0
$\begingroup$

The $t-V$ model is usually written as: $$H = -t\sum_{i}^{L}c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i + V\sum_i n_i n_{i+1}$$ where $c_i^\dagger(c_i)$ are creation (annahilation) operators and $n_i$ is number operator.

At some places (i.e. this), the same model is written as $$H = -t\sum_{i}^{L}c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i + V\sum_i \bigg(n_i-\frac{1}{2}\bigg) \bigg(n_{i+1}-\frac{1}{2}\bigg)$$ I have two question:

  1. What is the difference between these two?
  2. The latter is usually used when you want to apply Jordan-Wigner transformation but, if I apply Jordan-Wigner transformation on the first Hamiltonain then what will be the difference in ground-state?
$\endgroup$

1 Answer 1

2
$\begingroup$

The two expressions you wrote differ by the following term:

$$ V\sum_i (-\frac{1}{2}n_i-\frac{1}{2}n_{i+1}+\frac{1}{4})=V\sum_i(\frac{1}{4}-n_i) $$

The first term, $\sum_i \frac{1}{4}$, is just a constant. Adding a constant to the Hamiltonian corresponds to adding a constant to your potential energy. As you probably know, adding a constant to a potential energy does not affect the physics.

The second term, $\sum_i n_i$, is just the operator that returns the total number of particles in the system. Depending on what you're studying, this term might matter! However, in most cases, you assume you are working with some FIXED number of particles in the system. For example, you might say that you are working at half filling, or whatever. If the number of particles in the system is fixed, then, this operator is ALSO a constant. So it also does not affect the physics, and in particular does not affect the ground state.

However, both of these terms affect the ground state energy. The first term raises the ground state energy by $V\sum_i\frac{1}{4}$, and the second term lowers it by $VN$, where $N$ is the total number of particles in your system. This shouldn't surprise you: when you add constants to a potential energy, you change the zero of your energy scale.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your reply, Sir. It is very helpful. $\endgroup$ Commented Jan 25, 2019 at 20:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.