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Notes on Hermitian Matrices and Vector Spaces
- Hermitian matrices
Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. So, for example, if
M =
1 i 0 2 1 − i 1 + i
then its Hermitian conjugate M †^ is
M †^ =
1 0 1 + i −i 2 1 − i
In terms of matrix elements, [M †]ij = ([M ]ji)∗^.
Note that for any matrix (A†)†^ = A.
Thus, the conjugate of the conjugate is the matrix itself.
Defn: A square matrix M is said to be Hermitian (or self-adjoint) if it is equal to its own Hermitian conjugate, i.e. M †^ = M.
For example, the following matrices are Hermitian:
( 1 i −i 1
Note that a real symmetric matrix (the second example) is a special case of a Hermitian matrix.
Theorem: The Hermitian conjugate of the product of two matrices is the product of their conjugates taken in reverse order, i.e.
(AB)†^ = B†A†^.
Proof:
[LHS]ij = ([AB]ji)∗^ = (
k
[A]jk[B]ki)∗^ =
k
([A]jk)∗([B]ki)∗
k
([B]ki)∗([A]jk)∗^ =
k
[B†]ik[A†]kj = [A†B†]ij = [RHS]ij.
Exercise: Check this result explicitly for the matrices
A =
, B =
0 i i 0
Theorem: Hermitian Matrices have real eigenvalues.
Proof
Ax = λx
so x†Ax = λx†x. (1)
Take the complex conjugate of each side:
(x†Ax)†^ = λ∗(x†x)†^.
Now use the last theorem about the product of matrices and the fact that A is Hermitian (A†^ = A), giving x†A†x = x†Ax = λ∗x†x. (2)
Subtracting (1), (2), we have (λ − λ∗)x†x = 0.
Since x†x 6 = 0, we have λ − λ∗^ = 0, i.e. λ = λ∗. So λ is real. (QED).
Theorem: Eigenvectors of Hermitian matrices corresponding to different eigenvalues are orthogonal.
Proof Suppose x and y are eigenvectors of the hermitian matrix A corresponding to eigen- values λ 1 and λ 2 (where λ 1 6 = λ 2 ). So we have
Ax = λ 1 x, Ay = λ 2 y.
Hence y†Ax = λ 1 y†x, (3)
x†Ay = λ 2 x†y. (4)
Taking the Hermitian conjugate of (4), we have
(x†Ay)†^ = y†Ax = λ∗ 2 (x†y)∗^ = λ 2 y†x,
where we have used the facts that A is Hermitian and that λ 2 is real. So we have
y†Ax = λ 2 y†x. (5)
This matrix equation corresponds to 3 simple linear equations (of which only 2 are inde- pendent). We easily find z/x = i , y = 0.
So we can take
X 1 = n 1
i
where n 1 is a normalisation constant. For λ 2 = 2, (^)
− 3 0 − 2 i 0 0 0 2 i 0 − 3
x y z
This gives 3x = − 2 iz and 2ix = 3z so, x = z = 0 and y is arbitrary. So we can take
X 2 = n 2
Finally, for λ 3 = −3, (^)
2 0 − 2 i 0 5 0 2 i 0 2
x y z
From this we deduce x = iz and y = 0, so we can take
X 3 = n 3
i 0 1
(ii) The norm or magnitude of X 1 is
X 1 † X 1. Now
X 1 † X 1 = n^21 ( 1 0 −i )
i
(^) = n^21 (1 + 0 + 1) = 2n^21 ,
so we can take n 1 = 1/
2 and get the normalised form of the first eigenvector
X 1 =
i
Similarly we can take
X 2 =
X 3 = √^1
i 0 1
(iii) Form the matrix
P =
1 0 i 0
i 0 1
from the normalised eigenvectors. Then
P †AP =
1 0 −i 0
−i 0 1
− 1 0 − 2 i 0 2 0 2 i 0 − 1
1 0 i 0
i 0 1
1 0 −i 0
−i 0 1
1 0 − 3 i 0 2
i 0 − 3
Note that the corresponding eigenvalues appear as the diagonal elements. Note also that
P †P =
= I.
