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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:

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The Jordan-Wigner mappings read

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From this [4] source, the answer is given to be

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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.

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

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