- 36

- 0

- Summary
- How do I go about showing that the total Spin component commutes with the hamiltonian

Let there be four spin (1/2) particles at the corners of a tetrahedron, coupled such that the total hamiltonian is given by $$H=\sum_{i\neq j} S_{i} \cdot S_{J}$$.

How would I go about showing that each component of the total spin $$\sum_{i} S_{i}$$ commutes with the hamiltonian.

Work so far:

I know

$$\sum_{i} S_{i}^{x}=S_{1}^{x}+S_{2}^{x}+S_{3}^{x}+S_{4}^{x}$$

so $$[\sum_{i} S_{i}^{x},H]=[S_{1}^{x},H]+[S_{2}^{x},H]+[S_{3}^{x},H]+[S_{4}^{x},H]$$

From there, I can use $$H=S_{1}S_{2}+S_{1}S_{3}+S_{1}S_{4}+S_{2}S_{3}+S_{2}S_{4}+S_{3}S_{4}$$

I can also use $$S_{i}\cdot S_{j}=S_{i}^{x}S_{j}^{x}+S_{i}^{y}S_{j}^{y}+S_{i}^{z}S_{j}^{z}$$

As you can see, it gets rather messy. I just want to know if my reasoning is correct thus far and or if I am over complicating my solution.

Thank you.

How would I go about showing that each component of the total spin $$\sum_{i} S_{i}$$ commutes with the hamiltonian.

Work so far:

I know

$$\sum_{i} S_{i}^{x}=S_{1}^{x}+S_{2}^{x}+S_{3}^{x}+S_{4}^{x}$$

so $$[\sum_{i} S_{i}^{x},H]=[S_{1}^{x},H]+[S_{2}^{x},H]+[S_{3}^{x},H]+[S_{4}^{x},H]$$

From there, I can use $$H=S_{1}S_{2}+S_{1}S_{3}+S_{1}S_{4}+S_{2}S_{3}+S_{2}S_{4}+S_{3}S_{4}$$

I can also use $$S_{i}\cdot S_{j}=S_{i}^{x}S_{j}^{x}+S_{i}^{y}S_{j}^{y}+S_{i}^{z}S_{j}^{z}$$

As you can see, it gets rather messy. I just want to know if my reasoning is correct thus far and or if I am over complicating my solution.

Thank you.