# 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:  From this  source, the answer is given to be Could you please show a step by step instruction on how this was done? Even one or two terms would suffice the example, no need to do the entire Hamiltonian. Edit: The question was flagged as a homework question. It is not one. The concepts may be clear but an example application often clarifies what is left. The selected answer does exactly that.

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: $$b_1^\dagger b_2+b_2^\dagger b_1 = I\otimes \sigma^x \otimes \sigma^x + I \otimes \sigma^y \otimes \sigma^y.$$