Physics 3 for Electrical Engineering Ben Gurion University of the Negev Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 9. Quantum mechanics – angular momentum operators commutation relations eigenvalues and eigenvectors of Sources: Merzbacher (2 nd edition) Chap. 9; Merzbacher (3 rd edition) Chap. 11.

The time-independent Schrödinger equation for a particle in three dimensions is where r = (x,y,z), and A special, but important, class of potentials is the class of central potentials, which depend only on r = (x 2 + y 2 + z 2 ) 1/2, i.e.

The time-independent Schrödinger equation for a particle in three dimensions is where r = (x,y,z), and A special, but important, class of potentials is the class of central potentials, which depend only on r = (x 2 + y 2 + z 2 ) 1/2, i.e.

The time-independent Schrödinger equation for a particle in three dimensions is where r = (x,y,z), and A special, but important, class of potentials is the class of central potentials, which depend only on r = (x 2 + y 2 + z 2 ) 1/2, i.e.

If the only force acting is a central force, we know from classical physics about an important conserved quantity: angular momentum. Angular momentum L = r × p is conserved because F(r) and r are always parallel:

If the only force acting is a central force, we know from classical physics about an important conserved quantity: angular momentum. Angular momentum L = r × p is conserved because F(r) and r are always parallel: ? © George CoghillGeorge Coghill

If the only force acting is a central force, we know from classical physics about an important conserved quantity: angular momentum. Angular momentum L = r × p is conserved because F(r) and r are always parallel:

If the only force acting is a central force, we know from classical physics about an important conserved quantity: angular momentum. Angular momentum L = r × p is conserved because F(r) and r are always parallel: Is there a quantum operator for angular momentum? Is it conserved?

Angular momentum operators In classical mechanics, angular momentum is defined as L = r × p. So in quantum mechanics, is angular momentum defined as

Angular momentum operators In classical mechanics, angular momentum is defined as L = r × p. So in quantum mechanics, is angular momentum defined as (Isn ’ t there a problem with the ordering of and ?) © George CoghillGeorge Coghill ?

Angular momentum operators In classical mechanics, angular momentum is defined as L = r × p. So in quantum mechanics, is angular momentum defined as

Angular momentum operators In classical mechanics, angular momentum is defined as L = r × p. So in quantum mechanics, angular momentum is defined as

Angular momentum operators In classical mechanics, angular momentum is defined as L = r × p. So in quantum mechanics, angular momentum is defined as Is conserved?

Theorem: If a Hermitian operator commutes with, i.e., then we can find eigenstates of that are also eigenstates of. Proof: If commutes with an observable, then for any state,. So if is an eigenvector of with eigenvalue E n, i.e., then so is : If E n is nondegenerate, then for some number a, so is an eigenstate of. If E n is degenerate, then eigenstates of form a basis for the subspace of eigenvectors of with eigenvalue E n.

Corollary: If a Hermitian operator commutes with, i.e., then the expectation value of in any state is constant in time, i.e. Proof: Define the state such that and Write and then does not depend on time.

So is conserved? That is, do and commute? Let ’ s see:

So is conserved? That is, do and commute? Let ’ s see: Remember: the derivatives act also on a wave function ψ. [ ψ] ψ ψ

So is conserved? That is, do and commute? Let ’ s see:

So is conserved? That is, do and commute? Let ’ s see:

So is conserved? That is, do and commute? Let ’ s see:

So is conserved? That is, do and commute? Let ’ s see: [ ψ] Remember: the derivatives act also on a wave function ψ. ψ ψ

So is conserved? That is, do and commute? Let ’ s see: [ ψ] Remember: the derivatives act also on a wave function ψ. ψ ψ

So is conserved? That is, do and commute? Let ’ s see:

So is conserved? That is, do and commute? Let ’ s see: And what holds for holds for and.

So is conserved! What are the possible values of ? Let ’ s calculate commutation relations for.

Commutation relations using

Commutation relations We can choose a basis of eigenstates of, or of, or of, but only one of these bases at a time! Also, from our generalized uncertainty principle, we conclude

Eigenvalues and eigenvectors of. Let ’ s prove that is a raising operator: Similarly, is a lowering operator: Suppose. Then

Eigenvalues and eigenvectors of. Let ’ s prove that is a raising operator: Similarly, is a lowering operator: Suppose. Then

Eigenvalues and eigenvectors of. Let ’ s prove that is a raising operator: Similarly, is a lowering operator: Suppose. Then so is an eigenvector of with eigenvalue. Similarly, is an eigenvector of with eigenvalue.

Eigenvalues and eigenvectors of. Eigenvalues of : but since L z is bounded by we must have for some m max and m min. Note since Note also

Eigenvalues and eigenvectors of. To make the rest of the calculations easier, we should change to spherical coordinates: Now so The eigenfunctions of are with eigenvalues so Therefore m max = –m min and, by convention, m max = l.

Eigenvalues and eigenvectors of : Summary: For a given value of l, the eigenstates of are with respective eigenvalues These 2l+1 eigenvectors of are also eigenvectors of with degenerate eigenvalue

Schrödinger’s equation for a central potential is A vector identity: What is ? Since it is hence

Schrödinger’s equation for a central potential is A vector identity: What is ? Since it is hence

Solving for we obtain and by comparing this with the expression for in spherical coordinates,

Solving for we obtain and by comparing this with the expression for in spherical coordinates,

Solving for we obtain and by comparing this with the expression for in spherical coordinates we conclude

Now back to Schrödinger’s equation in spherical coordinates: since the eigenvalues of are, we can solve this equation by expressing ψ(r,θ,φ) as a product of two functions: ψ(r,θ,φ) = R(r) Y l m (θ,φ), where

Here are the lowest eigenfunctions Y l m (θ,φ) of

HereHere’s how they look: