Particle)in)a)central)potential.)The)...

Particle in a central potential. The hydrogen atom Lecture notes 6 (based on CT, Sec4on 6)

Introduction

  Central poten4al: V(r) depends only on the distance r from the origin

  The Hamiltonian commutes with the three components of the orbital angular momentum L

  The eigenfunc4ons of H can be required to be eigenfunc4ons of L2 and Lz as well

  This defines their angular part; the eigenvalue equa4on for H is then a differen4al equa4on in the variable r only

  Any isolated system of two interac4ng par4cles can be treated as a central problem

Stationary states of a particle in a central potential

  Consider a spinless par4cle of mass μ , subject to a central force derived from the poten4al V(r) (the center of force is the origin)

  The eigenvalue equa4on of the Hamiltonian is:

  Spherical coordinates are more suitable to solve the problem:

  We look for eigenfunc4ons which are func4ons of r, θ, φ


  We recall that:

  From which we can rewrite the Hamiltonian as:

  The angular dependence is en4rely contained in the L2 term

  We therefore need to solve:


  The three components of L commute with L2 and with all operators which only act on the r dependence

  Therefore, the three components of L are constants of the mo4on:

[H,L]=0 and [H,L2]

  The components of L do not commute with each other: we can only use L2 and one of the L components: we choose L2 and Lz

  Since H, L2 and Lz commute, we can find a basis of the state space E of common eigenfunc4ons to these three observables


  We need to solve the system of differen4al equa4ons:

  We already know the eigenfunc4ons of L2 and Lz. The func4ons we are looking for, are the products of a func4on of r and the spherical harmonic

  We now need to find R(r), such that the above func4on is eigenfunc4on of H


  When we apply L2 to we get:

is a common factor on both sides, therefore we simplified it

  No4ce that the expression for the laplacian is not valid at r=0

  We have to make sure that the behavior of the solu4ons R(r) is regular

  We look for eigenfunc4ons of an operator Hl which depends on l


  Therefore, we fix l and m. The equa4on to be solved depends only on l: it is the same in the (2l+1) subspaces E(l,m)

  We denote the eigenvalue of Hl by Ek,l

  The index k represents the various eigenvalues associated with the same value of l

  We rewrite the equa4on as:


  We can simplify the equa4on by wri4ng:

  Mul4plying both sides by r we get:

  This equa4on can be seen as the one for a par4cle which moves in the effec4ve poten4al


  The new term is always posi4ve or zero: the force tends to repel the par4cle from the force center O

  It is called centrifugal poten4al

  For an a_rac4ve Coulomb poten4al, the poten4al becomes:


  We assume that V(r) approaches infinity less rapidly than 1/r when r approaches 0

  We assume that the eigenfunc4on of H behaves like rs at the origin:

  Subs4tu4ng it into the equa4on and seang the coefficient of the dominant term to zero we get:

from which:


  Therefore, the two solu4ons of this equa4on behave at the origin as rl or 1/rl+1

  The la_er needs to be rejected because it is not a solu4on of

for r=0.

  Therefore, the acceptable solu4ons go to zero at the origin

  We add the condi4on to the eigenvalue equa4on


  Therefore, the fact that the poten4al is central allows:

-‐  To require the eigenfunc4ons of H to be simultaneous eigenfunc4ons of L2 and Lz, which determines their angular dependence

-‐  To replace the eigenvalue equa4on of H by a differen4al equa4on involving only r and depending on the parameter l


  In principle, the func4ons must be square-‐integrable:

  We can separate the variables:

  The spherical harmonics are normalized: this reduces to:


  However, if the spectrum of H has a con4nuum part, we only require:

where k is a con4nuous index

  It is only because of the behavior of the wavefunc4on for r∞ that the normaliza4on integrals diverge if k=k’

  We call k the radial quantum number, l the azimuthal quantum number and m the magne4c quantum number


  The (2l+1) func4ons with k and l fixed and m from –l to l are eigenfunc4ons of H with eigenvalue Ek,l

  The level Ek,l is therefore (2l+1)-‐fold degenerate

  This degeneracy is due to the fact that H does not contain Lz and it is called essen4al degeneracy

  It is also possible that an eigenvalue Ek,l is the same as Ek’,l’

  These degeneracies are called accidental

  For a fixed value of l, the radial equa4on has at most one solu4on for each Ek,l


  This follows from the condi4on that

  Since the radial equa4on is a second-‐order differen4al equa4on, in principle it has two solu4ons but the above requirement eliminates one of them

  H, L2 and Lz cons4tute a complete set of commu4ng observables

Motion of the center of mass and relative motion

  Consider a system of two spinless par4cles of mass m1 and m2 and posi4ons r1 and r2

  We assume that the forces exerted on these par4cles are derived from a poten4al energy V(r1-‐r2) which depends only on r1-‐r2

  This is true for an isolated system

  This system can be reduced to a single par4cle placed in a poten4al V(r)


  The operators R1, P1, R2, P2 which describe the posi4ons and momenta of the two par4cles sa4sfy the commuta4on rela4ons:

with analogous expressions along y and z. All observables labeled by the index 1 commute with all those of index 2.

