An operator operating on the elements of the vector space V has certain kets, called eigenkets, on which its action is simply that of rescaling.  W|V>=w|V>.  |V> is an eigenket (eigenvector) of W, w is the corresponding eigenvalue.

Example:

If |V> is an eigenvector, so is a|V>.  The eigenvectors are only fixed to an overall scale factor.  If we require that <V|V>=1, the ambiguity is partially removed, a phase factor is still arbitrary.

Examples:

How do we find the eigenvalues and eigenvectors of an arbitrary linear operator W ?

W|y>=w|y>.   (W-w I)|y>=0.

We can transform this equation into a matrix equation by choosing a basis {|u_i>} such that

.

Then

.

This matrix equation has a nontrivial solution only if det(W-w I)=0.  (For a trivial solution all c_j are zero.)

This is the characteristic equation.

The determinant is

  for even permutations, for odd permutations, and for repeating indices.

Our characteristic equation therefore is a polynomial of order N in w, where N is the dimension of the vector space.  It has N roots.  These roots can be real or complex, distinct or identical.  These N roots are the eigenvalues.  They are found in a particular basis, but are independent of the choice of basis.  If the vector space V is defined over the space of complex numbers, then each linear operator W has at least one eigenvalue.  The characteristic equation has at least one root.

Example:

 .

The eigenvalues are found from

,

or

.

We have 

,   .  

For sinq ¹ 0 no real, but two complex solutions exist.

How do we find the eigenvectors associated with the eigenvalues?

Assume one of the eigenvalues is w.  We have to solve the system of N equations, , (i=1 to N) for the unknowns c_j.  If W is a linear operator and |y> is an eigenvector, then a|y> is also an eigenvector.  The c_j can therefore only be uniquely defined up to a multiplicative constant.  That means that at most N-1 of the above equations can be linearly independent.  (The determinant could not be zero otherwise.)  If N-1 of the equations are linearly independent, then the solution for the c_j is unique up to an arbitrary multiplicative constant.  If less than N-1 of the equations are linearly independent, then we can find more than one linearly independent solution for the c_j.  If this happens, the eigenvalue w is said to be degenerate, and the associated eigenvectors form a subspace V_w of V.

Example:

If the characteristic equation has N distinct roots, we can find N (up to a phase factor) unique, normalized, linearly independent eigenvectors which form a basis of V.  If the characteristic equation does not have N distinct roots, we may be able to find more than one linearly independent eigenvector associated with a multiple root and so still find N linearly independent eigenvalues.  But we may also find that only one eigenvector is associated with a multiple root and that the operator does not have a basis of eigenvectors.

Eigenvalues and eigenvectors of a Hermitian operator

The eigenvalues of a Hermitian operator are real.

Proof:
Let W|w>=w|w >, then  <w|W|w >=w<w|w >.
Take the adjoint. <w|W^T|w>=w^*<w|w >.
If W=W^T then    <w|W|w >=w^*<w|w>, (w-w^*)<w|w >=0.
Since |w >¹0, we have w=w^*.

Every Hermitian operator has at least one basis of orthonormal eigenvectors.  The matrix of the operator is diagonal in this basis and has its eigenvalues as its diagonal entries.

Proof:
The characteristic equation has at least one root, call it w₁.  At least one non zero eigenvector |w₁> corresponds to this eigenvalue.  Let V_^1^n-1 be the subspace of all vectors orthogonal to |w₁>.  As a basis for V we can now choose the normalized vector |w₁> and n-1 orthonormal vectors |V_^1¹>, |V_^1²>, |V_^1³>, ..., |V_^1^n-1>.  In this basis the matrix of W has the following form:

.

We use the fact that W is Hermitian when we set <w₁|W=w₁<w₁|.
The characteristic equation now takes the form "(w₁-w) times determinant of the square matrix denoted by X = 0", or , where P^n-1(w) is a polynomial of order n-1.
The polynomial P^n-1 must also generate at least one root, call it w₂, and a normalized eigenvector |w₂>.  Define a subspace V_^1,2^n-2 of vectors in V_^1^n-1 orthogonal to |w₂> and repeat the procedure.  Finally the matrix of W in the basis |w₁>, |w₂>, ..., |w_n> becomes

.

