5.73 Lecture #13
1.
13 - 1
End Matrix Solution of H-O, a + a† Operators
p2 1 2
starting from H =
+ kx
and [x,p] = ih
2m 2
2.
showed p nm = m x nm ( E m − E n )
ih
i
x nm =
p nm ( E m − E n )
hk
x nm 
∴ x nn = 0, p nn = 0 and 
 = 0 if E n = E m
p nm 
3.
x 2nm = −
1 2
p nm
km
E m − E n = ±ihω
ω = ( k m )1/2
∴ the only non-zero x and p elements are between states whose E’s differ by ±hω
4.
5.
(i)
combs of connected states, block diag. of H, x, p, x2, p2 E n = hωn + ε i
lowest index must exist because lowest E must exist. Call this index 0
2 h
x 01 = (km)−1/2
2
from phase choice
2 h
+1/2
p 01 = (km)
x 01 = +i(km)−1/2 p 01
2
Today
6.
Recursion Relationship
x nn+1
2
in terms of x nn-1
2
2
general matrix elements x nn+1 , p nn+1
7.
general x and p elements
8.
only blocks correspond to ε i =
2
1
hω
2
Dimensionless x, p, H and a (annihilation) and a† (creation) operators
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 2
phase ambiguity: we can specify absolute phase of x or p BUT NOT BOTH because that
would affect value of [x,p]
BY CONVENTION:
matrix elements of x are REAL
p are IMAGINARY
x 01 = +i( km )−1/2 p 01 and plug this into
x 01p*01 − p 01x*01 = ih
try
get
x 01
p 01
2
2
−1/ 2
h
= ( km )
2
=
+1/ 2
h
km
( )
2
If we had chosen x 01 = −i( km )−1/ 2 p 01 we would have

obtained x 2 = − h ( km )1/ 2 which is impossible!
01

2




check for self-consistency of seemingly arbitrary phase
choices at every opportunity: * Hermiticity
*
6.
Recursion Relation for x ii+1
2
≥0
2
start again with gerand equation derived in #3 above using the phase
choice that worked above
−1/2
x nn+1 = i(km)
p nn+1
going up
Hermiticity
c.c. of both sides
x*n+1n = i( km )−1/2 p*n+1n
x n +1n = − i( km )−1/ 2 p n +1n
going down
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 3
∴ x nn±1 = ±i( km )−1/2 p nn±1
now the arbitrary part of the phase ambiguity in the relationship between x
and p is eliminated
Apply this to the general term in [x,p] ⇒ algebra
NONLECTURE : from terms in [x,p] = ih
 ( km )1/2 * 
x nn+1p n+1n = x nn+1p*nn+1 = x nn+1  −
x nn+1
i


= x nn+1
2
(+i(km)1/2 )
 ( km )1/2

2
−p nn+1x n+1n = −
x nn+1 x*nn+1 = x nn+1 +i( km )1/2
i


(
(
)
)
 ( km )1/2 * 
x nn−1p n−1n = x nn−1p*nn−1 = x nn−1  +
x nn−1
i


= x nn−1
2
(−i(km)1/2 )
 ( km )1/2

2
−p nn−1x n−1n = − −
x nn−1 x*nn−1 = x nn−1 −i( km )1/2
i


(
[
(
)
2
∴ ih = 2i( km )1/2 x nn+1 − x nn−1
h(km)−1/2
2
+ x nn−1
2
2
2 h
but x 01 = x10 = ( km )−1/2
2
2
)
]
2
x nn+1 =
thus
h
2
x nn+1 = ( n + 1) ( km )−1/2
2
h
2
p nn+1 = ( n + 1) ( km )+1/2
2
recursion
relation
general
result
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 4
7. Magnitudes and Phases for xnn±1 and pnn±1
verify phase consistency and Hermiticity for x and p
in # 3 we derived x nn±1 = ±i( km )−1/2 p nn±1
one self1/2


consistent set is 
h
1/2
x nn+1 = +( n + 1) 
= +x n+1n


 2( km )1/2 
x real

1/2
and

h

1/2 
= +x nn−1
positive
x nn−1 = +( n )  2 km 1/2 
 ( ) 

( km)1/2 = mω
AND
1/2

 h( km )1/2 
1/2

p imaginary
p nn+1 = −i( n + 1) 
= −p n+1n

2


w/sign flip for 

up vs. down
1/2  1/2

1/2  h( km )
= −p n−1n
 p nn−1 = +i( n ) 

