Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Explanation of superfluidity in He-4

  1. Mar 12, 2012 #1
    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?
     
  2. jcsd
  3. Mar 13, 2012 #2

    DrDu

    User Avatar
    Science Advisor

    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.
     
  4. Mar 13, 2012 #3
    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?
     
  5. Mar 14, 2012 #4

    DrDu

    User Avatar
    Science Advisor

    Stop! thats 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.
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook