Linear Vector Space and Matrix Mechanics Chapter 1 Lecture 1.7 Dr. Arvind Kumar Physics Department NIT Jalandhar e.mail: iitd.arvind@gmail.com https://sites.google.com/site/karvindk2013/

Eigen values and Eigen vectors: Given any dynamical variable A (represented by operator), how we find and characterize those particular states of the system in which A may have a definite value ? If we measure A in large number of system which are identically prepared and each is represent by same state ψ, under what conditions all these systems will yield same sharp value? What kind of state of system corresponds to a probability distribution of the value of A i.e. peaked at a and has no spread?

If we have some such state, then the standard deviation of A in this state (also called determinate state) would be zero. So   ---------(1) Note that we take <A> = a in 1st line because every measurement of dynamical observable gives us the value a and therefore the average will also be a. Since is hermitian and thus is also hermitian Therefore we moved over to 1st term in the inner product.

Only function whose inner product with itself vanishes is 0. Thus from Eq. (1), we writ Or --------(2) The state vector is called the eigenvector and a is called the eigenvalue. Eq. (2) is known as eigenvalue equation or eigenvalue problem.

Eigenvalue probelm for unit operator with eigenvalue 1 Note the following: Thus,

Note: Eigenvalue is a number not an operator or function. The collection of all Eigen values of an operator is called the spectrum. If two or more linearly independent Eigen functions share same Eigen value then the spectrum is called degenerate.

Theorem 1: For a Hermitian operator all of its eigenvalues are real and the eigenvector corresponding to different Eigenvalues are orthogonal. Proof: Note the following two Eqns . --------(1) --------(2)

Subtract (2) from (1), using fact that is hermitian operator, We get We have now two cases

Eigenvectors of non-degenerate Hermitian operator are unique and form orthonormal basis. These eigenvectors are linearly independent. Theorem: The Eigenvectors corresponding to equal eigenvalues of a degenerate hermitian operator are not unique and also not necessarily orthogonal. However we can construct a orthogonal set of eigenvectors for degenerate Hermitian operator using the Schmidt orthonormalization procedure.

Theorem:If two hermitian operators and commute and if has no degenerate eigenvalue, then each eigenvector of is also an eigenvector of . In addition, we Can construct a common orthonormal basis that is made of Joint eigenvectors of and . Proof: is hermitian operator with no degenerate Eigenvalue. For each eigenvalue of there corresponds one eigenvector

Above eq shows that is an eigenvector of with Eigenvalue an . But if is only linearly eigenvector of with eigenvalue an, then must be differ from by a constant. So we can write So above Eq show that is also eigenvector of .

Let an is degenerate eigenvalue of will also be eigenvector of operator A with Eigenvalue an. But is not necessarily eigenvector of . Explain!

Theorem: The eigenvalues of a unitary operator are complex numbers of moduli equal to one. The eigenvectors of a unitary operator that has no degenrate eigenvalues are mutually Orthogonal. Proof:

Theorem: The eigenvalues of an anti-Hermitian operator are either purely imagninary or equal to zero. The eigenstates of a Hermititian operator define a complete set of mutually orthonormal basis states. The operator is diagonal in this Eigenbasis with its diagonal elements equal to the eigenvalues. The basis set is unique if the operator has no degenerate eigenvalues and not unique (in fact it is infinite) if there is any degeneracy.

Unitary Transformations: How ket, bra and operators transform under unitary Transformation? Ket and bra ------(1) Operator: Let Transform to ----------(2) Using (1), =

Multiplying both sides by

Properties of unitary transformations: If is hermitan then the transformed operator The Eigenvalues of and are same i.e. Proof:

3. Commutators which are equal to complex numbers remain unchanged under unitary transformations 4. Complex number remain unchanged Under unitary transformations.

Let , then from last Eq, we can write the scalar Products invariant under unitary transformations Norm of state is conserved Verify Discuss finite and infinite unitary transformations.

Infinitesimal unitary transformation Consider the transformation Generator of transformation Infinitely small Real parameter will be unitary if is Hermitian and is real:

Transformation of state vector under above Where For operator Note:

Finite unitary transformation Finite unitary transformation can be generated by successive infinitesimal unitary Transformation N is integer and α is some finite parameter. We write

Transformation of operator where we used Note: