PHYS 517: Quantum Mechanics II Homework #2 Prakash Gautam April 26, 2018 ( ) ( ) 0 −i α 1. (Sakurai 3.1) Find the eigenvectors of σ = . Suppose an electron is in spin state . If S is y i 0 β y measured, what is the probability of the result ~=2? Solution: Suppose the eignevalues of the matrix are λ. The characterstics equation for the matrix is (0 − λ)(0 − λ) − (−i · i) = 0 ) λ = 1 ( ) x Let the eigenvector be . Then the eigenvector corresponding to λ = 1 we have y ( )( ) ( ) 0 −i x x −iy = x x = 1 = λ ) ) i 0 y y ix = y y = i p p Normalizing this eivenvector we have the normalization factor 12 + 12 = 2. So the required normalized eigenvector corresponding to λ = 1 is ( ) p1 1 2 i Then the eigenvector corresponding to λ = −1 we have ( )( ) ( ) 0 −i x x −iy = −x x = 1 = λ ) ) i 0 y y ix = −y y = −i p p Normalizing this eivenvector we have the normalization factor 12 + 12 = 2. So the required normalized eigenvector corresponding to λ = 1 is ( ) p1 1 2 −i So the eigenvectors corresponding to each eigenvalues are ( ) ( ) 1 1 1 1 λ = 1 ! p λ = −1 ! p 2 i 2 −i ( ) α Let the arbitrary spin state be γ = . Since the matrix representation of the S operator is ~ σ . The β y 2 y ~ probability that the state be measure to be in Sy with eigenvalue 2 is ( )( ) ( ) ( ) ~ 0 −i α ~ ( ) −iβ i~ hγj~=2σ jγi = α∗ β∗ = α∗ β∗ = (−βα∗ + β∗α) 2 2 i 0 β 2 iα 2 j i i~ ∗ − ∗ □ So the probability of measuring the given state in Sy; + state is 2 (αβ α β). 1 2. (Sakurai 3.2) Find, by explicit construction using Pauli matrices, the eigenvalues for Hamiltonian −2µ · H = ~ S B 1 for a spin 2 particle in the presence a magnetic B = Bx^x + By ^y + Bz^z. Solution: The hamiltonian operator in the given magnetic field as −2µ H = ~ (SxBx + SyBy + SzBz) Since the spin operators Sx;Sy and Sz are the pauli matrices with a factor of ~=2 we can write the above expression as [( ) ( ) ( ) ] 2µ ~ 0 1 0 −i 1 0 H = − B + B + B ~ 2 1 0 x i 0 y 0 −1 z ( ) B B − iB = −µ z x y Bx + iBy −Bz The characterstics equation for the this matrix is − − − − − ) 2 − 2 − 2 2 ) | j ((Bz λ)( Bz λ) (Bx iBy)(Bx + iBy)) = 0 λ Bz (Bx + By ) = 0 λ = B So the eigenvalue of the Hamiltonian which is −µ times the matrix is −µ · λ = ∓µjBj. □ 3. (Sakurai 3.3) Consider 2 × 2 matrix defined by a + iσ · a U = 0 a0 − iσ · a where a0 is a real number and a is a three-dimensional vector with real components. (a) Prove that U is unitary and unimodular. Solution: Given matrix U and hermitian conjugate can be written as P P y a0 + i ajσj a0 − i ajσ U = Pj U y = Pj j a − i a σ − y 0 j j j a0 i j ajσj Multiplying these two to check for unitarity P y P a − i a σ a + i a σ y 0 Pj j j · 0 Pj j j U U = y − a0 − i ajσ a0 i j ajσj Pj j P P P 2 y y a + ia0 σjaj − ia0 σ aj + σ ajσkak = 0 Pj Pj j Pj Pk j 2 − y y a0 ia0 j σjaj + ia0 j σj aj + j k σj ajσkak y Since each pauli