Interaction Hamiltonian coupling question

  • A
  • Thread starter td21
  • Start date
  • #1
td21
Gold Member
178
8

Main Question or Discussion Point

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:

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

Answers and Replies

  • #2
18,086
7,510
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?
 
  • #3
262
86
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.
 

Related Threads on Interaction Hamiltonian coupling question

  • Last Post
Replies
3
Views
3K
Replies
5
Views
4K
Replies
7
Views
604
Replies
0
Views
1K
Replies
8
Views
749
Replies
2
Views
916
Replies
2
Views
2K
Replies
1
Views
169
  • Last Post
Replies
6
Views
271
  • Last Post
Replies
1
Views
1K
Top