Transcript Too Many to Count

Too Many to Count
Three Notations
The Three Notations of Quantum
Mechanics

There are three notations (dialects if you like)
commonly used in quantum mechanics
 Sometimes they can be used interchangeably
and sometimes not


Each has a strength and each has a weakness
They are named for the 3 “fathers” of quantum
mechanics



Schroedinger
Heisenberg
Dirac
How they compare
Notation
Name
Type
Example
Comments
Schroedinger
“Wavefunction”
Math
function
Y =f(x,y,z,t)
Great for
integrating, doe
not handle
groups
Heisenberg
Column
matrix
Matrix
Dirac
“ket”
Both matrix
and math
function
1
 
0
some label
x
a
Handles spin,
isospin, and
some flavors,
can’t integrate
easily
Very
ambiguous,
sometimes too
ambiguous
Postulate 1
Quantum Mechanical
States are described
by vectors in a linear
vector space

Linear vector space means a field of
scalars over which the space
 From section 4.4 of Liboff’s text
Actually this is nothing new
First, define a basis such as
1 0
0
  
 
x̂, ŷ, and ẑ or  0 ,  1 , and  0  or 1 , 2 , and 3
 0  0
1
  
 

Schroeding er Y (x, y, z)  a 1 1x̂  a 2 2 ŷ  a 3 3 ẑ
Dirac
a  a1 1  a 2 2  a 3 3
Heisenberg
 a1 
 
a2 
a 
 3
Postulate 2
A dual space exists
with the same
dimensionally as the
original vector space
AKA “dual continuum”
the existence of a dual vector space reflects in an abstract way
the relationship between row vectors (1×n) and column vectors
(n×1)
 Required to allow the inner product so that vectors can be
normalized


Dual Spaces for the notations

To transform a vector from one space to
another, a Hermitian conjugation is
performed.
For example, a " bra"
a  a


means Hermitian conjugate or transpo se conjugate
a
a   
 ib 
a  a
Obviously, a

 ib  if a and b are real
 a
In Schroeding er Notation
Y  Y * e.g. Y  a  ib  Y *  a  ib
Postulate 3
An inner product
exists.

Back in E&M, we called the inner product: “dot
product”

Inner product = dot product = scalar product
In the 3 notations
Schroedinger
Heisenberg
Dirac



a
a
*
1
b
d
 b1 
a  
 b2 
*
2

ab
If a b  0 then a  b
Postulate 4
The dual space is
linear and has the
following property
If a  a1 1  a2 2
then
a a 1 a 2
*
1
*
2
Postulate 5
ab  ba
*
Postulate 6
a a 0

Only 0 if a  0, the null vector
Postulate 7
Multiplying a ket by a
complex number
(different from zero) does
not change the physical
state to which the ket
corresponds
Postulate 7 is discussing
normalization
Let a a  a
2
We can form a normalized ket, a'
such that a' a'  1 by defining
a' 
1
a
2
a
a' and a still represent the same physical state
It is convenient for define an orthonormal basis
(and you’ve been doing it all your life!)
In Physics 350, you used xˆ, yˆ, zˆ such that
xˆ  xˆ  1 xˆ  yˆ  0 or generally, mˆ  nˆ  δmn
In Dirac Notation,
x , y , z where m n  δmn
Operators

A mathematical
operation on a vector
which changes that
vector into another


This is not mere
multiplication (like
Postulate 7) but we are
actually changing
something like its
direction or perhaps
other quantities.
Example: Let Q be the
differential operator
with respect to x
Direction of operation
Qˆ n  m
or

ˆ
nQ  m
Direction of operation
Postulate 8
Physical observables
(such as position or
momentum) are
represented by linear
Hermitian operators
What does linear mean?
Operator, Qˆ , is linear if it satisfies two conditions :
1) Qˆ c a   cQˆ a where c is a constant
2) Qˆ  a  b   Qˆ a  Qˆ b
What does Hermitian mean?
Operator, Qˆ , is Hermitian if :

ˆ
ˆ
a Qb  aQ b
  


ˆ
ˆ
QQ
called " self adjoint"
A special case for operators
Mostly
Qˆ a  b
but sometimes, we see the following

*
ˆ
ˆ
Q a  q a or a Q  q a
Called
“Eigenvector” or
“Eigenfunction” or
“Eigenket”
Called “eigenvalue”
What does an eigenvalue mean in
Schroedinger notation?

