Physics 130B
Section 722382
Fall 2011
Physics 130B Midterm Solutions
Eric L. Michelsen
Fall 2011
Version 5: Small clarifications.
1.
A spin-1/2 particle has a definite value of spin “up” along an axis tilted 20 deg from the +z axis,
toward the +x axis.
(a 5pt) What is the probability of measuring the spin up along the z-axis?


 cos 20o / 2  
o
   cos10 
  

o
 sin 20o / 2 ei 0   sin10 




2
Pr  measuring z    z  
2
 cos10o 
  cos 2 10o
 1 0 
 sin10o 


(b 5) What is the probability of measuring the spin down along the z-axis?
Pr  measuring z    1  Pr  z 
  1  cos 2 10o
or
2
Pr  measuring z    z  
Note that
2
 cos10o 
  sin 2 10 o
  0 1 
 sin10o 


Pr(|z+>) + Pr(|z–>) = 1
(c 5) What is the average (over many particles) z-component of spin?
sz 
or






Pr  z    Pr  z    cos 2 10o  sin 2 10o  cos 20o
2
2
2
2
0   cos10o 
/ 2


sz   sˆz   cos*10o , sin*10o  

  0  / 2  
o 
 sin10 

 cos10o  


  cos2 10o  sin 2 10o  cos 20o
cos10o , sin10o  


o


2
2
  sin10  2


(d 5) What is the classical z-component of spin for the given particle?

cos 20o , which must equal the quantum average, since the classical result is the
2
average of billions of quantized results.
s z ( classical ) 
(e 5) If the particle is tilted 20 deg toward +y (instead of +x), what is the probability of measuring spin up
along the z-axis?
By axial symmetry about the z-axis, this is the same as part (a)
2.
In computational quantum chemistry, the local energy of a trial wave-function is computed
numerically, to provide adjustments to the wave-function, which is then the starting point for a new
iteration. Given a trial wave-function:
 ( x)  N
1
2
x 1
,
N  normalization constant
(a 10) The potential is everywhere 0. Find the local energy, E(x).
11/14/2011 9:47:23
32
Physics 130B
Section 722382
E ( x) 
Fall 2011
Eric L. Michelsen
 2  ''( x )
1 pˆ 2 ( x )
Eˆ ( x )


,
 ( x ) 2 m  ( x)
2m  ( x )


 '   x2  1
2


 2 x  2 x x 2  1



 ''  2  x  2  x 2  1

3

2


 2 x  x2  1
   x2  1
2

 ''/  2  4 x 2 x 2  1

2

 2  4 x 2 x 2  1
E ( x)  
2m


N cancels, so we drop it:
2 


1 


   x  1
2
2


 2  3 x 2  1 
  m  2
2

 x 1 


1 


(b 5) Is the total energy for this particle finite, or infinite? Justify your answer.
E

 E ( x)  ( x)
2
dx
Since both terms in E(x) drop off faster than 1/x2, and integral 1/x2 is finite, and |ψ|2 is bounded,
the total energy is finite.
(c 10) What is the average momentum of this state?
By symmetry of ψ(x) about x = 0, <p> = 0.
Or,
p   pˆ  


*
 

 dx 
i x
i


*

 dx, but this must be real.
x
Since  is real , the only way to get rid of the 'i ' is if the int egral  0
Note this is true for any real function ψ(x).
Or,
For any real ψ(x), we can evaluate the integral by parts:



 ' dx   ' 

Or,
p   pˆ  

 ' dx
Using
 UV ' dx  UV    U 'V dx

 ' dx  0


*
 

 dx 
i x
i
  x

2

1
1


  1 x 2  1
2
 2 x)  dx,
but the integrand is odd, and we integrate over a symmetric interval, so the left and right halves (about 0)
cancel. Thus, <p> = 0.
3.
Consider a simple harmonic oscillator, in standard notation.
(a 5) What is <x> in the state |0>?
By symmetry of |ψ(x)|2 about x = 0, <x> = 0.
(b 5) What is <x> in the state |1>?
Ditto. Note that ψ(x) is anti-symmetric about x = 0.
(c 5) What is <x> in the state
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1
3
0 
1 ?
2
2
33
Physics 130B
Section 722382

3
 1 3


2m  2 2
2
(d 5) What is <x> in the state
x 


 a†  a  
2m
x   x 

Fall 2011
1

2 
Eric L. Michelsen

 1 3
31
0a1 
1 a† 0 

2m  2 2
2 2

3


2m 2
1
3
2 
1 ?
2
2

  31
1 3
1a 2 
2 a† 1  



2m  2 2
2 2


  31
1 3

6
2
2




2m  2 2
2 2
2

2
m

(e 5) What is <x2> in the state of (d)?
x2 
4.


 a†  a
2m

2
 



 a† a†  a† a  aa†  aa 
2m


  3 3
11
1 a† a  aa† 1 
2 a† a  aa† 2 

2m  2 2
22


1
7
 3
 14
 7



 1  2    2  3  
m
m
m
2m  4
4
2

4
2

2
4





For l = 1, we have
 0
1/ 2
0 


Lˆ x   1/ 2
0
1/ 2  ,


 0
1/ 2
0 

 0

Lˆ y    i / 2

 0

i / 2
0
i/ 2


i / 2 

0 
0
(a 10) What is the operator for angular momentum along an axis 30 deg CCW from the +x axis?
L-hat can be written as a superposition of L-hatx and L-haty, each of which projects a component onto Lhat:


Lˆ  cos30o Lˆx  sin 30o Lˆ y  cos Lˆx  sin Lˆ y
6
6
 0
 0
1/ 2
0 




  cos 1/ 2
0
1/ 2    sin  i / 2
6
6

 0

 0
1/
2
0




0


i
/
6

 e
/ 2


0

e i / 6 / 2
0
e
i / 6
/ 2
i / 2
0
i/ 2


i / 2 

0 
0


ei / 6 / 2 


0

0
(b 10) What are its eigenvalues?
From the isotropy of space, the eigenvalues of any l = 1 component measurement are ħ, 0, –ħ
(c 5) What is the eigenstate for measuring +ħ along this tilted axis?
Dropping the ħ, so the eigenvalues reduce to +1, 0, –1, we use the eigenvector equation:
11/14/2011 9:47:23
34
Physics 130B
Section 722382
ˆ     ,
where
Fall 2011
Eric L. Michelsen
 is the known eigenvalue. In this case:
1 1
   
ˆ
L  b   1 b 
c c
   
where we have assumed we can set a = 1. This is justified by getting a result below. Solving:

 1
0
0
e i / 6 / 2

  1
   

i

/
6

i

/
6
e
/ 2
0
/ 2 b  b
e


   

c  c 
0
0
e i / 6 / 2


From the top row :
be i / 6 / 2  1 
b  2e i / 6
From the 2nd row :
e i / 6 / 2  cei / 6 / 2  b  2e i / 6
e i / 6  ce i / 6  2e i / 6 , ce i / 6  e i / 6 , c  e i / 3
1




1
eigenvector   2e i / 6  , Normalize : mag 2  1  2  1  4, multiply by
4
 i / 3 
 e



1/ 2


eigenstate   1/ 2 e i / 6 


 1/ 2 e i / 3 

 


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
35