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Thermal Boundary Resistance of the Superfluid 3 He A-B Phase Interface D.I. Bradley S.N. Fisher A.M. Guénault R.P. Haley H. Martin G.R. Pickett J.E. Roberts.

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Presentation on theme: "Thermal Boundary Resistance of the Superfluid 3 He A-B Phase Interface D.I. Bradley S.N. Fisher A.M. Guénault R.P. Haley H. Martin G.R. Pickett J.E. Roberts."— Presentation transcript:

1 Thermal Boundary Resistance of the Superfluid 3 He A-B Phase Interface D.I. Bradley S.N. Fisher A.M. Guénault R.P. Haley H. Martin G.R. Pickett J.E. Roberts V. Tsepelin

2 Thermal Boundary Resistance of the Superfluid 3He A-B Phase Interface D.I. Bradley S.N. Fisher A.M. Guénault R.P. Haley H. Martin G.R. Pickett J.E. Roberts V. Tsepelin

3 Outline Helium Background Experiment Low Field B Phase Results A Phase Layer in Cell Distorted B Phase in Cell Conclusions – Kapitza Resistance, Thermal Conductivity

4 Helium 3 Phase Diagram 2 nd order transition through Tc P = O bar T = 130-200µK Critical Field ~ 340mT 1st order transition between A and B Superfluid 3He is a BCS condensate with “spin triplet p-wave pairing”

5 The A-B interface is the interface with the highest order, highest purity and in principle best- understood phase interface to which we have access. It’s a phase boundary between two quantum vacuum states. We find that we are able to measure the transport of quasiparticle excitations between these two order parameters. Why study the A-B interface?

6 A Phase has only parallel components

7   Anisotropic gap

8 B Phase has all 3 components:

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10  Pseudo-isotropic gap

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12 Opposite spins suppressed Parallel spins enhanced  Polar gap suppressed Equatorial gap enhanced Apply a magnetic field to the B phase – gap becomes distorted:   pp ee

13 We thus obtain the above complex structure across the interface. B- phase A- phase The quasiparticle motion.

14 Zeeman splitting decreases the energy of the down-spin qp’s, so the low energy ones are Andreev reflected. Any that reach the A-phase are high enough in energy to travel straight through. The energy of the up-spin qp’s is increased. Those with energy below the A- phase gap are Andreev reflected

15 Vibrating Wire Resonators Width Parameters W =  f 2 * T * E  Power Few mms

16 VWR Range of Measurement Critical Velocity

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18 The Experimental Cell

19 Do this to check the cell’s working as a BBR i.e. VWR damping is proportional to Power

20 LOW FIELD ISOTROPIC GAP B PHASE The cell appears to be hotter at the bottom than at the top! Why?

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22 Magnetic Field Profile used to Produce A Phase Layer

23 QUASIPARTICLE TRANSPORT A PHASE “SANDWICH”

24 QUASIPARTICLE TRANSPORT HIGH FIELD DISTORTED B PHASE

25 This extra resistance may be caused by a textural defect remaining after the A phase layer has been removed

26 Thermal Resistance of Cell

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28 The “Kapitza Resistance” of the A-B interface is: We can now calculate the thermal conductivity through the cell: Measured : R K (AB) = 0.3 µK/pW at 140µK Predicted by S.Yip 1 : R K (AB) = 2.6*10 -3 µK/pW 1 S. Yip. Phys Rev B 32, 2915 (1985)

29 Thermal Conductivity of Cell

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31 Summary Have we measured the “Kapitza resistance” of the A-B interface in superfluid Helium -3? Resistance decreases as temperature increases. The thermal conductivity appears to have an exponential dependence on temperature.  The thermal conductivity is dominated by the heat capacity of the helium 3.

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36 How do we get smoothly from the anisotropic A phase with gap nodes to...

37 .... the B phase with an isotropic (or nearly isotropic) gap?

38 We start in the A phase with nodes in the gap and the L- vector for both up and down spins pairs parallel to the nodal line.

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40 The up spin and down spin nodes (and L- vector directions) separate

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42 ..... and separate further.

43 The up spin and down spin nodes finally become antiparallel (making the topological charge of the nodes zero) and can then continuously fill to complete the transformation to the B phase.

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45 But think for a moment about the excitations!

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51 Why is the B-phase gap distorted? In zero magnetic field L and S are both zero. However, a small field breaks the symmetry between the  spins and the  spins, the energy gap becomes distorted and a small L and S appear.


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