# Rotating Bose gases, quantum vortices

1. Jul 8, 2008

### confused_man

Hello everyone,

4. Jul 8, 2008

### confused_man

Thanks Cthugha for pointing that out. I can't believe that was my mistake... In my defense, it was pretty late and I was tired! Since I can't edit my post anymore, here's the middle part:

This can be shown by using the continuity equation for the probability density $$n$$,

$$\frac{\partial n}{\partial t} + \nabla \cdot (n \mathbf{v}) = 0$$

where $$n = |\psi|^2$$ and

$$\mathbf{v} = \frac{-i\hbar}{2m}\frac{\psi^*\nabla\psi-\psi\nabla\psi^*}{|\psi|^2}.$$

If generic wavefunction written in polar form $$\psi=\psi_0 e^{i\phi}$$ is inserted into the velocity equation, then it is found that the velocity is the gradient of a scalar function,

$$\mathbf{v} = \frac{\hbar}{m}\nabla\phi$$

which means that the wavefunction is irrotational since the curl of the gradient of a scalar is zero ($$\nabla \times \mathbf{v} = 0$$). However, the wavefunction can have angular momentum if it possesses a singularity in the form of a vortex.

Last edited: Jul 8, 2008
5. Jul 10, 2008

### Creator

Don't exactly see what you're asking.

That's somewhat like asking "Does a superconductor need to be in superconducting state in order to have vortex flux lines?". How would you recognize the flux 'lines' if the entire material was penetrated by magnetic flux? ...(which is what happens if it is not a condensed superfluid).

Likewise in a Bose condensate there is no need to talk of quantum vortices of angular momentum if the entire fluid is in a normal 'rotational' state (above T(c))....Like the SC case, the vortices are the 'defects' that allow the angular momentum to 'penetrate' through an otherwise irrotational (Bose) state.
At least that's the way I see it.

6. Jul 10, 2008

### confused_man

Thanks for your reply. I guess what I'm asking is when does a Bose gas cease to be irrotational. For example, at around 2 K, 4He is a superfluid but only around 7 percent of the atoms are Bose condensed.

Suppose that you had a Bose gas that was slightly above the transition temperature Tc, then it might occupy a few different energy levels instead of the one. Wouldn't this gas also be irrotational?

7. Jul 10, 2008

### Creator

Hello, CM.

If it were a Bose condensate then it is already IN the (lowest) ground state throughout, (at least throughout that portion that is superfluid) ....which requires it to be < T(c). ..that's my understanding....at least for helium.

London identified the 'lambda' transition (2.17 * K. for helium) with Bose condensaton.....where there is long range ordering as a result of the single valued wavefunction....which is required for superfluidity.

It is the single valued wavefunction that results in superfluidity that accounts for Helium Irrotation since , as you have shown, the superfluid velocity is proportional to the gradient of the phase of the ground state wavefunction, the curl of which = 0.

By taking the line integral (of your last equation) around any (simply connected) closed path in the superfluid, you will arrive at the conclusion that velocity can only develop when there is a change in the phase of the wavefunction around the path.

But by definition, in a Bose condensed state the phase change is zero since it overlaps uniquely (in a simply connected system) every 2(pi) ....so it should only rotate with angular velocity (macroscopically) when the circulation of the wavefunction is 'out of phase'...And Above the 'lambda' point it is all 'out of phase'.
This means that the superfluid moment of inertia (below T(c)) must be different from that of 'rigid' rotation of a normal fluid (above T(lambda)).

However, this irrotation can usually only be observed at low velocities, particularly, below a certain 'critical velocity'....(due to the development of the vortices which 'dissapate' the angular momentum at higher velocities).

As a practical matter the irrotational angular velocity value, omega(0), given by Hess, below which a container of helium of radius R can show irrotational response is given by the radius dependent formula:

omega(0) = h/m(R^2); where h is actually h(bar).

Below this critical angular velocity the portion of the helium that is superfluid cannot rotate ...according to your last equation....Above that velocity the angular momentum is quantized ...all of which has been shown experimentally; (see Hess & Fairbank, 1967 ). ....and quantum vortices develop, starting at around 15 times the critical angular velocity, I believe. - (See Feynman). This quantization is analogous to quantization of magnetic flux threading a (simply connected) superconductor....and the vortices are the analog of flux lines pinned in a superconductor.

