Quantum Mechanics A Paradigms Approach 1st Edition McIntyre Solutions Manual

Page 1

and write the S x eigenvalue equations in matrix notation

which yields

Solve by adding and subtracting the equations to get

Hence the matrix representing S

in the S

basis is

by adding and subtracting the equations to get

Ch. 2 Solutions 3/20/19 2-1 2.1 Let S x a b c d      
a b c d       1 2 1 1         2 1 2 1 1       a b c d       1 2 1 1        2 1 2 1 1      
a  b   2 c  d   2 a b  2 c d   2
a  0 b  2 c  2 d  0
x
z
S x 2 0 1 1 0       Let S y a b c d      
a b c d       1 2 1 i         2 1 2 1 i       a b c d       1 2 1 i        2 1 2 1 i      
a  ib   2 c  id  i 2 a ib  2 c id  i 2
a  0 b  i 2 c  i 2 d  0
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and write the S y eigenvalue equations in matrix notation
which yields
Solve
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Hence the matrix representing S y in the S z basis is

which was to be expected, because we know that the only possible results of a measurement of any spin component are  2. Find the eigenvectors. For the positive eigenvalue:

2
3/20/19 2-2
Ch.
Solutions
S y 2 0 i i 0      
Solve
equation det S x I  0  2 2   0 Solve to find the eigenvalues 2 2       2  0 2  2       2   2
2.2
the secular
2 0 1 1 0       a b         2 a b       yields b  a The normalization condition yields a 2  a 2  1 a 2  1 2
a to be real and positive, resulting in a  1 2 b  1 2
Choose

so the eigenvector corresponding to the positive eigenvalue is

Likewise, the eigenvector for the negative eigenvalue is

The normalization condition yields

Choose a to be real and positive, resulting in

so the eigenvector corresponding to the negative eigenvalue is

2.3 From Eq. (1.37), we know the S z eigenstates in the S x basis:

Let the representation of S z in the S

basis be

and write the S z eigenvalue equations in matrix notation

Ch. 2 Solutions 3/20/19 2-3
 x 1 2 1 1      
2 0 1 1 0       a b        2 a b       b  a
a 2  a 2  1 a 2  1 2
a  1 2 b  1 2
x 1 2 1 1      
  1 2  x  x    1 2  x x  
S z a b c d      
a b c d       1 2 1 1         2 1 2 1 1       a b c d       1 2 1 1        2 1 2 1 1      
x

These yield

Solve by adding and subtracting the equations to get

Hence the matrix representing

in the

basis is

Now diagonalize:

as expected. Find the eigenvectors:

Ch. 2 Solutions 3/20/19 2-4
a  b   2 c  d   2 a b  2 c d   2
a  0 b  2 c  2 d  0
S z
S x
S z Sx 2 0 1 1 0      
 2 2   0  2 2 2  0    2
2 0 1 1 0       a b         2 a b        b  a yielding  Sx 1 2 1 1       Likewise 2 0 1 1 0       a b        2 a b        b  a Sx 1 2 1 1      
the eigenvalue equations are 2 0 1 1 0       1 2 1 1         2 1 2 1 1        S z    2  OK 2 0 1 1 0       1 2 1 1        2 1 2 1 1        S z  2 OK
Hence

2.4

Ch. 2 Solutions 3/20/19 2-5
The
matrix is A a b c d       The matrix elements are  A   1 0   a b c d       1 0        1 0   a c        a  A  1 0   a b c d       0 1        1 0   b d        b A   0 1   a b c d       1 0        0 1   a c        c A  0 1   a b c d       0 1        0 1   b d        d Hence we get A  A   A A  A         2.5 The commutators are [S x ,S y ] 2 0 1 1 0       2 0 i i 0       2 0 i i 0       2 0 1 1 0       2       2 i 0 0 i       i 0 0 i               2       2 2i 0 0 2i       i 2       1 0 0 1        i S z [S y ,S z ] 2 0 i i 0       2 1 0 0 1       2 1 0 0 1       2 0 i i 0       2       2 0 i i 0       0 i i 0               2       2 0 2i 2i 0       i 2       0 1 1 0        i S x
general

The spin component operator

expected. Find the eigenvectors:

Ch. 2 Solutions 3/20/19 2-6 [S z ,S x ] 2 1 0 0 1       2 0 1 1 0       2 0 1 1 0       2 1 0 0 1       2       2 0 1 1 0       0 1 1 0               2       2 0 2 2 0       i 2       0 i i 0        i S y 2.6
Sn is S n  S ˆ n  S x sincos S y sinsin S z cos Using the matrix representations for Sx, Sy, and Sz gives S n 2 0 1 1 0       sincos 2 0 i i 0       sinsin 2 1 0 0 1       cos 2 cos sincos isinsin sincos isinsin cos       2 cos sine i sine i cos       2.7
S n 2 cos sine i sine i cos       Now diagonalize: 2 cos  2 sine i 2 sine i 2 cos   0 2 2 2 cos 2  2 2 sin2  0  2 2 2  0    2
Diagonalize Sn:
as
Ch. 2 Solutions 3/20/19 2-7 2 cos sine i sine i cos       a b         2 a b       acos bsine i  a  b  ae i 1 cos sin The normalization condition yields a 2  a 2 1 cos sin       2  1 a 2  sin2  2 2cos  4sin2  2 cos 2  2 4sin2  2  cos 2  2 yielding  n  cos  2   e i sin  2 Likewise 2 cos sine i sine i cos       a b        2 a b       acos bsine i  a  b  ae i 1  cos sin The normalization condition yields a 2  a 2 1  cos sin       2  1 a 2  sin2  2  2cos  4sin2  2 cos 2  2 4cos2  2  sin2  2 yielding n  sin  2  e i cos  2 2.8 The  n eigenstate is  n  cos  2   e i sin  2  cos  8   e i5 3 sin  8
probabilities are
The

2.9 The expectation value of S

is easy to do in Dirac notation:

The expectation values of S x and S y are easier in matrix notation:

To find the uncertainties, we need the expectation values of the squares:

Ch. 2 Solutions 3/20/19 2-8 P y  y   n 2  1 2  i 2  cos  8   e i5 3 sin  8  2  1 2 cos  8 iei5 3 sin  8 2  1 2 cos  8  sin  8 sin 5 3 isin  8 cos 5 3 2  1 2 cos 2  8  sin2  8 sin2 5 3  sin2  8 cos 2 5 3  2cos  8 sin  8 sin 5 3    1 2 1 2cos  8 sin  8 sin 5 3   0.194 P y  y  n 2  1 2   i 2  cos  8   e i5 3 sin  8  2  1 2 cos  8  iei5 3 sin  8 2  1 2 cos  8 sin  8 sin 5 3  isin  8 cos 5 3 2  1 2 cos 2  8  sin2  8 sin2 5 3  sin2  8 cos 2 5 3 2cos  8 sin  8 sin 5 3    1 2 1 2cos  8 sin  8 sin 5 3   0 806
z
S z   S z    2   2    2
S x  1 0  2 0 1 1 0       1 0        2 1 0   0 1        0 S y  1 0  2 0 i i 0       1 0        2 1 0   0 i        0
S z 2   S z 2    2 2   2       2    2       2 S x 2  1 0  2 0 1 1 0       2 0 1 1 0       1 0        2       2 1 0   1 0        2       2 S y 2  1 0  2 0 i i 0       2 0 i i 0       1 0        2       2 1 0   1 0        2       2 The uncertainties are

2.10 These expectation values are easier in matrix notation:

To find the uncertainties, we need the expectation values of the squares:

Ch. 2 Solutions 3/20/19 2-9 S z  S z 2 S z 2  2       2 2       2  0 S x  S x 2 S x 2  2       2 0  2       S y  S y 2 S y 2  2       2 0  2      
S x  1 2 1 i  2 0 1 1 0       1 2 1 i        4 1 i   i 1        0 S y  1 2 1 i  2 0 i i 0       1 2 1 i        4 1 i   1 i        2 S z  1 2 1 i  2 1 0 0 1       1 2 1 i        4 1 i   1 i        0
S x 2  1 2 1 i  2 0 1 1 0       2 0 1 1 0       1 2 1 i        1 2 2       2 1 i   1 i        2       2 S y 2  1 2 1 i  2 0 i i 0       2 0 i i 0       1 2 1 i        1 2 2       2 1 i   1 i        2       2 S z 2  1 2 1 i  2 1 0 0 1       2 1 0 0 1       1 2 1 i        1 2 2       2 1 i   1 i        2       2
are S x  S x 2 S x 2  2       2 0  2       S y  S y 2 S y 2  2       2 2       2  0 S z  S z 2 S z 2  2       2 0  2      
The uncertainties

In the vector model, shown below, the spin is precessing around the y-axis at a constant angle such the y-component of the spin is constant and x- and z-components oscillate about zero.

