Where does the phase difference come from in a Josephson Junction?

When you separate two superconductors by a thin insulating film, a current $I(t)=I_0 \sin{\theta(t)}$ flows between the superconductors, where $\theta$ is the phase difference between the superconducting order parameters. Where does this phase difference come from? Does it come from initial conditions?

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@JohnRennie As I said in my previous comment I could not understand the Latex. If $\theta$ is the phase diff. then $I(t)=I_0\sinθ(t)$ seems dimensionaly incorrect. –  user31782 Apr 6 '14 at 8:56
@Anupam: why is it dimensionally incorrect? The dimensions seem fine to me. $\sin\theta$ is dimensionless and the dimensions of $I$ and $I_0$ obviously match. –  John Rennie Apr 6 '14 at 9:02
Is it $(\sin \theta)t$ or $\sin(\theta\times t)$. –  user31782 Apr 6 '14 at 9:04
@Anupam: $\sin\theta(t)$ is a perfectly standard way of specifying $\sin(\theta(t))$. We use this notation to save multiple brackets, just as we often write $\sin\theta$ instead of $\sin(\theta)$. If that is the reason for your downvote it is unjustified. –  John Rennie Apr 6 '14 at 9:09

The short answer is it comes from the current and/or voltage difference applied across the junction.

A longer answer should be of interest I believe, so here it goes.

The phase in the superconductor circuit stands for a redundant degree of freedom that you can never measure. This redundancy is important: it somehow means that a voltage is meaningless whereas a voltage-drop can be measured, and that a flux requires a closed loop to be measured as well. So it is just a way to discuss charges dynamics at the quantum level if you wish.

NB: I introduce the phase as a redundancy ; electrical engineers prefer to refer to flux node and voltage loop to discus the duality between current and voltage in circuit (along the duality between loop and node in 2D networks), see e.g. Quantum fluctuations in electrical circuits by M. Devoret (1997, free version easily found by Googling a bit) or the first years lectures by M. Devoret at Collège de France (in french). These nodes and loops are as unphysical as a redundant degree of freedom: at the end of the day, you measure voltage-drops and flux-around-loops, and that's it.

Now let's have a look at the current-voltage relation of a Josephson junction, picture taken from this website.

You see that there is a region of zero voltage and still a finite current in the diagram (plain curve). Of course this corresponds to the superconducting current violating the Ohm's law $V=RI$ (represented as the dashed line on the figure), since $R=0$ in superconductors. So what's happening there ? Well, the phase adapts itself such that the current flows. So we may prefer to write that the phase is given by the current-phase relation $$\varphi=\arcsin\dfrac{I}{I_{c}}$$ instead of the more common Josephson (first-)relation $I=I_{c}\sin\varphi$. These two rewritings are strictly equivalent as long as $I\leq I_{c}$.

Next the question: what's happening for currents larger that the critical one: $I\geq I_{c}$ ? Clearly, the phase $\varphi\in\left[0,2\pi\right]$ is no more a correct solution of the above relation. Nevertheless, this just means that we forgot to take into account the normal (non-superconducting) component of the current. We can nevertheless guess that a more complete current-phase-voltage relation should be $$I=I_{c}\Theta\left(I+I_{c}\right)\sin\varphi+\dfrac{V}{R}\Theta\left(I-I_{c}\right)$$ with $\Theta$ the step function. This is the current-phase-voltage relation you see on the picture: no voltage below the critical current since the phase redundancy allows for superconducting currents up to the critical current, and no supercurrent above the critical current when Ohm's law is verified.

Unfortunately this relation is never verified experimentally, since there are some non-linear effects for $I\geq I_{c}$ in real junctions. But the main picture remains: below a critical value (corresponding to the maximum number of Cooper pairs that can be formed across the tunnelling barrier) there is a super current flowing without resistance, and at higher current intensity you start opening normal conducting channels having some resistance. A even larger voltage you recover the linear Ohm's law.

