Lecture 4 - Quantum mechanics of a two-level system

by Leonid Pryadko

Initialization

Definition of Pauli matrices

(Subscript[σ, x])={{0,1},{1,0}};(Subscript[σ, y])={{0,-I},{I,0}};(Subscript[σ, z])={{1,0},{0,-1}};(Subscript[σ, 0])={{1,0},{0,1}};

(Subscript[σ, y])//MatrixForm

WF will be a two-component spinor

ψ={y1[t],y2[t]};

Equation generator from Lecture 3

DoEq[x_,psi_,psi0_]:=Block[{i,y={}},
For[i=1,i≤Length[x],i++,y=Append[y,(x[[i]]==0)]];
For[i=1,i≤Length[psi],i++,y=Append[y,(psi[[i]]==psi0[[i]])]];
y]

Hamiltonian

Most general Hermitian 2x2 matrix

(Subscript[H, 0])={{(Subscript[a, 0])+(Subscript[a, 3]),(Subscript[a, 1])-I (Subscript[a, 2])},{(Subscript[a, 1])+I (Subscript[a, 2]),(Subscript[a, 0])- (Subscript[a, 3])}};(Subscript[H, 0])//MatrixForm

Same can be written in terms of Pauli Matrices:

H=(Subscript[a, 0])(Subscript[σ, 0])+(Subscript[a, 1])(Subscript[σ, x])+(Subscript[a, 2])(Subscript[σ, y])+(Subscript[a, 3])(Subscript[σ, z]);H//MatrixForm

Standart interpretation: spin-1/2.  s=1/2 σ  is the spin operator.  In this representation two-component wavefunction ψ gives the amplitudes of spin-up and spin-down states.  Vector {(Subscript[a, 1]),(Subscript[a, 2]),(Subscript[a, 3])} is proportional to the magnetic field, a=B/2.  The value (Subscript[a, 0]) defines the overall energy origin.

Basis change

Eigenvalues[H]

This just tells us that it is the total magnetic field that is relevant.  We might want to rotate our coordinates properly.

How to make a rotation?  Method 1: use eigenvectors of the Hamiltonian for the new basis.

vecs=Eigenvectors[H]//FullSimplify

{{(((Subscript[a, 3])-(Sqrt[(Subscript[a, 1]^2)+(Subscript[a, 2]^2)+(Subscript[a, 3]^2)]))/((Subscript[a, 1])+I (Subscript[a, 2]))),1},{(((Subscript[a, 3])+(Sqrt[(Subscript[a, 1]^2)+(Subscript[a, 2]^2)+(Subscript[a, 3]^2)]))/((Subscript[a, 1])+I (Subscript[a, 2]))),1}}

ss=FullSimplify[ConjugateTranspose[Eigenvectors[H]],Assumptions->{(Subscript[a, 1])>0,(Subscript[a, 2])>0,(Subscript[a, 3])>0}]

FullSimplify[ConjugateTranspose[ss].ss,Assumptions->{(Subscript[a, 1])>0,(Subscript[a, 2])>0,(Subscript[a, 3])>0}]//MatrixForm

Need to normalize the Eigenvectors.

HQ=FullSimplify[H/.{(Subscript[a, 1])->Sin[θ]Cos[ϕ],(Subscript[a, 2])->Sin[θ] Sin[ϕ],(Subscript[a, 3])->Cos[θ]}];HQ//MatrixForm

Eigenvalues[HQ]//Simplify

Eigenvectors[HQ]//Simplify

v1=Transpose[ssQ][[1]];v1=FullSimplify[v1/Norm[v1],Assumptions-> {θ>0,ϕ>0,θ<π}]

v2=Transpose[ssQ][[2]];v2=FullSimplify[v2/Norm[v2],Assumptions-> {θ>0,ϕ>0,θ<π}]

Correctly normalized matrix is

UQ=FullSimplify[ConjugateTranspose[{v1,v2}],Assumptions-> {θ>0,ϕ>0,θ<π}];UQ//MatrixForm

FullSimplify[ConjugateTranspose[UQ].HQ.UQ,Assumptions-> {θ>0,ϕ>0}]

General property of any Hermitian matrix -- it can be diagonalized.

