Lecture 4 Outline - Eigenbits
•Q: does matrix with orthogonal vectors imply U?
•ze eigenvalue problem [Section 1.8]
•Hermitian matrix eigenvalues [Section 1.8]
•Diagonalisation [Section 1.8]
•an eigen example [Section 1.8]
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Eigenvalues and Eigenvectors of Hermitian Matrices: Real and Orthogonal - Prof. M. W. Brom, Study notes of Quantum Mechanics
The properties of hermitian matrices, focusing on their real eigenvalues and orthogonal eigenvectors. The text derives these properties through mathematical reasoning and provides examples to illustrate the concepts. It also discusses the diagonalization of hermitian matrices and their relation to unitary matrices.
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Lecture 4 Outline - Eigenbits
Q: does matrix with orthogonal vectors imply
U
ze eigenvalue problem [Section 1.8]
Hermitian matrix eigenvalues [Section 1.8]
Diagonalisation [Section 1.8]
an eigen example [Section 1.8]
Hermitian matrix fundamentals
- Eigenvalues of a Hermitian matrix (operator) are
real
- Eigenvectors of Hermitian matrices (operators) are
orthogonal
(if they correspond to distinct eigenvalues)
- Eigenvectors of Hermitian matrices (operators)
span the space
(ie. matrix can be diagonalised)
Firstly: let
Ω | ω 〉 = ω | ω 〉
where
†
and dot away
〈 ω | Ω | ω 〉 = ω 〈 ω | ω 〉
and taking ajoint
〈 ω | Ω † | ω 〉 = ω ∗ 〈 ω | ω 〉
subtracting:
ω
ω
ω
ω ∗ 〈 ω | ω 〉
ω − ω ∗ ) 〈 ω | ω 〉
thus
ω
ω
∗
)
so
ω
ω
∗
ie. real asreqd.
Hermitian fundamentals 3
Eigenvalues
real
. Eigenvectors
orthogonal
and
span
Thirdly: (again assuming non-degenerate
ω
i ’s)
eigenvectors span the space, use them as basis!
Remembering that defined
ij
= 〈 i | Ω | j 〉
means that
ii′
= 〈 ω i | Ω | ω i 〉 = ω i 〈 ω i | ω i 〉 = ω i
and also that
i ′ 6 =
j = 〈 ω i | Ω | ω j
ω
j (^) 〈
ω
i | ω
j (^) 〉
Normal
matrices (
[
N
(^) †
, N
] =
) have eigenvectors that
span the space and are thus diagonalisable
Unitary matrices and Diagonalisation
- Eigenvalues of a Unitary matrix (
U
(^) †
U
(^) −
1 ) are
complex
numbers of modulus 1 (ie.
ω
e
iφ
- Eigenvectors of a Unitary matrix are
orthogonal
(again if they correspond to distinct eigenvalues)
- If
is Hermitian there exists a Unitary matrix
U
built
from the eigenvectors of
such that
U
(^) †
Ω
U
is diagonal.
(with elements = eigenvalues)
Really is a
rotation
transform:
| ω i 〉 = U
i 〉
(eg.unitary transform preserves trace and det
Tr(
U
(^) †
Ω
U
(^) ) = Tr(Ω)
Example Diagonalisation contd.
Example Diagonalisation contd.
Eigenproblem contd 1...
Choice of basis: as unity displacement of each mass
where
was represented in this basis:
so
| x ( t ) 〉 =
x
1
x
2
= x 1 | 1 〉 + x 2 | 2 〉
Want a basis in which
is diagonal:
Ω | I 〉 = − ω
I 2 (^) | I
〉
and
II
ω
II 2
II
Solving eigenproblem
ω
I
mk
and
ω
II
3 k
m
I
and
II
which span space
| x ( t ) 〉 = x 1 | 1 〉 + x 2 | 2 〉 = x I
I
x
II
II
Eigenproblem contd 2...
Given
| x ( t ) 〉 = x I
I
x
II
II
we can rewrite the original
¨x
( t ) 〉
| x ( t ) 〉
eqn as
¨x
I
¨x
II
ω
I 2
ω
II 2
x
I
x
II
ie.
¨x
I
ω
I 2 (^) x
I
which has solns
x
I (^) (
t ) =
x
I (^) (0) cos
ω
I (^) t
| x ( t ) 〉 = x I
(^) (0) cos
ω
I (^) t
| I
〉
x
II
(^) (0) cos
ω
II
(^) t
| II
I
I
x
(0)
cos
ω
I (^) t
II
II
x
(0)
cos
ω
II
(^) t
Remember: Unitary transform preserves inner products
I
x
(0)
x
1 (0)
x
2
(0)
x
1 (0) +
x
2 (0)
Simultaneous Diagonalisation
If
and
are two Hermitian operators that commute
[Ω
Λ] =
, then there is a basis of common eigenvectors
that diagonalises them both.
Assuming non-degeneracy, let
Ω | ω i 〉 = ω i | ω i 〉
and
| ω i 〉 = ω i Λ | ω i 〉
ω
i 〉
| ω i 〉 = ω i Λ | ω i 〉
So
ω
i 〉
is also an eigenvector of
with eigenvalue
ω
i .
BUT
ω
i
eigenvalue has one eigenvector (up to a scale)
Λ | ω i 〉 = λ i | ω i 〉
thus
| ω i 〉 = ω i λ i | ω i 〉
So every eigenvector of
is an eigenvector of
and thus
ω
i
diagonalises both
and
Function of Operators
f In analogy with a power series expansion of functions
(^) (
x
) =
∑ n a n x n
Can define equivalent functions of operators to be
f
(^) (Ω) =
∞
n ∑
a
n
Ω
n
eg
e
Ω
∞
n ∑
n
n
!
makes sense if the sum sensibly converges
eg. choosing the operator in it’s eigenbasis:
m
ω
1 m
ω
(^2) m
ω
nm
e
Ω
ω
1 m
m
!
ω
2 m
m
!
ω
nm
m
!