Transcription of Spin Eigenstates - Review
1 spin Eigenstates - ReviewDr. R. HermanPhysics & Physical Oceanography, UNCWS eptember 20, 2019SG Devices Measure SpinIOrient device in directionnIThe representation of| intheSn-basis for spin12:| n=In| ,whereIn=|+n +n|+| n n|| n=|+n +n| +| n n| =a+|+n +a | n ( +n| n| )IProb(|+n ) =| +n| |2 Representation of OperatorsMatrix Representation of AinSn-basis A An=( +n| A|+n +n| A| n n| A|+n n| A| n )Matrix Representations A An=S AzS,whereS=( +z|+n +z| n z|+n z| n )andAz=( +z| A|+z +z| A| z z| A|+z z| A| z )Change of Basis (zton) - StatesTransform Kets| z=|+z +z| +| z z| = (|+n +n|+| n n|)|+z +z| + (|+n +n|+| n n|)
2 | z z| = [ +n|+z +z| + +n| z z| ]|+n + [ n|+z +z| + n| z z| ]| n Matrix Representation| n= S | z( +n| n| )=( +n|+z +n| z n|+z n| z ) Components ofzstates( +z| z| ) S ( +z| z| )Change of Basis (zton) - OperatorsBegin with States| n= S | z,n |=z | | S,whereS =( +n|+z +n| z n|+z n| z )Relate | A| inz-basis to value inn-basis, using S S = S S= | Az| z=z | S S Az S S | z=n | S Az S | n=n | An| n(1)If we define Anew= Anand Aold= Az,then Anew= S Aold SAngular Momentum OperatorsRotations and Generators ( Jnis Hermitian) R( n) =e i Jn /~Commutation Relations[ Jx, Jy] =i~ Jz,[ Jy, Jz] =i~ Jx,[ Jz, Jx] =i~ J2= J2x+ J2y+ J2z, J = Jx i JyEigenstates (2j+ 1) for j, j+ 1.
3 ,j 1,j J2|j,m =j(j+ 1)~2|j,m Jz|j,m =m~|j,m J |j,m = j(j+ 1) m(m 1)~|j,m 1 Spin12 RepresentationsIOperators S2, Sz, S IStates|12,12 ,|12, 12 ,IRepresentations S ~2 S+=~(0 10 0),S =~(0 01 0)Sx=S++S 2,Sy=S+ S 2iSx=~2(0 11 0),Sy=~2(0 ii0),Sz=~2(100 1)IExpectation Values Sx = | Sx| =( +z| z| )~2(0 11 0)( +z| z| ) spin 1 RepresentationsIOperators S2, Sz, S IStates|1,1 ,|1,0 ,|1, 1 ,IRepresentations Sz=~ 1 000 000 0 1 , S+= 2~ 0 1 00 0 10 0 0 Sx=S++S 2,Sy=S+ S 2iIRepresentation of A.
4 A 1,1| A|1,1 1,1| A|1,0 1,1| A|1, 1 1,0| A|1,1 1,0| A|1,0 1,0| A|1, 1 1, 1| A|1,1 1, 1| A|1,0 1, 1| A|1, 1 Representation of Sx- spin 1 CaseNoting that Sx=12( S++ S ),Aii= 0,i= 1,2,3,andA12= 1,1| Sx|1,0 =A 21=12 1,1| S++ S |1,0 =12( 2 1,1|1,1 + 2 1,1||1, 1 )=1 1,1| Sx|1, 1 =A 31=12 1,1| S++ S |1, 1 = 1,0| Sx|1, 1 =A 32=12 1,0| S++ S |1, 1 =1 final representation is .. Eigenstates of Sx- spin 1 CaseFind the Eigenstates ofSxinSz-basis~ 2 0 1 01 0 10 1 0 abc = ~ abc Eigenvalues [ = 1,0,1] 1/ 201/ 2 1/ 201/ 2 = 0 ( 2 12)+12 = 0 Eigenvectors for = 11 2 0 1 01 0 10 1 0 abc = abc b= 2a= 2c,a+c= 2b, |1, 1 x Sz12 1 21 spin 1 Particles - SG DevicesSend spin 1 particles through 3 Stern Gerlach devices.
5 | SGz|1, 1 z|1,0 z|1,1 zSGx|1, 1 x|1,0 x|1,1 xSGz|1, 1 z|1,0 z|1,1 z=?Probability [to find| in state| ] =| | |2,|z 1,1| |2,|x 1, 1|1,1 z|2,|z 1,1|1, 1 x|2,Example:Evaluate|x 1, 1|1,1 z| that|1, 1 xis an eigenstate ofSx:Therefore,|1, 1 x Sz12 1 21 |x 1, 1|1,1 z|2= EvolutionThe Schr odinger Equationi~ddt| (t) = H| (t) For Htime-independent,| (t) = U| (0) , U=e i H t/~Energy Eigenstates H|E =E|E ,Time Evolution: Initial State| (0) = n|En En| (0) | (t) = ne i Ent/~|En En| (0) Expectation Values:i~ddt A = (t)|[ H, A] + A t| (t) Precession in a Magnetic FieldConstant Magnetic FieldB=B0k H= B=ge2mcB0Sz 0 SzEvolution of States H| z = ~ 02| z | (t) =e i H t/~|+x =1 2(e iE+t/~|+z +e iE t/~| z )=1 2(e i 0t/2|+z +ei 0t/2| z )Expectation values Sz = 0, Sx =~2cos 0t, Sy =~2sin 0tUncertaintyFor Hermitian matrices A, B,[ A, B] =i C, A B | C |2 Example.
6 Jx Jy ~2| Jz |Recall A2= A2 A 2and[ A, B] = A B B spin SystemsBasis States for spin -12 Particles|+z,+z =|1212,1212 |+x,+z =1 2|+z,+z +1 2| z,+z Hyperfine Splitting H=2A~2 S1 S2 H=A~2( S1+ S2 + S1 S2++ 2 S1z S2z) A20000 A2A00A A20000A2 (2)Eigenvalue ProblemSeek Eigenvalues and Energy Eigenstates H|E =E|E . A20000 A2A00A A20000A2 abcd =E abcd .Eigenvalue Equation0 = A2 E0000 A2 EA00A A2 E0000A2 E =(A2 E)2[(E+A2)2 A2]Energy EigenstatesFor eigenvaluesE=A2,we get the triplet 1000 ,1 2 0110 , 0001 ,or|1,1 =|+z,+z ,|1,0 =1 2|+z, z +1 2| z,+z ,|1, 1 =| z, z.
7 For eigenvaluesE= 3A2,we get the singlet1 2 01 10 ,or|0,0 =1 2|+z, z 1 2| z,+z .EPR Paradox -|0,0 DecayzxSpin121 SGAliceSpin122 SGBobA spin -0 particle decays into two spin -12particles.|0,0 =1 2|+z, z 1 2| z,+z =1 2|+z 1| z 2 1 2| z 1|+z do Alice and Bob measure?
