Does a Massive Photon Make a Superconductor Heavier?
Appendix to the Standard Model lecture
Does a Massive Photon Make a Superconductor Heavier?
The Paradox
Consider a normal conductor placed in a magnetic field.
Now cool it below the critical temperature. The conductor becomes a superconductor.
There are two languages for the same phenomenon.
Current language:
a superconducting screening current appears near the surface and expels the magnetic field.Field language:
the photon inside the superconductor becomes massive; the electromagnetic field acquires a finite penetration length.Question:
are these two descriptions adding two different energies, or are they the same energy written twice?
Static Free Energy
Use units
\[ \hbar=c=\mu_0=\varepsilon_0=1. \]
For a nonrelativistic superconductor the Ginzburg–Landau free energy is
\[ F[\psi,A] = \int d^3x \left[ \frac{1}{2}B^2 + \frac{1}{2m_*} \left| (-i\nabla-qA)\psi \right|^2 + \alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4 \right]. \]
Here
\[ B=\nabla\times A, \]
\[ q=2e, \qquad m_*=2m_e \]
for Cooper pairs.
The first term is magnetic-field energy.
The second term is the kinetic energy of the superconducting condensate.
The last two terms determine whether the condensate exists.
London Limit
Deep in the superconducting phase the condensate amplitude is nearly fixed:
\[ \psi(x)=\sqrt{n_s}\,e^{i\theta(x)}. \]
Then
\[ (-i\nabla-qA)\psi = \sqrt{n_s}\,e^{i\theta} \left(\nabla\theta-qA\right). \]
Therefore the relevant part of the free energy becomes
\[ F_{\rm L} = \int d^3x \left[ \frac{1}{2}B^2 + \frac{n_s}{2m_*} \left(\nabla\theta-qA\right)^2 \right]. \]
Define the London penetration depth by
\[ \lambda_L^{-2}=\frac{q^2n_s}{m_*}. \]
Then
\[ F_{\rm L} = \int d^3x \left[ \frac{1}{2}B^2 + \frac{1}{2\lambda_L^2} \left(A-\frac{1}{q}\nabla\theta\right)^2 \right]. \]
This expression is gauge invariant.
The gauge-invariant object is not \(A\) alone, but
\[ A-\frac{1}{q}\nabla\theta. \]
Unitary Gauge: The Photon Mass Term
Choose the gauge in which the condensate phase is constant:
\[ \theta=0. \]
Then
\[ F_{\rm L} = \int d^3x \left[ \frac{1}{2}B^2 + \frac{1}{2\lambda_L^2}A^2 \right]. \]
The second term has the form of a mass term for the vector field:
\[ \frac{1}{2}m_\gamma^2A^2, \qquad m_\gamma=\frac{1}{\lambda_L}. \]
Thus the photon is massive inside the superconductor in the precise sense that the electromagnetic field has a mass gap and a finite penetration length.
The Same Term as Supercurrent Energy
The superconducting current follows from the free energy:
\[ j_s = -\frac{1}{\lambda_L^2} \left(A-\frac{1}{q}\nabla\theta\right). \]
In the gauge \(\theta=0\),
\[ j_s=-\frac{1}{\lambda_L^2}A. \]
Therefore
\[ \frac{1}{2\lambda_L^2}A^2 = \frac{\lambda_L^2}{2}j_s^2. \]
So the photon mass term is not an additional energy on top of the current energy.
It is the kinetic energy of the condensate current written in the gauge-field language.
\[ \boxed{ \text{photon mass term} = \text{supercurrent kinetic energy} } \]
Meissner Equation from Free Energy Minimization
Minimize
\[ F_{\rm L} = \int d^3x \left[ \frac{1}{2}(\nabla\times A)^2 + \frac{1}{2\lambda_L^2}A^2 \right] \]
with respect to \(A\), in the gauge
\[ \nabla\cdot A=0. \]
The variation gives
\[ -\nabla^2 A+\frac{1}{\lambda_L^2}A=0. \]
Therefore
\[ \left(\nabla^2-\lambda_L^{-2}\right)A=0. \]
This is the static Proca equation inside the superconductor.
The magnetic field also obeys
\[ \left(\nabla^2-\lambda_L^{-2}\right)B=0. \]
Hence magnetic fields decay exponentially inside the superconductor.
Plane Boundary Example
Let the superconductor occupy the half-space
\[ x>0. \]
Let the magnetic field be parallel to the surface:
\[ B(x)=B_0e^{-x/\lambda_L}. \]
Choose
\[ A(x)=-\lambda_L B_0e^{-x/\lambda_L}. \]
Then
\[ B=\partial_x A. \]
The magnetic energy per unit area is
\[ \frac{U_B}{S} = \int_0^\infty \frac{1}{2}B_0^2e^{-2x/\lambda_L}\,dx = \frac{B_0^2\lambda_L}{4}. \]
The mass-term energy per unit area is
\[ \frac{U_m}{S} = \int_0^\infty \frac{1}{2\lambda_L^2} A^2\,dx = \int_0^\infty \frac{1}{2}B_0^2e^{-2x/\lambda_L}\,dx = \frac{B_0^2\lambda_L}{4}. \]
Thus
\[ \frac{U_B}{S} = \frac{U_m}{S}. \]
The total electromagnetic-plus-condensate energy stored in the penetration layer is
\[ \frac{U_{\rm layer}}{S} = \frac{B_0^2\lambda_L}{2}. \]
Half of it is magnetic-field energy.
Half of it is condensate kinetic energy, or equivalently the photon mass term.