ˆ
Let Q 
x
Qˆ   q
and   e ikx
 ikx
e  ikeikx  ik
x
ik  q
ik  q,
ik is the eigenvalue for this function
What does an eigenvalue mean in
Heisenberg notation?
0 1
ˆ
 and
Let Q  
1 0
Qˆ a  q a
1
a   
1
 0 1 1  0  1 1

   
     a
 1 0 1 1  0  1
1a q a
1  q,
or 1 a
1 is the eigenvalue for this function
1
What if a    ?
  1
Theorem 1
Eigenvalues of a
Hermitian operator
are real
i.e.
If Q+=Q then q*=q
Proof of Thm 1
Let Qˆ a  q a
so
 
and
a Qˆ a  a q a
a Qˆ a  q a a
a Qˆ a  q
a a 1
Let a Qˆ   a q *
so

 a Qˆ  a   a q  a

*
a Qˆ  a  q * a a
but Qˆ   Qˆ
a Qˆ a  q *
q  q * , QED
Theorem 2
Eigenvectors of a
Hermitian operator
are orthogonal if they
belong to different
eigenvalues
Proof of Thm 2
Let Qˆ a  q1 a
and
Qˆ b  q2 b
q1  q2  q1  q2  0
so
a Qˆ   q1* a
Since Hermitian, a Qˆ  q1 a
 
a Qˆ b  q1 a  b
a Qˆ b  q1 a b
 
a Qˆ b  a q2 b
a Qˆ b  q2 a b
Obviously
q1 a b  q2 a b
or q1  q2  a b  0
Since q1  q2  0 then a b  0  a  b
QED

Note: An operator may
have a set of
eigenvalues of which 2
or more are equal; this is
called degeneracy
Projection operators

Graphically,
the inner
product
represents the
project of a
onto b or in
Dirac notation
|a> onto |b>
|b>
|a>
<a|b>
If |a> is considered a unit vector, then the
vector which represents projection of |b>
onto |a> is written <a|b>|a> or |a><a|b>
Theorem 3
A projection operator
is idempotent i.e.
Q2 =Q
Proof :
Let a a  1 and Qˆ  a a
Qˆ 2   a a
 a
Qˆ 2  a a a a
Qˆ 2  a a  Qˆ
Qˆ 2  Qˆ QED
a

Theorem 4
If 1 , 2 , 3 , k form an orthogonal basis, then
k
k
k  Iˆ, the identity matrix
Proof of Thm 4
Let a be an arbitraril y chosen vec tor and can be written in terms of a set
of basis vectors :
a  cj j
j
Have a projection operator, k k , operate on a
k k a   c j k k j   c j k  k , j ck k
j
j
So
k
k a   ck k
k
k
But k has same dimension of j so  ck k  a
k
k
k a  a
k
k
k
k 1
Creating a set of orthogonal vectors from a set of
normalized linear independent kets
Let |a>, |b>, and |c> be a set of
normalized linear independent kets
 We are going to create a new set of kets
(|1>, |2>, |3>) from these which will be
orthogonal to one another i.e. <1|2>=0,
<1|3>=0 and <2|3>=0
 First, pick one of the original set and
build the rest of the set around it
 |1>=|a>

Constructing |2>
Geometrically
|2>=|b>-|1><1|b>
|b>
|b>-|1><1|b>
|1><1|b>
-|1><1|b>
|1>
Test that |2> is orthogonal to |1>
2  b 1 1b
1 2  1b  1 1 b

1 2  1 b  11 1 b
12  1b  1b
QED
Normalizing |2>
2  b  1 1 b and 2  b  b 1 1

1b
2 2   b  b 1 1  b  1 1 b
2 2  b  b  1 1 b  b 1
1 1b

2 2  b b  b1 1b  b1 1b  b1 11 1b
2 2  1 b 1 1 b  b 1 1 b  b 1 1 b
2 2  1 b 1
2
Normalized 2 is
2 
b 1 1b
1 b 1
2
|3>
You may almost guess the structure of 3
3  c  1 1 c  2 2 c (unormaliz ed)
with a normalizat ion constant of 1  1 c
3 
c 1 1c  2 2c
1 1 c
2
 2c
2
k 1
Generalizi ng
k 
j  L L j
L 1
k 1
1  L j
L 1
2
2
 2c
2
Postulate 9
Eigenvalues are the
only possible
outcome of physical
measurements

If physical observables are represented by
Hermitian operators and these have real
eigenvalues, it is reasonable to assume that
there is a connection between their eigenvalues
and the results of experiments.
Theorem 5
Operators
representing
simultaneously
observable quantities
commute
Proof of Thm 5
Let
Qˆ a  q a


