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Suppose that an ion having s = 1, l
= 0 is surrounded by an array of charges.
Its Hamiltonian is H = D[S_{z}^{2} - 1/3 s(s + 1)I] + G(S_{x}^{2}
- S_{y}^{2})
. Assuming D/3 > G > 0.

(a) Construct the energy matrix for this Hamiltonian.

(b) Solve exactly for the energy eigenvalues.

(c) Sketch the energy levels.

Given: I is the identity matrix.

In this problem the units are chosen such that ħ = 1.

Solution

- (a) You can solve this problem without knowing anything about the physical quantities
represented by the operators S
^{2}, L^{2}, S_{x}, S_{y}, and S_{z}. L^{2}and S^{2}represent the operators for the square of the orbital angular momentum and the spin angular momentum of a spin 1 particle and S_{x}, S_{y}, and S_{z}represent the operators for the Cartesian components of the spin. The eigenvalues of S^{2}are s(s + 1)ħ^{2}and the eigenvalues of L^{2}are l(l + 1)ħ^{2}. In this problem the units are chosen such that ħ = 1.

S_{z}is diagonal. The basis picked is an eigenbasis of S_{z}. The eigenvalues of S_{z}are 1,0,-1.

H = D[S_{z}^{2}- 1/3 s(s + 1) I] + G(S_{x}^{2}- S_{y}^{2}). D and G are numbers, S_{x}^{2}, S_{y}^{2}, and S_{z}^{2}are matrices and s(s + 1) is a number.

s = 1, s(s + 1) = 2, (1/3)s(s + 1) = 2/3.

(b)

-(D/3 - β)^{2}(2D/3 + β) + G^{2}(2D/3 + β) = 0.

Either (2D/3 + β) = 0, β_{1}= -2D/3, or D/3 - β = ąG, β_{2}= D/3 + G, β_{3}= D/3 - G.

The eigenvalues are β_{1}, β_{2}, and β_{3}.

The eigenvector corresponding to the eigenvalue β_{1}, β_{2}, and β_{3}are

.

(c) We have found the eigenbasis of H.

The eigenvalues are E_{1}= -2D/3 < 0, E_{2}= D/3 + G > 0, E_{3}= D/3 - G > 0.

The angular momentum operators {J_{x}, J_{y}, J_{z}} are central to quantum theory.
States are classified according to the eigenvalues of these operators when
**J** is conserved by the respective Hamiltonian H.

(a) What condition(s) is (are) necessary for all eigenstates of H to be
eigenstates of **J**?

An eigenstate of **J** is usually specified by |j,m_{z}>,

where J^{2}|j,m_{z}> = j(j + 1)ħ^{2}|j,m_{z}>
and J_{z}|j,m_{z}> = m_{z}ħ|j,m_{z}>.

(b) We can substitute J_{x} or J_{y} for J_{z}
in (a). However a state cannot be simultaneously an eigenstate of J_{z}
and J_{x}. Derive the commutation relation for the angular
momentum operators J_{x} and J_{z},
(i.e. [J_{x},J_{z}] = -iħJ_{y})
from the definition of the linear momentum operator.

(c) Prove that it is indeed possible for a state to be simultaneously an eigenstate of
J^{2 }= J_{x}^{2 }+ J_{y}^{2 }+J_{z}^{2}
and J_{z}.

Solution:

- Concepts::

Commuting operators - Reasoning:

Two commuting observables can be measured simultaneously, i.e. the measurement of one does not cause loss of information obtained in the measurement of the other. If we measure a complete set of commuting observables (C.S.C.O.), then the state of the system after the measurement is one element of an unique eigenbasis. The results of the measurement specify the state completely. - Details of the calculation:

(a) It is possible to specify a common eigenbasis of two operators if they commute. If one of the operators is non degenerate, then all of its eigenvectors are also eigenvectors of the other operator. In order for all eigenstates of H to be eigenstates of J^{2}and J_{z}we need [J^{2},H] = 0 and [J_{z},H] = 0 and H is non degenerate. (States with different j or m_{z}must have different energy eigenvalues E.)

b) Assume the classical relationship**J**=**r**×**p**.

J_{x}= yp_{z}- zp_{y}, J_{y}= zp_{x}- xp_{z}, J_{z}= xp_{y}- yp_{x}.

For the quantum mechanical operators we then have

J_{x}= YP_{z}- ZP_{y}, J_{y}= ZP_{x}- XP_{z}, J_{z}= XP_{y}- YP_{x}.

[J_{x},J_{z}] = [YP_{z}- ZP_{y},XP_{y}- YP_{x}] = [YP_{z},XP_{y}] + [ZP_{y},YP_{x}] - [YP_{z},YP_{x}] - [ZP_{y},XP_{y}]

0 0

= [YP_{z},X]P_{y}+ X[YP_{z},P_{y}] + [ZP_{y},Y]P_{x}+ Y[ZP_{y},P_{x}]

0 0

= Z[P_{y},Y]P_{x}+ X[Y,P_{y}]P_{z}= -iħ(ZP_{x}- XP_{z}) = -iħJ_{y}.

(In general [J_{i},J_{j}] = iħε_{ijk}J_{k}.)

(c) J^{2}= J_{x}^{2}+ J_{y}^{2}+ J_{z}^{2}. If [J^{2},J_{z}] = 0 then it is possible to find simultaneous eigenstates.

[J^{2},J_{z}] = [J_{x}^{2},J_{z}] + [J_{y}^{2},J_{z}] + [J_{z}^{2},J_{z}] = J_{x}[J_{x},J_{z}] + [J_{x},J_{z}]J_{x}+ J_{y}[J_{y},J_{z}] + [J_{y},J_{z}]J_{y }= -iħ(J_{x}J_{y}+ J_{y}J_{x}) + iħ(J_{x}J_{y}+ J_{y}J_{x}) =0.

The matrix σ_{x}
is defined by

.

Prove the relation exp(iασ_{x}) = I cosα + i σ_{x}
sinα, where I
is the 2 x 2 identity matrix.

- Solution:

Functions of operators

Let σ_{x}be the matrix of the Hermitian operator S_{x}in the {|1>,|2>} basis, <i|S_{x}|j> = (σ_{x})_{ij}.

exp(iαS_{x}) = cos(αS_{x}) + i sin(αS_{x}) = ∑_{n=even}f_{n}α^{n}S_{x}^{n}+ i ∑_{n=odd}f_{n}α^{n}S_{x}^{n}.

<i|S_{x}^{2}|j> = ∑_{k=1}^{2}<i|S_{x}|k><k|S_{x}|j> = ∑_{k=1}^{2}(σ_{x})_{ik}(σ_{x})_{kj}= δ_{ij}.

Therefore S_{x}^{2 }= I, S_{x}^{n }(n=even) = I, S_{x}^{n }(n = odd) = IS_{x }= S_{x}.

exp(iαS_{x}) = ∑_{n=even}f_{n}α^{n}I + i ∑_{n=odd}f_{n}α^{n}S_{x}= I cosα + i S_{x}sinα,

exp(iασ_{x}) = I cosα + i σ_{x}sinα.