Schrodinger equation with arbitrary initial conditions

Solve equations for the wavefunction

ψ

This is for IC {1,0} (spin up):

eqs1=DoEq[I D[ψ,t]-H.ψ,ψ/.t->0,{1,0}]/.(Subscript[a, 0])->0

sol1=DSolve[eqs1,ψ,t];sol1=FullSimplify[sol1,Assumptions->(Subscript[a, 1])>0&&(Subscript[a, 2])>0 && (Subscript[a, 3])>0]

This is for IC {0,1} (spin down):

eqs2=DoEq[I D[ψ,t]-H.ψ,ψ/.t->0,{0,1}]/.(Subscript[a, 0])->0

sol2=DSolve[eqs2,ψ,t];sol2=FullSimplify[sol2,Assumptions->(Subscript[a, 1])>0&&(Subscript[a, 2])>0 && (Subscript[a, 3])>0]

General solution is a linear combination

Cup ({y1[t],y2[t]}/. sol1 [[1]])+Cdn {y1[t],y2[t]}/. sol2 [[1]]

FullSimplify[%/.{Cup->1/Sqrt[2],Cdn->1/Sqrt[2]}]

Instead, we can solve equations for the unitary matrix

U={{u1[t],u2[t]},{u3[t],u4[t]}};

For[i=1;ans={},i≤9,i++,ans=Append[ans,ToExpression["u"<>ToString[i]<>"[t]"]]];Partition[ans,3]

lhs=I D[U,t]-H.U

Flatten[lhs]

Flatten[{{a,b},{c,d}}]

eqs=DoEq[Flatten[lhs],Flatten[U]/.t->0,IdentityMatrix[2]//Flatten]

eqs=eqs/.{(Subscript[a, 0])->0,(Subscript[a, 3])->0,(Subscript[a, 2])->0}

{-(Subscript[a, 1]) u3[t]+I u1'[t]==0,-(Subscript[a, 1]) u4[t]+I u2'[t]==0,-(Subscript[a, 1]) u1[t]+I u3'[t]==0,-(Subscript[a, 1]) u2[t]+I u4'[t]==0,u1[0]==1,u2[0]==0,u3[0]==0,u4[0]==1}

sol=DSolve[eqs,Flatten[U],t]

MatrixForm[U/.sol[[1]]]

Spin precession (Heisenberg representation)

Consider what happens to the components of the average spin

FullSimplify[(ConjugateTranspose[U].(Subscript[σ, x]).U)/.sol[[1]],Assumptions->t>0&&(Subscript[a, 1])>0]//MatrixForm

FullSimplify[(ConjugateTranspose[U].(Subscript[σ, y]).U)/.sol[[1]],Assumptions->t>0&&(Subscript[a, 1])>0]//MatrixForm

FullSimplify[(ConjugateTranspose[U].(Subscript[σ, z]).U)/.sol[[1]],Assumptions->t>0&&(Subscript[a, 1])>0]//MatrixForm

Evolution of the density matrix via U

write the initial density matrix as

ρ[0]=1/2((Subscript[σ, 0])+(Subscript[R, x])(Subscript[σ, x])+(Subscript[R, y])(Subscript[σ, y])+(Subscript[R, z])(Subscript[σ, z]));ρ[0]//MatrixForm

ρ[t]=FullSimplify[U.ρ[0].ConjugateTranspose[U]/.sol[[1]],Assumptions->t>0&&(Subscript[a, 1])>0]

Average spin components

FullSimplify[{Tr[ρ[t].(Subscript[σ, x])],Tr[ρ[t].(Subscript[σ, y])],Tr[ρ[t].(Subscript[σ, z])]}]

Precession around the field B.

Evolution equation for the density matrix

RR=1/2((Subscript[σ, 0])+(Subscript[R, x])[t](Subscript[σ, x])+(Subscript[R, y])[t](Subscript[σ, y])+(Subscript[R, z])[t](Subscript[σ, z]))

Simplify[H.RR-RR.H]//MatrixForm

I D[RR,t]-(H.RR-RR.H)//FullSimplify

Alternative set

-I{Tr[(H.RR-RR.H).(Subscript[σ, x])],Tr[(H.RR-RR.H).(Subscript[σ, y])],Tr[(H.RR-RR.H).(Subscript[σ, z])]}//FullSimplify

Rotation around the vector a