2



Phase is a recurrent problem in matrix
mechanics because we never look at
This is the usual phase convention wavefunctions or evaluate integrals explicitly.
8. Possible existence of noncommunicating blocks along diagonal of H, x, p
k  1/2

you show that H nm = ( n + 1 / 2)h
δ nm
 m
 note that x 2 and p2 have non - zero ∆n = ±2 elements but

2
1

p
 kx 2 +
has cancelling contributions in ∆n = ±2 locations
2

2m
This result implies
all of the possibly independent blocks in x, p, H are
identical
* εi = (1/2)hω for all i
* degeneracy of all En? all same, but can’t prove that it
is 1.
*
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 5
Creation and Annihilation Operators (CTDL pages 488-508)
*
*
*
*
*
Dimensionless operators
simple operator algebra rather than complicated real algebra
matrices arranged according to “selection rules”
matrix elements calculated by extremely simple rules
automatic generation of any basis function by repeated operations on
lowest (nodeless) basis state
get rid of system-specific factors of k, µ, ω and also h.
ω = ( k /m)
1/2
1/2
dimensionless
 h  2
x2 = 
x
 mω  ~
mω 
x~ ≡ 
x
 h 
regular
−1/2
p
p ≡ (hmω )
p2 = hmω p2
1
1  2 2
=
H
=
x~ + p
H
~
~ 

hω
2


1 2 p2 1
= hω  x~ 2 + p2 
H = kx +
~ 
2
2m 2

~
We choose these
factors to make
everything come
out dimensionless.
~
1/2
1
   mω 1 
x
,
p
x,
p
=
=
(ih) = i
[
]
 ~ ~   h hmω 
h
from results for x, p, H
[
i[( n + 1)
x~ mn = 2 −1/ 2 ( n + 1) δ mn +1 + n1/ 2δ mn −1
1/ 2
p mn = 2 −1/ 2
~
1/ 2
]
]
square root of
larger q.n.
δ mn +1 − n1/ 2δ mn −1
H
~ mn = ( n + 1 / 2 )δ mn
note the negative sign
diagonal
Kronecker - δ’s specify selection rules for nonzero matrix elements
now define something new
more!
a, a† to clean things up even
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 6
( )
p = 2 −1/2 i(a† − a )
~


a = 2 −1/2 x~ + i p
~



a† = 2 −1/2 x~ − i p
~

x~ = 2 −1/2 a + a†
Let’s examine the matrix elements of a and a†


a mn = 2 −1/2 x~mn + 2 −1/2 i p mn 
~


1
1
1
1
=  ( n + 1)1/2 δ mn+1 − ( n + 1)1/2 δ mn+1 + n1/2δ mn−1 + n1/2δ mn−1 
2
2
2
2

group according to
“selection rule”
x~
ip
x~
~
~
add
cancel
a mn = n1/2δ mn−1
ip
first index is one smaller than second
column
a mn = m a n
row
similarly
a †mn = ( n + 1)1/2 δ mn+1
n1/2 n − 1
a is lowering or “annihilation” operator
first index is one larger than second
a† is a “creation” operator
updated 9/27/02 9:22 AM
5.73 Lecture #13
0
0 0
1 11/ 2

a † =2  0
3 0

4 0
13 - 7
1
2
3
0
0
21/ 2
0
0
0
0
0
31/ 2
0
0
0
0
0
41/ 2
4
0
0

0
0

0
square root of integers always
only one step below main
diagonal. a, a† are obviously not
Hermitian
e.g. 3 a † 2 = 31/ 2
a † raises
 0 11/ 2
0 0

a = 0 0
0 0

0 0
0
21/ 2
0
0
0
31/ 2
0
0
0
0
0 
0 

0 
41/ 2 

0 
square root of integers
always only one step
above main diagonal
e.g. 3 a 4 = 4 1/ 2
a lowers
What is so great about a, a†?
a n = n1/2 n − 1
a† n = ( n + 1)1/2 n + 1
( )
annihilates 1 quantum
creates 1 quantum
n
n = [ n!]−1/2 a† 0
generate any state from lowest one 0
needed to normalize. Each application of a† gives next
integer. Do it n times, get n!
more tricks: look at aa† and a†a
is aa† Hermitian?
(aa† )
†
[(AB)† = B†A†] definition of hermitian
= a††a† = aa†
∴ aa† and a†a are Hermitian — to what “observable” quantity do they
correspond? We will see that one of these is the “number operator.”
updated 9/27/02 9:22 AM
5.73 Lecture #13
aa † =
13 - 8

