More problems

Consider the Helium atom.
(a) Write down the Hamiltonian in the central field approximation and the dominant correction term.
(b) What is the configuration of the ground, first excited, and second excited state?
(c) What type of splitting is introduced by the non-central part of the electron-electron interaction?  What is the degeneracy of each sublevel?
(d) What type of additional splitting is introduced by the spin-orbit interaction?

Solution:

(a) H0=p12/(2m)+p22/(2m)-2e2/r1-2e2/r2+Vc(r1)+Vc(r2)
is the Hamiltonian in the central field approximation.
W =e2/r12-Vc(r1)-Vc(r2)
is the dominant correction term.  It is due to the non-central part of the electron-electron interaction.

(b) The eigenvalues of H0 depend on n1, l1 and n2, l2.
Configuration of the ground state: 1s2
Configuration of the first excited state: 1s2s
Configuration of the second excited state: 1s2p
(c) The electron-electron repulsion term in the Hamiltonian commutes with L and S.
[W, L1]¹0, [W, L2]¹0, but [W, L]=[e2/r12, L]=0. [W, S]=0, where L=L1+L2, S=S1+S2.
L and S are constants of motion.
The anti-symmetric eigenfunctions of L2, Lz, S2, and Sz form an orthonormal basis for the subspace EA(n1l1,n2l2).  The matrix elements of W between those basis vectors are diagonal and do not depend on ML and MS.  (W commutes with L± and S±.)  The eigenvalues of H therefore are E0(n1l1,n2l2)+E1(L,S). They are (2L+1)(2S+1)-fold degenerate.

(d) The perturbation due to the spin orbit interaction may be written as

.

Li is a vector operator in orbital space and Si is a vector operator in spin space.  Within the subspace EA(n1l1,n2l2,L) LiµL by the Wigner-Eckart theorem, and within the subspace E(s1,s2,S) SiµS.
In the LS coupling scheme we therefore have WsoµL×S.
L×S=(1/2)(J2-L2-S2)
Inside each subspace EA(n1l1,n2l2,LS) the eigenvalues of H depend on J and are (2J+1)-fold degenerate.

Consider a carbon atom whose electrons are in the configuration (1s)2 (2s)2 2p 3p. List all expected terms on the basis of the LS (Russell-Sanders) coupling scheme.
Solution:

(1s)2 couples to L=0, S=0. (Pauli exclusion principle)
(2s)2 couples to L=0, S=0. (Pauli exclusion principle)
The 2p and 3p electrons both have l=1, s=1/2. This can lead to L=2,1,0, S=0,1.

We therefore can form the following terms:
1D2, 3D123, 1P1, 3P012, 1S0, 3S1.

Scandium has a ground state configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d.  Consider a transition from the ground state 4s2 3d to the excited state 3d 4s 4p.

(a) Assuming LS coupling find L, S, and J values for the terms derived from the 4s2 3d and 3d 4s 4p configurations.  Write down each term in standard notation.

(b) Order the quadriplet terms of the 3d 4s 4p in order of increasing energy.  Order the levels of the 4D multiplet in order of increasing energy and indicate the relative spacing between levels of the multiplet.

(c) Write down dipole selection rules for LS coupling and indicate the allowed transitions.

(d) Which additional intersystem transitions are likely to occur?

Solution:

(a)
  4s2 3d
L 0 2
S 0 1/2

LS coupling: L=2; S=1/2

Possible terms: 2D5/2, 2D3/2

  3d 4s 4p
L 2 0 1
S 1/2 1/2 1/2

LS coupling: L=3,2,1; S=3/2,1/2

Possible terms: 4F9/2,7/2,5/2,3/2; 2F7/2,5/2,; 4D7/2,5/2,3/2,1/2; 2D5/2,3/2; 4P5/2,3/2,1/2; 2P3/2,1/2.

(b)  Hund's rule: Emax to Emin
2P, 2D, 2F, 4P, 4D, 4F

For the 4D multiplet: L×S=(1/2)(J2-L2-S2)=(J(J+1)-6-15/4)=(J(J+1)-39/4).

  J=7/2 J=5/2 J=3/2 J=1/2
(J(J+1)-39/4) 6 -1 -6 -9
 

(c) LS coupling selection rule for allowed transitions:
DL=±1, DJ=0,±1, DMJ=0,±1, DS=0.

Ground state ß   à Excited state
2D   2F
2D   2P
 

(d)  The next most likely transitions are magnetic dipole and electric quadrupole transitions.

Magnetic dipole transitions:
DL=0, DJ=0,±1, DMJ=0,±1, DS=0, no J=0 to J=0 transitions.

Ground state ß à Excited state
2D   2D

Electric quadrupole transitions:
DL=0,2, DJ=0,±1,±2, DMJ=0,±1,±2, DS=0.

Ground state ß à Excited state
2D   2D
 

The zeroth order Hamiltonian in jj coupling is given by

.

(a) What are the quantum numbers of the one-electron functions?

(b) What are the three jj coupling states that arise from (np)2?

(c) What are the possible j values for each of these states?

(d) What are the expectation values of the spin-orbit energy for each of these states expressed in terms of <np|a(r)|np>?

(e) What is the effect of introducing the electron-electron repulsion as a perturbation?  Sketch the transition from jj to LS coupling.

Solution:

(a) n,l,s,j,m

(b) Configuration np2

    possible terms
np1/2 , np1/2 m1=1/2, m2=-1/2 (1/2,1/2)0
np1/2 , np3/2 m1= ±1/2, m2= ±1/2, ±3/2 (1/2,3/2)2,1
np3/2 , np3/2 m1=3/2, m2=1/2
m1=3/2, m2=-1/2
m1=3/2, m2=-3/2
m1=1/2, m2=-1/2		 (3/2,3/2)2,0

(c)  The possible values of J are

0 for (1/2,1/2),
2 and 1 for (1/2,3/2),
2 and 0 for (3/2,3/2).

