Quantum Mechanics for Scientists and Engineerstera.yonsei.ac.kr/class/2016_2_2/lecture/Lesson15...

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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

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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

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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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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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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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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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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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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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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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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

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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

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Thank You for Your Attention,

Do You Have Any Questions?