# Should a superconductor act as a perfect mirror?

I have been told that metals are good reflectors because they are good conductors. Since an electric field in conductors cause the electrons to move until they cancel out the field, there really can't be electric fields in a conductor.

Since light is an EM wave, the wave cannot enter the conductor, and it's energy is conserved by being reflected (I'm guessing in a similar fashion to a mechanical wave being reflected when it reaches a medium it cannot travel through, like a wave on a rope tied to a wall for example).

I imagine then that more conductive materials are better reflectors of light. Wouldn't a perfect conductor then, such as a superconductor, be a perfect reflector of light? (Or reach some sort of reflective limit?)

• Good reasoning, good question. I also would like to know the answer! There are metals like Al and Zn that when reduced to near zero Kelvin temperatures become superconducting. They start as good reflectors (over some range of wavelengths), so when they are superconducting is this improved? What about the so-called high temp superconducting ceramics like YBCO? I do know at visible light frequencies they are not very good reflectors at superconducting temps. – docscience Oct 24 '15 at 0:55
• For this to be true, the material would have to be superconducting at its surface, rather than just at some depth within the body of the material. Perhaps not all superconductors meet this criteria. – Daniel Griscom Oct 24 '15 at 11:48
• I believe this is somewhat linked to skin effects in superconductor. What I'm saying is that the primary characteristic of a superconductor is to avoid letting a magnetic field inside the conductor; this is true in depth of material, while there is a typical distance in which you can find magnetic field in a conductor with temperature inferior than its critical temperature for superconducting. – Klopmint Oct 25 '15 at 12:12
• look at the answers of this question – user46925 Oct 25 '15 at 15:18
• It's worth mentioning that the gaps of superconductors are in the THz, so they aren't very good mirrors for visible/UV/IR light, but are very good for RF radiation. That's why they are used as RF cavities – KF Gauss Sep 12 '19 at 3:11

## 2 Answers

Yes and no. Below the superconducting gap a superconductor is a near perfect reflector and superconductivity has its say in it.

Reflectivity at normal incidence is given by the equation.

$$R = \left| \frac{1-\sqrt{\varepsilon}}{1+\sqrt{\varepsilon}} \right|^2$$

where $\varepsilon$ is the complex-valued frequency-dependent dielectric function of the reflective material. Let's look at the dielectric function of a superconductor above and below the superconducting transition temperature:

This is a plot of the real part of the optical conductivity (in arbitrary units) in the normal state (blue) and the superconducting state (orange). The relationship between the real part of the optical conductivity and the imaginary part of the dielectric function is given by $\varepsilon_0 \mathrm{Im}(\varepsilon) \omega = \mathrm{Re}(\sigma)$

The area under the curve must be conserved, therefore the missing part of the area is hidden in a delta-function at zero frequency (we must take it into account to perform a Kramers-Kronig transformation properly). This is important, because the delta function in the conductivity (that's the manifestation of dissipationless dc current!) leads to a $-a/\omega^2$ term in the real part of the dielectric function. Large-by-magnitude values of the dielectric function give a good coefficient of reflection.

The other part of the dielectric function is $\mathrm{Re}(\varepsilon)$ and is obtained by doing a Kramers-Kronig transformation:

Now this can be plugged into the expression for reflectivity:

As you can see, and this is due to the vastly negative real part of the dielectric function, reflectivity below the gap is near 100%.

EDIT - actually, since the real part of $\varepsilon$ is negative, and within the superconducting gap the imaginary part is exactly zero (with caveats, such as s-wave vs. d-wave superconductors) $R$ would be exactly 100%.

Above the gap the reflectivity is actually slightly worse. Now because the superconducting gap lies at energies far lower than those of visible light, the reflectivity for visible lightly is barely affected. As superconductors are often bad conductors in their normal state, their visible light reflectivity leaves much to be desired. Stick to silver.

• Nice answer! However the permeability for superconductor which asuumed to be unitary may not be reasonable for quite a lot cases. Let's apply more constrain,how about when incident light is close to the gap energy; and close to the magnetic(fero,antifero,para..)/superconductor phase coundary? – Wilson Ko Dec 13 '17 at 8:01
• @WilsonKo The above is an ideal situation for s-wave superconductors. For example, d-wave superconductors are not absorption-free even at zero temperature. Being close to the gap energy is just another way of being far from the ideal scenario of zero absorption. Your remark about assuming $\mu=1$ is interesting, but 20 minutes of googling didn't help me find anything really conclusive. Intuitively, I suspect that the complex frequency-dependent $\mu^{-1}$ behaves much like $\varepsilon$. – LLlAMnYP Dec 13 '17 at 10:59
• For superconductor , it repel static magnetic field completely, thus the relative permeability is zero, I guess it's small for small frequency EM waves, but can not tell if frequency is comparable to the gap's energy. – Wilson Ko Dec 19 '17 at 2:28
• Does this hold true for high temperature superconductors? Are not their gaps much higher in energy and capable of reflecting visible light? – Mathews24 Aug 20 '18 at 14:21

The above explanation by LLlAMnYP holds fully in the limit case $$T=0$$, where the $$R$$ would indeed be exactly $$1$$ for $$\hbar \omega<2\Delta(0)$$. For $$T>0$$, unpaired thermally excited quasiparticles, whose density goes to zero as $$T$$ approaches $$0$$, absorb and dissipate energy even at frequencies $$\hbar \omega<2\Delta(T)$$. A more complete and rigorous treatment requires Mattis-Bardeen theory. Ref. Tinkham 3.9.3 (Fig. 3.10 in particular, case II ).