Linear Operators for Quantum Mechanics

Thomas F. Jordan

• Publisher: Dover Publications
• Year: 2012
• ISBN:
  9780486140544
• Topic: Naturwissenschaften
• Subtopic: Physik

This compact treatment highlights the logic and simplicity of the mathematical structure of quantum mechanics. Suitable for advanced undergraduates and graduate students, it treats the language of quantum mechanics as expressed in the mathematics of linear operators.
Originally oriented toward atomic physics, quantum mechanics became a basic language for solid-state, nuclear, and particle physics. Its grammar consists of the mathematics of linear operators, and with this text, students will find it easier to understand and use the language of physics. Topics include linear spaces and linear functionals; linear operators; diagonalizing operators; operator algebras; states; equations of motion; and representation of space-time transformations. The text concludes with exercises and applications.

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DIAGONALIZING OPERATORS

To use linear operators in quantum mechanics we should know how to diagonalize them. We will see that putting operators in diagonal form makes working with them almost as easy as working with numbers. Some things work out better for complex vector spaces than for real vector spaces. From now on we consider operators defined on a complex separable Hilbert space.

12.EIGENVALUES AND EIGENVECTORS

Let B be a linear operator. If ψ is a nonzero vector and b is a scalar such that Bψ = bψ), then we say that ψ is an eigenvector of B and b is an eigenvalue of B. What is required for a number to be an eigenvalue of a linear operator?

Theorem 12.1. If b is an eigenvalue of a linear operator B, then the operator B – b has no inverse.

Proof. If B – b has an inverse, then there is no nonzero vector ф such that (B – b)ф = 0.¹

For a finite-dimensional space this is all that is required.

Theorem 12.2. If B is a linear operator on a finite-dimensional space, then each of the following is a necessary and sufficient condition for a scalar b to be an eigenvalue of B.

(i)The operator B – b has no inverse.

(ii)The determinant of the matrix corresponding to B – b is zero.

Proof. The operator B – b has an inverse if and only if there is no nonzero vector ф such that (B – b)ф = 0 (Theorem 7.2). Also B – b has an inverse if and only if the determinant of the matrix corresponding to B – b is not zero (Theorem 7.2).

Condition (ii) gives us a way to calculate eigenvalues. If the space is n-dimensional, we get an n x n matrix for B – b. Setting the determinant of this matrix equal to zero gives us an n^th-order ...