskxsinkxdx =

+

sin2x

2|+

= 0

Thus < px >= 0

This should not surprise you from the discussion in Lecture 7

since coskx =1

2(e ikx + e-ikx )

(c) = ex2

,ddx

= 2xex2

*ddx

+ dx = - 2

+ xex

2

dx = 0 [by symmetry this integral is zero since xe

is an odd function ]


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4.

Evaluate the expectation values of r and r 2 for ahydrogen atom with the wavefunctions given by

( a ) : = 132a o3

(2 - rao

)e r / 2 ao

(b) : = rsincose r / 2 ao

(a)

< r > =1

32a o3

*r d [ d = r2 sindd]

=1

32ao3

r(2 r

a o ) 2 e

r / ao dr4 [4 = sindo

d

0

2 ]

=1

8a o3

(4r3

o

4r4

a o

+r

5

ao2

)er / ao dr

=1

8ao3

(4 3!a o4 4 4!a o

4 + 5!a o4 ) = 6ao [ x

n eax dx =n!

a n+1 ]

< r2 >=1

8ao3

(4r 4

o

4r 5

a o+

r 6

ao2

)e r / ao dr = 42a o

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Chem 373- Problems related to expectation values and superposition of states (Lecture-7)