Homework assignment 1 - Solution - s...
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Homework assignment 1 - Solution 1. Show that the Pauli matrices σ z ,σ x and σ y correspond to operators measuring the polarisation state of light in the horizontal-vertical, diagonal- antidiagonal and right-lefthand circular basis, respectively. Hint: Find the eigenvectors of the Pauli matrices and compare them to the Jones vectors of polarised light. eigenvalues of σ x : Using characteristic equation det(σ x -λI ) = 0, where I is the identity matrix. So we have: det -λ 1 1 -λ ! = 0. Hence λ 2 - 1=0 ⇒ λ = ±1. Eigenvector for λ 1 = 1: σ x x y ! =1 x y ! It follows that x = y and y = x. we chose x = 1 and have the following eigenvector: ν 1 = 1 1 ! . To normalise the eigenvector we demand ν 2 1 = 1. The normalised eigenvector for λ 1 is hence ν 1 = 1 √ 2 1 1 ! . For eigenvector ν 2 with eigenvalue λ 2 we find: x = -y and y = -x. The normalised eigenvector is hence ν 2 = 1 √ 2 1 -1 ! . Similarly for σ y we have: λ 1 = 1 and λ 2 = -1 with normalised eigen- vectors: ν 1 = 1 √ 2 1 i ! and ν 2 = 1 √ 2 1 -i ! . Similarly for σ z we have: λ 1 = 1 and λ 2 = -1 with normalised eigen- vectors: ν 1 = 1 0 ! and ν 2 = 0 1 ! . From the lecture we know that a general polarisation state can be writ- ten as a Jones vector: - → J = cos(α) sin(α) e iΔφ ! , where α is the angle of electric field vector from the horizontal and Δφ is the phase difference
Transcript of Homework assignment 1 - Solution - s...
Homework assignment 1 - Solution
1. Show that the Pauli matrices σz, σx and σy correspond to operatorsmeasuring the polarisation state of light in the horizontal-vertical, diagonal-antidiagonal and right-lefthand circular basis, respectively. Hint: Findthe eigenvectors of the Pauli matrices and compare them to the Jonesvectors of polarised light.eigenvalues of σx : Using characteristic equation det(σx−λI) = 0, where
I is the identity matrix. So we have: det
(−λ 11 −λ
)= 0. Hence
λ2 − 1 = 0⇒ λ = ±1. Eigenvector for λ1 = 1: σx
(xy
)= 1
(xy
)It follows that x = y and y = x. we chose x = 1 and have the following
eigenvector: ν1 =
(11
). To normalise the eigenvector we demand
ν21 = 1. The normalised eigenvector for λ1 is hence ν1 = 1√
2
(11
). For
eigenvector ν2 with eigenvalue λ2 we find: x = −y and y = −x. The
normalised eigenvector is hence ν2 = 1√2
(1−1
).
Similarly for σy we have: λ1 = 1 and λ2 = −1 with normalised eigen-
vectors: ν1 = 1√2
(1i
)and ν2 = 1√
2
(1−i
).
Similarly for σz we have: λ1 = 1 and λ2 = −1 with normalised eigen-
vectors: ν1 =
(10
)and ν2 =
(01
).
From the lecture we know that a general polarisation state can be writ-
ten as a Jones vector:−→J =
(cos(α)
sin(α) ei∆φ
), where α is the angle of
electric field vector from the horizontal and ∆φ is the phase difference
between the horizontal and vertical component.For α = 0 we have horizontally polarised light represented by
−→JH =(
10
), similarly for α = 90◦ we have vertically polarised light repre-
sented by−→JV =
(01
). Hence σz corresponds to measurements in the
horizontal-vertical basis.Linear polarised light at +45◦ has α = 45◦ and ∆φ = 0 resulting in
the following Jones vector:−→JP = 1√
2
(11
). For linear polarised light
at −45◦ we get:−→JM = 1√
2
(1−1
). Hence σx corresponds to mea-
surements in the diagonal-antidiagonal basis. Finally a right-handedcircular polarised light beam has α = 45◦ and ∆φ = π
2resulting in:
−→JR = 1√
2
(1i
). For left-handed circular light we get:
−→JL = 1√
2
(1−i
).
Hence σy corresponds to measurements in the right and left circularbasis.
2. Calculate the expectation values of σz, σx and σy for the following twostates:
(a) A pure state polarised along +45◦ degree. |P 〉 = 1√2
(|H〉+ |V 〉)
< σx >= 〈P |σx|P 〉. In vector form 〈P 〉 is written as 1√2
(11
).
So, < σx >= 1√2
(1, 1) ·(
0 11 0
)· 1√
2
(11
)= 1
2(1, 1) ·
(11
)= 1.
Similarly for:< σy >= 〈P |σy|P 〉 = 0 and< σz >= 〈P |σz|P 〉 = 0.
(b) A mixed state from a source emitting randomly states with hor-izontal and vertical polarisation. Hint: use the density matrix|ρ〉 = 1
2|H〉〈H|+ 1
2|V 〉〈V |.
For a mixed state we have:< σx >= tr(ρ σx).
The density matrix for the state is given by:ρ = 12
(1, 0) ·(
10
)+
2
12
(0, 1) ·(
01
)= 1
2
(1 00 1
). So,
< σx >= tr
(12
(1 00 1
)·(
0 11 0
))= 0.
Similarly for:< σy >= tr(ρ σy) = 0 and< σz >= tr(ρ σz) = 0.
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