Superfluid Helium-4
Noah Miller
Columbia University
March 6, 2018
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 1 / 143 Why is helium a liquid at low temperatures? Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 2 / 143 Why is helium a liquid at low temperatures?
Helium is small inert
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 3 / 143 Why is helium a liquid at low temperatures?
Helium atoms are neutral, but...
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 4 / 143 Why is helium a liquid at low temperatures?
they still have weak inter-atomic interactions when you account for the forces between all individual particles
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 5 / 143 Why is helium a liquid at low temperatures?
This creates the London dispersion force. Using (second order) perturbation theory, with the extra interactions taken as the perturbing Hamiltonian, we get
3 IAIB αAαB EAB = − 6 2 IA + IB R where
EAB is the energy between atoms A and B
IA and IB are the first ionization energies
αA and αB are the dipole polarizabilities R is the inter-atomic distance
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 6 / 143 Why is helium a liquid at low temperatures?
Helium is very small.
Therefore its polarizability is very small.
Therefore its inter-atomic attraction is very small.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 7 / 143 Why is helium a liquid at low temperatures?
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 8 / 143 Why is helium a liquid at low temperatures?
But how can it remain a liquid at absolute zero, when all motion stops? Shouldn’t that small attraction still be enough to make it a solid?
http://www.columbia.edu/∼vjd1/solid-liquid-gas.htm
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 9 / 143 Why is helium a liquid at low temperatures?
Answer: the atoms have some momentum from the uncertainty principle!
Because helium is so light, this momentum is large enough to keep it a liquid.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 10 / 143 Why is helium a liquid at low temperatures?
Lindemann criterion: crystals liquefy when the amplitude of vibration δx is about 5% − 15% the inter-atomic spacing ∆x
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 11 / 143 Why is helium a liquid at low temperatures?
We can approximate the inter-atomic energy using the Lennard-Jones potential. ! σ 12 σ 6 E = ε − 2 r r
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 12 / 143 Why is helium a liquid at low temperatures?
At equilibrium, the potential is approximately harmonic 1 V ≈ k(δx)2 2 with ε k = 36 σ2
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 13 / 143 Why is helium a liquid at low temperatures?
From the uncertainty principle
∆x∆p ≈ ~ we can approximate the kinetic energy:
p2 2 KE = ≈ ~ 2m 2m(∆x)2
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 14 / 143 Why is helium a liquid at low temperatures?
We can find δx via the conservation of energy. ε V = KE =⇒ 18 (δx)2 = KE σ2 Assuming σ ≈ ∆x and ignoring constants, we get r δx KE ≈ ∆x ε
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 15 / 143 Why is helium a liquid at low temperatures?
For helium, ε = 3.9 × 10−3eV σ = 2.2A˚ ∆x = 3.7A˚≈ σ m = 6.6 × 10−27kg =⇒ KE = 3.8 × 10−5eV
r s δx KE 3.8 × 10−5eV ≈ = = 10% ∆x ε 3.9 × 10−3eV So it is plausible that helium is a liquid at 0K!
values from http://www.cengage.com/resource uploads/downloads/1111988617 420953.pdf
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 16 / 143 Transition to superfluid Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 17 / 143 Transition to superfluid
Helium becomes a liquid at 4K.
But at 2.17K it becomes a superfluid!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 18 / 143 Transition to superfluid
Leggett, p.71
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 19 / 143 Transition to superfluid
https://en.wikipedia.org/wiki/Lambda point
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 20 / 143 Transition to superfluid
Superfluid helium flows without viscosity under a critical velocity of about 20 cm/s.
It just keeps going!
This produces some strange phenomena.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 21 / 143 Transition to superfluid
It can leak through cracks far too small for other liquids to fit through
https://www.youtube.com/watch?v=2Z6UJbwxBZI
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 22 / 143 Transition to superfluid
It can climb up walls via microscopic films
https://en.wikipedia.org/wiki/Superfluidity
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 23 / 143 Transition to superfluid
(The reason for this doesn’t seem understood. The common explanation is that, like all liquids, there is a microscopic layer of helium on the walls of the container called the Rollin film, which is attracted to the atoms on the wall. Because helium can flow with no resistance, it can flow right up and over the wall. This is probably incorrect.)
