Download Quantum Mech. II HW 6: Particle Continuity, Dirac to Nonrel. Schrodinger, Rel. Charged - P and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics II: HW 6 Solutions Leo Radzihovsky April 23, 2008 1 Particle Continuity Equation in Dirac Equation The number density in the Dirac equation is given by n = ψ†ψ, where ψ is a Lorentz spinor satisfying the Dirac equation i~∂tψ = (cα · p + βmc2)ψ. (1) Thus, to determine the continuity equation and the corresponding (spatial) components of the current density, j, we differentiate n with respect to time and use the Dirac equation to obtain: ∂tn = (∂tψ†)ψ + ψ†(∂tψ), (2) = i~−1[(cα · p + βmc2)ψ]†ψ − ψ†i~−1[(cα · p + βmc2)ψ], (3) = ∇ · j, (4) where we can then identify the current density to be j = cψ†αψ, (5) and in above we used p = −i~∇, and hermiticity of α and β. 2 From Dirac to Nonrelativistic Schrodinger Equation, Spin and Zeeman interaction We start out with time-independent Dirac equation in a magnetic field: Eψ = (cα ·Π + βmc2)ψ, (6) where the gauge invariant momentum operator is given by Π = p− qA/c, (7) and Lorentz spinor ψ = (χ, φ). We eliminate the small spinor component φ in favor of the large component χ, obtaining in the nonrelativistic limit Esχ = 1 2m (σ ·Π)(σ ·Π)χ, (8) 1 where Schrodinger energy eigenvalue is given by Es = E −mc2, and σi are the three Pauli matrices. To simplify the right hand side of the resulting Schrodinger equation we use the identity (σ ·C)(σ ·D) = C ·D + iσ · (C×D), (9) arising from the anticommutation relations of Pauli matrices. Using C = D = Π we find (σ ·Π)(σ ·Π) = Π2 + iijkΠiΠjσk, (10) = Π2 − i q c ijk(piAj +Aipj)σk, (11) = Π2 − q~ c ijk∇iAjσk, (12) = Π2 − q~ c σ · ∇ ×A, (13) which leads to the nonrelativistic Pauli Hamiltonian with Zeeman energy accounting for spin magnetic moment interaction with the B field: HPauli = 1 2m (p− qA/c)2 − gµB ~ S ·B, (14) where g = 2, µB = q~/(2mc) is the Bohr magneton, and B is the magnetic field. 3 Relativistic charged particle in a constant magnetic field Following the same procedure of eliminating the small component, φ of the Lorentz spinor and getting the equation for the large component, χ, we obtain (E −mc2)χ = c 2 E +mc2 (σ ·Π)(σ ·Π)χ. (15) This then readily gives the following relativistically exact equation for χ: (E2 −m2ct)χ = c2(σ ·Π)(σ ·Π)χ, (16) (E2 −m2ct)χ = [ c2(p− qA/c)2 − q~c σ ·B ] χ, (17) where B = B0ẑ is the constant magnetic field. Now, clearly, for a constant magnetic field the above equation is no more difficult to solve than the ordinary Schrodinger’s equation. Recall from the nonrelativistic theory that in 3d we have 1 2m (p− qA/c)2ψn,ky,kz = Ẽn,kzψn,ky,kz . (18) where in the Landau gauge we have Ẽn,kz = ~2k2z/2m + ~ωc(n + 1/2), with the cyclotron frequency ωc = qB0/(mc), and the standard degeneracy with respect to the quantum number ky. Using this result we can easily diagonalize the first term of the right hand side of the above equation for χ, obtaining (E2 −m2c4)χn,ky,kz = [ c22mẼn,kz − q~ c σ ·B ] χn,ky,kz , (19) 2
