PH 401
Dr. Cecilia Vogel
Review
Resuscitating Schrödinger's cat
Pauli Exclusion Principle
EPR Paradox
Outline
Sx, Sy, Sz eigenstates
spinors, matrix representation
states and operators as matrices
multiplying them
Eigenstates of Spin Components
Let the eigenstates of Sz be represented by
|+>
and |->
Sz|+> = /2 |+>
Sz|-> = -/2 |->
Then the eigenstates of Sx and Sy
are not eigenstates of Sz
but rather linear combinations of |+> and |->
Eigenstates of Sx and Sy
The eigenstates of Sx and Sy
are linear combinations of |+> and |->
1
|Sx=+/2> = 2 |    |  
|Sx=-/2> = 12 |    |  
1
|Sy=+/2> = 2 |   i |  
|Sy=-/2> = 12 |   i |  
Note that the amplitude of |+> and |-> in these
equations for the eigenstates, when we take the
absolute square, gives us the probability …
that we will find that value of Sz, given that the
particle is in that eigenstate of Sx (or Sy)
Matrix notation
Instead of writing all that out,
we can write each as a matrix, listing the
amplitudes in the linear combo:
1
|+> =  0 
|-> =  10 
|Sx=+/2> = 12 11
 
|Sx=-/2> =
|Sy=+/2> =
|Sy=-/2> =
 1 


 1
1  1
 
2  i 
1
2
1
2
 1 


i
Overlap
The overlap of two states |a> and |b>
is written <b|a>
bra…ket
tells you how much state |a> is like state |b>
can be calculated by multiplying matrices
ket matrix is the column matrix we just saw
bra matrix is the transpose (turn row into
column) and complex conjugate
The absolute square of the overlap
is the probability of finding state |b> when
observing a particle known to be in state |a>
Overlap
Let’s calculate the overlap of |Sy=+/2> and |->
 0
1
1
i



1

i
1
  2 (0  i)  2
 < Sy=+/2 |->= 2
 
note complex conj
If we take the absolute square of this, we get the
probability that a particle known to have Sy=+/2
will be found to have Sz=-/2. That prob=1/2.
Likewise the probability that a particle known to
have Sx=+/2 will be found to have Sy=-/2 can be
calculated:
1 1
1
1 i


1 i 

(
1

i
)

< Sy=-/2 | Sx=+/2 >= 2 1 2
2
Prob = 1  i   1  i  1  i   1
2
 2  2  2
(Note: you’ll get ½ any time the two dir’s are ┴ )
2
Overlap
Let’s calculate the overlap of |Sy=+/2> and
| Sy=+/2 >
1 1
1

1  i 
i 
  (1  i )  1
 < Sy=+/2 | Sy=+/2 >= 2
  2
2
If we take the absolute square of this, we get
the probability that a particle known to have
Sy=+/2 will be found to have Sy=+/2.
That prob better be 1, or 100%
If you know it has Sy=+/2, then it is in an
eigenstate of Sy, and you will find that value
100% of the time
generally: <a|a> = 1
if state is normalized
Matrix notation
In matrix notation
states are written as column vectors,
operators are written as square matrices.
Operating with an operator on a state
means multiplying a square matrix by a
column matrix
… the result is a column matrix,
as it should be:
when you operate on a state, you should get
a (probably un-normalized) state
Matrix notation
In matrix notation
the observable operators corresponding to each
component of the spin are given by these
matrices:
 0 1
Sx = 2  1 0 
0  i
Sy = 2  i 0 
0 
Sz =   10 1

2

Eigenstates
We can verify that the eigenstates ?I gave earlier
are indeed eigenstates with the stated eigenvalue
For example, is |+> and eigenstate of Sz?
0  1    1  0    1 
 1


Sz |+>= 2  0 1 0   2  0  0   2  0 
=/2 times the original state
so it is an eigenstate with eigenvalue /2
Similarly Sy | Sy=-/2>
= 2  2  0i 0i  1i   2  2  i001  2  2  i1  2 2  1i 
=- /2 times the original state
so it is an eigenstate with eigenvalue -/2
PAL