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 24 / 143 Transition to superfluid
And many other things too.
We will give a qualitative first-principles explanation of some properties of superfluid helium.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 25 / 143 Multi-particle wave-function and phonons Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 26 / 143 Multi-particle wave-function and phonons
A helium-4 atom contains two protons two neutrons two electrons It is a spin 0 boson.
(Helium-3 is a fermion but we will not discuss it.)
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 27 / 143 Multi-particle wave-function and phonons
The quantum state of a collection of atoms is a multi-particle wave-function.
Ψ(~x1, ~x2, ~x3,..., ~xN )
Each x~i is the position of atom i.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 28 / 143 Multi-particle wave-function and phonons
As the particles are bosons, it must be completely symmetric under any permutation of atoms.
Ψ(~x3, ~x69, ~x420,..., ~x1) = Ψ(~x1, ~x2, ~x3,..., ~xN )
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 29 / 143 Multi-particle wave-function and phonons
So a wave-function Ψ assigns a complex number to each possible configuration.
The bigger |Ψ|2 is for a configuration, the more likely it is.
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Let us call the ground state wave-function Ψ0.
What is Ψ0 like qualitatively?
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N X 2 1 X Hˆ = − ~ ∇2 + V (|x~ − x~ |) 2m i 2 i j i=1 i6=j
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 32 / 143 Multi-particle wave-function and phonons
Basic facts about definite energy states: |Ψ|2 is large when potential energy of configuration is small Ψ has higher energy when the gradient is large
Basic facts about ground states: Completely real (up to global phase) Never zero, no “nodes”
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 33 / 143 Multi-particle wave-function and phonons
So we take Ψ0 to be real and positive everywhere.
It will be very small when atoms overlap, and gradually grow as they get further apart.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 34 / 143 Multi-particle wave-function and phonons
Value of Ψ0 per configuration:
Ψ0 large Ψ0 small Ψ0 minuscule
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Ψ0 is large when the atoms are evenly spaced out, and small when there are density variations and clumps.
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That is the case for the lowest energy state Ψ0.
What about the states with slightly higher energy?
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 37 / 143 Multi-particle wave-function and phonons
The fluid will have a small energy and the multi-particle wave-function will have a few “nodes.”
(Energy eigenstates can still be taken to be real.)
Call our state ΨE .
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 38 / 143 Multi-particle wave-function and phonons
Because we want ΨE to have a small energy, we want to minimize its gradient as much as possible.
Therefore, it should change slowly as we vary a configuration.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 39 / 143 Multi-particle wave-function and phonons
ΨE will attain its maximum positive value for some configuration of atoms (configuration A), and its maximum negative value for a different configuration (configuration B).
As the atoms are slowly varied from configuration A to configuration B, ΨE must pass through 0.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 40 / 143 Multi-particle wave-function and phonons
Imagine that configuration A and configuration B were both comprised of evenly spaced out atoms, as would seem natural for a low-energy state. (We will soon show that this is impossible.)
Config A,ΨE very positive Config B,ΨE very negative
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 41 / 143 Multi-particle wave-function and phonons
Vitally, all atoms are identical! As the atoms in both configurations are evenly spaced out, we can change one configuration to the other by varying the positions of the atoms only very slightly (with average distance on the order of the inter-atomic spacing).
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 42 / 143 Multi-particle wave-function and phonons
This would mean that a tiny change in the configuration resulted in a big change in ΨE , as ΨE shoots from its maximum value to its minimum value. The gradient of ΨE would be very large.
Therefore, configuration A and B cannot be evenly spaced out if ΨE has a low energy.
They must instead contain density variations to keep the gradient (and therefore energy) low.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 43 / 143 Multi-particle wave-function and phonons
Example configurations:
Config A,ΨE very positive Config B,ΨE very negative
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 44 / 143 Multi-particle wave-function and phonons
Configurations A and B represent the most likely configurations of ΨE .