Rˆ Qˆ a  Rˆ q a
Rˆ a  r a
 

Rˆ Qˆ a  qRˆ a  qr a

Qˆ Rˆ a  Qˆ r a


Qˆ Rˆ a  rQˆ a  rq a
Rˆ Qˆ a  Qˆ Rˆ a  Rˆ Qˆ  Qˆ Rˆ a
qr a  rq a  qr  rq  a  0


Rˆ , Qˆ   0 therefore they commute
Rˆ Qˆ  Qˆ Rˆ a  0
Commutator Brackets
[a,b]=(ab-ba)
If [a,b]=0 then a and b
commute
QM analog of Poisson
brackets
An Example of non-commuting
operators
 
Consider  , x 
 x 
Use a dummy function,  , to evaluate

 
 
 x , x   x  x   x x  

 
 
 x , x     x x   x x  
 
 x , x   
 
 x , x   1


Postulate 10
The average value in
the state |a> of an
observable
represented by an
operator Q, is
Qˆ 
ˆ
aQa
a a
Called an
“expectation value”
or called the
“mean”
In Schroedinger Notation
Average value of position

x   x  dx
*

In Heisenberg Notation
0 1
1 1
 and a 
 
Let  x  
2 1
1 0
Find the expectatio n value of  x in state a
1
1 1
a 
2
  0 1  1 1 
1
 
  
1 1 
a x a 
2
  1 0  2 1 
1
1
a  x a  1 1   1
2
1
Defining Standard Deviation

Let Q= operator
 DQ=
standard deviation of measurement of
Q
 (DQ)2= variance of that measurement
Sometimes called mean square deviation
from the mean
 (DQ)2 =<(Q-<Q>)2>
 Or, more compactly
 (DQ)2 =<Q2>-<Q>2

The Uncertainty Principle




If two observables are represented by commuting
operators then you can measure the physical
observables simultaneously
If the operators DO NOT COMMUTE then a
SIMULTANEOUS measurement will NOT BE
EXACTLY REPEATABLE
There will be a spread in the measurement such that
the product of the standard deviations will exceed a
minimum value; the size of the minimum depends on
the observable
To calculate this, we first have to build some
mathematical machinery.
Theorem 6
Schwartz’s Inequality
j j k k  j k
2
Proof of Thm 6
Construct a ket
f  j k j  j j k
f  j k j  k j j
f f  j k j j k j  j k j j j k
 k j j j k j  k j j j j k
2
f f  j k
f f  j k
f f  j j
j j  j k
2
2
j j  j j
j j
j j  j k
2
k k  j k
j j k k  j k
QED
j j  j j
k k
2

f f  0 and j j  0  j j k k  j k
2
2
2
0
2
k k
Theorem 7
Let a = A-<A> and
b =B -<B> then
[a,b] =[A,B]
Proof
α,β   A  A B 
α,β   AB  B A 
B   B  B A  A
A B A B
 BA  A B  B A  A B
α,β   AB  BA  A,B QED

Derivation of the Uncertainty
Principle for any Operator
Using the definition s in Thm 7,
2
DA 
 A  A
2
DB
 B  B
2

 a
2

 b2
2
Let' s calculate DA   DB
2
DA   DB 
DA 2  DB2 
2
2
2
a2 b2
aa a a b a
2
2
Derivation of the Uncertainty
Principle … page 2
DA 2  DB2 
a a2 a a b2 a
Let
aa  j
so
and
ba  k
DA 2  DB2   a a a a  a b b a 
2
2
2
DA   DB  j j k k  j k
Need more power!

Now the absolute square of any complex
number, z, can be written as


|z|2 = (Re(z))2 +(Im(z))2
Of course, |z|2  (Im(z))2
DA   DB
 j k
DA   DB
 
2
2
2
2
2
 Re j k
  Im
2
j k

2 2
An Aside
Let
z  a  ib
then
zz
a  ib  a  ib  2ib
Im (z) 


b
2i
2i
2i
*
So we can now start having fun…
DA   DB
 j k
DA   DB
 
2
2
2
2
2
 Re j k
  Im
2
j k

2
Since sum of Real and Imaginary greater th an its
parts,
DA   DB
2
2
 
 Im j k

2
 j k  k j 

DA   DB  

2
i


2
2
2
The final slide
 j k  k j 

DA   DB  

2
i


2
2
2
 a ab a  a ba a 

DA   DB  

2
i


2
2
 a , b 
DA   DB  
 2i
2
2
 A, B 
DA   DB  
 2i
2
2




2




2
2
Does it work?
 j k  k j 

DA   DB  

2i


 
Let A 
 px
i x
Bx
2
2
2
  p , x 
Dp x   Dx    x 
 2i 
 

 
Recall  , x   1  , x    p x , x  
i  x 
i
 x 
2
2
2

 
2
2
Dp x   Dx    i 
 2i 
 
 
2
Dp x 2  Dx 2  
4
Eliminatin g squares

Dp x Dx 
2
2