6_eigenvectors_QM.md

title: Eigenvalues and eigenvectors

Eigenvalues and eigenvectors

The lecture on eigenvalues and eigenvectors consists of the following parts:

• Eigenvalue equations in linear algebra

• Eigenvalue equations in quantum mechanics

and at the end of the lecture one can find the corresponding Problems

The contents of this lecture are summarised in the following videos:

• 6_eigenvectors_QM_video1

In the previous lecture we discussed a number of operator equations, which are equations of the form

$$\hat{A}|\psi\rangle=|\varphi\rangle \, ,$$

where

$|\psi\rangle$

and

$|\varphi\rangle$

are state vectors belonging to the Hilbert space of the system

$\mathcal{H}$

.

A specific class of operator equations, that appear frequently in quantum mechanics, are equations of the form

$$\hat{A}|\psi\rangle= \lambda_{\psi}|\psi\rangle \, ,$$

where

$\lambda_{\psi}$

is a scalar (in general complex). These are equations where the action of the operator

$\hat{A}$

on the state vector

$|\psi\rangle$

returns the same state vector multiplied by the scalar

$\lambda_{\psi}$

. This type of operator equations are known as eigenvalue equations, and are of great importance for the description of quantum systems.

In this lecture we present the main ingredients of these equations and how we can apply them to quantum systems.

Eigenvalue equations in linear algebra

First of all let us review eigenvalue equations in linear algebra. Assume that we have a (square) matrix

$A$

with dimensions

$n\times n$

and

$\vec{v}$

is a column vector in

$n$

dimensions. The corresponding eigenvalue equation will be of form

$$A \vec{v} =\lambda \vec{v} .$$

with

$\lambda$

being a scalar number (real or complex, depending on the type of vector space). We can express the previous equation in terms of its components, assuming as usual some specific choice of basis, by using the rules of matrix multiplication,

$$\sum_{j=1}^n A_{ij} v_j = \lambda v_i \, .$$

The scalar

$\lambda$

is known as the eigenvalue of the equation, while the vector

$\vec{v}$

is known as the associated eigenvector.

The key feature of such equations is that applying a matrix

$A$

to the vector

$\vec{v}$

returns the original vector up to an overall rescaling,

$\lambda \vec{v}$

. In general there will be multiple solutions to the eigenvalue equation

$A \vec{v} =\lambda \vec{v}$

, each one characterised by an specific eigenvalue and eigenvectors. Note that in some cases one has degenerate solutions, whereby a given matrix has two or more eigenvectors that are equal.

In order to determine the eigenvalues of the matrix

$A$

, we need to evaluate the solutions of the so-called characteristic equation of the matrix

$A$

, defined as

$${\rm det}\left( A-\lambda \mathbb{I} \right)=0 \, ,$$

where

$\mathbb{I}$

is the identity matrix of dimensions

$n\times n$

, and

${\rm det}$

is the determinant.

This relations follows from the eigenvalue equation in terms of components

$$\sum_{j=1}^n A_{ij} v_j = \lambda v_i \, ,\quad \to \quad \sum_{j=1}^n A_{ij} v_j - \sum_{j=1}^n\lambda \delta_{ij} v_j =0 \, ,\quad \to \quad \sum_{j=1}^n\left( A_{ij} - \lambda \delta_{ij}\right) v_j =0 \, .$$

Therefore the eigenvalue condition can be written as a set of coupled linear equations

$$\sum_{j=1}^n\left( A_{ij} - \lambda \delta_{ij}\right) v_j =0 \, , \qquad i=1,2,\ldots,n\, ,$$

which only admit non-trivial solutions if the determinant of the matrix

$A-\lambda\mathbb{I}$

vanishes (the so-called Cramer condition), thus leading to the characteristic equation.

Once we have solved the characteristic equation, we end up with

$n$

eigenvalues

$\lambda_k$

,

$k=1,\ldots,n$

.

We can then determine the corresponding eigenvector

$$\vec{v}_k = \left( \begin{array}{c} v_{k,1} \\ v_{k,2} \\ \vdots \\ v_{k,n} \end{array} \right) \, ,$$

by solving the corresponding system of linear equations

$$\sum_{j=1}^n\left( A_{ij} - \lambda_k \delta_{ij}\right) v_{k,j} =0 \, , \qquad i=1,2,\ldots,n\, ,$$

