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In the previous
chapter, we defined the Hermitian
conjugate of a matrix as its transpose conjugate: T† = T*. The more fundamental definition for Hermitian
conjugate of a linear transformation for all vectors |α> and |β> is that,
when we applied transformation $\hat{T}$† to the first member of an
inner product, it gives the same result as if $\hat{T}$ itself had been applied
to the second vector:
<
$\hat{T}$†α|β > = < α|$\hat{T}$β > (18.1)
In particular, the
notation |$\hat{T}\beta $> means $\hat{T}$|$\beta $>, and <$\hat{T}$†α|β>
means the in-product of the vector $\hat{T}$†|α> with vector |β>.
Please notice also for any scalar c
that:
<α|cβ> = c<α|β> (18.2)
But,
<cα|β> = c*<α|β>
(18.3)
If we’re working in an
orthonormal basis the Hermitian conjugate of a linear transformation is
represented by the Hermitian conjugate of the corresponding matrix, so that:
<α|$\hat{T}\beta $> = a†Tb = (T†a)†b
= <$\hat{T}$†α|β> (18.4)
The Hermitian
transformations ($\hat{T}$† = $\hat{T}$) play a fundamental role in
quantum mechanics. The eigenvectors and eigenvalues of a Hermitian
transformation have three important properties:
1.
The eigenvalues of a Hermitian
transformation are real.
This
property can be proven as follows: Let λ be
an eigenvalue of$\hat{T}$ and |α>
≠ |0>: $\hat{T}$|α> = λ|α>,
then <α|$\hat{T}\alpha $> =
<α|λα> = λ<α|α>. Meanwhile, if $\hat{T}$is
Hermitian then: <α|$\hat{T}$α>
= <$\hat{T}$α|α> = <λα|α> = λ*<α|α>.
But <α|α> ≠ 0 so that λ = λ* and hence λ is real.
2.
The eigenvectors of a Hermitian
transformation belonging to distinct eigenvalues are orthogonal.
Suppose
$\hat{T}$|α > = λ|α>
and $\hat{T}$|β> = μ|β> with μ ≠ λ then:
<α|$\hat{T}$β> = <α|μβ>
= μ<α|β>
And
if $\hat{T}$is Hermitian,
<α|$\hat{T}$β> = <$\hat{T}$α|β> = <λα|β> = λ*<α|β>
Since
λ = λ* (from property 1), and λ ≠ μ by assumption, thus <α|β> = 0.
3.
The eigenvectors of a Hermitian
transformation span the space.
If
all the n-roots of the characteristic
equation are distinct, then by property 2 we have n mutually orthogonal eigenvectors, so that they span the space.
Suppose there are m-fold λ duplicate roots (degenerate eigenvalues), any linear combination of two eigenvectors
belonging to the same eigenvalue is still an eigenvector, it can be shown that
there are m linearly independent
eigenvectors with eigenvalues λ.
These eigenvectors can be orthogonalized by the Gram-Schmidt procedure. So, the eigenvectors of a Hermitian transformation
can always be taken to constitute an orthonormal basis. It follows, that any
Hermitian matrix can be diagonalized by a similarity transformation, with S
unitary.