To ensure that electromagnetism is not violated in quantum circuit, one needs also to verify $$\dfrac{\hbar}{2e}\dfrac{d\varphi}{dt}=V$$ which is nothing more than the Faraday's equation applied to the phase-difference of Cooper pairs, hence the charge $2e$ appearing (recall the engineers jargon of node-flux? here it has some interests). It is sometimes called the second-Josephson relation. So a voltage also can change the phase in a dynamic way. The previous discussion about current is valid also in the static / quasi-static situation, when the phase adapts itself instantaneously to the current change.

Of course, once again, everything is more complicated at the microscopic level, but I think I gave you the raw picture.

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Please see also the answer physics.stackexchange.com/a/107129/16689 overlaps mine for the dynamics contribution due to the second-Josephson-relation. I completed it with the discussion fo current and quasi-static effects. –  FraSchelle Apr 7 '14 at 9:39
Thanks. This was well explained. Why must there be a current at $V=0$? Why not $I=0$? –  ChickenGod Apr 7 '14 at 22:31
It's not that a current must be at $V=0$, it's just that, if you current-bias the junction, there will be no voltage drop, whereas the phase changes so that $I=I_{c}\sin\varphi$ is verified, $I$ being the bias current in that case. A flowing current without associated voltage drop is the hallmark of superconductivity electromagnetism (with the Meißner effect). If you choose to voltage-bias the junction, then you induce dynamics according to the second-Josephson relation $\dot{\varphi}=2eV/\hbar$. The associated dc-voltage is still zero then. –  FraSchelle Apr 8 '14 at 8:22
I am beginning to understand now. So, if we tie both superconductors to a common voltage $V$, there will be no current? And from $\dot{\phi}=2eV/\hbar$, $\phi =2eVt/\hbar+C$. What determines the integration constant? –  ChickenGod Apr 9 '14 at 9:38
The integration constant is determined by the value of the phase at $t=0$, as usual. You can choose $C=0$ for superconductors in equilibrium at $t=0$. I'm not sure you get it correctly: applying a voltage generates currents, but applying current does not necessarily generates a voltage: look at the current-voltage characteristic one more time :-) So finite voltage means current (the second Josephson relation tells you that this current is oscillating / ac-current), but finite dc-current $I\leq I_{c}$ means no voltage. Michael Tinkham's book on superconductivity should give you further details. –  FraSchelle Apr 9 '14 at 12:22

When an isolated superconductor goes through the superconducting transition, $T_c$, a symmetry is broken. The electrons above $T_c$ have an assortment of different phases, whereas below they "choose" one macroscopic phase. This "choice" is analogous to that of a magnetic material going through its Curie temperature or the 'ball on a sombrero' example. This choice undoubtedly depends upon the properties of the individual elements before going through the transition, but not in an obvious way and for all intents and purposes the choice is arbitrary.

The two superconducting elements comprising a Josephson junction can initially be thought of as two isolated superconductors each randomly selecting a phase. It is therefore highly unlikely that they would both select the same. Once in the superconducting state the two systems become coupled and will influence one another (The free energy of a standard junction is minimized when the phase difference between the two superconducting elements is zero). Again this can be thought of as analogous to two magnetic elements being brought together whereby they will tend to align and fall into an energy minimum.

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Thanks. What I meant to ask was: what determines the phase that the superconductor starts with? Is it completely random? If I take a superconductor, break it into two, will the Josephson current between them be zero at $V=0$ because they have the same phase? –  ChickenGod Apr 7 '14 at 22:33

Perhaps it is worth asking, "Why would the order parameters have the same phase?" But the phase difference $\phi$ across a Josephson junction can be forced to change by the application of a voltage across the junction,

$$U(t) = \frac{\hbar}{2 e} \frac{\partial \phi}{\partial t}$$

where $U(t)$ is the applied bias. The picture I have in my head is - the condensate on one side of the junction now has a higher energy, and so its phase evolves more rapidly, just as we find for higher energy solutions to the Schrödinger equation, in, say, a square well. Any other way of forcing the sides of the junction to have an energy difference will have the same effect.

As always, the wiki page is informative, and a useful device to test your understanding of this physics would be, of course, the SQUID.

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I would have assumed that if we prepared two superconductors in the exact same way at the exact same time (of course, experimentally impossible), they would have the same phase. Or is that wrong? Is there something else that determines the initial phase? –  ChickenGod Apr 7 '14 at 22:35