Let us remind ourselves that in

$n=2$

dimensions the determinant of a matrix is evaluated as

$${\rm det}\left( A \right) = \left| \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right| = A_{11}A_{22} - A_{12}A_{21} \, ,$$

while the corresponding expression for a matrix belonging to a vector space in

$n=3$

dimensions will be given in terms of the previous expression $$ {\rm det}\left( A \right) = \left| \begin{array}{ccc} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} \end{array} \right| = A_{11} \left| \begin{array}{cc} A_{22} & A_{23} \ A_{32} & A_{33} \end{array} \right|

• A_{12} \left| \begin{array}{cc} A_{21} & A_{23} \ A_{31} & A_{33} \end{array} \right|

• A_{13} \left| \begin{array}{cc} A_{21} & A_{22} \ A_{31} & A_{32} \end{array} \right| $$

Let us illustrate how to compute eigenvalues and eigenvectors by considering a

$n=2$

vector space. Consider the following matrix

$$A = \left( \begin{array}{cc} 1 & 2 \\ -1 & 4 \end{array} \right) \, ,$$

which has associated the following characteristic equation

$${\rm det}\left( A-\lambda\cdot I \right) = \left| \begin{array}{cc} 1-\lambda & 2 \\ -1 & 4-\lambda \end{array} \right| = (1-\lambda)(4-\lambda)+2 = \lambda^2 -5\lambda + 6=0 \, .$$

This is a quadratic equation which we know how to solve exactly, and we find that the two eigenvalues are

$\lambda_1=3$

and

$\lambda_2=2$

.

Next we can determine the associated eigenvectors

$\vec{v}_1$

and

$\vec{v}_2$

. For the first one the equation that needs to be solved is

$$\left( \begin{array}{cc} 1 & 2 \\ -1 & 4 \end{array} \right) \left( \begin{array}{c} v_{1,1} \\ v_{1,2} \end{array} \right)=\lambda_1 \left( \begin{array}{c} v_{1,1} \\ v_{1,2} \end{array} \right) = 3 \left( \begin{array}{c} v_{1,1} \\ v_{1,2} \end{array} \right)$$

from where we find the condition that

$v_{1,1}=v_{1,2}$

: an important property of eigenvalue equations is that the eigenvectors are only fixed up to an overall normalisation condition. This should be clear from its definition: if a vector

$\vec{v}$

satisfies $A\vec{v}=\lambda\vec{v} $, then the vector

$\vec{v}'=c \vec{v}$

with c some constant will also satisfy the same equation. So then we find that the eigenvalue \lambda_1 has associated an eigenvector \vec{v}_1 = \left( \begin{array}{c} 1 \\ 1 \end{array} \right) \, , and indeed one can check that A\vec{v}_1 = \left( \begin{array}{cc} 1 & 2 \\ -1 & 4 \end{array} \right) \left( \begin{array}{c} 1 \\ 1 \end{array} \right) = \left( \begin{array}{c} 3 \\ 3 \end{array} \right)= 3 \vec{v}_1 \, , as we wanted to demonstrate. As an exercise, you can try to obtain the expression of the eigenvector corresponding to the second eigenvalue \lambda_2=2.

Eigenvalue equations in quantum mechanics

We can now extend the ideas of eigenvalue equations from linear algebra to the case of quantum mechanics. The starting point is the eigenvalue equation for the operator \hat{A}, \hat{A}|\psi\rangle= \lambda_{\psi}|\psi\rangle \, , where the vector state |\psi\rangle is the eigenvector of the equation and $ \lambda_{\psi}$ is the corresponding eigenvalue, in general a complex scalar.

In general this equation will have multiple solutions, which for a Hilbert space \mathcal{H} with n dimensions can be labelled as \hat{A}|\psi_k\rangle= \lambda_{\psi_k}|\psi_k\rangle \, , \quad k =1,\ldots, n \, .

In order to determine the eigenvalues and eigenvectors of a given operator \hat{A} we will have to solve the corresponding eigenvalue problem for this operator, what we called above the characteristic equation. This is most efficiently done in the matrix representation of this operation, where we have that the previous operator equation can be expressed in terms of its components as \begin{pmatrix} A_{11} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}= \lambda_{\psi_k}\begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix} \, , \quad k=1,\ldots,n \, .

As discussed above, this condition is identical to solving a set of linear equations for the form \begin{pmatrix} A_{11}- \lambda_{\psi_k} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi_k} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi_k} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix} \begin{pmatrix} \psi_{k,1}\\\psi_{k,2}\\\psi_{k,3} \\\vdots\end{pmatrix}=0 \, , \quad k=1,\ldots,n \, . This set of linear equations only has a non-trivial set of solutions provided that the determinant of the matrix vanishes, as follows from the Cramer condition: {\rm det} \begin{pmatrix} A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{pmatrix}= \left| \begin{array}{cccc}A_{11}- \lambda_{\psi} & A_{12} & A_{13} & \ldots \\ A_{21} & A_{22}- \lambda_{\psi} & A_{23} & \ldots\\A_{31} & A_{32} & A_{33}- \lambda_{\psi} & \ldots \\\vdots & \vdots & \vdots & \end{array} \right| = 0 which in general will have n independent solutions, which we label as \lambda_{\psi,k}.

