Interaction Hamiltonian coupling question

[td21]

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:

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

 

[Truecrimson]

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.