Let E_{s}(1) denote the two-dimensional state space of particle 1 and E_{s}(2)
the two-dimensional state space of particle 2. E_{s}= E_{s}(1)ÄE_{s}(2) then is the state space of the system of the two
particles. E_{s} is four-dimensional.

The vectors {*|i:+>,|i:->*} form a basis for the two-dimensional state space
of each particle. They are eigenvectors of *S _{iz} *and

,

.

The vectors {*|++>,|+->,|-+>,|-->*}, (where* |++>=|*1*:+>*Ä*|*2*:+>* etc. ), form a basis for E_{s}.
Every vector in E_{s} can be written as a linear combination of these basis
vectors.

Let e denote + or -. The we express orthogonality through

,

and completeness through

.

The most general normalized state in E_{s} is

,

with .

In the four dimensional state space of the two particles *S _{iz} *and

etc.

The matrix of any product operator *A(1)*Ä*B(2)*
in a basis of tensor product vectors {*|i(1)>*Ä*|j(2)>=|i,j>*}
is

,

where the *A _{ij}* are the matrix elements of

.

The matrix of *S _{1z}* in the {

.

The matrix of *S _{2z}* is

.

The matrix of *S _{1}^{2}* and

.

**What do we see inspecting these matrices?**

The basis vectors {*|++>,|+->,|-+>,|-->*} are eigenvectors of *S _{1z},
S_{2z}, S_{1}^{2}*, and

**The total spin of the two particles is S=S_{1}+S_{2}.
What are the eigenvectors of S^{2} and S_{z}?**

.

Any linear combination of basis vectors are eigenvectors of* S _{1}^{2}*
and

.

In the {*|++>,|+->,|-+>,|-->*} basis we have

.

.

.

What do we see by inspecting these matrices?

The basis vectors are not eigenvectors of *S*_{ix}, *S*_{iy},
, and *S*^{2}.

The eigenvalues of *S ^{2}* are 0 and . The eigenvalue is 3 fold degenerate.

Any linear combination of the normalized vectors |++>, is an eigenvector with eigenvalue .

The eigenvector with eigenvalue 0 is .

The matrix of *S _{z}=S_{1z}+S_{2z}* is

in the {*|++>,|+->,|-+>,|-->*} basis.

Common eigenvectors of *S ^{2}* and

,

**triplet states**

and

.

**singlet state**

*S*^{2} and *S*_{z}
form a C.S.C.O. for the 4-dimensional state space E_{s}.

(a) Write the Hamiltonian for the system.

(b) Work out the matrix form of *H*, defining the basis states explicitly.

(c) Find the **exact** expressions for the energies of the eigenstates.

(d) Calculate the state energies using perturbation theory and compare these approximate results with your exact expressions from part c).

- Solution:
No orbital angular momentum is mentioned, we assume that the total angular momentum of each particle is due to its spin

.**I**_{1}=**S**_{1},**I**_{2}=**S**_{2}.

- (a) .
- (b) In the {
*|++>,|+->,|-+>,|-->*} basis*S*and_{1z}*S*are diagonal, but is not. In the {_{2z}*|S,S*} basis is diagonal but_{z}>*S*and_{1z}*S*are not. Therefore the matrix of_{2z}*H*is not diagonal in either basis.Let us choose the {

*|++>,|+->,|-+>,|-->*} basis. Then.

(c) . The

*l*are the eigenvalues of*H*. We have (since the matrix is block-diagonal)´

=0,

where . One of the factors in this product has to be zero.

The eigenvalues are

,

and from

.

- (d) If
*H=H*, where_{0}+W*W*is a small perturbation compared to*H*, and the set {*|f*} is a non degenerate eigenbasis of_{n}>*H*with eigenvalues {_{0}*E*} then to first order in_{n}^{0}*W*the eigenvalues of*H*are {*E*}, where_{n}*E*._{n}=E_{n}^{0}+<f_{n}|W|f_{n}>.

.

.

.

A system of two distinguishable spin ½
particles (**S**_{1} and **S**_{2}) are in some
triplet state of the total spin, with energy *E _{0}.* Find the energies of
the states, as a function of

- Solution:
In the {

*|S,S*}={|1,1>,|1,0>,|1,-1>,|0,0>} basis the matrix of the Hamiltonian_{z}>*H*is_{0}.

We have

*H=H*._{0}+V.

In the {

*|S,S*} the matrix of_{z}>*V*is diagonal, since the matrices of are diagonal..

For the triplet state .

For the singlet state .

,

.

Therefore we have

.

,

.

The energies into which the triplet state splits are and . The value is twofold degenerate.