Ch. 2 Solutions 3/20/19 2-10
[S2 ,S x ] 3 2 4 1 0 0 1       2 0 1 1 0       2 0 1 1 0       3 2 4 1 0 0 1       3 3 8 0 1 1 0       0 1 1 0                0 [S2 ,S y ] 3 2 4 1 0 0 1       2 0 i i 0       2 0 i i 0       3 2 4 1 0 0 1       3 3 8 0 i i 0       0 i i 0                0 [S2 ,S z ] 3 2 4 1 0 0 1       2 1 0 0 1       2 1 0 0 1       3 2 4 1 0 0 1       3 3 8 1 0 0 1       1 0 0 1                0
2.11 The commutators in matrix notation are

Ch. 2 Solutions

In abstract notation, the commutators are

x 2S y S y S x

S z 2S y S y S z

S x S x S y S y S x S x

i S z

S y 3 S y 3

S z S z S y S y S z S z  S x S y S x

S x S y i S z

S x  0  S z S y S z i S x 

S z S y  i S x

 S x S y S x  i S x S z S x S y S x  i S z S x  S z S y S z i S z S x S z S y S z i S x S z

 0 [S2 ,S z ]

S x 2  S y 2  S z 2 ,S z

S y 2S z S z S y 2

 S x S x S z S z S x S x

S z 2S z S z S z 2

S y S y S z S z

 0 2.12 For S x the diagonalization yields the eigenvalues

i S y S x S y S z S y  i S x S y

S z

3/20/19 2-11
[S2 ,S x ]  S x 2  S y 2  S z 2 ,S x      S x 2 ,S x      S y 2 ,S x      S z 2 ,S x      S x 2S x S x S x 2  S y 2S x S x S y 2  S z 2S x S x S z 2  S x 3 S x 3  S y S y S x S x S y S y  S z S z S x S x S z S z  0  S y S x S y i S z   S y S x  i S z  S y  S z S x S z  i S y   S z S x i S y  S z  S y S x S y i S y S z S y S x S y i S z S y  S z S x S z  i S z S y S z S x S z  i S y S z  0 [S2 ,S y ]  S x 2  S y 2  S z 2 ,S y      S x 2 ,S y      S y 2 ,S y      S z 2 ,S y      S
 
 
2  S y 2S y S y S y 2
2
 
 
    
