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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?

enter image description here

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  • $\begingroup$ In 1+1 dimensions, when the theory is either conformal or has two Fermi surfaces (points), left-moving and right-moving particles are always separated to independent. After all, the space of $\vec p$ in $D=1$ that obey $|\vec p|=p_F$ isn't connected - it's two points. So they're surely different. They don't even have to have the same properties. It's surely wrong to say that they have "the same wave function". If they had the same wave function, they would be absolutely the same situations, right? They are not. $\endgroup$ Jun 1 '16 at 14:55
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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.

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