Every |w_i> is chosen from a subspace that is orthogonal to the previous one, therefore the basis {|w_i>} is orthogonal.

Note: We did not assume that the eigenvalues are all distinct (non degenerate).

If the eigenvalues are degenerate, then there exist many bases of eigenvectors that diagonalize W.

Assume W|w₁>=w|w₁> and W |w₂>=w|w₂>.  Then W(a₁|w₁>+a₂|w₂>)=w(a₁|w₁>+a₂|w₂>) for any a₁, a₂ in F.  There exists a whole subspace spanned by |w₁> and |w₂>, whose elements are eigenvectors of W with eigenvalue w.

Summary

We have proofed that in a finite dimensional vector space a basis of eigenvectors of a Hermitian operator can be found.  We assume that this is also true for an infinite dimensional "physical Hilbert space", where we define as an observable any Hermitian operator whose eigenvectors can form a basis.

Links: 

Problem:

 
The eigenvector associated with is with.
The eigenvector associated with is with .

(b) The projector onto .  The matrix elements are

matrix of P₁: , matrix of P₂:

The eigenvalues of a unitary operator are complex numbers of unit magnitude.  Eigenvectors with different eigenvalues are mutually orthogonal.

Proof:
Let U|u_i>=u_i|u_i>, U|u_j>=u_j|u_j>.  
Let i, j denote eigenvectors with different eigenvalues if i¹j.  
Then <u_j|U^TU|u_i>=u_j^*u_i<u_j|u_i>.  
But U^TU=I, therefore <u_j|u_i>=u_j^*u_i<u_j|u_i>, or (1-u_j^*u_i)<u_j|u_i>=0.  
If i=j then <u_i|u_i>¹0, therefore u_i^*u_i=1. 
If i¹j then 1-u_j^*u_i¹0. 
(u_j¹u_i Þ u_j^*u_i ¹ u_i^*u_i ¹ u_j^*u_j ¹ 1.)  
Therefore <u_j|u_i>=0.

Commuting observables

Let A and B be two commuting Hermitian operators.  A=A^T, B=B^T, [A,B]=0.

BA|a_i>=AB|a_i> since A and B commute;

a_iB|a_i>=AB|a_i> since |a_i> is an eigenvector of A.

This implies B|a_i>=b_i|a_i>, since B|a_i> is an eigenvector of A with eigenvalue a_i and a unique normalized basis vector corresponds to every eigenvalue.  Therefore the {|a_i>} are also eigenvectors of B.  Exactly one basis of eigenvectors common to A and B exists if at least one of the operators is non degenerate.  This basis diagonalizes both A and B.

<a_i|A|a_j>=a_id_ij, <a_i|B|a_j>=b_id_ij.

.

This basis, however, is not unique.

The eigensubspace corresponding to every degenerate eigenvalue has an infinity of bases.

How does B appear in this basis?

AB|a_i^a>=BA|a_i^a>=a_iB|a_i^a>, where a=1, ..., g_n and g_i denotes the degree of degeneracy of the eigenvalue a_i.  B|a_i^a> lies in the eigensubspace associated with the eigenvalue a_i.  Vectors from different eigensubspaces are orthogonal.  The matrix of B is therefore a block diagonal matrix.

.

The matrix of B in the eigensubspace is Hermitian and therefore can be diagonalized by trading the basis {|a_i>} for an eigenbasis of B_i.  The matrix of A remains diagonal, since we are choosing another orthonormal basis in a degenerate eigenspace.  If B is not degenerate in a given subspace, we will end up with a unique orthonormal basis of eigenvectors of both A and B.  If B is degenerate in any given subspace, the basis we find is not unique.  A third operator C, which commutes with A and B, ([A,C]=0, [B,C]=0) may still not be diagonal in this basis, and we may have to diagonalize the matrix of C in the eigensubspaces which belong to both A and B.  It is however always possible to find a set of operators {A, B, C, ...,} which commute by pairs and uniquely define a common eigenbasis.  Such a set of operators is called a complete set of commuting observables (C.S.C.O.).  It is generally understood that C.S.C.O. refers to the minimal set.

Summary

If two Hermitian operators commute a common eigenbasis can be found.  If they do not commute, then no common eigenbasis exists.  A complete set of commuting observables is the minimal set of Hermitian operators with a unique common eigenbasis.