A matrix satisfying this condition is said to be unitary.
- Vector spaces
The vectors described above are actually simple examples of more general objects which live in something called a Vector Space. (*) Any objects which one can add, multiply by a scalar (complex number) and get a result of a similar type are also referred to as vectors. Defn: A set of objects (vectors) x, y,... is said to form a linear vector space V if: (i) The set is closed under addition (usual commutative and associative properties apply), i.e. if x, y ∈ V, then the ‘sum’ x + y ∈ V. (ii) The set is closed under multiplication by a scalar, i.e. if λ is a scalar and x ∈ V, then λx ∈ V. The usual associative and distributive properties must also apply. (iii) There exists a null or zero vector 0 such that x + 0 = x. (iv) Multiplication by unity (λ = 1) leaves the vector unchanged, i.e. 1 × x = x. (v) every vector x has a negative vector −x such that x + (−x) = 0.
[Note that, in the above, as is often the case, I have not bothered to denote vectors by bold face type. It is usually clear within the context, what is a vector and what is a scalar!]
(*) An excellent Mathematics text book covering all of the tools we need is the book ‘Mathematical Methods for Physics and Engineering’ by Riley, Hobson and Bence, CUP
- Chapter 7 contains material on vectors, vector spaces, matrices, eigenvalues etc. Chapter 15 has more on the application to functions which we describe in what follows.
A vector space of this form, with an inner product, is sometimes referred to as a Hilbert Space (e.g. Mandl, Chapter 12).
Properties of Hermitian linear operators We can now generalise the above Theorems about Hermitian (or self-adjoint) matrices, which act on ordinary vectors, to corresponding statements about Hermitian (or self- adjoint) linear operators which act in a Hilbert space, e.g. the space of wave functions in Quantum Mechanics. For completeness, I rewrite the above Theorems (and Proofs) using the more general notation to describe scalar products. In this context, one often talks about eigenfunctions rather that eigenvectors if the ‘vectors’ happen to be functions or, specifically in our case, wave functions.
Defn: The Hermitian conjugate A†^ of a linear operator A is defined by
〈y|Ax〉∗^ = 〈x|A†y〉.
An equivalent definition is given by
〈Ax|y〉 = 〈x|A†y〉.
This follows using property (2) above of the inner product.
Thus, if A is a Hermitian operator (A†^ = A), we must have
〈Ax|y〉 = 〈x|Ay〉.
Exercise: Check that this definition agrees with that given when A is a complex matrix.
Theorem: The Hermitian conjugate of the product of two linear operators is the product of their conjugates taken in reverse order, i.e.
(AB)†^ = B†A†^.
Proof: For any two vectors x and y,
〈y|ABx〉 = 〈(AB)†y|x〉.
But also 〈y|ABx〉 = 〈A†y|Bx〉 = 〈B†A†y|x〉.
So we must have (AB)†^ = B†A†^.
Theorem: Hermitian (linear) operators have real eigenvalues.
Proof
Ax = λx
so 〈x|Ax〉 = λ〈x|x〉. (1)
Take complex conjugate: 〈x|Ax〉∗^ = λ∗〈x|x〉∗
so, using the definition of Hermitian conjugation (see above)
〈x|A†x〉 = λ∗〈x|x〉.
Now A is hermitian, so this means
〈x|Ax〉 = λ∗〈x|x〉. (2)
Subtracting (1), (2), we have (λ − λ∗)〈x|x〉 = 0.
Since 〈x|x〉 6 = 0, we have λ − λ∗^ = 0, i.e. λ = λ∗. So λ is real.
Theorem: Eigenvectors (eigenfunctions) of Hermitian operators corresponding to different eigenvalues are orthogonal.
Proof Suppose x and y are eigenvectors (eigenfunctions) of the hermitian operator A corre- sponding to eigenvalues λ 1 and λ 2 (where λ 1 6 = λ 2 ). So we have
Ax = λ 1 x, Ay = λ 2 y.
