all Cooper pairs must behave in the same way LIQUID HELIUM-THREE Helium is the only element that remains liquid down to absolute zero temperature. The “common” isotope of helium, helium-4, becomes “superfluid” at about 2 degrees absolute (2K) The “rare” isotope, helium-3 (3He) behaves normally above 3/1000 of a degree (3mK) Theoretically, the atoms of 3He are thought to be “the same kind of particles” as the electrons in a metal. — and experimentally, above 3mK liquid 3He behaves like the electrons in a “normal” (nonsuperconducting) metal. But at low temperatures, many metals become superconducting: this is because the electrons form Cooper pairs (a sort of giant di-electronic molecules) and once formed, according to the BCS (Bardeen-Cooper-Schrieffer) theory, all Cooper pairs must behave in the same way Question: could something similar happen in liquid 3He? Much theoretical work 1959–1972: strongly suspected Cooper pairs, if formed, would (unlike those in metals) have one or more orientational degrees of freedom (“anisotropic superfluid”)

COMPLETELY UNEXPECTED! THE EXPERIMENTS OF NOV. 1971–MAY 1972 Doug Osheroff, graduate student at Cornell, cools a mixture of liquid and solid 3He, measuring pressure (P) as a function of temperature (T). Goes down into region below 3 mK where few if any had ventured before. . . — sees two small glitches in the curve (“A” and “B”). (Nowadays, liquid at temperatures between A and B said to be in “A phase,” below B in “B phase”). — data originally interpreted (incorrectly) as evidence for something happening in solid 3He. Subsequently (spring 1972) Cornell group performs nuclear magnetic resonance (NMR) experiment to measure resonance frequency (res) as function of temperature (T). Sees anomalous behavior clearly identified as coming from liquid: res 3mK 2mK A phase B phase normal phase temperature COMPLETELY UNEXPECTED!

NMR: THE PROBLEM In traditional theory of NMR, resonance frequency is a direct measure of the “average total magnetic field” acting on the system (3He nucleus). If we interpret the data in this way, then in the normal (N) phase, and also in the B phase, the total magnetic field is just the externally applied one (no problem!). But in the A phase, there seems to be an extra “internal” field which is not simply proportional to the external one. In fact, it looks as if i.e. Hint is perpendicular to Hext! Problem: There is no obvious source of a large enough Hint! (electrons are all paired off in the atom, and field of other nuclei is far too weak) ? ? First evidence of a breakdown of standard laws of physics under extreme conditions? Hint Hext Htot

THE SOLUTION—STEP 1 The whole idea of an “average total field” is just wrong for liquid 3He (too “classical”: correlations in 3He are quantum-mechanical). Rather, we can explain the data if we assume that in equilibrium, pairs of nuclear spins always have the orientation which minimizes the nuclear dipole energy. 3He nuclei are just like tiny bar magnets: So, all nearby pairs should have the “good” orientation. But this raises a new problem: difference in energy between “good” and “bad” (pair) is only ~10–7K, so much smaller than thermal energy (~10–3K)  thermal disorder should make orientation almost random! ? ? + – + – – + – + bad good

Difference in energy is not pair but N pair! THE SOLUTION—STEP 2 (JULY 1972) Problem is similar to behavior of electrons in ferromagnetic material (e.g. iron). There, we have For a single electron, difference between “good” and “bad” (el) is much less than thermal energy (Eth), yet all electron spins point along field! The reason: other strong (“exchange”) forces, having nothing to do with external field, force all spins to point parallel  competition is between The energy difference between these two configurations is not el but N el, i.e. much larger than Eth! Similarly, in the A phase of liquid 3He, let’s suppose some effect nothing to do with the nuclear dipole interaction forces all pairs to behave in the same way. Then competition is between + – – + external field good bad      . . . . and      . . . . good bad      . . . and       . .         bad good Difference in energy is not pair but N pair!

should we consider it is not an anisotropic superfluid? THE SOLUTION—STEP 2 (cont.) So, in the A phase some effect must be forcing all pairs of nuclear spins to behave in the same way. What could it be? Most obvious guess: Cooper pairing (note: Cooper pairs must behave identically not just in their center-of-mass motion, but also in their orientational state). So, reasonable hypothesis: A phase of liquid 3He is a particular kind of anisotropic Cooper-paired state (as previously conjectured). But, what about the B phase? Since NMR behavior seems to be the same as in normal state . . . B phase A phase normal phase res temperature should we consider it is not an anisotropic superfluid?

(use “Born-Oppenheimer” approximation) THE SOLUTION—STEP 3 (APRIL 1973) Proper “microscopic” calculation. Variables: (1) total nuclear spin (2) “spin of Cooper pairs” (use “Born-Oppenheimer” approximation) Result: NMR behavior can be used as fingerprint of anisotropic superfluid phases! (each possible phase gives different signature in NMR) Most spectacular prediction: one particular (theoretically favored) anisotropic superfluid phase should show (like the experimentally observed B phase) no anomalous behavior in a traditional NMR experiment, but in a novel type of experiment (rf field potential parallel to external field) should show a sharp nonzero-frequency resonance (whose frequency can be calculated) Experiment (July 1974): B phase does just that!  A, B phases are both anisotropic superfluid phases, unambiguously identified by their NMR behavior.