Explanation of superfluidity in He-4

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In summary, the authors claim that there is mass transfer in a superfluid due to the quasi-momentum of the particles. However, they do not provide a clear explanation of how this works. Their equation for the total momentum operator does not match the expectation for a harmonic oscillator.
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sam_bell
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Hi. I was just reading the explanation of superfluidity in He-4 (from the beginning of QFT Methods in Statistical Physics by Abrikosov et. al.). There is something I don't understand. At finite temperatures there is a "gas of excitations", which they take to be moving at an average velocity v relative to the stationary liquid. They then derive that the quasi-momentum of this gas (per unit volume) is P = (const.) v. They claim this constant represents a mass and therefore there is mass transfer and that this part of the liquid is "normal". The rest of the mass is taken to be in the ground state superfluid. OK, my question: If we are talking about *quasi-*momentum, how can we be sure that there is really mass transfer? After all, a single quasi-particle has quasi-momentum, but this doesn't correspond to mass transfer as a drift of He-4 atoms.

I suppose this is related to diffraction experiments, where is deflection of photon has a conversation law written in terms of quasi-momentum.

Helpful thoughts?
 
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I would say that the quasi-momentum and the true momentum coincide in this case.
A liquid does not break translation invariance whence momentum is well defined.
 
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There is still a consistency here that I can't follow. For simplicity, I imagine the case of a linear chain of oscillators. In this case the total momentum operator is P = sum(i = 1..N, P(i)) where P(i) is the momentum operator of the ith body in the chain. Expading P(i) in terms of normal modes gives = sum(n = 1..N, sum(all k, f(k) (a(k) exp(inb) - a(k)* exp(-inb)) where f(k) ~ 1/sqrt(energy) and b is the periodicity of the lattice. This doesn't look like the crystal momentum operator P = sum(all k, k a*(k) a(k)). Nevertheless, since b --> 0, if we excite a phonon of crystal momentum hk, then the external environment loses "real" momentum hk. Alternately, this means the linear chain gains a "real" momentum hk. But calculating <k|P|k> = 0 because none of the P(i) conserves phonon number. In going from |0> to |k> the real momentum of the chain didn't change?
 
  • #4
sam_bell said:
There is still a consistency here that I can't follow. For simplicity, I imagine the case of a linear chain of oscillators. In this case the total momentum operator is P = sum(i = 1..N, P(i)) where P(i) is the momentum operator of the ith body in the chain. Expading P(i) in terms of normal modes gives = sum(n = 1..N, sum(all k, f(k) (a(k) exp(inb) - a(k)* exp(-inb))
Stop! that's not correct. There should be a k in the exponents.
The sum over n then gives a delta function in k and only the k=0 components are left.
Either in a or in a^* the k should read -k.
. But calculating <k|P|k> = 0 because none of the P(i) conserves phonon number. In going from |0> to |k> the real momentum of the chain didn't change?
Of course in true harmonic oscillator eigenstates the expectation of momentum always vanishes.
However in the limit k=0 (and omega=0!), coherent states with unsharp number of quanta become alternative true eigenstates and have non-vanishing momentum. You may replace a(0) by its expectation value on these states. (You are not forced to do so. A state with fixed number of quanta would correspond to a macroscopic superposition of states with opposite momenta.)
Due to the f(k) factor it diverges (an infinite long moving chain will have an infinite momentum) and you should divide by sqrt(L) to calculate the finite momentum per length. Note the strong analogy to your previous thread. Here, we have an example how a finite momentum density breaks symmetry (Galilean symmetry).
I am not totally sure how the crystal momentum enters. I think we need to take coupling to the lattice into account to describe state with like momentum but unlike velocity.
 

FAQ: Explanation of superfluidity in He-4

1. What is superfluidity?

Superfluidity is a state of matter where a substance, such as helium-4, has zero viscosity and can flow without any resistance. This phenomenon occurs at extremely low temperatures, close to absolute zero, and is characterized by unique properties such as a lack of friction and the ability to climb up walls.

2. How is superfluidity explained in helium-4?

The explanation of superfluidity in helium-4 is based on the concept of Bose-Einstein condensation. At very low temperatures, the individual helium atoms undergo a phase transition and form a single quantum state, resulting in a collective behavior with unique properties. This is known as a Bose-Einstein condensate and is responsible for the superfluidity observed in helium-4.

3. What are some key properties of superfluid helium-4?

Superfluid helium-4 exhibits several interesting properties, including zero viscosity, the ability to flow through extremely narrow spaces without any resistance, and the ability to climb up walls. It also has a very high thermal conductivity and a unique response to external forces, such as rotation and magnetic fields.

4. How is superfluid helium-4 different from regular liquid helium?

Regular liquid helium is a normal fluid that follows classical laws of physics, whereas superfluid helium-4 exhibits behavior that can only be explained by quantum mechanics. Additionally, regular liquid helium has a finite viscosity and cannot flow without resistance, while superfluid helium-4 has zero viscosity and can flow without any hindrance.

5. What are some practical applications of superfluid helium-4?

Superfluid helium-4 has several important applications, including its use in cryogenic systems and cooling devices. It is also used in research and development of superconductors, as well as in the study of quantum mechanics and low-temperature physics. It has also been proposed for use in precision instruments, such as gyroscopes and accelerometers.

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