We have therefore proven that the lowest energy states must be states with density variations.
These are just sound waves, A.K.A. phonons!
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Conversely, it must take a lot of energy to “stir” the atoms, i.e. move them around without varying the density.
This means that at low temperatures, when there isn’t much energy available, the atoms won’t get “stirred about,” qualitatively explaining why superfluid helium has no viscosity.
Much later we will see exactly how the fluid can get “stirred.”
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 46 / 143 Multi-particle wave-function and phonons
Strangely, the boson statistics have made our fluid act quite rigid, like a crystal.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 47 / 143 Dispersion curve Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 48 / 143 Dispersion curve
Let us now investigate those higher energy states, the ones with no density variation, more carefully.
Recall that ΨE should reach its maximum value on configuration A and its minimum value on configuration B.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 49 / 143 Dispersion curve
The A sites should be in between the B sites, and vice versa.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 50 / 143 Dispersion curve
The wave-function should vary smoothly as more and more atoms are moved from A sites to the B.
This can be accomplished with a function f (~x) that is +1 on A sites and −1 on B sites, and varies smoothly in between (and doesn’t have to be real).
In order to retain the indistinguishability of the bosons, we should sum f (~x) over all coordinates ~xi . X F = f (~xi ) i
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 51 / 143 Dispersion curve
However, we still want our wave-function to be small if atoms overlap. We can accomplish this by multiplying F by the ground state Ψ0.
ΨE = F Ψ0 This is an educated guess at the form our wave-function.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 52 / 143 Dispersion curve
Our job now is to solve for f (~x).
Because our state is to have a low energy, we can solve for f (~x) using the variational method.
The energy functional to minimize is
R ∗ ˆ N E ΨE HΨE d V E = = R ∗ N I ΨE ΨE d V
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 53 / 143 Dispersion curve
The Hamiltonian is
N X 2 Hˆ = − ~ ∇2 + V − E 2m i 0 i=1 shifted so that
HˆΨ0 = 0
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 54 / 143 Dispersion curve
After a bit of work, we get
2 X Z E = ~ (∇ F )∗ · (∇ F )|Ψ |2dN V 2m i i 0 i and Z X ∗ 2 N I = F F |Ψ0| d V i
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 55 / 143 Dispersion curve
After substituting in the form of F , we get
2 X Z E = ~ (∇ f (~x )∗ · ∇ f (~x ))|Ψ |2dN V 2m i i i i 0 i and carrying out the N − 1 easy integrals, we get
2 Z E = ρ ~ (∇f )∗ · (∇f )dV 2m where ρ is the number density of atoms.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 56 / 143 Dispersion curve
I on the other hand is Z X X ∗ 2 N I = f (~xi ) f (~xj )|Ψ0| d V i j which, from the indistiguishability of the particles, can be integrated over the N − 2 easy integrals and rewritten as Z ∗ I = f (~x1) f (~x2)ρ(~x1, ~x2)dV1dV2 where ρ(~x1, ~x2) is the probability density in the ground state that there is an atom at ~x2 given that there is an atom at ~x1. From the isotropy of space,
ρ(~x1, ~x2) = ρp(|~x1 − ~x2|) where ρ is the number density of atoms and p is the “correlation function.”
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 57 / 143 Dispersion curve
To recap:
2 Z E = ρ ~ (∇f )∗ · (∇f )dV 2m Z ∗ I = ρ f (~x1) f (~x2)p(|~x1 − ~x2|)dV1dV2
We want an f that minimizes
E E = I
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Varying f ∗ by δf ∗, E is minimized when f satisfies
Z 2 E p(|~x − ~x |)f (~x )dV = − ~ ∇2f (~x ) 1 2 2 2 2m 1 This is satisfied by
f (~x) = exp(i~k · ~x)
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 59 / 143 Dispersion curve
Likewise, the energy is
2k2 E = ~ 2mS(k) where S(k) is the Fourier transform of the correlation function p. Z S(k) = p(~x) exp(i~k · ~x)dV
S(k) is sometimes called the “form factor.”
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 60 / 143 Dispersion curve
Recall the condition that f needs to be +1 on A sites and −1 on B sites. This implies that k should be on the order of 2π k ≈ ∆x where ∆x is the average inter-atomic distance.
If k is far from that value, then our analysis is unjustified.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 61 / 143 Dispersion curve
However, I will now argue that our expression for the wave-function is also 2π justified if k is much smaller than ∆x , that is, for larger wavelengths!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 62 / 143 Dispersion curve
If the inter-atomic interaction were 0,
V = 0 then our system would be an ideal gas. Rewriting our wave-function as a function of the Fourier modes of the density, qk , our Hamiltonian would just be a set of many independent harmonic oscillators, one for each mode.
The excitations of our harmonic oscillators would be phonons.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 63 / 143 Dispersion curve
Recall that the first excited state of a 1D harmonic oscillator is just proportional to x times the ground state.
Therefore, the state with one excited phonon of momentum k is
ideal ∝ qk Ψ0 ideal where Ψ0 is the ground state of our ideal boson gas.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 64 / 143 Dispersion curve
In the non-ideal interacting case, that wave-function does not contain enough small-distance detail. We still don’t want atoms to overlap.
We can solve this by replacing the ideal ground state with the true ground state, which should be okay as long as k is small (wavelength long) and therefore does not affect the small-distance detail of our state.
Therefore, the one-phonon state of our interacting gas ought to be
X ~ ∝ qk Ψ0 = exp(ik · ~xi ) Ψ0 i
(where we substituted in the expression for qk ).
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 65 / 143 Dispersion curve
Wow! Our phonon wave-function has the exact same form as our previous wave-function! They are both
X ~ exp(ik · ~xi ) Ψ0. i The only difference is that for the phonon, k is small, and for the highly excited state, k is about 2π/∆x.
This suggests that this form for the wave-function should be roughly valid 2π for all momenta in between k = 0 and k = ∆x !
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 66 / 143 Dispersion curve
(Interesting note:
If you try to make a wave-function where only one atom is moving, and the rest are in the ground state, you get ~ exp(ik · ~xi )Ψ0.
However, the atoms of this wave-function are not symmetric. Blindly symmetrizing, you get X ~ exp(ik · ~xi ) Ψ0 i which is just our phonon state. Therefore, one atom cannot move around by itself. If one moves, they all move (as a phonon).
This is another way to intuitively see that bosons cannot get “mixed around” at low energies.)
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 67 / 143 Dispersion curve
We now return to our (now justified) expression for the phonon energy:
2k2 E = ~ 2mS(k) where Z S(k) = p(~x) exp(i~k · ~x)dV
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Intuitively, what does p(x) look like?
It is the probability that there is an atom at x given that there is one at the origin.
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It will start at 0, then reach a maximum at ∆x, before dropping again. It will attain decreasing maxima for the multiples of ∆x as it asymptotes to 1.
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2π S(k) will therefore peak at ∆x , fall again, and asymptote to 1 as well. It will also grow linearly for small k.
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2 2 ~ k E = 2mS(k) will therefore grow linearly at small k, have a minimum at 2π about ∆x , and continue to grow afterwards.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 72 / 143 Dispersion curve
This is indeed the case! Our variational method does over-estimate the energy by a considerable amount (as it is known to do) but it is qualitatively correct.
The initial linear growth, maximum, and minimum, are all there.
Feynman, 1972, p.331
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The initial linear growth means the heat capacity grows like that of a crystal for low temperatures.
Indeed, the heat capacity curve follows the T 3 Debye law.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 74 / 143 Dispersion curve
Usually, only the small k modes are called “phonons,” where the ones at the maximum are called “maxons” and the ones in the minimum are called “rotons.” Note that maxons and rotons have 0 group velocity.
We have seen that that the shape of the energy-momentum dispersion curve reflects the existence of a “local order” to the distribution of atoms in the liquid.
A dilute boson gas does not have the minimum.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 75 / 143 Dispersion curve The roton minimum is responsible for the deviation in the Debye T 3 law around 1K. The Boltzmann factor e−E/kT suppresses the probability for independent modes to excite. When the factor is appreciable for the roton energy gap, a small increase in temperature results in a big increase in energy as the roton states fill, causing the heat capacity curve to grow steeper.
At the phase transition, however, the modes are far from independent, and more sophisticated techniques need to be used.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 76 / 143 Critical velocity (first attempt) Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 77 / 143 Critical velocity (first attempt)
The linear dispersion relation for small momenta gives a clue as to why there is a critical velocity.
For simplicity let’s take
E = cs p where cs is necessarily the speed of sound in the material.
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Imagine there is a large mass of fluid flowing down a tube with velocity v.
1 E = Mv 2 2
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Suddenly, a tiny phonon excitation is randomly produced with momentum p. The fluid recoils with a change in momentum −p due to the conservation of momentum.
p δv = − M
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The energy of the configuration after the excitation can not exceed the initial energy. 1 1 0 ≤ δE = M(v − p )2 + c p − Mv 2 2 M s 2 0 ≤ −vp + cs p
v ≤ cs Therefore, we see that there can be no excitation unless the velocity of the fluid is at least the speed of sound.
This is the critical velocity.
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Interestingly, the speed of sound in helium is a few orders of magnitude higher than the measured critical velocity. (According to this argument, the critical velocity would be 70m/s when it is really about 20cm/s.)
Our argument fails because the entire liquid need not move in unison. One part of the liquid can recoil without other parts doing so.
Later, we will come back and refine our understanding of the critical velocity.
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1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 83 / 143 Irrotational flow
We will now investigate how helium “flows.”
A flowing liquid has a lot of kinetic energy, but it is ordered motion, not disordered.
Therefore, helium can still flow at 0K.
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If we move the center-of-mass of the ground state Ψ0 by ~v, our wave-function will be X exp i m ~v · ~x Ψ . ~ i 0 i Note that X X exp i m ~v · ~x Ψ 6= exp(i m ~v · ~x ) Ψ ~ i 0 ~ i 0 i i so this is not a phonon. The phase is global, and all the atoms are moving together. The energy is also a lot larger.
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The energy density can be found by integrating 1 Ψ∗HˆΨ = mv 2|Ψ |2 0 0 2 0 over N − 1 configuration variables. The result is 1 ρv 2 2 which is just what you would expect classically.
If ~v doesn’t change much on atomic distances, the above statement is still true.
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Imagine a likely configuration in the ground state.
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Assume that the velocity vectors ~v do not vary much microscopically.
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Consider any loop of atoms. The wave-function will not change if the atoms are permuted. (The loop can be any size.)
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Therefore, if we “slide” each atom into the next one on the loop, the wave-function must end up the same as when we started.
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If there is enough space to slide the atoms without colliding them, there will be no density variation, and Ψ0 will not change.
We can also briefly move neighboring atoms out of the way of the loop if we have to.
Therefore, Ψ0 will not change during this sliding, and the only change in the wave-function during the sliding will be in the phase.
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Say the positions of the atoms before the sliding are ~xi , the positions after 0 0 the sliding are ~xi , and ~xi − ~xi = ∆~xi .
The change in the phase is X X ∆θ = m ~v · ~x0 − m ~v · ~x ~ i ~ i i i X ∆θ = m ~v · ∆~x ~ i i I ∆θ = m ~v · d~x ~
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As eiθ cannot change under a permutation, we must have I 2πn = m ~v · d~x ~ for some integer n.
The line integral of velocity around any loop is quantized.
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This has some strange effects.
Imagine getting the fluid flowing around an annulus (perhaps by spinning the container when T > 2.17K and then lowering the temperature).
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The integer n cannot drop continuously!
