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The fact that the Pauli matrices, along with the identity matrix I, form an orthogonal basis for the Hilbert space of all 2 × 2 complex matrices , over , means that we can express any 2 × 2 complex matrix M as = + where c is a complex number, and a is a 3-component, complex vector.
Given an n × n square matrix A of real or complex numbers, an eigenvalue λ and its associated generalized eigenvector v are a pair obeying the relation [1] =,where v is a nonzero n × 1 column vector, I is the n × n identity matrix, k is a positive integer, and both λ and v are allowed to be complex even when A is real.l When k = 1, the vector is called simply an eigenvector, and the pair ...
The eigenvectors may be found by the usual methods of linear algebra, but a convenient trick is to note that a Pauli spin matrix is an involutory matrix, that is, the square of the above matrix is the identity matrix. Thus a (matrix) solution to the eigenvector problem with eigenvalues of ±1 is simply 1 ± S u. That is,
Suppose there is a spin 1/2 particle in a state = [].To determine the probability of finding the particle in a spin up state, we simply multiply the state of the particle by the adjoint of the eigenspinor matrix representing spin up, and square the result.
In the QR algorithm for a Hermitian matrix (or any normal matrix), the orthonormal eigenvectors are obtained as a product of the Q matrices from the steps in the algorithm. [11] (For more general matrices, the QR algorithm yields the Schur decomposition first, from which the eigenvectors can be obtained by a backsubstitution procedure. [13])
This method of generalizing the Pauli matrices refers to a generalization from a single 2-level system to multiple such systems. In particular, the generalized Pauli matrices for a group of N {\displaystyle N} qubits is just the set of matrices generated by all possible products of Pauli matrices on any of the qubits.
An n × n matrix A is diagonalizable if there is a matrix V and a diagonal matrix D such that A = VDV −1. This happens if and only if A has n eigenvectors which constitute a basis for C n. In this case, V can be chosen to be the matrix with the n eigenvectors as columns, and thus a square root of A is = ,
Rellich draws the following important consequence. << Since in general the individual eigenvectors do not depend continuously on the perturbation parameter even though the operator () does, it is necessary to work, not with an eigenvector, but rather with the space spanned by all the eigenvectors belonging to the same eigenvalue. >>