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.

17. Formalism of Quantum Mechanics: Eigenvectors and Eigenvalues

Consider a complex vector space; every linear transformation $\hat{T}$ has vectors |α>, which are transformed into simple multiples of themselves:

$\hat{T}|\alpha >=\lambda |\alpha >$                                     (17.1)

The vectors |α> are the eigenvectors of the transformation and the complex number λ is the eigenvalue. The null vectors, are ‘trivial solutions’ of Equation (17.1) and doesn’t count. Therefore, any nonzero multiple of eigenvector is still an eigenvector with the same eigenvalue.

The eigenvector equation, with respect to a particular basis, in matrix form is given by:

Ta = λa                                                                    (17.2)

Or:

(T – λ1)a = 0                                                               (17.3)

Here, 0 is the zero matrix. By the assumption that a is nonzero, the matrix (T – λ1) must in fact be singular, which means that its determinant is equal to zero:

det (T – λ1) = 0                                                            (17.4)

An algebraic equation for λ can be obtained by expansion of the determinant:

${{C}_{n}}{{\lambda }^{n}}+{{C}_{n-1}}{{\lambda }^{n-1}}+$ …$+{{C}_{1}}\lambda +{{C}_{0}}=0$                                          (17.5)

This is called the characteristic equation for the matrix, where the coefficients C_i depend on the elements of T and its solutions determine the eigenvalues. Equation (17.5) is an n^th-order equation that has n (complex) roots. The corresponding eigenvectors can be constructed by plugging each λ back into Equation (17.2) and solve for the components of a. We are going to show you how it goes by mean of an example.

QM Solved Problems #2

1.      A mono-energetic bundle of particles is moving along the x-axis to the right in a potential field V:

V = 0, x < 0;

V = V₀ > 0, x > 0;

      The energy E of the particles is greater than V₀.

a.       What are the solutions of the time-independent Schrödinger equation for x < 0 and x > 0?

b.      At x = 0, an incoming bundle from left, a reflected bundle and a transmitted bundle correspond to the boundary conditions. Calculate the reflection coefficient (the absolute value of the square of the amplitude of reflected and incoming bundles).

c.       If the energy of the particles is smaller than V₀, what are the solutions of the time-independent Schrödinger equation for x > 0 and the reflection coefficient?

Solution:

2.      A one dimensional rectangular potential field with energy E < 0, is given by:

V(x) = 0 for |x| > a;

V(x) = -V₀ < 0 for |x| ≤ a;

      Write the general solution of time independent Schrödinger equation for each area:

      x < -a, -a < x < a, x > a. Use the following definitions:

κ² = 2m|E|/ħ²;

q = 2m(V₀ - |E|)/ħ²

      Solution:

3.      Quantum mechanical states are vectors in a Hilbert space. A Hilbert space is a complex vector space with a Hermitian inner product. The properties of this inner product are that it is linear:

$<\phi |a{{\psi }_{1}}+b{{\psi }_{2}}>=a<\phi |{{\psi }_{1}}>+b<\phi |{{\psi }_{2}}>$

$<\phi |\psi >=<\psi |\phi >*$

$<\psi |\psi >\ge 0$

a.       Prove the Schwarz inequality: $|<\psi |\phi >{{|}^{2}}\le <\psi |\psi ><\phi |\phi >$. Hint: consider $<\phi +\lambda \psi |\phi +\lambda \psi >\ge 0$ and choose for λ the value which minimizes this expression.

b.      Prove the triangle inequality: $\sqrt{<\psi +\phi |\psi +\phi >}\le \sqrt{<\psi |\psi >}+\sqrt{<\phi |\phi >}$  a physical observable corresponds to a Hermitian operator.

c.       What is the definition of a Hermitian operator?

d.      Prove that the expectation value of a Hermitian operator is always real.

A Hermitian operator is called positive definite if for every ψ it holds that $<\psi |A|\psi >\ge 0$.

e.       A projection operator P is a Hermitian operator which satisfies P² = P. Show that the projection operator is always positive definite.

f.       Show that if A is a positive definite operator, $|<\psi |A|\phi >{{|}^{2}}\le <\psi |A|\psi ><\phi |A|\phi >$. Use the fact that you can expand in a basis of eigen-states and that the eigen-values of A are positive.

Solution: