# How do you perform a Jordan-Wigner transformation step-by-step? [closed]

I have understood the transformation equations to some extent, but I am unable to perform even a rather easy transformation. Here is an example Hamiltonian:

Well, you only need to know some properties of the tensor product: $$(A_1 \otimes B_1)\cdot(A_2 \otimes B_2) = (A_1 \cdot A_2)\otimes(B_1 \cdot B_2),$$ where I left explicit that $$a \cdot b$$ is the normal matrix multiplication between $$a$$ and $$b$$. It is understood that $$A_1$$ and $$A_2$$ live in the same space, i.e. it could be $$A_1,A_2 \in \mathbb{C_{2 \times 2}}$$. (Same applies for $$B_1$$ and $$B_2$$, i.e. they belong to the same space). Note $$A_1$$ and $$B_1$$ may not belong to the same space.
Lets take the first term: $$b_1^\dagger b_2 = (I \otimes I \otimes \sigma^+)(I \otimes \sigma^- \otimes \sigma^z)=(I \otimes \sigma^-\otimes \sigma^+ \sigma^z) \equiv -(I \otimes \sigma^-\otimes \sigma^+),$$ cause $$\sigma^+ \sigma^z = -\sigma^+.$$ Do that for all terms and use the distributive property $$A_1\otimes(B_1 + B_2) = A_1 \otimes B_1 + A_1 \otimes B_2, \tag{*}$$ remembering that $$\sigma^+ = (\sigma^x+i\sigma^y)/2$$ so $$b_1^\dagger b_2$$ really is $$b_1^\dagger b_2 = -(I \otimes \sigma^-\otimes \sigma^+) = - (I \otimes (\sigma^x-i\sigma^y)/2\otimes (\sigma^x+i\sigma^y)/2),$$ and I'll let you expand that one out. Spoiler alert: $$b_1^\dagger b_2+b_2^\dagger b_1 = I\otimes \sigma^x \otimes \sigma^x + I \otimes \sigma^y \otimes \sigma^y.$$