I am learning Luttinger liquid now. It is a very basic question, I think.
Look at the figure. For each $k $, there is a state for the left mover and a state for the right mover, right? They have the same wave function, right?
Then, why are their energies different generally?
Are the left movers and right movers idential particles, or, indistinguishable particles?
Luttinger liquids are a case of bosonization. Two fermions $\psi$ and $\psi'$ are written according to a boson field $\phi$ $$ \psi^\dagger\psi' = exp(i\phi), \psi'^\dagger\psi = exp(-i\phi^\dagger), $$ where the "field" $\phi$ serves as a phase and is real valued. For the fermion operator $c_j$ the number operator $n_j = c_j^\dagger c_j$ permits us to replace the fermion operator with $$ c_j \rightarrow exp\left(i\pi \sum_{j<i}n_i\right)b_j, $$ where $b_j$ is a boson operator that may be expressed according to a field $\phi$ that acts as a phase $b_j\rightarrow \sqrt{n_j}e^{i\phi_j}$. For two different fermions we then have the product of $c_i'c_j$ bosonized into a product $$ {c'_i}^\dagger c_j \rightarrow exp\left(-i\pi \sum_{k<i}n_k\right) exp\left(i\pi \sum_{l<i}n_l\right)b_i^\dagger b_j. $$ Now write the boson operator according to the phase and the separation into the two sets of states, here the $i, j$ standing in for right and left movers, is apparent.
