18. Formalism of Quantum Mechanics: Hermitian Transformations

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.