Problem 5 a Before we solve the problem in the presence of a finite vector

Problem 5 a before we solve the problem in the

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Problem 5:(a.) Before we solve the problem in the presence of a finite vector potential, let us consider thecase of~A= 0.In this case, it is straightforward to write the wavefunction for the electronicground-state. It is given by,ψ0(~r) =Ne-|~r-~R|,(22)whereσ=aH/Zand the the Bohr-radius,aH=~2/(Zme2). In the above,Nis a normalizationcondition, chosen to be real, that can be computed easily.Let us now consider a finiteBandchoose the circular gauge:~A=B2(-y, x,0).(23)The electronic contribution to the Hamiltonian then becomes,Hel=|~p+ec~A(~r)|22m-2e2|~r-~R|=~p22m-2e2|~r-~R|+eB2mcLz+e2B28mc2(x2+y2).(24)Now if the nucleus was fixed at the origin, so that~R= 0, then~r2=x2+y2a2Hand we couldignore the last term in the Hamiltonian above for smallB. The ground state ofHelis then also aneigenstate of the angular momentum withLz=~L2= 0 and there is no change inψ0to first orderinB.However, if the nucleus is far away from the origin (which is what we have been told in theproblem), then these arguments don’t apply.They would have applied if we used the gauge,
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5~A0= (B/2)(Y-y, X-x,0) (i.e.working in the relative coordinates), thus replacingLzandx2+y2by the angular momentum and distance relative to the nucleus at~R. In this gauge, theelectronic ground state would be given by the wavefunctionψ0to first order inB. However, thegauge one needs to use is~A, which is related to~A0by,~A(~r) =~A0(~r) +χ(~r),where(25)χ(~r) =~A(~R)·~r,or,χ(~r) = (B/2)(-Y, X,0).(26)The electronic ground state in the gauge~A(~r) then becomes,ψ0(~r,~R) =Ne-|~r-~R|e-ieχ/(~c),where(27)χ=B2(Xy-Y x).(28)(b.) In order to compute the effective HamiltonianHeff, we need to compute the Berry connection,~Aeff(~R), where,~Aeff(~R) =i~ψel∂ψel~R=-~Imψel∂ψel~R.(29)Let us consider theXcomponent of the quantity:∂ψel∂X=-ie~cBy2ψel+e-ieχ/(~c)N∂X(e-|~r-~R|).(30)Now notice that in order to compute the imaginary part of the matrix element in Eqn.29, thesecond term above drops out. Therefore, we are left withZψel*∂ψel∂Xd3r=-ieB2c~Zy|ψel|2d3r+ real pieces =--Be2~cY+ real pieces.(31)Therefore, we find,Axeff=eBY2c=-ecAx(~R),and similarlyAyeff=-eBX2c=-ecAy(~R), Azeff= 0.(32)Finally, we have,Heff=12M|~P-2ec~A(~R)-~Aeff(~R)|2=12M|~P-ec~A|2.(33)Notice that this is the Hamiltonian for a particle with chargeein an applied magnetic field,B,as is expected for an ion with total chargee! It is important to note that we have assumed theelectron mass to be much smaller than the nuclear mass and ignored higher order terms inm/M.
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6If one were to include these terms, the nuclear mass would be replaced by the total mass,M+min the effective Hamiltonian.
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