Eigenvalues and Eigenvectors

As a general rule, an Operator \(A:\mathcal{V}\to\mathcal{V}\) changes a vector to another vector, \(A|\alpha\rangle = |\beta\rangle\). I.e. \(|\beta\rangle \not\propto |\alpha\rangle\).

For eigenvectors of \(A\), \(A\) scales the vector, \(A|a_i\rangle = a_i|a_i\rangle\).


An eigenvector of \(A\) is a vector \(|v\rangle\) such that \(A|v\rangle = \alpha|v\rangle\). \(\alpha\) is called the eigenvalue.

The set of eigenvalues, \(\Sigma(A)\) is the spectrum of \(A\).


Angular momentum tensor: \(\vec{L}=I\cdot \vec{\omega}\).

Scale Factor

Every eigenvector has an arbitrary scale factor \(c=N\exp(i\theta)\), which comprises an arbitrary scale and phase.

Phase Factor

All eigenvectors have an arbitrary phase, \(\exp(i\theta)\). The phase is generally set so that the first non-zero component is a positive real number.


\(A|a\rangle = a\mathbb{I}|a\rangle = (A-a\mathbb{I})|a\rangle = 0\). Finding \(a\) such that \(\det(A-a\mathbb{I})=0\). So, the non-trivial solutions of \(|a\rangle\) only exist when \(A-a\mathbb{I}\) is not invertible. Thus, it must have a zero determinant.

When \(a\) s are obtained, Solve for vectors that solve \(A-a\mathbb{I}=0\)


If \(a\) is degenerate, find any linearly independent set and use \(Gram-Schmidt\) to orthogonalize.


For any Hermitian operator on a finite dimensional vector space,

  1. The eigenvalues of \(H\) are real
  2. Eigenvectors that belong to distinct eigenvalues are orthogonal
  3. The eigenvectors can always be chosent to be an Orthonormal basis for the vector space (for \(\mathbb{C}\))
  4. There exists an orthonormal basis of simultaneous eigenvectors of two Hermitian operators \(A,B\) iff \([A,B]=AB-BA=0\)

Proof of 4

Assume \([A,B]=0\). Let \(|a\rangle\) be an eigenvector for \(A\). Then \(BA|a\rangle = a(B|a\rangle) = A(B|a\rangle)\). Thus, \(B|a\rangle\) is also an eigenvector for \(A\). Vice-versa for \(B\).

Assume \(A\) and \(B\) have simultaneous eigenvectors. Then let \(|a,b\rangle\) be any eigenvector for both \(A\) and \(B\). So, \(AB|a,b\rangle = BA|a,b\rangle = ab|a,b\rangle\), thus, \((AB-BA)|a,b\rangle = 0\). Then, any linear combination of eigenvectors gives zero. Thus, \(AB-BA=0\).

Spectral Theorem for Hermitian Operators

Given a Hermitian operator \(H\) on a finite-dimensional, \(n=\dim\mathcal{V}\), vector space, \(\mathcal{V}\), choose an orthonormal basis of \(H\)’s eigenvectors: \(H|h_i\rangle = h_i|h_i\rangle, \langle h_i|h_j\rangle = \delta^i_j\). Then the matrix of \(H\) in the basis \(\{|h_i\rangle\}\) is diagonal, \(H^i_j=\langle h_i|H|h_j\rangle=h_i\delta^i_j\).

If we are in another basis, when we express the eigenvectors in this basis we can construct the unitary matrix with the column vectors \(|h_i\rangle\), \(U=(|h_1\rangle\:|h_2\rangle\:\cdots\:|h_n\rangle)\). So, \(H=\sum_i h_iP_i\) with \(P_i=|h_i\rangle\langle h_i|\). This is the spectral theorem.

Then the matrix of \(H\) is given by \(H=U^\dagger H_DU\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:24