 1 2
1
2
 x~ + i p~   x~ − i p~  =  x~ + i p~ x~ − i x~ p~ + p~ 
2

 2




1
 1
=  x~ 2 + p 2 − i[ x~ , p] =  x~ 2 + p 2 + 1
~
~
~
2

{  2


i 

1
similarly a †a =  x~ 2 + p 2 − 1
~
2

(
1 †
†
∴ H
=
~ 2 a a + aa
H = a †a + 1 / 2
~
(
)
and
[a,a ] = 1 simple form for H
†


number operator + 1/2
†
H = hω H
~ = hω a a + 1 / 2
~
)
# of
quanta
a †a n = n n
[aa
†
n = (n + 1) n
a †a is "number operator"
]
not as useful
What have we done? We have exposed all of the “symmetry” and universality of the
H–O basis set. We can now trivially work out what the matrix for any XnPm operator
looks like and organize it according to selection rules.
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 9
What about X3?
mω 
X =
 h 
3
−3/2
When you multiply this out,
preserve the order of a and a†
factors.
3
X
~
( )( )
= (2 −3/2 )a 3 + (a†aa + aa†a + aaa† ) + (aa†a† + a†aa† + a†a†a ) + a† 


X3 = 2 −3/2 a + a†
~
3
3
∆n =
–3,
–1,
+1,
+3
(# of † minus
# of non-†)
Simplify each group using commutation properties so
that it has form
a a †a n
or
a † a †a n
[ ]
1/2
n
[ ]
⇓
⇓
1/2
( n + 1)
n n −1
n n +1
NONLECTURE: Simplify the ∆n = –1 terms.
†
a†aa = aa†a−aa
a†3
aa = aa†a − a
14
4a2+44
†
†
[ ]
add and subtract term
needed to reverse order
a † ,a a=−a
†
†
†
aaa = aa a−aa
+44
aaa
144a2
3 = aa a + a
[ ]
a a,a † =a
†
[a†aa + aa†a + aaa ] = 3aa†a
try to put everthing into aa†a order
NONLECTURE: Simplify the ∆n = +1 terms.
† †
aa†a† = a†aa† −a
aa 2+44
aa†3
a† = a†aa† + a†
144
[a,a† ]a† =a†
† †
†
† †
†
a †aa† = a†a†a−a
a a2
+44
a†aa
144
3 = a a a+a
[ ]
a † a,a † =a †
† †
† †
[aa†a† + a†aa† + a a a] = 3a a a + 3a†
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 10
∆n = ±3
1/2
a 3 n = [ n( n − 1)( n − 2)]
n−3
1/2
a†3 n = [( n + 1)( n + 2)( n + 3)]
n+3
∆n = ±1
[a
†
] ( )
+ a a a ] n = 3[(n + 1)
aa + aa †a + aaa † n = 3 n 3/ 2 n − 1
[aa a
† †
+ a †aa †
1/ 2
† †
= 3(n + 1)
1/ 2
= 3(n + 1)
3/ 2
n n + 1 + (n + 1)
1/ 2
n +1
(n + 1) n + 1
]
n +1
no need to do matrix multiplication. Just play with a, a† and commutation
rule and a†a number operator
“Second Quantization”
h 
x 3mn = 
 2mω 
3/2
[δ
same as |n+3⟩⟨n|
mn+3
(( n + 1)( n + 2)( n + 3))1/2
+ δ mn+1 3( n + 1)3/2
+ δ mn−1 3n 3/2
|n+1⟩⟨n|
1/2
+ δ mn−3 ( n( n − 1)( n − 2))
]
|n⟩⟨n–3|
|n⟩⟨n–1|
simple! x3 is arranged into four
separate terms, each with its
own explicit selection rule.
1 2
3
4
kx + ax
+ bx
142
4
3
2
anharmonic terms → perturbation theory
*
V(x) =
*
IR transition intensities ∝ n x n + 1
2
* Survival and transfer probabilities of initially prepared pure
harmonic oscillator non-eigenstate in anharmonic potential.
* Expectation values of any function of x and p.
updated 9/27/02 9:22 AM
5.73 Lecture #13
13 - 11
Universitality: all k,m (system-specific) constants are removed until we put
them back in at the end of the calculation.
e.g., What is ∆x
2
[
= x2 − x
2
]
1 (
† )2 
x=
x =
a +a 

mω ~ mω  2
2
∆x
2
h
h
=
(a + a )
[
2mω
h
† 2
− a + a†
2
]?
pure numbers in [ ]
updated 9/27/02 9:22 AM