# Superconductor question

1. Sep 25, 2005

### sniffer

in deriving quantization of flux in superconductor ring, the momentum of cooper pair p:
$$p=\hbar\nabla\theta=e^*(\Lambda J_s + A)$$
then integrate around the ring,
$$\hbar\oint\nabla\theta dl=e^*\oint(\Lambda J_s + A)dl$$
using stoke's theorem and integrate sufficiently deep in the ring where current density is very small, the RHS becomes
$$RHS=e^*\Phi_s$$
and the left hand side,becomes hn where n is integer.
So the quantized flux is
$$\Phi_s=nh/e^*$$
e star is the effective cooper pair charge which is -2e.

i got confused here,why the left hand side integral becomes hn?

The argument used by the book (Van Duzer, superconductivity page 116) is that because theta is unique or differ by a multiple of 2 Pi at each point, so the integral
$$\oint\nabla\theta dl=2\pi n$$
why?? where does n come from? theta is a scalar function of r.

thanks.

Last edited: Sep 25, 2005
2. Sep 25, 2005

### Gokul43201

Staff Emeritus
$\theta$ is the phase of the order parameter $\psi = |\psi|e^{i \theta}$. Since $\psi$ must be single valued at some specific angular position $\phi + 2k \pi$ on the ring, we need, $\psi(\phi ) = \psi(\phi + 2m \pi )$, or we need $e^{i \theta ( \phi )} = e^{i \theta ( \phi + 2m \pi)}$, for all m. For the phase factor to remain unchanged over integral number of traversals of the loop, the phase $\theta$ must itself change by only an integer multiple of $2 \pi$ (since $e^{2in \pi} = 1$).

So
$$\theta ( \phi + 2m \pi ) = \theta ( \phi) + 2n \pi$$.

In other words
$$\hbar\oint\nabla\theta dl = \hbar \Delta\theta = \hbar (\theta ( \phi + 2 \pi ) - \theta ( \phi)) = 2 \pi n \hbar = nh$$

Last edited: Sep 25, 2005
3. Sep 26, 2005

### Creator

In your opinion, from an experimental point of view, what would be a simple method to effect a change in the phase of the wavefunction of the supercurrent while remaining below T(c)?

Creator

Last edited: Sep 26, 2005
4. Sep 26, 2005

### sniffer

change the radius of the ring.

5. Sep 27, 2005

### ZapperZ

Staff Emeritus
Have a magnetic flux through the loop. That, after all, is how a SQUID works.

Zz.

6. Sep 28, 2005

### Creator

True, in theory; but by what method are you going to change the length of a brittle superconducting wire while it is in the superconducting state?

7. Sep 28, 2005

### Creator

Of course; externally applying B thru the loop. I should be more specific. I guess I am referring to a non-electromagnetic method of altering the quantum phase.

Creator

Last edited: Sep 28, 2005