General Structure of Wave Mechanics (Ch. 5) • • • • Sections 5-1 to 5-3 review items covered previously use Hermitian operators to represent observables (H,p,x) eigenvalues.
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Transcript General Structure of Wave Mechanics (Ch. 5) • • • • Sections 5-1 to 5-3 review items covered previously use Hermitian operators to represent observables (H,p,x) eigenvalues.
General Structure of Wave Mechanics (Ch. 5)
•
•
•
•
Sections 5-1 to 5-3 review items covered previously
use Hermitian operators to represent observables (H,p,x)
eigenvalues of Hermitian operators are real and give the expectation values
eigenvectors for different eigenvalues are orthogonal and form a complete set of
states
• any function in the space can be formed from a linear series of the eigenfunctions
• some variables are conjugate (position, momentum) and one can transform from one
to the other and solve the problem in either’s “space”
eigenfunctions : ui ( x )
i | j ui | u j
( x, t )
C
N
*
u
i
u j dx ij dot product
uN ( x )e iEt /
p * pop dx * ( i
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) d x
x
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Notation
•
•
•
•
there is a very compact format (Dirac notation) that is often used
|i> = |ui> = eigenfunction
<c|f is a dot product between 2 function
|i><j| is an “outer” product (a matrix). For example a rotation between two different
basis
• if an index is repeated there is an implied sum
eigenfunctions : ui ( x )
i | j ui | u j
( x)
C
N
u u dx
*
i
uN ( x )
j
ij
dot product
CN n n n
n n projectionoperator
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Degeneracy (Ch. 5-4)
• If two different eigenfunctions have the same eigenvalue they are degenerate
(related to density of states)
• any linear combination will have the same eigenvalue
Ou i aui
Ou j au j
O (ui u j ) a (ui u j )
• usually pick two linear combinations which are orthogonal
• can be other operators which have only some specific linear combinations being
eigenfunctions. Choice may depend on this (or on what may break the degeneracy)
• example from V=0
sin kx, coskx, e
ikx
,e
ikx
2k 2
E
2m
sin kx, coskx orthogonal: sin kx coskxdx 0
eikx , e ikx
orthogonal: ( e ikx )*eikxdx 0
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Degeneracy (Ch. 5-4)
• Parity and momentum operators do not commute
( x )
( x )
) i
x
x
( x )
pop ( P ( x )) i
x
[ P, pop ] Ppop pop P 2i
P
x
P ( pop ( x )) P ( i
• and so can’t have the same eigenfunction
• two different choices then depend on whether you want an eigenfunction of Parity or
of momentum
sin kx, coskx
eikx , eikx
Parity
m om entum
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Uncertainty Relations
(Supplement 5-A)
• If two operators do not commute then their uncertainty product is greater then 0
• if they do commute 0
• start from definition of rms and allow shift so the functions have <U>=0
( A) 2 A2 A 2
A2 * A2 dx
U A A U 0, ( U ) 2 ( A) 2
• define a function with 2 Hermitian operators A and B U and V and l real
f ( x ) (U ilV ) ( x )
U A A
V B B
• because it is positive definite
I (l ) f *f d x 0
• can calculate I in terms of U and V and [U,V]
I
((U ilV ) )
*
(U ilV ) dx
* (U dag ilV dag )(U ilV ) dx
U dag U T * U
Herm itian
* (U ilV )(U ilV ) dx
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Uncertainty Relations
• rearrange
I
*
*
(Supplement 5-A)
(U ilV )(U ilV ) dx
(U 2 il (UV VU ) l2V 2 ) dx
• But just the expectation values
I U 2 il ( UV VU ) l2 V 2
U 2 il [U ,V ] l2 V 2
• can ask what is the minimum of this quantity
dI
i [U ,V ]
0 lmin
dl
2 V2
• use this “uncertainty” relationship from operators alone
I (lmin ) U 2 ilmin [U ,V ] l2min V 2 0
1
i[U ,V ] 2
4
1
i[ A, B ] 2
4
U 2 V 2
( A) 2 ( B ) 2
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Uncertainty Relations -- Example
• take momentum and position operators
• in position space
xop x
pop i
x
px i
x i ix
x
x
x
px xp [ p, x ] i
xp ix
• that x and p don’t commute, and the value of the commutator, tells us directly the
uncertainty on their expectation values
( A) 2 ( B ) 2
px
1
i[ A, B ] 2
4
1
1
i[ p, x ] 2
4
2
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i ( i) 2
2
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Time Dependence of Operators
• the Hamiltonian tells us how the expectation value for an operator changes with time
A t
*
( x, t ) A ( x, t )dx
d
A
A t * ( x , t )
( x, t )dx
dt
t
*
*
[
(
x
,
t
)]
A
(
x
,
t
)
dx
(
x
,
t
)
A
[
( x, t )]dx
t
t
• but know Scrod. Eq.
1
*
1
i
H
(
H )* H * *
t
i
t
i
H *T H ( H )* A *HA
• and the H is Hermitian
• and so can rewrite the expectation value
d
A
A t
dt
t
d
A
A t
dt
t
i
i
*
*
HA
dx
AHdx
i
H , A
t
t
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Time Dependence of Operators
II
• so in some sense just by looking at the operators (and not necessarily solving S.Eq.)
we can see how the expectation values changes.
d
A
i
H , A
A t
t
dt
t
• if A doesn’t depend on t and [H,A]=0 <A> doesn’t change and its observable is a
constant of the motion
• homework has H(t); let’s first look at H without t-dependence
• and look at the t-dependence of the x expectation value
d
i
x
dt
H , x
V ( x ), x 0
d
x
dt
p
2
i
p2
H
V ( x)
2m
p2
2m V ( x ), x
, x p p , x p , x p
2
p
i
p
m
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Time Dependence of Operators
III
• and look at the t-dependence of the p expectation value
d
i
p
dt
H , p
i
p2
2m V ( x ), p
i
p ,V ( x )
( pV Vp ) p (V ) V ( p )
d
d
dV
(V ) V
i dx
i dx
i dx
• rearrange giving
p ,V ( x )
and
dV
d
i dx
dt
d2
m
dt2
x
p
dV ( x )
dt
dV ( x )
dt
• like you would see in classical physics
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