# Interaction Hamiltonian coupling question

• A
Gold Member
System is composed of two qubits and the bath is one bath qubit.

The interaction Hamiltonian is:

$$\sigma_1^x\otimes B_1 + \sigma_2^x\otimes B_2$$ where $$B_i$$ is a 2 by 2 matrix.

I try to interpret and understand this, is it the same as:
$$(\sigma_1^x\otimes B_1)\otimes I_2 + I_1\otimes(\sigma_2^x\otimes B_2)~?$$

If the situation is that two system qubits are coupled to the bath qubit in the same way, such that $$B_1= B_2 = B$$, may I write the the interaction hamiltonian as:

$$(\sigma_1^x\otimes B + \sigma_2^x\otimes B)\stackrel{?}{=} (\sigma_1^x\otimes B)\otimes I_2 + I_1\otimes(\sigma_2^x\otimes B)\stackrel{?}{=}(\sigma_1^x + \sigma_2^x)\otimes B$$

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

is it the same as:
$$(\sigma_1^x\otimes B_1)\otimes I_2 + I_1\otimes(\sigma_2^x\otimes B_2)~?$$

Yes.

If the situation is that two system qubits are coupled to the bath qubit in the same way, such that $$B_1= B_2 = B$$, may I write the the interaction hamiltonian as:

$$(\sigma_1^x\otimes B + \sigma_2^x\otimes B)\stackrel{?}{=} (\sigma_1^x\otimes B)\otimes I_2 + I_1\otimes(\sigma_2^x\otimes B)\stackrel{?}{=}(\sigma_1^x + \sigma_2^x)\otimes B$$

Yes. Tensor product is distributive.