13:29
I hope there's somebody Lieb Liniger model savvy in here -- Simple conceptual question:
Suppose one is in the ground state, and now one goes to the excited state
In the ground state the set of quantum numbers is described by a sequence $\xi_j = 1/L (j-(N+1)/2)$, with $j$ some index $\xi$ the quantum number over system length $L$ and $N$ the number of particles. So now suppose we have an excited state
In the excited state we choose to remove a subset of what the above relation generates for $1<j<N$, so far so good
The associated rapidities to me appear well defined in the ground state case
Now in the excited state, we may choose to replace a few of those quantum numbers with other quantum numbers, suppose we choose them all to be on the right side of the right "Fermi point"
Where the Fermi point is essentially the last quantum number on the line as seen from the left
Again so far all is clear
Now the way I always did calculations is by considering a couple of sets of quantum numbers and associated sets of rapidities
And for definiteness I just consider a left to right ordering, so the first particle to the right side of the Fermi point also corresponds to the first hole as seen from the left Fermi point in the Fermi sea
So I have a few coinciding normalised quantum numbers $\xi$, call them $\xi^0$, and I guess a few numbers that are holes in the excited state so $\xi^-$, and particles $\xi^+$
This all seems clear I guess, and we may associate say, rapidities in the GS as $\lambda^0$, $\lambda^-$ etc., but we don't have $\mu^-$. These don't have a corresponding quantum number since it's absent from the set $\xi$ for the excited state
Now I found a paper that seems to use this rapidity anyway, how can that be?
What is the rapidity at the hole in the excited state? It's zero I would argue?
And similarly I found a paper that has $\lambda^+$, but there is no notion of that additional particle in the ground state
I suppose I could extend the set $\{\xi^0\}$ to have cardinality $N+n$ with $n$ the number of excitations in the chosen excited state configuration
But this all seems rather moot