## Addition of two one half spins

Posted by peeterjoot on March 10, 2013

In class an example of interacting spin was given where the Hamiltonian included a two spins dot product

The energy eigenvalues for this Hamiltonian were derived by using the trick to rewrite this in terms of just squared spin operators

For each of these terms we can calculate the total energy eigenvalues from

where takes on the values of the total spin for the (possibly composite) spin operator. Thinking about the spin operators in their matrix representation, it’s not obvious to me that we can just add the total spins, so that if and are the spin operators for two respective particle, then the total system has a spin operator (really , since the respective spin operators only act on their respective particles).

Let’s develop a bit of intuition on this, by calculating the energy eigenvalues of using Pauli matrices.

First lets look at how each of the Pauli matrices operate on the eigenvectors

Summarizing, these are

For convienience let’s avoid any sort of direct product notation, with the composite operations defined implicitly by

Now let’s compute all the various operations

Tabulating first the action of the sum of the and operators we have

so that

Now we are set to write out the Hamiltonian matrix. Doing this with respect to the basis , we have

Two of the eigenvalues we can read off by inspection, and for the other two need to solve

or

These are the last of the triplet energy eigenvalues and the singlet value that we expected from the spin addition method. The eigenvectors for the eigenvalue is given by the solution of

So the eigenvector is

For our eigenvalue we seek

So the eigenvector is

An orthonormal basis with respective eigenvalues is thus given by

**Confirmation of spin additivity.**

Let’s use this to confirm that for , the two spin particles have a combined spin given by

With

we have for the energy eigenstate of

and for the energy eigenstate of

We get the and eigenvalues respectively as expected.

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