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ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Quantum Mechanics for
Scientists and Engineers
Sangin KimAdvanced Computational Electromagnetics Lab
redkid@yonsei.ac.kr
Nov. 14th, 2016
ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Outline
Quantum Mechanics for Scientists and Engineers
• Bilinear expansion of linear operators
• The identity operator
• Inverse and Unitary operators
2
ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Bilinear expansion of linear operators
3
By acting with a specific operator
Expand functions in a basis set
n n
nnnn xcxforxcxf )()()()(
A
fAg ˆ
Expand g and f on the basis set i
i
iidg j
jjcf
From our matrix representation of fAg ˆ
j
jiji cAd we know that fc jj j
jiji fAd So,
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Bilinear expansion of linear operators
4
Substituting back into
Expand functions in a basis set
Remember that is simply a number
i
iidg
So, switched multiplicative order
fc jj
j
jiji fAd
i
ji
jij fAg ,
fAfAg ji
ji
ijji
ji
ij
,,
ji
ji
ijAA ,
ˆ
ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Bilinear expansion of linear operators
5
This form
Expand functions in a basis set
is referred to as a “bilinear expansion” of the operator
on the basis and is analogous to the linear
expansion of a vector on a basis
Though the Dirac notation is more general 11)(ˆ)( dxxfAxg
ji
ji
ijAA ,
ˆ
A
i
)()(ˆ1
*
,
xxAA ji
ji
ij
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Outer product
6
An expression of the form
Bilinear expansion form
Contains an outer product of two vectors
An inner product expression of the form
ji
ji
ijAA ,
ˆ
fg
An outer product expression of the form fg
Complex number
Matrix
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ELECTROMAGNETICS LAB
Outer product
7
The specific summation
is actually, then, a sum of matrices
ji
ji
ijAA ,
ˆ
In the matrix ji
the element in the ith row and the jth column is 1, another's are 0
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ELECTROMAGNETICS LAB
The identity operator
8
In matrix form, the identity operator is
The identity operator
In bra-ket form
where the
The identity operator can be written
I
i
iiI ˆ
i
form a complete basis for the space
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ELECTROMAGNETICS LAB
The identity operator
9
For an arbitrary function , we know
Proof
And, the multiplication each side of
So
i
iiI ˆ
i
ii ff
f
i
ii
i
ii fffI ˆ
By using, . So, i
ii ff ffI ˆ
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ELECTROMAGNETICS LAB
The identity operator
10
The statement
Similarly, we can obtain
i
iiI ˆ
0
0
1
1 So that
000
000
001
0010
0
1
11
100
010
001
ˆ
i
iiI
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Proof that the trace is independent of the
basis
11
Consider the sum,
of the diagonal elements of an operator
S
on some complete orthonormal basis
i
ii AS ˆ
i
And, suppose some other complete orthonormal basis
A
i
m
mmI ˆ
In , we can insert an identity operator just before i
ii AS ˆ A
AAI ˆˆˆ
i
i
m
mmi
i
ii AAIS ˆˆˆ
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Proof that the trace is independent of the
basis
12
Rearranging
i
i
m
mmi
i
ii AAIS ˆˆˆ
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ELECTROMAGNETICS LAB
Proof that the trace is independent of the
basis
13
So, with now
Using the
m
mm
i
ii IAAS ˆˆˆ
Hence the trace of an operator the sum of the diagonal elements
is independent of the basis used to represent the operator
AIA ˆˆˆ
m
mm
i
ii AAS ˆˆ
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ELECTROMAGNETICS LAB
Inverse and projection operator
14
Inverse operator
For an operator operating on an arbitrary function
Using the inverse operator
A
1ˆ A
f
The inverse operator
fAAf ˆˆ 1
IAA ˆˆˆ 1
Projection operator
For example,
the one corresponding to the specific vector
ffP ˆ In general has no inverse
f
because it projects all input vectors onto only one axis in the space
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ELECTROMAGNETICS LAB
Unitary operator
15
Unitary operator
One for which UU ˆˆ 1
That is, its inverse is its Hermitian adjoint
U
The Hermitian adjoint is formed by reflecting on a -45
degree line and taking the complex conjugate
ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Unitary operator
16
Conservation of length for unitary operators
For two matrices and A
That is, the Hermitian adjoint of the product is the
“flipped round” product of the Hermitian adjoints
B
Consider the unitary operator and vectorsU oldfoldg
oldnew fUf ˆ Using the operatoroldnew gUg ˆ
Then, So,Ugg oldnewˆ
The unitary operation does not change the inner product
The length of a vector is not changed by a unitary operator
ADVANCED COMPUTATIONAL
ELECTROMAGNETICS LAB
Thank You for Your Attention,
Do You Have Any Questions?
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