faqs for module8

7/30/2019 Faqs for Module8

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FAQs & their solutions for Module 8:

Angular Momentum-II

Question1: The spin angular momentum operator for electron is given by

z z y y x x s s s 2

1,

2

1,

2

1 (1)

Where x , y and z are Pauli spin matrices and are given by

10

01,

0

0,

01

10 z y x

i

i

Write the eigenvalues and eigenvectors of , and . x y z s s s

Solution1: We first determine the eigenvalues of the x matrix which are determined from the

following equation

01

1

or

`012

implying

1 (2)

Thus the eigenvalues of the x matrix are 1 and therefore the eigenvalues of s x are 2

1 . Thus

if we measure s x [i.e., the x component of the spin angular momentum of a spin2

1 particle like

electron, proton or neutron] then we will get only one of the two possible (eigen) values 2

1 or

2

1

. The corresponding eigenfunctions are easy to determine; e.g., for the eigenvalue 2

1

, the

eigenvalue equation is written as

b

a

b

a1

01

10

giving

ab

Thus the eigenfunction is given by

11a



Thus

i

1

2

1^y (8)

would represent the normalized “ y-down” state. Similarly

i

1

2

1^y (9)

would represent the normalized “ y-up” state.

Question2: A spin half particle is in the “ z -up” state. On that particle, if we make a

measurement of s x then what are the values that we will obtain and what will be their

probabilities.

Solution2: The spin half particle is in the “ z -up” state. On that particle, if we make a

measurement of s x then we will get one of the two eigenvalues of s x. In order to determine their

probabilities we have to express the (normalized) “ z -up” state as a linear combination of the

(normalized) “ x-up” and “ x-down” states:

^^^ |

2

1|

2

1

0

1| xxz (10)

Thus, if we make a measurement of s x then the probability of obtaining a “ x-up” state [i.e., the

probability of obtaining the eigenvalue 2

1 for s x] is

2

1 and the probability of obtaining a “ x-

down” state is also2

1 .

Question3: The magnetic moment of the neutral Ag-atom is the same as that of an electron and

is given by

q

m μ s (11)

whereq and m represent the charge and mass of the electron and

z z y y x x s s s 2

1,

2

1,

2

1 (12)

Such a particle is placed in a static magnetic field given by

^zB 0 B (13)

Obtain the eigenvalues and eigenfunctionsof the energy associated with magnetic field.



Solution3: The magnetic moment of the neutral Ag-atom is the same as that of an electron and

is given by

q

m

μ s (14)

where z z y y x x s s s 2

1,

2

1,

2

1 (15)

where x , y and z are Pauli spin matrices. If such a particle is placed in a static magnetic field

given by

^zB 0 B (16)

then the potential energy associated with magnetic field would be given by

z H 0021 B.μ (17)

where

00

0

2 B

m

qB B (18)

and

J/T10274.9~2

24m

q B

(19)

represents the Bohr magneton. Since the eigenvalues of z are +1 and -1, the solution of the

eigenvalue equation

2,1;||0 nn E n H n (20)

would be given by

0

1|1|

2

1^

01z E (21)

1

0

|2|2

1^02 z E (22)

Question4: Write the most general solution of the time dependent Schrodinger equation

)(|)(| 0 t H t

t i (23)

where

0 0

1

2 z

H



Solution4 : The most general solution of the time dependent Schrödinger equation

0 0

1| ( ) | ( ) | ( )

2z i t H t t

t

(24)

would be

2|1|

|)(|

2/2/

/

0021

2

1

t it i

t iE

eC eC

neC t n

nn

(25)

where

0

1|1|

2

1^

01 z E (26)

10|2|

21 ^

02 z E (27)

Further, the coefficients C 1 and C 2 are to be determined from the knowledge of |(t = 0) >:

011 C

and

021 C

Of course, if the system is initially in the | z > or | z > states, then it will remain in those

states for all times to come; these are the stationary states of the problem.

Question5: In continuation of the previous problem, we assume that at t = 0, the atom is in

the x state. Obtain the time evolution of the state and calculate the expectation values of

, and . x y z s s s

Solution5:

At t = 0, the atom is in the | x > state. Since

1

1

2

1^x

we readily get

2

1

1

1)01(

2

1| ^^

1

xzC

Similarly C 2 = 1/ 2 . Thus



1

0

2

1

0

1

2

1)(

iieet

or

i

i

e

et

21)( (28)

where

t 02

1 (29)

Equation (21)(28) describes the time evolution of the state. Further,

i

i

ii

x x

eeee

t t s

0110

41

)()(2

1

or

t s x 0cos2

1 (30)

Similarly

t s y 0sin2

1 (31)

and

0 z s (32)

The above equations physically imply that the direction of the spin angular momentum vector

rotates about the z -axis with angular velocity 0 .

Question6: Next consider the more general case when

^^

2sin

2cos)0( zz

Obtain the time evolution of the state.

Solution6:

We have

^^

2sin

2cos)0( zz

(33)

[when /2, we obtain the results of the previous problem]. Obviously

2cos1

C and 2sin2

C



so that

i

i

ii

e

e

eet

2sin

2

cos

2sin

2cos)( ^^ zz

Thus

)()(2

1t t s x x

or

t s x 0cossin

2

1 (34)

Similarly

t s y 0sinsin2

1 (35)

and

cos2

1 z s (36)

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