Gauge-Invariant Meissner Response and Superfluid Weight

The Meissner effect is the equilibrium transverse electromagnetic response of charged paired matter. It appears only after the paramagnetic current correlator and the diamagnetic contact term are combined in a gauge-consistent convention. A finite conductivity delta function or a mean-field gap alone does not establish magnetic screening: the static transverse limit and penetration depth must be obtained explicitly.

Required background. Gauge redundancy and observables, the gauge-invariant Higgs mechanism, and Kubo response supply the field and limit conventions.

Helpful background. Nambu–Gor’kov propagators supplies the paired current bubble and Ward identity.

Electromagnetic kernel

For particles of charge $q$ with quadratic dispersion, minimal coupling gives a current

$$\mathbf j=\mathbf j_{\mathrm p}-\frac{nq^2}{m}\mathbf A.$$

Linear response may be written

$$j_i(q)=-K_{ij}(q)A_j(q), \qquad K_{ij}=\frac{nq^2}{m}\delta_{ij}+\Pi_{ij}^R,$$

where $\Pi^R$ is the retarded paramagnetic correlator in this sign convention. Gauge invariance requires the longitudinal Ward identity $q_\mu K^{\mu\nu}=0$. In the normal continuum state the static uniform transverse paramagnetic term cancels $nq^2/m$. In a superconductor the cancellation is incomplete:

$$D_{s,ij}=\lim_{\mathbf q\to0}K_{ij}(\mathbf q,0), \qquad i,j\perp\mathbf q.$$

$D_s$ is the charged superfluid weight. A bare bubble with a dressed propagator generally violates the Ward identity unless the matching vertex is included. This is why “add the diamagnetic term” is necessary but not always sufficient.

Nambu 1960, §§III–IV shows how the gauge vertex and collective response restore the conservation constraints of the paired state.

For a clean, Galilean-invariant, fully gapped BCS state at $T=0$, $D_s=nq^2/m$. On a lattice, band curvature replaces $n/m$ and interactions can renormalize the result. At finite temperature quasiparticles reduce $D_s$; its low-$T$ behavior can help diagnose a full gap versus nodes, but disorder and multiband structure must be included in the comparison.

Penetration depth and the order of limits

Combine $\mathbf j=-D_s\mathbf A_T$ with static Maxwell theory. In SI units,

$$\boldsymbol\nabla^2\mathbf B=\lambda_L^{-2}\mathbf B, \qquad \lambda_L^{-2}=\mu_0D_s.$$

Magnetic field therefore decays over $\lambda_L$ inside a bulk sample. Geometry and nonlocality matter when the coherence length or mean free path is not small compared with $\lambda_L$ or the sample dimensions.

Tinkham 2004, chs. 2–3 develops the London and nonlocal screening limits in the same charge convention.

The optical conductivity uses a different limit:

$$\operatorname{Re}\sigma(\omega) =\pi D_s\delta(\omega)+\sigma_{\mathrm{reg}}(\omega)$$

in an ideal superconductor. A translation-invariant normal metal can also have a Drude delta function because momentum does not relax. Only the equilibrium transverse kernel diagnoses Meissner screening. Scalapino, White, and Zhang 1993, §§II–IV makes this order-of-limits distinction explicit.

Phase-only interpretation

At long wavelength, the gauge-invariant free energy is

$$F=\frac{\rho_s}{2}\int\mathrm d^d x\, (\boldsymbol\nabla\theta-q_{\mathrm p}\mathbf A)^2,$$

where $q_{\mathrm p}=2e$ in magnitude for electronic Cooper pairs and $\rho_s$ uses the pair-phase convention. Expanding in transverse $\mathbf A$ gives $D_s=q_{\mathrm p}^2\rho_s$. This connection is powerful but conditional: it assumes the phase-only description, correct charge normalization, and an equilibrium static limit.

The structure map highlights the response gate between Nambu calculations and magnetic screening.

Meissner screening is licensed only by the gauge-consistent static transverse kernel. The conductivity delta function, phase stiffness, and penetration depth are related only after their charges and limit orders are translated. Original schematic, not to scale.

See the paired-matter claim test matrix for the independent checks.

Exercise

Derive London screening. In transverse gauge, combine $\mathbf j=-D_s\mathbf A$ with $\boldsymbol\nabla\times\mathbf B=\mu_0\mathbf j$ and $\mathbf B=\boldsymbol\nabla\times\mathbf A$.

Solution

Taking a curl of Ampère’s law gives $\nabla\times(\nabla\times\mathbf B)=\mu_0\nabla\times\mathbf j=-\mu_0D_s\mathbf B$. Since $\nabla\cdot\mathbf B=0$, the left side is $-\nabla^2\mathbf B$. Hence $\nabla^2\mathbf B=\mu_0D_s\mathbf B=\lambda_L^{-2}\mathbf B$. Positivity of $D_s$ is required for exponential decay; a sign error in the response convention would predict an instability instead of screening.

References

• Nambu, Y. (1960). “Quasi-particles and gauge invariance in the theory of superconductivity.” Physical Review 117, 648–663. doi:10.1103/PhysRev.117.648.
• Scalapino, D. J., White, S. R., and Zhang, S. C. (1993). “Insulator, metal, or superconductor: The criteria.” Physical Review B 47, 7995–8007. doi:10.1103/PhysRevB.47.7995.
• Tinkham, M. (2004). Introduction to Superconductivity, 2nd ed. Dover. Publisher record.