matrices are Hermitian, for each i we have σi = σi. This makes the numerator the exact same as the denominator. Thus they cancel out P P P P 2 a + ia0 σjaj − ia0 σjaj + σjajσkak U yU = 0 Pj Pj Pj Pk = 1 2 − a0 ia0 j σjaj + ia0 j σjaj + j k σjajσkak This shows that this matrix is unitary. Expanding out the matrix in terms of the pauli matrices we get a0 + ia3 ia1 + a2 2 2 2 2 ia1 − a2 a0 − ia3 (a + ia )(a − ia ) − (ia + a )(ia − a ) a + a + a + a det U = = 0 3 0 3 1 2 1 2 = 0 1 2 3 = 1 − − − − − − − 2 2 2 2 a0 ia3 ia1 + a2 (a0 ia3)(a0 + ia3) ( ia1 + a2)( ia1 a2) a0 + a1 + a2 + a3 −ia1 − a2 a0 + ia3 This shows that the matrix is unimodular. □ 2 (b) In general, a 2 × 2 unitary unimodular matrix represents a rotation in three dimensions. Find the axis and the angle of rotation appropriate for U in terms of a0; a1; a2 and a3. Solution: The matrix can be rewritten as ( ) 1 a − a2 + 2ia a 2a a + 2ia a U = 0 0 3 0 2 0 1 2 2 − − 2 − a0 + a 2a0a2 + 2ia0a1 a0 a 2ia0a3 ( ) a b Since the most general unimodular matrix of the form represent a rotaion through an −b∗ a∗ angle ϕ through the direction ^n = nx^x + ny ^y + nz^z related as ( ) ( ) ϕ ϕ Re(a) = cos ; Im(a) = −n sin (1) 2 z 2 ( ) ( ) ϕ ϕ Re(b) = −n sin ; Im(b) = −n sin (2) y 2 x 2 Making these comparision in this matrix we get ( ) ( ) 2 − 2 2 − 2 ϕ a0 a ) a0 a cos = 2 2 ϕ = 2 acos 2 2 2 a0 + a a0 + a And similarly we get a a a n = − 1 n = − 2 n = − 3 x jaj y jaj z jaj This gives the rotation angle and the direction of rotation for this given unimodular matrix. □ 4. (Sakurai 3.9) Consider a sequence of rotations represented by ( ) ( ) ( ) ( ) −iσ α −iσ β −iσ γ e−i(α+γ)=2 cos β −e−i(α+γ)=2 cos β D(1=2) 3 2 3 2 2 (α; β; γ) = exp exp exp = −i(α−γ)=2 β i(α+γ)=2 β 2 2 2 e sin 2 e cos 2 Solution: Again this final matrix can be written as a coplex form as 2( ( ) ( )) ( ) ( ( ) ( )) ( )3 cos α+γ + i sin α+γ cos β − cos α+γ − i sin α+γ cos β D1=2(α; β; γ) = 4( ( 2 ) ( 2 )) ( 2 ) ( ( 2 ) ( 2 )) ( 2 )5 α+γ α+γ β − α+γ − α+γ β cos 2 + i sin 2 cos 2 cos 2 i sin 2 cos 2 Let ϕ be the angle of rotation represented by this final rotation matrix. Using again the equations (1) we get ( ) ( ) ( ) [ ( ) ( )] ϕ α + γ β α + γ β cos = cos cos ) ϕ = 2 cos−1 cos cos 2 2 2 2 2 This gives the angular rotation value for this matrix. The direction of rotation can similarly be found by using (1) to calculate the directions. □ 5. (Sakurai 3.15a) Let J be angular momentum. Using the fact that Jx;Jy;Jz and J ≡ Jx Jy satisfy the usual angular-momentum commutation relations, prove 2 2 − ~ J = Jz + J+J− Jz . Solution: 3 Multiplying out J+ and J− we get J+J− = (Jx + iJy)(Jx − iJy) 2 − 2 = Jx iJxJy + iJyJx + Jy 2 − 2 = Jx i[Jx;Jy] + Jy 2 2 − ~ = Jx + Jy i(i Jz) 2 − 2 ~ = J Jz + Jz 2 2 − ~ □ Rearranging above expression gives J = Jz + J+J− Jz which completes the proof. 4.