This makes a “superflow” which can persist indefinitely.
The origin of this conservation law comes from topology (not symmetry).
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However, locally our fluid is simply connected, and n must be 0.
By Stoke’s theorem,
∇~ × ~v = 0. Our fluid has no “vorticity.”
This is called “irrotational flow.”
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 96 / 143 Vortex lines Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 97 / 143 Vortex lines
But if this is true, how can our fluid ever have angular momentum when the container doesn’t have a hole in the middle?
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 98 / 143 Vortex lines
Imagine we have a cylinder of helium at 0K, subjected to great pressure so it is a solid, rigid, crystal.
Then spin the cylinder.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 99 / 143 Vortex lines
Then, slowly relieve pressure from the cylinder until it liquifies.
Angular momentum is conserved, so what happens?
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 100 / 143 Vortex lines
From the condition I 2πn = m ~v · d~x ~ we can see that our fluid can circulate only if it has holes in it.
If there isn’t one big hole in the middle of the container, then the fluid will make many thin holes, anyway!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 101 / 143 Vortex lines
Vortex lines will form in our fluid, where the wave-function is 0.
The phase and velocity are undefined on a vortex line.
These lines are thinner than an atom, so the “hole” isn’t real.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 102 / 143 Vortex lines
Vortex lines will form in the cylinder along the axis of rotation.
They will be evenly spaced out.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 103 / 143 Vortex lines
Proof that energy is lowest when vortex lines are evenly distributed: Assume circular symmetry, dA = 2πrdr. R is the radius of the cylinder, H is the height, m is the mass of an atom, ρ is the mass density. l(r) is the number of vortex lines per area element at radius r, v(r) is velocity of the atoms at radius r, L(r) is the angular momentum of the atoms within the inner cylinder of radius r. Note L(0) = 0 and L(R), the total angular momentum, is held fixed. We will vary L(r), keeping the boundary conditions fixed, trying to minimize KE.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 104 / 143 Vortex lines
The line integral of v around the circle of radius r is equal to 2π~/m times the number of vortex lines in the circle. Therefore, I h v(r)dr = (number of lines) m h Z r (2πr)v = l(r 0)2πr 0dr 0 m 0 d h (2πrv) = 2π l(r)r dr m 1 d h (rv) = l(r) r dr m If we can show that KE is minimized when l(r) above is constant, we are done.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 105 / 143 Vortex lines
The angular momentum within the cylinder of radius r is Z r L(r) = (ρH)vr 02πr 0dr 0 0 This allows us to solve for v(r) in terms of L(r):
dL = L0 = (2πρH)vr 2 dr 1 L0 v(r) = 2πρH r 2
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 106 / 143 Vortex lines
The total kinetic energy, which we wish to minimize, is
Z R Z R 02 ρH 2 0 0 1 L KE = v 2πr dr = 3 dr 0 2 4πρH 0 r The Euler-Lagrange equation tells us this is minimized when
d ∂ L02 d L0 = 0 =⇒ 2 = 0 dr ∂L0 r 3 dr r 3 L0 =⇒ = const r 3
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 107 / 143 Vortex lines
Therefore, rv = r 2 · const h 1 d 1 d l(r) = (rv) = (r 2 · const) m r dr r dr and l(r) is constant, as desired. This implies the number of vortex lines per unit area is constant, and the vortex lines are evenly spaced out.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 108 / 143 Vortex lines
After being predicted by Feynman in 1955, this has been experimentally verified! We have pictures from injecting solid hydrogen into the liquid, which get drawn into the cores of the vortex lines:
https://www.aps.org/units/dfd/pressroom/papers/gaff09.cfm
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 109 / 143 Vortex lines
There aren’t even that many lines.
If the cylinder is spinning at ω rad/sec, there will be
lines 2 × 103 ω . cm2 At ω = 1, the lines will be 0.2mm apart.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 110 / 143 Vortex lines
The energy of a configuration of vortex lines can be found by calculating
Z 1 ρv 2dV 2 over the whole liquid, estimating the “core” of the line to be of atomic thickness.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 111 / 143 Vortex lines
Intuitively, we can see that takes energy to push two parallel vortex lines together.