(d) Both particles have the same n, l=1, and s=1/2. The allowed values for j1, j2 are (1/2,1/2), (1/2,3/2), (3/2,3/2). For indistinguishable particles the state vector must be anti-symmetric.

|yA>=(2)-1/2(|j1m1;j2m2>-|j2m2;j1m1>)

Wso=a(r1)2l1×s1/+ a(r2)2l2×s2/Image5165.gif (871 bytes)=W'(1)+W'(2).

The matrix elements of W' are
< jmj|W'| jmj'>= <nlsjmj| a(r)(j2-l2-s2)| nlsjmj'>=
(j(j+1)-2-3/4) )<nlsjmj| a(r)| nlsjmj'>.
The matrix elements are zero if mj¹mj'.

We therefore have
<yA|Wso|yA>=(1/2)(< j1m1|W'| j1m1>+< j2m2|W'| j2m2>)2.
All the cross terms are zero since either j1¹j2 or m1¹m2.

We have

<yA|Wso|yA>=2<np|a(r)|np> for (3/2,3/2),
<yA|Wso|yA>=-1<np|a(r)|np> for (1/2,3/2),
<yA|Wso|yA>=-4<np|a(r)|np> for (1/2,1/2).

(d)  The operator for the electron-electron interaction, is a scalar operator.  It does not commute with ji=li+si but commutes with j=åji.  The non-zero matrix elements of any scalar operator A, <kjmj|A|k'j'm'j> must satisfy j=j' and mj=mj' and do not depend on mj.

The matrix elements of in the subspace of vectors (j1,j2) are diagonal between eigenvectors of J2, (j1,j2)J.

dia.gif (8965 bytes)

In the transition, we match levels with the same value of J.

Consider a system of 4 identical particles.  Each particle has 3 possible eigenvalues, E1, E2, E3, of some observable.

(a)    Use Young’s tableaux or any other method to find the number of possible states that are (i) symmetric, (ii) anti symmetric, and (iii) of mixed symmetry.

(b)    Write down the normalized symmetric wave function in which two of the particles have eigenvalue E1, one has eigenvalue E2, and one has eigenvalue E3.

Solution:

Background :

Young’s tableaux is a method of counting the number of states which are totally symmetric or anti-symmetric under the exchange of any two particles and the number of states with mixed symmetry.  States with mixed symmetry are symmetric or anti-symmetric under the exchange of some of the particles, but not all of the particles.

Method:

Each indistinguishable particle is represented by a box ([ ]), each of the possible states is represented by a number in that box (1,2,3,… ).

Example:
A spin ½ particle can be in one of two possible eigenstates of Sz.  It is represented either by [1] (for spin up) or [2] (for spin down).

A symmetric state of N particles is represented by a row of N boxes, an anti-symmetric state by a column of N boxes.

For a row of N boxes, the numbers in the boxes cannot decrease when going from left to right, (ni ³ ni-1).  For a column of boxes , the numbers in the boxes must increase when going from top to bottom (nj>nj-1).

Example:

For two spin ½ particles the possible symmetric states are [1][1], [1][2], [2][2], i.e. there are three symmetric states.  There is only one possible anti-symmetric state, .

A tableaux for a mixed state has the following form:

i.e., the total number of boxes is N, and the number of boxes in row i must be less or equal the number of boxes in row i-1.  If the entries in the boxes of the tableaux T are labeled Tij, the we need Tij £ Tij+1, Tij < Ti+1j. (Rows are labeled by the index i, columns by the index j.)

Why are we interested in mixed states?

Assume the state space E=E(1)ÄE(2) is a tensor product space and we want to find the number of totally anti-symmetric states in E.  Product states which are products of a state symmetric in E(1) and a state anti-symmetric in E(2) or a state symmetric in E(2) and a state anti-symmetric in E(1) are obviously totally anti-symmetric in E.  But we also can find states totally anti-symmetric in E by combining mixed tableauxs.

(a)  To find the number of symmetric states we count the number of rows we can form with four boxes (particles) and three numbers (states).

Table 1

P1

P2

P3

P4

 

 

 

1

 

 

1

2

 

 

 

3

 

1

2

2

 

 

 

3

 

 

3

3

1

 

2

2

 

2

 

3

 

 

3

3

 

3

3

3

 

 

2

2

 

2

 

3

2

 

3

3

 

3

3

3

3

3

3

3

We can form 15 symmetric states.

There are no anti-symmetric states since we cannot form a column with four boxes (particles) and three numbers (states).

We can form a mixed state by adding a box to a column with 3 entries
(3 possibilities for [x]).

or adding a box to a row with 3 entries.

Referring to table 1, we have 6 rows beginning with 1.  For these rows y can be 2 or 3.  We have 3 rows starting with 2.  For these rows y can only be 3.  We therefore have 6+6+3=15 possibilities.

We can also form a mixed state by forming a square.  There are 6 possibilities to form a square.

The total number of mixed states therefore is 24.

(b)  We find the sum of all possible permutations and divide by the square root of the number of permutations.
|y>=[ |1,1,2,3>+|1,1,3,2>+|1,2,1,3>+|1,2,3,1>

                   +|1,3,2,1>+|1,3,1,2>+|2,1,1,3>+|2,1,3,1>

                   +|2,3,1,1>+|3,1,1,2>+|3,1,2,1>+|3,2,1,1>]/(12)½ .