z     
S
   
   S x 2 ,S z
S y 2 ,S
S z 2 ,
z
 S x 2S z S z S x 2
S y S y  S z 3 S z 3  S x S z S x i S y   S x S z  i S y  S x  S y S z S y  i S x   S y S z i S x  S y  0  S x S z S x i S x S y S x S z S x i S y S x  S y S z S y 
S x 2 0 1 0 1 0 1 0 1 0          2 0 2  2 0 2               0   2 2 2   2  2   0  2 2   0   1 ,0, 1 and the eigenvectors
3/20/19 2-12 2 0 1 0 1 0 1 0 1 0         a b c          1 a b c          b  a 2 a  c  b 2 b  c 2 a 2  b 2  c 2  1  b 2 1 2  1 1 2   1  b  1 2 ,a  1 2 ,c  1 2 1 x  1 2 1  1 2 0  1 2 1 2 0 1 0 1 0 1 0 1 0         a b c          0 a b c          b  0 a  c  0 b  0 a 2  b 2  c 2  1  a 2 1  1   1  a  1 2 ,b  0,c  1 2 0 x  1 2 1 1 2 1 2 0 1 0 1 0 1 0 1 0         a b c          1 a b c          b  a 2 a  c  b 2 b  c 2 a 2  b 2  c 2  1  b 2 1 2  1 1 2   1  b  1 2 ,a  1 2 ,c  1 2 1 x  1 2 1 1 2 0  1 2 1 For S y the diagonalization yields S y 2 0 i 0 i 0 i 0 i 0          i 2 0 i 2  i 2 0 i 2               0   2 2 2   i 2  i 2   0  2 2   0   ,0, and the eigenvectors 2 0 i 0 i 0 i 0 i 0         a b c          1 a b c          ib  a 2 ia ic  b 2 ib  c 2 a 2  b 2  c 2  1  b 2 1 2  1 1 2   1  b  i 2 ,a  1 2 ,c  1 2 1 y  1 2 1  i 2 0 1 2 1
Ch. 2 Solutions
Ch. 2 Solutions 3/20/19 2-13 2 0 i 0 i 0 i 0 i 0         a b c          0 a b c          ib  0 ia ic  0 ib  0 a 2  b 2  c 2  1  a 2 1  1   1  a  1 2 ,b  0,c  1 2 0 y  1 2 1  1 2 1 2 0 i 0 i 0 i 0 i 0         a b c          1 a b c          ib  a 2 ia ic  b 2 ib  c 2 a 2  b 2  c 2  1  b 2 1 2  1 1 2   1  b  i 2 ,a  1 2 ,c  1 2 1 y  1 2 1 i 2 0 1 2 1 2.13 The commutators are [S x ,S y ] 2 0 1 0 1 0 1 0 1 0         2 0 i 0 i 0 i 0 i 0         2 0 i 0 i 0 i 0 i 0         2 0 1 0 1 0 1 0 1 0         2 2 i 0 i 0 0 0 i 0 i         i 0 i 0 0 0 i 0 i                   2 2 2i 0 0 0 0 0 0 0 2i         i 1 0 0 0 0 0 0 0 1          i S z [S y ,S z ] 2 0 i 0 i 0 i 0 i 0         1 0 0 0 0 0 0 0 1         1 0 0 0 0 0 0 0 1         2 0 i 0 i 0 i 0 i 0         2 2 0 0 0 i 0 i 0 0 0         0 i 0 0 0 0 0 i 0                   2 2 0 i 0 i 0 i 0 i 0         i 2 0 1 0 1 0 1 0 1 0          i S x

The eigenvalue equation is

For spin-1 this is

Hence the S2 operator must be 2 2 times the identity matrix:

Ch. 2 Solutions 3/20/19 2-14 [S z ,S x ] 1 0 0 0 0 0 0 0 1         2 0 1 0 1 0 1 0 1 0         2 0 1 0 1 0 1 0 1 0         1 0 0 0 0 0 0 0 1         2 2 0 1 0 0 0 0 0 1 0         0 0 0 1 0 1 0 0 0                   2 2 0 1 0 1 0 1 0 1 0         i 2 0 i 0 i 0 i 0 i 0          i S y 2.14 Using the component matrices we find S2  S x 2  S y 2  S z 2 2 0 1 0 1 0 1 0 1 0         2 0 1 0 1 0 1 0 1 0          2 0 i 0 i 0 i 0 i 0         2 0 i 0 i 0 i 0 i 0          1 0 0 0 0 0 0 0 1         1 0 0 0 0 0 0 0 1         2 2 1 0 0 0 2 0 0 0 1          2 2 1 0 0 0 2 0 0 0 1          1 0 0 0 0 0 0 0 1         2 2 1 0 0 0 1 0 0 0 1        
S2 sm  s s  1   2 sm
S2 1m  2 2 1m
S2 2 2 1 0 0 0 1 0 0 0 1        

2.15 a) The possible results of a measurement of the spin component S z are always 1 , 0 , 1 for a spin-1 particle. The probabilities are

The three probabilities add to unity, as they must.

b) The possible results of a measurement of the spin component S x are always 1 , 0 , 1 for a spin-1 particle. The probabilities are

The three probabilities add to unity, as they must.

c) For the first measurement, the expectation value is

For the second measurement, the expectation value is

The histograms are shown below.