Therefore, |n| > 1 vortex lines are unstable and break apart.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 112 / 143 Vortex lines
Meanwhile, anti-parallel vortex lines lose energy by coming together.
Therefore, they like to annihilate each other.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 113 / 143 Vortex lines
There are all sorts of interesting vortex line dynamics.
For example, the vortex lines themselves move with the velocity of the background fluid (i.e., the component of the velocity not caused by the vortex line itself).
Therefore, anti-parallel vortex lines will orbit each other, while parallel vortex lines will not.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 114 / 143 Vortex rings Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 115 / 143 Vortex rings
Vortex lines can also close into little loops called “vortex rings.”
Leggett, p.94
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 116 / 143 Vortex rings
Vortex rings have their own interesting dynamics. They can grow, shrink, become disordered, interact with walls, vortex lines, each other, etc.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 117 / 143 Vortex rings
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 118 / 143 Vortex rings
Were rotons really vortex rings this whole time?
Feynman, 1972, p.333
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 119 / 143 Vortex rings
No!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 120 / 143 Vortex rings
This is from 2014???
Boy, I hope someone got fired for this!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 121 / 143 Critical velocity (second attempt) Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 122 / 143 Critical velocity (second attempt)
Let us return to the phenomenon of resistanceless flow below a critical velocity.
Experimentally, the actual “critical velocity” is not well defined, and depends upon the geometry of the container.
Therefore, we expect for the true cause to be dynamical in nature.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 123 / 143 Critical velocity (second attempt)
In a regular fluid, energy is dissipated by turbulence, when the flow lines get mixed up on geometry of the pipe, pointing in different directions. This is a very chaotic process.
http://www.croberts.com/erosion-corrosion.htm www.bg.ic.ac.uk/research/k.parker/homepage/Mechanics of the Circulation/Chap 05/ Chapter 05.htm
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 124 / 143 Critical velocity (second attempt)
To see why this does not happen in a superfluid flowing under a critical velocity, we will consider just one particular set up:
fluid flowing into a big container from a thin pipe.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 125 / 143 Critical velocity (second attempt)
In a regular fluid, the liquid coming out of the pipe will be quite turbulent, and the vorticity will be high.
Feynman, 1955, p.46
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 126 / 143 Critical velocity (second attempt)
Turbulence requires vorticity.
https://en.wikipedia.org/wiki/Turbulence
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 127 / 143 Critical velocity (second attempt)
Therefore, if the flow is irrotational, it can’t become turbulent, and energy will not be dissipated.
(This is called potential flow.)
Feynman, 1955, p.45
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 128 / 143 Critical velocity (second attempt)
But, if the flow is fast enough, then vortex rings can be created on the corner of the pipe.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 129 / 143 Critical velocity (second attempt)
The pattern of vortex rings would look something like this.
The fluid velocity inside the rings would be about v. The fluid velocity outside the rings would be about 0.
Feynman, 1955, p.46
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 130 / 143 Critical velocity (second attempt)
The vortex rings will eventually get all jumbled up, and the energy of the system will be found in disordered velocity vectors pointing in different directions.
This explains the phenomenon of critical velocity: if the velocity isn’t large enough to make these vortex rings, there will be no turbulence.
We have now properly characterized the states that “mix around” the atoms in the fluid, which we speculated about in the beginning.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 131 / 143 Critical velocity (second attempt)
We could estimate the critical velocity by first estimating the energy of a vortex ring, found by calculating
Z 1 ρv 2dV 2 and estimating the ”core” of the ring to be of atomic thickness.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 132 / 143 Critical velocity (second attempt)
Vortex rings can only form if the rate of kinetic energy flowing into our system matches the energy it takes to continuously form rings.
The critical velocity calculated through this estimate matches the observed critical velocity within an order of magnitude, much better than our last attempt.
