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The Nuclear and Electron Spin Barnett Effects B.K. Tenn 1 , E.L. Hahn 2 , and M.P. Augustine 1 1 Department of Chemistry, University of California - Davis 2 Department of Physics, University of California - Berkeley Abstract In 1914 Barnett [1] detected the magnetization due to the polarization of electron spins caused by the rotation of a cylinder of unmagnetized iron. One year later in 1915 Einstein and de-Haas [2] accomplished the inverse experiment, namely the mechanical rotation of an iron cylinder due to magnetization developed by application of a magnetic field. Prior to the development of magnetic resonance, these two experiments were the basis of measuring magnetic g-factors in ferromagnetic and paramagnetic solids. To date, the Barnett effect has only been mentioned in connection with NMR by Purcell in 1979 [3], where he postulates that the origin of weakly polarized starlight in interstellar space is due to the alignment “suprathermal” grains by rotation induced Barnett magnetization. The focus of this project is to understand the mechanism of the Barnett effect and its manifestation in nuclear and electron spin systems. Acknowledgements Dr. Alex Pines, Dr. Maurice Goldman, Dr. John Waugh, Dr. Jean Jeener, Dr. Eugene Commins, Dr. Carlos Meriles, Dr. Demitrius Sakellari, Dr. Andreas Trabesinger, Dr. Jamie Walls, and The Augustine Group Definitions References [1] S.J. Barnett, Phys. Rev., 6, 239 (1915). [2] A. Einstein and W.J. de Haas, Verhandl. Deut. Phsik. Ges., 17, 152 (1915). [3] E.M. Purcell, Astrophys. J., 231, 404 (1979). [4] D.K. Sodickson and J.S. Waugh, Phys. Rev. B, 52, 6467-6469 (1995). [5] R.L. Strombotne and E.L. Hahn, Phys. Rev. 133, A1616 (1964). [6] G. Whitfield and A.G. Redfield, Phys. Rev., 106, 918 (1957). [7] A.A. Abragam. Principles of Nuclear Magnetism, Oxford, 2002. Magnetization by Rotation We will now discuss the dynamics of the nuclear Barnett effect - nuclear spin polarization due to sample spinning (taken along the z-axis for simplicity). In the absence of an externally applied magnetic field we find that implying that any change in the lattice angular momentum will result in a corresponding change in the spin angular momentum, i.e. sample magnetization. Note that in the absence of sample rotation, In the presence of rotation, the previously static dipolar coupling, H D , becomes time dependent in φ(t) and upon rotation of the spin components, the time dependence can be effectively eliminated as shown: The rotating frame Hamiltonian, in combination with the Liouville - von Neumann equation during sample spinning can be used to determine both the formation of magnetization <J> and the change in spin rate due to this magnetization in spin systems. Note the formation of a "ghost" field due to the rotation, Simulations of the effect of rotation about the z-axis at ω r /2π = 10kHz on an isolated dipole- dipole coupled two spin-1/2 system (θ = π/2, φ = 0), a powder of non-interacting dipole-dipole coupled two spin-1/2 systems, and an eight spin-1/2 dipole-dipole coupled cube intially at thermal equilibirum are presented (ω D /2π = 10kHz). Rotation by Magnetization Now it is natural to ask if one could cause an ensemble of spins at thermal equilibrium in zero field to rotate by instantaneously jumping a real DC magnetic field. In the presence of a real DC magnetic field, one finds that implying that the application of an DC magnetic field represents an additional source of angular momentum. However we note that the rotor only exchanges angular momentum with the spins through H D , the dipolar interaction, and not through the direct coupling with H z . The applied field, H z , represents an energy-momentum source from a solenoid or an oscillator not directly coupled to the rotor. When an applied field is jumped from 0 to H 0 , the source of angular momentum flowing into J z is the solenoid. The solenoid now plays the role of the rotor because of the application of an electromotive force which launches a current and therefore momentum into the applied H 0 field. The resulting exchange of order between the dipolar and Zeeman reservoirs is that of Strombotne and Hahn [5]. Simulations of the dynamics resulting from the application of a 10kHz DC magnetic field along the z-axis are shown (ω D /2π = 10kHz). Effects upon NMR It is instructive to note that (1) explains why neither the Barnett effect nor its inverse are seen in high field solid state NMR experiments. At high magnetic field, the Zeeman interaction scales the non-secular, q ≠ 0, part of the dipolar interaction to roughly, ω D 2 /ω 0 << ω D << ω 0 . To zeroth order in perturbation theory, the NMR spectrum is governed by the Zeeman interaction, H z and 〈[H D ,J]〉 = 0, thus removing any coupling between the nuclear spin and mechanical angular momenta. Extending this argument to first order in perturbation theory also does not yield any useful coupling between ω r and 〈J〉 when the applied DC field is parallel to the sample spinning axis because the secular term of the dipolar Hamiltonian does not have any ω r dependence. It is the time dependence imparted on the q ≠ 0 parts of H D (t) that translate into the generation of magnetization or the onset of sample rotation in zero and non-zero magnetic fields, respectively. + ω r t φ 0 φ 0 θ x y z r r' ω r : internuclear vector : angurlar velocity : frequency of applied DC magnetic field : angle from z-axis : angle from x-axis Total Spin Angular Momentum: Total Hamiltonian: Second Rank Irreducible Spherical Tensor Operators: Liouville-von Neumann Equation: () =- () i H, Ehrenfest's Theorem: = i H , Angular Momentum Operators: L d d d d x = + i sin cot cos φ θ θ φ φ L d d d d y = - + i cos cot sin φ θ θ φ φ L z =- i Angular Momentum While investigating nuclear spin diffusion in zero field, Sodickson and Waugh [4] introduced an interesting relation dealing with momentum exchange between the spins and the lattice. Starting with the classical definition of angular momentum, L = r x p, in connection with Ehrenfest's Theorem one can show that We notice that the above reformulated Bloch's equation demonstrates that the total angular momentum of our system, spins plus lattice, is conserved. Provided there are no real magnetic fields present, 〈J〉 = 0 at all times, therefore, verification (of above) requires showing that 〈L〉 = 0. The calculation necessary to calculate 〈L〉 requires a rigorous quantum mechanical treatment of our system, which in turn, necessitates the quantization of the rotational motion of our lattice in direct analogy to the rotational level structure in molecular hydrogen gas. This approach requires some knowledge of the partition function corresponding to rotation in addition to the calculation of the lattice coefficient's matrix elements from the spherical harmonic basis elements. In the case of molecular hydrogen gas, however, the moment of inertia is low, molecular rotation is fast, and the temperature is low, so we can safely truncate our basis functions to a finite number of rotational energy levels. In a real macroscopic sample at room temperature, however, our moment of inertia is large and sample rotation is slow. Therefore one finds a large number of occupied, closely spaced energy levels will be populated at thermal equilibrium. However, the inability to define a tractable density matrix for this case can be circumvented by recognizing that the expectation value for the lattice angular momentum 〈L〉, must reduce to the classical value, Iωr,for a macroscopic object. In this case, we find, which, upon expansion in the principle axis system of the moment of inertia tensor yields: r ω , d dt t t t d dt J J xx rx yy ry zz rz x y I I I ω ω ω , , , () () () =- 〈 〉 〈 〉 〈 〉 + - - - J z tot z tot y tot z tot x 0 0 ω ω ω , , () () () () tot y tot x x y z t t J J J , , () () ω 0 g =- Iω r t r = D = i H , J = [ ] + i H i H d dt Z L J J =- +i H , D H =- J M M H H T z SS Redfield barn r = + - 1 2 1 1 ω sin Einstein-de Haas Effect Barnett Effect 0 150 50 100 0 150 50 100 0 30 10 20 1.0 .75 .50 .25 0 0 100 60 80 40 20 time (μs) time (μs) time (μs) ω r /2π (mHz) ω 0 /2π (kHz) Δω r /2π (mHz) 0 -20 -40 -60 -80 -100 0.5 0.2 0 0.1 0.3 0.4 0 -20 -30 -40 -50 -10 1.0 0.8 0.6 0.4 0.2 0 0 -.1 -.2 -.3 0.1 0 0.2 0.3 0 150 50 100 0 -20 -40 -60 -80 -100 1.0 0.8 0.6 0.4 0.2 0 0 150 50 100 time (μs) 0.5 0.2 0 0.1 0.3 0.4 0 -20 -30 -40 -50 -10 0 30 10 20 0.1 0 0.2 0.3 1.0 .75 .50 .25 0 0 100 60 80 40 20 0 -.1 -.2 -.3 time (μs) time (μs) Electron Spin Barnett Effect Bloch's equations can be reformulated so that the relaxation terms make the magnetization approach its equilibrium value with respect to the instantaneous applied magnetic field instead of the fixed field [6,7]. Solution of these modified Bloch's equations reveal that in the presence of a circularly polarized RF field, even in the absence of an applied magnetic field, a magnetization perpendicular to the plane of polarization is developed. In the case of the electron spin Barnett effect, we wish to investigate the effect of applying a circularly polarized RF field in addition to the, albeit small, Barnett magnetization of diphenyl picryl hydrazil (DPPH) upon the overall polarization of the electron spins. Accounting for both the circularly polarized RF and the developing oscillatory Barnett magnetization in our sample, we find that the Bloch's equations become: In the limit that γ H 1 << χ 0 /T and H barn << H 1 , at steady state, we find In the absence of the Barnett field, we reproduce the results of Whitfield and Redfield [6]. Introduction of a Barnett field will introduce oscillations in the steady state magnetization at the frequency ω r -ω. The numerical solution to this problem is shown where H 1 = 0.1, χ 0 /T = 1000, H barn = 0, 10 -2 , 10 -3 . We find that in the absence of the Barnett field, the typical steady state magnetization is acheived. In the presence of a Barnett field oscillations in the steady state magnetization are induced at approximately the difference frequency described above. t H tM M H t barn r bar + - + + ω 1 0 0 1 sin sin co os t H tM H t H tM M barn r x barn r y =- + + + - ω ω sin sin z t H tM M H t x barn r bar + ( γ ω ω 1 0 0 1 co os si in ) <Mz> (10 -8 ) H barn /H 1 = .01 H barn /H 1 = .1 H barn /H 1 = 0 0 5 10 15 0 1 3 4 5 6 7 2 Numerical Solution of the Modified Bloch's Equation time (μs) (1) & M M H M H 0 = × - - γ χ T 〈J〉 〈J z 〉 〈J z 〉 〈J z 〉 〈J z 〉 〈J z 〉 〈J z 〉 〈J z 〉 〈J z 〉 T I I I I I I n m nzmz 0 2 3 () , , , ⋅ T I I I I I I n m n mz nzm ± ( ) 1 2 () , , , , , nm nm nm nm ± ± ( ) = 2 2 () , , in , θ φ θ φ i nm nm nm 0 1 3 , , , os θ Δω r /2π (mHz) Δω r /2π (mHz) ω r /2π (mHz) ω r /2π (mHz) ω r /2π (mHz)