Once we have solved the n eigenvalues ${ \lambda_{\psi,k} } $, we can insert each of them in the original evolution equation and determine the components of each of the eigenvectors, which we can express as columns vectors |\psi_1\rangle = \begin{pmatrix} \psi_{1,1} \\ \psi_{1,2} \\ \psi_{1,3} \\ \vdots \end{pmatrix} \,, \quad |\psi_2\rangle = \begin{pmatrix} \psi_{2,1} \\ \psi_{2,2} \\ \psi_{2,3} \\ \vdots \end{pmatrix} \,, \quad \ldots \, , |\psi_n\rangle = \begin{pmatrix} \psi_{n,1} \\ \psi_{n,2} \\ \psi_{n,3} \\ \vdots \end{pmatrix} \, .

An important property of eigenvalue equations is that if you have two eigenvectors $ |\psi_i\rangle$ and $ |\psi_j\rangle$ that have associated different eigenvalues, $\lambda_{\psi_i} \ne \lambda_{\psi_j} $, then these two eigenvectors are orthogonal to each other, that is \langle \psi_j | \psi_i\rangle =0 \, \quad {\rm for} \quad {i \ne j} \, . This property is extremely important, since it suggest that we could use the eigenvectors of an eigenvalue equation as a set of basis elements for this Hilbert space.

Recall from the discussions of eigenvalue equations in linear algebra that the eigenvectors |\psi_i\rangle are defined up to an overall normalisation constant. Clearly, if |\psi_i\rangle is a solution of \hat{A}|\psi_i\rangle = \lambda_{\psi_i}|\psi_i\rangle then c|\psi_i\rangle will also be a solution, with c some constant. In the context of quantum mechanics, we need to choose this overall rescaling constant to ensure that the eigenvectors are normalised, that is, that they satisfy \langle \psi_i | \psi_i\rangle = 1 \, \quad {\rm for~all}~i \, . With such a choice of normalisation, one says that the set of eigenvectors are orthogonal among them.

The set of all the eigenvalues of an operator is called eigenvalue spectrum of the operator. Note that different eigenvectors can also have the same eigenvalue. If this is the case the eigenvalue is said to be degenerate.

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##Problems

1) Eigenvalues and Eigenvectors

Find the characteristic polynomial and eigenvalues for each of the following matrices,

A=\begin{pmatrix} 5&3\\2&10 \end{pmatrix}\, \quad B=\begin{pmatrix} 7i&-1\\2&6i \end{pmatrix} \, \quad C=\begin{pmatrix} 2&0&-1\\0&3&1\\1&0&4 \end{pmatrix}

2) The Hamiltonian for a two-state system is given by H=\begin{pmatrix} \omega_1&\omega_2\\ \omega_2&\omega_1\end{pmatrix} A basis for this system is |{0}\rangle=\begin{pmatrix}1\\0 \end{pmatrix}\, ,\quad|{1}\rangle=\begin{pmatrix}0\\1 \end{pmatrix}

Find the eigenvalues and eigenvectors of the Hamiltonian H, and express the eigenvectors in terms of \{|0 \rangle,|1\rangle \}

3) Find the eigenvalues and eigenvectors of the matrices

A=\begin{pmatrix} -2&-1&-1\\6&3&2\\0&0&1 \end{pmatrix}\, \quad B=\begin{pmatrix} 1&1&2\\2&2&2\\-1&-1&-1 \end{pmatrix} .

4) The Hadamard gate

In one of the problems of the previous section we discussed that an important operator used in quantum computation is the Hadamard gate, which is represented by the matrix: \hat{H}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix} \, . Determine the eigenvalues and eigenvectors of this operator.

5) Show that the Hermitian matrix

\begin{pmatrix} 0&0&i\\0&1&0\\-i&0&0 \end{pmatrix}

has only two real eigenvalues and find and orthonormal set of three eigenvectors.

6) Confirm, by explicit calculation, that the eigenvalues of the real, symmetric matrix

\begin{pmatrix} 2&1&2\\1&2&2\\2&2&1 \end{pmatrix}

are real, and its eigenvectors are orthogonal.