Basically, below a certain value, angular momentum is 'repelled' by the superfluid He 4 which is analogous to the repulsion of magnetic field by a superconductor below t(c); (i.e., Meissner effect). I see it as an 'inertial' Meissner effect. ;)

Creator

Last edited: Jul 10, 2008
8. Jul 11, 2008

### confused_man

Hi Creator,

That's a nice detailed post. Thanks! If I correctly understand what you are saying, the superfluid properties of He4 is due to it Bose condensing at temperatures below the lambda point (T = 2.17 K). It it then clear how the quantized vortices come about.

In reality however, the superfluid transition temperature and the Bose condensation temperature are not the same thing. Like you said, in the 1930's London proposed that the lambda point corresponds to the Bose condensation temperature, and this is what is taught in most textbooks. However, the reality is that the strong interactions between particles in He4 prevent most of the liquid from condensing. Below 1 Kelvin, only about 7% of the He4 liquid is a Bose-Einstein condensate, but 100% of the liquid is a superfluid. Vice-versa, an ideal BEC (where the particles don't interact), is not a superfluid (although it is still irrotational and can support quantized vortices). The reason for this is that superfluidity requires an energy gap against the creation of excitations, which is not present without the presence of interactions. So an ideal BEC would not support frictionless flow.

What I'm having problems with is this: if He4 is not a BEC, then what kind of state is it in? How can there be phase coherence so that a wavefunction $$\psi = \psi_0 e^{i\phi}$$ be used to describe the whole liquid? If I can use $$\psi = \psi_0 e^{i\phi}$$, then I can understand irrotationality, vortices, etc.

This might be dumb, but what exactly are the requirements for phase coherence (which I understand as meaning a single wavefunction can be used to describe the whole material)?

Thanks for your help!

9. Jul 11, 2008

### confused_man

Here's a better question: how can 100% of He4 atoms be in a superfluid state when T~0 K, while < 10% are Bose condensed? What makes the other 90% superfluid?

10. Jul 16, 2008

### Creator

Sorry for the late response, CM.

That's a tough question. I don't think anyone has a complete answer yet. And the answer somewhat hinges on your definition of a Bose condensed state in helium.

It is compunded by the fact that, the atomic interaction in He 4 makes the exact criterion for definition of a Bose condensate somewhat ambiguous. It is far easier (not to produce, but to describe) a dilute Bose gas...and 100% Bose can be cleanly distinguished. With the atomic interaction of He 4 things get very complicated, as you have pointed out.

Your 7 % Bose at 1 *K in He 4 is clearly only a QM theoretically determined value based on a phenomenological model and not experimental.(BTW, You should include the reference).

That is the first point.
Secondly, superfluidity (in He 4) is usually an experimentally measured attribute, and you should realize that its determination is dependent upon the methodology of viscosity measurement.

In the capillary method, He 4 (below lambda point) flows frictionlessly through tiny capillaries, the velocity of which is dependent upon the tube thickness....as you are probably aware.

However, In the rotaing disk method ... Keesom & MacWood, (or the oscillating torsion cylinder method- see Hollis), where the disk is rotated in the helium (below lambda pt.), we do measure viscous drag. BTW, helium viscosity measurements between 4* and 5 * K by Bowers, et al ...revealed that the coefficient decreases over the whole range (this is above the lambda point.)

This conundrum developed into the "two fluid' model over the years, with various explainations of how a superfluid fraction and a normal fraction can co-exist; (thus I used the commonly accepted phase in my previous posts 'the fraction of helium that is superfluid' , etc.) Remember this is a phenomenological model that appears to correctly describe the results. But since it is impossible to physically separate the two 'fluids' we need to recognize it is based on assumptions systemic to our model and there may be other reasonable explaination....and we certainly are far from a complete understanding....all we can say is that it is 'closely connected' to the "Bose condensed state'.

Making experimental superfluid (viscosity)measurements of a dilute Bose gase is very complicated due to the containment issues etc, and I haven't seen a comparable method to that of He 4 for which to compare. However, there have been some recent empirical results showing that the newer non-interacting Bose gases do indeed show superfluid behavior.

see for ex.; http://www.aip.org/pnu/1999/split/pnu449-1.htm

And Lieb et al rigoriously 'prove' that dilute Bose gases can support superfluidity....here...http://prola.aps.org/abstract/PRB/v66/i13/e134529

So I think your conjecture that Bose gases cannot be superfluid will be shown to be erroneous .

Creator