Ch. 2 Solutions 3/20/19 2-15
P1  1 in 2  1 2 29 1  3i 29 0 4 29 1     2  2 29 11  3i 29 1 0 4 29 1 1 2  2 29 2  4 29 P0  0 in 2  0 2 29 1  3i 29 0 4 29 1     2  2 29 0 1  3i 29 0 0 4 29 0 1 2  3i 29 2  9 29 P 1  1 in 2  1 2 29 1  3i 29 0 4 29 1     2  2 29 11  3i 29 1 0 4 29 1 1 2  4 29 2  16 29
P1x  x 1 in 2  1 2 1  1 2 0  1 2 1   2 29 1  3i 29 0 4 29 1  2  1 29  3i 2 29 2 29 2  1 58 2  3i 2  11 58 P0x  x 0 in 2  1 2 1 1 2 1   2 29 1  3i 29 0 4 29 1  2  2 2 29  4 2 29 2  36 58 P 1x  x 1 in 2  1 2 1 1 2 0  1 2 1   2 29 1  3i 29 0 4 29 1  2  1 29 3i 2 29 2 29 2  1 58 2 3i 2  11 58
S z  m P m m   1 4 29  0 9 29  1   16 29  12 29
S x  m P mx m   1 11 58  0 36 58  1   11 58  0

2.16 a) The possible results of a measurement of the spin component S z are always 1 , 0 , 1 for a spin-1 particle. The probabilities are

The three probabilities add to unity, as they must.

b) The possible results of a measurement of the spin component S y are always

1 , 0 , 1 for a spin-1 particle. The probabilities are

Ch. 2 Solutions 3/20/19 2-16
P1  1 in 2  1 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 11 y  3i 29 1 0 y 4 29 1 1 y 2  2 29 1 2  3i 29 1 2 4 29 1 2 2  1 58 2  3i 2  11 58 P0  0 in 2  0 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 0 1 y  3i 29 0 0 y 4 29 0 1 y 2  2 29 i 2  3i 29 0 4 29 i 2 2  1 58 i 2  4  2  36 58 P 1  1 in 2  1 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 11 y  3i 29 1 0 y 4 29 1 1 y 2  2 29 1 2  3i 29 1 2 4 29 1 2 2  1 58 2  3i 2  11 58
P1y  y 1 in 2  y 1 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 y 11 y  3i 29 y 1 0 y 4 29 y 1 1 y 2  2 29 2  4 29 P0y  y 0 in 2  y 0 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 y 0 1 y  3i 29 y 0 0 y 4 29 y 0 1 y 2  3i 29 2  9 29

The three probabilities add to unity, as they must.

c) For the first measurement, the expectation value is

For the second measurement, the expectation value is

The histograms are shown below.

2.17 a) The possible results of a measurement of the spin component S z are always

1 , 0 , 1 for a spin-1 particle. The probabilities are

The expectation value of S z is

b) The expectation value of S x is

Ch. 2 Solutions 3/20/19 2-17 P 1y  y 1 in 2  y 1 2 29 1 y  3i 29 0 y 4 29 1 y     2  2 29 y 11 y  3i 29 y 1 0 y 4 29 y 1 1 y 2  4 29 2  16 29
S z  m P m m   1 11 58  0 36 58  1   11 58  0
S y  m P my m   1 4 29  0 9 29  1   16 29  12 29
P  1  2  1 0 0   1 30 1 2 5i         2  1 30 1 2  1 30 P0  0  2  0 1 0   1 30 1 2 5i         2  1 30 2 2  4 30 P  -1  2  0 0 1   1 30 1 2 5i         2  1 30 5i 2  25 30
S z  P  P0 0  P   1 30  4 30 0  25 30   24 30  4 5

2.18

a) The possible results of a measurement of the spin component S z are always 1 , 0 , 1 for a spin-1 particle. The

b) After the S z measurement, the system is in the state

measurement of the spin component

c) Schematic of experiment.

Ch. 2 Solutions 3/20/19 2-18 S x   S x   1 30 1 2 5i   2 0 1 0 1 0 1 0 1 0         1 30 1 2 5i          1 30 2 1 2 5i   2 1 5i 2          1 30 2 2  2 1 5i   5i  2   2 15
probabilities are P1  1 in 2  1 1 14 1 3 14 0  2i 14 1     2  1 14 11 3 14 1 0  2i 14 1 1 2  1 14 2  1 14 P0  0 in 2  0 1 14 1 3 14 0  2i 14 1     2  1 14 0 1 3 14 0 0  2i 14 0 1 2  3 14 2  9 14 P 1  1 in 2  1 1 14 1 3 14 0  2i 14 1     2  1 14 11 3 14 1 0  2i 14 1 1 2  2i 14 2  4 14
1 . The possible results
a
S x are always 1 , 0 , 1 for a spin-1
The probabilities are P1x  x 1 in 2  1 2 1  1 2 0  1 2 1   1 2  1 2 2  1 4 P0x  x 0 in 2  1 2 1 1 2 1   1 2  1 2 2  1 2 P 1x  x 1 in 2  1 2 1 1 2 0  1 2 1   1 2  1 2 2  1 4
of
particle.