  We define the observables (center of mass and rela4ve mo4on):


  The commutators of the new observables are:

  With analogous expressions along y and x. All other commutators are 0

  Therefore, R, P, RG, PG sa4sfy canonical commuta4on rela4ons

  We can interpret {R, P}, {RG, PG} as posi4on and momentum of two fic44ous par4cles


  The Hamiltonian is:

  Or equivalently

  Therefore, the Hamiltonian is the sum of two terms:

  with


  HG and Hr commute with each other and with H

  There exists a basis of eigenvectors of H which are eigenvectors of HG and Hr

  We look for solu4ons of the system:

which implies with


  We consider the {|rG,r>} representa4on, whose basis vectors are the eigenvectors common to RG and R

  In this representa4on, a state is characterized by a wavefunc4on

  In this representa4on, RG and R correspond to mul4plying the wavefunc4on by rG and r. PG and P become the differen4al operators

  The state space E is the tensor product of E rG and Er


  Therefore we can find a basis in the form

  with


  In the {|rG>} and {|r>} representa4ons we have:

  The par4cle associated with the center of mass is free:

  EG can take any posi4ve value or zero


  The second equa4on concerns the “rela4ve” par4cle

  It is subject to a central poten4al V(r)

  The total angular momentum is

or equivalently:

where

The hydrogen atom

  The hydrogen atom consists of a proton:

and an electron:

q= -‐ 1.6 x 10-‐19 Coulomb

  Their interac4on is electrosta4c

  We study the system in the center of mass frame

The center of mass coincides with the proton, the rela4ve par4cle is the electron

The hydrogen atom

  The spectrum of H includes a discrete part (nega4ve eigenvalues) and a con4nuous part (posi4ve eigenvalues)

  For E>0, the classical mo4on is not bounded in space: the eigenvalue equa4on has solu4ons for any value of E. The wavefunc4ons are not square integrable

  For E<0, the classical mo4on is bounded: the eigenvalue equa4on has solu4ons only for discrete values of E. The wavefunc4ons are square-‐integrable

The hydrogen atom

  We have to find eigenvalues and eigenfunc4ons of the Hamiltonian. We consider E<0. In the {|r>} representa4on it is:

  The poten4al is central: the eigenfunc4ons are of the form

  We write the radial wavefunc4on as:

The hydrogen atom

  The func4on χ has to sa4sfy the following equa4on:

  It is useful to introduce a new (dimensionless) variable η=αr

with:

  Besides, we define:

  The equa4on becomes:

The hydrogen atom

  The eigenvalues of H will be iden4fied by values of λ corresponding to physically acceptable solu4ons

  In terms of λ they read:

  We start by inves4ga4ng the asympto4c behavior of χEl

  In the limit η∞ the eigenvalue equa4on reduces to:

  The only acceptable solu4on has the form:

The hydrogen atom

  In order for the solu4on to be acceptable, QEl(η) has to diverge at most as a power of η

  The eigenvalue equa4on for QEl(η) is:

  The solu4on to the above equa4on has to behave as a power of η and it has to vanish for η=0

  For the above equa4on, the origin is a regular singular point. Its characteris4c exponents are l+1 and –l. Only l+1 is acceptable

  Therefore:

The hydrogen atom

  The series has to be truncated, in order for the func4on to have the desired asympto4c behavior

  Therefore, all coefficients ck are zero for k>n’, with n’ arbitrary integer number

  Therefore, for η∞ the func4on behaves as:

  Replacing into the eigenvalue equa4on, and keeping only the dominant terms we have:

The hydrogen atom

  This equa4on has solu4ons only for λ=n’+l+1

  Conven4onally we call n=n’+l+1 the “main quantum number”

  Once we fix n, we have l≤(n-‐1)

  The quan4za4on law for hydrogen atoms is therefore: λ=n and the quan4zed energies are:

  We can write the solu4on as:

The hydrogen atom

  Replacing into the eigenvalue equa4on, we find that the coefficients have to sa4sfy:

  These are the coefficients of Laguerre polynomials. The radial func4on can therefore be wri_en as:

  The Laguerre polynomials are given by:

The hydrogen atom

  The following quan44es are fundamental for the hydrogen atom:

  Indeed the first wavefunc4ons are:

  The Compton wavelength of the electron is:

  a0 is ≃100 4mes the Compton wavelength of the electron

The hydrogen atom

  The degeneracy of the states goes as follows:

  For every value of n, there are n-‐1 possible values for l, and for each of them there are 2l+1 values for m. The degeneracy therefore is:

  The various energy levels are:

The hydrogen atom

Behavior of the angular func4ons:

The hydrogen atom

Behavior of the radial func4ons:

The hydrogen atom

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