How to Derive the Second Hamiltonian Using the Jordan-Wigner Transformation?

[yukawa]

How to go from this Hamiltonian :

$$H= \sum^{N}_{j=1}[\sigma^{+}_{j}\sigma^{-}_{j+1}+\sigma^{-}_{j}\sigma^{+}_{j+1}]$$

to the following Hamiltonain:

$$H = \sum^{N-1}_{j=1}[c^{*}_{j}c_{j+1}+c^{*}_{j+1}c_{j}]-(c^{*}_{1}c_{N}+c^{*}_{N}c_{1})exp[i\pi\sum^{N}_{j=1}c^{*}_{j}c_{j}]$$

(where c^* is the Hermitian conjugate of c)

using the Jordan-Wigner transformation:
$$\sigma^{+}_{j} = exp[i\pi\sum^{j-1}_{n=1}c^{*}_{n}c_{n}] c_{j}$$

$$\sigma^{-}_{j} = exp[-i\pi\sum^{j-1}_{n=1}c^{*}_{n}c_{n}] c^{*}_{j}$$

where are fermionic operators.

The following is what I have calculated, please correct my mistakes if there are.

$$\sigma^{+}_{j}\sigma^{-}_{j+1}=c_{j}exp[-i\pi c^{*}_{j}c_{j}]c^{*}_{j+1}=c_{j}(1-2c^{*}_{j}c_{j})c^{*}_{j+1}$$

$$\sigma^{-}_{j}\sigma^{+}_{j+1}=c^{*}_{j}exp[i\pi c^{*}_{j}c_{j}]c_{j+1}=c^{*}_{j}(1-2c^{*}_{j}c_{j})c_{j+1}$$

putting them into the first Hamiltonian yeilds:

$$\sum^{N}_{j=1}c_{j}(1-2c^{*}_{j}c_{j})c^{*}_{j+1}+c^{*}_{j}(1-2c^{*}_{j}c_{j})c_{j+1}$$

how to proceed further in order to arrive at the second Hamiltonian?

 

[eljose79]

You are on the right track! To proceed further, we can use the Jordan-Wigner transformation once again on the terms c^{*}_{j}c_{j+1} and c^{*}_{j+1}c_{j}. This will result in the following expressions:

c^{*}_{j}c_{j+1} = exp[i\pi\sum^{j-1}_{n=1}c^{*}_{n}c_{n}] c^{*}_{j}c_{j+1} = c^{*}_{j}(1-2c^{*}_{j}c_{j})c_{j+1}

c^{*}_{j+1}c_{j} = exp[i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}] c^{*}_{j+1}c_{j} = c^{*}_{j+1}(1-2c^{*}_{j}c_{j})c_{j}

Substituting these expressions into the original Hamiltonian and rearranging the terms, we get:

H = \sum^{N}_{j=1}c_{j}(1-2c^{*}_{j}c_{j})c^{*}_{j+1}+c^{*}_{j+1}(1-2c^{*}_{j}c_{j})c_{j} - (c^{*}_{1}c_{N}+c^{*}_{N}c_{1})exp[i\pi\sum^{N}_{j=1}c^{*}_{j}c_{j}]

This is almost the same as the desired Hamiltonian, except for the exponent term. To get rid of this term, we can use the identity exp[i\pi\sum^{N}_{j=1}c^{*}_{j}c_{j}] = (-1)^{\sum^{N}_{j=1}c^{*}_{j}c_{j}}. This identity holds because the fermionic operators c^{*}_{j} and c_{j} anticommute, which means that (c^{*}_{j})^{2} = (c_{j})^{2} = 0. Therefore, the exponent term becomes:

(c^{*}_{1}c_{N}+c^{*}_{N}c_{1})(-1)^{\sum^{N}_{j=1

 

[denian]

Your calculations so far seem correct. To proceed further, we can use the fact that fermionic operators anticommute, meaning that c^{*}_{j}c_{j+1} = -c_{j+1}c^{*}_{j}. This allows us to rewrite the first Hamiltonian as:

\sum^{N}_{j=1}c_{j}(1+2c_{j}c^{*}_{j+1})c^{*}_{j+1}+c^{*}_{j}(1+2c^{*}_{j}c_{j+1})c_{j+1}

Next, we can use the Jordan-Wigner transformation to rewrite the terms in the parentheses as:

1+2c_{j}c^{*}_{j+1} = exp[i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}]

1+2c^{*}_{j}c_{j+1} = exp[-i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}]

Substituting these into the Hamiltonian and simplifying, we get:

\sum^{N}_{j=1}c_{j}exp[i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}]c^{*}_{j+1}+c^{*}_{j}exp[-i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}]c_{j+1}

We can then use the identity exp[i\pi\sum^{j}_{n=1}c^{*}_{n}c_{n}]c^{*}_{j+1} = c^{*}_{j+1}exp[i\pi\sum^{j-1}_{n=1}c^{*}_{n}c_{n}] to rewrite the first term as c^{*}_{j+1}c_{j} and similarly for the second term. This gives us the desired form of the second Hamiltonian:

\sum^{N-1}_{j=1}c^{*}_{j}c_{j+1}+c^{*}_{j+1}c_{j}-(c^{*}_{1}c_{N}+c^{*}_{N}c_{1})exp[i\pi\sum^{N}_{j