Hence 〈y|Ax〉 = λ 1 〈y|x〉, (3)
〈x|Ay〉 = λ 2 〈x|y〉. (4)
Taking the complex conjugate of (4), we have
〈x|Ay〉∗^ = 〈y|A†x〉 = λ∗ 2 〈x|y〉∗^ = λ 2 〈y|x〉,
where we have used the facts that 〈x|y〉∗^ = 〈y|x〉, and that λ 2 is real. Then, since A is hermitian, it follows that 〈y|Ax〉 = λ 2 〈y|x〉. (5)
Subtracting (5) from (3), we have
(λ 1 − λ 2 )〈y|x〉 = 0,
The boundary conditions then imply
f (L) = 0 ⇒ A cos(
λL) + B sin(
λL) = 0 f (−L) = 0 ⇒ A cos(
λL) − B sin(
λL) = 0.
There are two types of eigenfunctions odd: A = 0 and sin(
λL) = 0, so √ λL = nπ, (n = 0, 1 , 2.. .)
giving eigenvalues and eigenfunctions of K as
λ =
n^2 π^2 L^2
, f (^) n(o )(x) = Bn sin
( (^) nπx L
, (n = 0, 1 , 2.. .)
even: B = 0 and cos(
λL) = 0, so √ λL = (n + 1/2)π, (n = 0, 1 , 2.. .)
giving eigenvalues and eigenfunctions of K as
λ =
(n + 1/2)^2 π^2 L^2
, f (^) n(e )(x) = An cos
( (^) (n + 1/2)πx L
, (n = 0, 1 , 2.. .)
If you are feeling really keen, you could check that all these eigenfunctions are indeed orthogonal, i.e. 〈f (^) n(p )|f (^) m(q )〉 = 0
whenever m 6 = n, for p = o, e and q = o, e.
Basis vectors In dealing with any vector space V, it is very convenient to choose a set of basis vectors in terms of which any of the vectors in V can be written. The number of these needed is called the dimension of the vectors space (maximum number of linearly independent vectors in V). (Recall the discussion of such things in MATH102!) For example, in ordinary 3D (Euclidean) space, we are all used to adopting i, j and k as basis vectors. There are, of course, 3 of them and they are clearly linearly independent. Actually, they are orthogonal and unit (≡ orthonormal). We don’t have to use these as a basis. They don’t even have to be orthogonal! For example we could change basis to
e 1 = i + j , e 2 = i − j , e 3 = i + j + k.
The important thing about a basis is that an arbitrary vector can always be written as a linear combination of the basis vectors. Thus if ψ ∈ V and {φi}, i = 1, 2... N is a basis of V, we can write
ψ =
∑^ N
i=
ciφi.
It is often convenient to choose the basis vectors to be the eigenvectors of some linear operator which acts on the space. In Quantum Mechanics, we will do this all the time!
Example The matrix
A =
− 1 0 − 2 i 0 2 0 2 i 0 − 1
(see page 3) acts on the 3D space of column vectors of the form
x y z
Construct a basis for this 3D space consisting of the eigenvectors of A. How do you know the 3 vectors are linearly independent? Express the vector
v =
a b c
in terms of these basis vectors.
Solution See example on page 3, for the extraction of eigenvectors of A:
X 1 =
i
, X 2 =
, X 3 = √^1
i 0 1
Since there are 3 of them and they are linearly independent, they form a basis. We know they are linearly independent because they are orthogonal (correspond to different eigenvalues). Now let
v =
a b c
∑^3
i=
ciXi =
c 1 √ 2
i
(^) + c 2
(^) + √c^3 2
i 0 1
There are two ways to solve this (a) Equate components on each side of this vector equation, so finding 3 linear simulta- neous equations in c 1 , c 2 and c 3. Solve for these. (b) Use the fact that the Xi are orthonormal to show that
cj = X j† v.
