Quantum Mechanics, A Graduate Course

Quantum Mechanics, A Graduate Course Solutions manual

HORATIU NASTASE 1 / 4

Chapter 0 1)Doing the integral overBin the formula forP, we nd P= 2h c 2 Z 1

d

3 e h k B T 1 2 Z

sindjcosj: (0.1)

Deningx=h=(kBT), we nd P= 2k 4 B h 3 c 2 T 4 Z 1

dx x 3 e x 1 4 Z 1

d(cos)(cos) = 2k 4 B h 3 c 2 T 4

(4)(4)1 =

2k 4 B

4 h 3 c 2 15 T 4

:

(0.2) 2)The photon has energyEph=h, and is split between the kinetic energy of the photoelectric electron, and its binding energyW. By the kinetic theory of gases (applied, by assumption, to the electron), mv 2 2 = 3 2 kBT. Thus, Eph= mv 2 2 +W= 3 2 kBT+W) T= hW 3kB=2

h 3kB

;(0.3)

where in the lastwe used the assumption ofWof the same order ash. But visible4810 14

Hz ; h'6:610

34 m 2

Kg=s ; kB'1:410

23 m 2 Kg=(s 2 K); (0.4) so T 48 3

6:6

1:4

1000K7141000K: (0.5)

3)The photon has momentump=h=and energyE=hc=, while the electron has initially negligible momentum and (rest) energymec 2 , and after has momentump

eand energy p (p

ec) 2

• (mec

2 ) 2 .Conservation of energy is then E+mec 2 =E

+ p (p

ec) 2

• (mec

2 ) 2 )(p

ec) 2 =

hc

hc

+mec 2

2 (mec 2 ) 2 ; (0.6) and conservation of momentum is ~p

e=~p~p

)p 02 ec 2 = (pc) 2

• (p

c) 2 2pp

c 2

cos: (0.7)

Identifying the two ways of calculatingp

e, we nd 1

1

= h mec 1

(1cos))

= h mec

• sin

2

=2: (0.8)

3 2 / 4

4Chapter 0 4)For a relativistic wave,!= 2andk= 2== 2=c. If the two-slit experiment has a distanceabetween the slits and is at a distancedafrom the screen, and we consider interference at a distancexfrom the midpoint on the screen, we see that the dierence in distance traversed by the the two waves is l2l1= l'a, but thenis approximately the same as the angle made by the interference point from the slits, i.e.,'x=d. In between two maxima or minima, the dierence in distance travelled is of, so x=l d a = c

d a

:(0.9)

5)Room temperature is aboutTroom300K. An electron at room temperature would have, by kinetic gas theory, me v 2 2 = 3 2 kBT)= h mev = h p 3kBTroomme

: (0.10)

Sinceme'910 31 Kg, we nd '

6:610

34 p

31:410

23

300910

31

m'0:610

8

m: (0.11)

On the other hand, visible light, with4810 14 Hz, has = c

3 48 10 6

m:(0.12)

But since the accuracy of the microscope is of the order of, we see that a microscope with room temperature electrons is still better than one with visible light (has a better accuracy).6)We could indeed apply Bohr-Sommerfeld quantization, since we have a peri- odic motion in phase space. Dening asthe angle around the circle traced by the circular pendulum as it moves, we would have Z

pd=lh:(0.13)

However, this is not very useful, since the circular pendulum is a classical object, so the angular momentum is very large, solis well approximated by being contin- uous. If however we have a quantum system of the same kind, like for instance a precession motion in a quantum system, then it would be useful.7)Light is a relativistic system, so we would need to use Quantum Field Theory.Since we have Quantum Mechanics, in some sense yes, we should use a form of the Schrodinger equation, but really only the Quantum Field Theory formalism does it. 3 / 4

1Chapter 1 1)The triangle inequality for~c=~a+ ~ bis~c 2 ~a 2 + ~ b 2 . Generalizing to vectors jvi=jwi+jui, we nd

hvjvi hwjwi+hujui:(1.1)

2)Considering~a ~ b6= 0,~a~c6= 0, ~ b~c6= 0, we rst normalize~a, ~a

= ~a p ~a 02

:(1.2)

Then we orthogonalize ~ bto it, ~ ~ b= ~ b(~a

~ b) ~a

~a 02

;(1.3)

so that~a

~ ~ b= 0, then normalize it, ~ b

= ~ ~ b q ~ ~ b 2

:(1.4)

Then we orthogonalize~cto~a

, ~ ~c=~c(~a

~c) ~a

~a 02

;(1.5)

so that ~ ~c~a

= 0, then orthogonalize it to ~ b

also, ~~ ~c= ~ ~c( ~ b 0~ ~c) ~ b

~ b 02

;(1.6)

so that ~~ ~c ~ b

= 0, as well as ~~ ~c~a

= 0, and nally normalize it, ~c

= ~~ ~c q ~~ ~c 2

:(1.7)

3)If ^ A= X a;a

jaiha

j; ^ B= X b;b

jbihb

j;(1.8) then ^ A ^ B= X a;a

;b;b

jaiha

jbihb

j= X b

@ X a;a

;b ha

jbijai 1 Ahb

j: (1.9)

• / 4