2.19 The probability is

2.20 Spin 1 unknowns. Follow the solution method given in the lab handout. (i) For unknown number 1, the measured probabilities are

3/20/19 2-19
Ch. 2 Solutions
P  f   f i 2  1 i 7 y 1  2 7 y 0  i 7 y 1   1 6 1 2 6 0  i 3 6 1  2  1 i 7 1 6 y 11 1 i 7 2 6 y 1 0  1 i 7 i 3 6 y 1 1  2 7 1 6 y 0 1 2 7 2 6 y 0 0  2 7 i 3 6 y 0 1   i 7 1 6 y 11 i 7 2 6 y 1 0  i 7 i 3 6 y 1 1 2  1 i 7 1 6 1 2  1 i 7 2 6 i 2 1 i 7 i 3 6 1 2  2 7 1 6 1 2 2 7 2 6 0  2 7 i 3 6 1 2   i 7 1 6 1 2 i 7 2 6 i 2 i 7 i 3 6 1 2 2  1 168 1 i  2  2i 3 i 3  2 2 0  i2 6  i  2  3 2  1 168 5  2 2  2i i 3  i2 6 2  1 168 5  2 2   2  2 3  2 6   2    1 168 64  8 2 4 3  8 6   0 524 or in matrix notation P  f   f i 2  1 i 7 1 2 i 2 1 2   2 7 1 2 0 1 2   i 7 y 1 2 i 2 1 2     1 6 1 2 i 3           2  1 168 1  2 2 2 2 i 2 1  2 2   1 2 i 3           2  1 168 1  2 2  4  2i i 3  i2 6 2  1 168 64  8 2 4 3  8 6   0.524
P1  1 4 P1x  1 4 P1y  1 P0  1 2 P0x  1 2 P0y  0 P 1  1 4 P 1x  1 4 P 1y  0 Write the unknown state as

Equating the predicted S z probabilities and the experimental results gives

allowing for possible relative phases. So now the unknown state is

Equating the predicted S x probabilities and the experimental results gives

Equating the predicted S y probabilities and the experimental results gives

Ch. 2 Solutions 3/20/19 2-20 1  a 1  b 0  c 1
P1  1 1 2  1 a 1  b 0  c 1  2  a 2  1 4  a  1 2 P0  0 1 2  0 a 1  b 0  c 1  2  b 2  1 2  b  1 2 e i P 1  1 1 2  1 a 1  b 0  c 1  2  c 2  1 4  c  1 4 e i
1  1 2 1  1 2 e i 0  1 2 e i 1
P0 x  x 0 1 2  1 2 1 1   1 2 1  1 2 e i 0  1 2 e i 1  2  1 2 2 1 e i  2  1 8 1 e i  1 e i   1 8 1 1 e i e i   1 4 1 cos   1 2  cos 1    Giving the state 1  1 2 1  1 2 e i 0 1 2 1
P1y  y 1 1 2  1 2 1 i 2 0 1 2 1   1 2 1  1 2 e i 0 1 2 1  2  1 4 i 2 e i  1 4 2  1 4 1 iei  1 ie i   1 4 1 1 iei  ie i   1 2 1  sin   1  sin  1     2 Hence the unknown state is 1  1 2 1  1 2 e i  2 0 1 2 1  1 2 1  i 2 0 1 2 1  1 y
For unknown number 2, the measured probabilities are P1  1 4 P1x  9 16 P1y  0 870 P0  1 2 P0x  3 8 P0y  0 125 P 1  1 4 P 1x  1 16 P 1y  0.005 Write the unknown state as 2  a 1  b 0  c 1
the predicted S z probabilities and the experimental results gives Quantum Mechanics A Paradigms Approach 1st Edition McIntyre Solutions Manual Full Download: http://testbanktip.com/download/quantum-mechanics-a-paradigms-approach-1st-edition-mcintyre-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
(ii)
Equating

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