PHYS 517: 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 state . If S is y i 0 β y measured, what is the probability of the result ℏ/2? Solution: Suppose the eignevalues of the 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 √ √ Normalizing this eivenvector we have the normalization factor 12 + 12 = 2. So the required normalized eigenvector corresponding to λ = 1 is ( ) √1 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 √ √ Normalizing this eivenvector we have the normalization factor 12 + 12 = 2. So the required normalized eigenvector corresponding to λ = 1 is ( ) √1 1 2 −i So the eigenvectors corresponding to each eigenvalues are ( ) ( ) 1 1 1 1 λ = 1 → √ λ = −1 → √ 2 i 2 −i ( ) α Let the arbitrary spin state be γ = . Since the matrix representation of the S is ℏ σ . The β y 2 y ℏ probability that the state be measure to be in Sy with eigenvalue 2 is ( )( ) ( ) ( ) ℏ 0 −i α ℏ ( ) −iβ iℏ ⟨γ|ℏ/2σ |γ⟩ = α∗ β∗ = α∗ β∗ = (−βα∗ + β∗α) 2 2 i 0 β 2 iα 2

| ⟩ iℏ ∗ − ∗ □ 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 , the eigenvalues for Hamiltonian

−2µ · H = ℏ S B 1 for a spin 2 particle in the presence a magnetic B = Bxˆx + By ˆy + Bzˆz. Solution: The hamiltonian operator in the given magnetic field as −2µ H = ℏ (SxBx + SyBy + SzBz)

Since the spin operators Sx,Sy and Sz are the pauli matrices with a factor of ℏ/2 we can write the above expression as [( ) ( ) ( ) ] 2µ ℏ 0 1 0 −i 1 0 H = − B + B + B ℏ 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 ⇒ | | ((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 = ∓µ|B|. □

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 ∑ ∑ † a0 + i ajσj a0 − i ajσ U = ∑j U † = ∑j j a − i a σ − † 0 j j j a0 i j ajσj Multiplying these two to check for unitarity ∑ † ∑ a − i a σ a + i a σ † 0 ∑j j j · 0 ∑j j j U U = † − a0 − i ajσ a0 i j ajσj ∑j j ∑ ∑ ∑ 2 † † a + ia0 σjaj − ia0 σ aj + σ ajσkak = 0 ∑j ∑j j ∑j ∑k j 2 − † † a0 ia0 j σjaj + ia0 j σj aj + j k σj ajσkak † 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 ∑ ∑ ∑ ∑ 2 a + ia0 σjaj − ia0 σjaj + σjajσkak U †U = 0 ∑j ∑j ∑j ∑k = 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 represents a 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 ˆn = nxˆx + ny ˆy + nzˆz 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 |a| y |a| z |a|

This gives the rotation angle and the direction of rotation for this given unimodular matrix. □

4. (Sakurai 3.9) Consider a sequence of 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 ( ( ) ( )) ( ) ( ( ) ( )) ( ) cos α+γ + i sin α+γ cos β − cos α+γ − i sin α+γ cos β D1/2(α, β, γ) = ( ( 2 ) ( 2 )) ( 2 ) ( ( 2 ) ( 2 )) ( 2 ) 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 . 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 . Using the fact that Jx,Jy,Jz and J ≡ Jx  Jy satisfy the usual angular-momentum commutation relations, prove

2 2 − ℏ 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 − ℏ = Jx + Jy i(i Jz) 2 − 2 ℏ = J Jz + Jz

2 2 − ℏ □ Rearranging above expression gives J = Jz + J+J− Jz which completes the proof.

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