It also suggests avenues for improvement.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 133 / 143 Critical velocity (second attempt)
Critical velocities can arise in different geometries through different processes involving vortex lines.
For example, in our annulus, if the temperature is high enough and there is enough energy, a vortex line can form on one wall and terminate on another, changing n and disrupting the superflow.
Likewise, a vortex ring can form in the middle of a fluid and terminate on the wall, also disrupting superflow.
These processes will only occur near 2.17K.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 134 / 143 Critical velocity (second attempt)
Conclusion:
Superfluid helium is pretty dank.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 135 / 143 Critical velocity (second attempt)
Thank you!
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 136 / 143 References Table of Contents
1 Why is helium a liquid at low temperatures?
2 Transition to superfluid
3 Multi-particle wave-function and phonons
4 Dispersion curve
5 Critical velocity (first attempt)
6 Irrotational flow
7 Vortex lines
8 Vortex rings
9 Critical velocity (second attempt)
10 References
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 137 / 143 References
Much of this talk was taken from Feynman’s original work on the subject (which carries the unfortunate belief that rotons are vortex rings).
Much of his work is in his statistical mechanics textbook: Richard Feynman, 1972. Statistical Mechanics: A Set Of Lectures. Reading, Mass.: W.A. Benjamin. The main paper to consult on vortex lines is Richard Feynman, 1955. Application of Quantum Mechanics to Liquid Helium. Progress in low temperature physics (Vol. 1, p. 17). North-Holland, Amsterdam.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 138 / 143 References
Feynman’s work on using the variational method to find the energy-momentum dispersion relation can be found in Richard Feynman., 1954. Atomic theory of the two-fluid model of liquid helium. Physical Review, 94(2), p.262. and further attempts to find better curves can be found in Richard Feynman, Michael Cohen, 1956. Energy spectrum of the excitations in liquid helium. Physical Review, 102(5), p.1189.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 139 / 143 References
Feynman also tried to understand the nature of the λ-transition in Richard Feynman, 1953. Atomic theory of the λ transition in Helium. Physical Review, 91(6), p.1291. David Ceperley later used Feynman’s idea, aided by powerful numerical techniques, to better understand the transition in David Ceperley, 1995. Path integrals in the theory of condensed helium. Reviews of Modern Physics, 67(2), p.279.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 140 / 143 References
A friendly and modern exposition on superfluid helium is given by Anthony Leggett, 2006. Quantum Liquids: Bose condensation and Cooper pairing in condensed-matter systems. New York, NY: Oxford University Press. The above book also describes how superflow can decay.
Another very helpful book, which also describes the history of theories on liquid helium in great detail, is Allan Griffin, 1993. Excitations in a Bose-condensed Liquid. Cambridge, UK: Cambridge University Press. A very short and pleasant article on the history of liquid helium with nice figures is Russell Donnelly, 2009. The two-fluid theory and second sound in liquid helium. Phys. Today, 62(10), pp.34-39.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 141 / 143 References
An insanely deep work on the subject is Grigory Volovik, 2009. The Universe in a Helium Droplet. New York, NY: Oxford University Press. Landau’s original phenomenological paper on liquid helium (still a fun read) is Lev Landau, 1941. Theory of the Superfluidity of Helium II. Physical Review, 60(4), p.356. The book I made fun of, but which has a good discussion of the boson gas and Bogoliubov transformations, is Tom Lancaster, Stephen Blundell, 2014. Quantum Field Theory for the Gifted Amateur. New York: Oxford University Press.
Noah Miller (Columbia University) Superfluid Helium-4 March 6, 2018 142 / 143 References
In depth discussion on quantized vortices can be found in Russell Donnelly, 1991. Quantized Vortices in Helium II. Cambridge, UK: Cambridge University Press. The pictures of the vortices in helium were taken from the following press release: https://www.aps.org/units/dfd/pressroom/papers/gaff09.cfm Lennard-Jones coefficients for helium are taken from http://www.cengage.com/resource uploads/ downloads/1111988617 420953.pdf
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