Physics Asked by jan0155 on October 20, 2020

I am currently reading S. Sachdevs Book on Quantum Phase Transitions focusing on the Bose-Hubbard Model (Chapter 9) and especially the Dilute-Boson Field Theory (Chapter 16).

When describing the fluid phase of the one dimensional model Sachdev says that this phase has quasi-long range order in the superfluid order parameter, intermediate boson occupation number and a non-zero superfluid stiffness.

I could not find any definition of a superfluid stiffness in the entire book and also doing some research on the internet I was not able to find a clean definition of superfluid stiffness in this context (Most likely because of my incapability 😀 to find something).

Therefore my question:

- Could somebody provide a definition of a superfluid stiffness in the context of the Bose-Hubbard Model?
- Any further explaination of this quantity in the "quasi long-range ordered" phase of the XY-chain would also be very kind?

Thank you all in advance.

There are probably different conventions that lead to definitions that differ by some numberical factor or factors of the mass density, but essentially the *superfluid stiffness* is the coefficient $alpha$ in the expressions for the energy density

$$
E[theta]= int d^3 xfrac 12 alpha |nabla theta|^2,
$$
where $theta$ is the phase of the superfluid order parameter. A non-zero $alpha$ means that it costs energy to have a space-varying phase, hence "stiffness". The superfluid particle-number current is then
$$
rho_s{bf v}_s = alpha nabla theta,
$$
where $rho_s$ is the *superfluid (number) density.*
As
$${bf v_s}=frac 1 m nablatheta
$$ where mass is $m$ of the superfluid particle one often writes
$$
E= int d^3 xfrac {rho_s}{2m} |nabla theta|^2,
$$
so $alpha= rho_s/m$. At finite temperature, the "energy" should be understood to be a local *free energy* $F=E-TS$.

Correct answer by mike stone on October 20, 2020

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