Uitgewerkte opdrachten - Quantum Engineering & Applications ( TN2306) - Minor Modern Physics

TN2306 UITWERKINGEN WERKCOLLEGES




Quantum Engineering and
Applications
TN2306

Uitwerkingen opgaven




Pagina 1 van 33

, TN2306 UITWERKINGEN WERKCOLLEGES


Inhoudsopgave
Inhoudsopgave 2
College 1 De qubit 3
College 2 2 qubits 9
College 3 Kwantumcircuits 14
College 4 Kwantumcomputer 18
College 5 Kwantumencryptie 21
College 6 Kwantumhardware I 30
College 7 Kwantumalgoritmen 33




Pagina 2 van 33

, TN2306 UITWERKINGEN WERKCOLLEGES


College 1 De qubit
Problem 1
a) Calculate the eigenvalues and eigenstates of the following matrices:
⎛ 1 0 ⎞ ⎛ 0 1 ⎞ ⎛ 0 −i ⎞
σz = ⎜ σx = ⎜ σy = ⎜
⎝ 0 −1 ⎟⎠ ⎝ 1 0 ⎟⎠ ⎝ i 0 ⎟⎠
b) Show that the three matrices above are hermitian and unitary.
c) Calculate the commutator between the three matrices above

A.
σz
⎧λ1 = 1
(1− λ )( −1− λ ) = 0 → ⎨
⎩λ2 = −1
⎛ 1 0 ⎞⎛ α ⎞ ⎛ α ⎞ ⎧α = α ⎛ 1 ⎞
⎜⎝ 0 −1 ⎟⎠ ⎜ β ⎟ = 1• ⎜ β ⎟ → ⎨− β = β → ν1 = ⎜ 0 ⎟ = 0
⎝ ⎠ ⎝ ⎠ ⎩ ⎝ ⎠
⎛ 1 0 ⎞⎛ α ⎞ ⎛ α ⎞ ⎧α = −α ⎛ 0 ⎞
⎜
⎜⎝ 0 −1 ⎟⎠ β ⎟ = −1• ⎜ ⎟ → ⎨ → ν = ⎜⎝ 1 ⎟⎠ = 1
⎝ β ⎠ ⎩− β = − β
2
⎝ ⎠

σx
⎧λ1 = 1
λ2 −1 = 0 → ⎨
⎩λ2 = −1
⎛ 0 1 ⎞⎛ α ⎞ ⎛ α ⎞ ⎧β = α 1 ⎛ 1 ⎞
⎜⎝ 1 0 ⎟⎠ β⎜ ⎟ = 1• ⎜ ⎟ → ⎨ → ν = ⎜ ⎟ =
1
(0 +1 )
⎝ β ⎠ ⎩α = β
1
⎝ ⎠ 2⎝ 1 ⎠ 2
⎛ 0 1 ⎞⎛ α ⎞ ⎛ α ⎞ ⎧β = −α ⎛ 1 ⎞
⎜⎝ 1 0 ⎟⎠ ⎜ β ⎟ = −1• ⎜ β ⎟ → ⎨a = − β → ν 2 = ⎜⎝ −1 ⎟⎠ = 2 ( 0 − 1 )
1
⎝ ⎠ ⎝ ⎠ ⎩

σy
⎧λ1 = 1
λ 2 − ( −i ) • i = λ 2 − 1 = 0 → ⎨
⎩λ2 = −1
⎛ 0 −i ⎞ ⎛ α ⎞ ⎛ α ⎞ ⎧−iβ = α 1 ⎛ i ⎞
⎜⎝ i 0 ⎟⎠ ⎜ β ⎟ = 1• ⎜ β ⎟ → ⎨iα = β → ν1 = ⎜ ⎟
2 ⎝ −1 ⎠
=
1
(i 0 − 1 )
⎝ ⎠ ⎝ ⎠ ⎩ 2
⎛ 0 −i ⎞ ⎛ α ⎞ ⎛ α ⎞ ⎧−iβ = −α 1 ⎛ 1 ⎞ 1 ⎛ i ⎞
⎜
⎜⎝ i 0 ⎟⎠ β ⎟ = −1• ⎜ β ⎟ → ⎨ → ν = ⎜ ⎟ ∼ ⎜ ⎟ =
1
(i 0 + 1 )
⎩iα = − β 2 ⎝ −i ⎠
2
⎝ ⎠ ⎝ ⎠ 2⎝ 1 ⎠ 2

B.
Hermitisch
⎛ 1 0 ⎞ ⎛ 1 0 ⎞ ⎛ 0 1 ⎞ ⎛ 0 1 ⎞
σ zT = ⎜ ⎟ → σ z† = ⎜ =σz σ xT = ⎜ → σ x† = ⎜ = σ x en
⎝ 0 −1 ⎠ ⎝ 0 −1 ⎟⎠ ⎟
⎝ 1 0 ⎠ ⎝ 1 0 ⎟⎠
⎛ 0 i ⎞ ⎛ 0 −i ⎞
σ yT = ⎜ ⎟ → σ y† = ⎜ =σy
⎝ −i 0 ⎠ ⎝ i 0 ⎟⎠

Pagina 3 van 33

, TN2306 UITWERKINGEN WERKCOLLEGES
Eenheid
⎛ 1 0 ⎞⎛ 1 0 ⎞ ⎛ 1 0 ⎞ ⎛ 0 1 ⎞⎛ 0 1 ⎞ ⎛ 1 0 ⎞
σ zσ z† = ⎜ = = I2 σ xσ x† = ⎜ = = I2
⎝ 0 −1 ⎟⎠ ⎜⎝ 0 −1 ⎟⎠ ⎜⎝ 0 1 ⎟⎠ ⎝ 1 0 ⎟⎠ ⎜⎝ 1 0 ⎟⎠ ⎜⎝ 0 1 ⎟⎠
⎛ 0 −i ⎞ ⎛ 0 −i ⎞ ⎛ 1 0 ⎞
σ yσ y† = ⎜ = = I2
⎝ i 0 ⎟⎠ ⎜⎝ i 0 ⎟⎠ ⎜⎝ 0 1 ⎟⎠

C.
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎡⎣σ x , σ y ⎤⎦ = ⎜ 0 1 ⎟ ⎜ 0 −i ⎟ − ⎜ 0 −i ⎟ ⎜ 0 1 ⎟ = ⎜ i 0 ⎟ − ⎜ −i 0 ⎟ = ⎜ 2i 0 ⎟ = 2i ⎜ 1 0 ⎟ = 2iσ z
⎝ 1 0 ⎠ ⎝ i 0 ⎠ ⎝ i 0 ⎠ ⎝ 1 0 ⎠ ⎝ 0 −i ⎠ ⎝ 0 i ⎠ ⎝ 0 −2i ⎠ ⎝ 0 −1 ⎠
⎛ 0 1 ⎞ ⎛ 1 0 ⎞ ⎛ 1 0 ⎞ ⎛ 0 1 ⎞ ⎛ 0 −1 ⎞ ⎛ 0 1 ⎞ ⎛ 0 −2 ⎞ ⎛ 0 −i ⎞
⎡⎣σ x , σ z ⎤⎦ = ⎜ − = − = = −2i ⎜ = −2iσ y
⎝ 1 0 ⎟⎠ ⎜⎝ 0 −1 ⎟⎠ ⎜⎝ 0 −1 ⎟⎠ ⎜⎝ 1 0 ⎟⎠ ⎜⎝ 1 0 ⎟⎠ ⎜⎝ −1 0 ⎟⎠ ⎜⎝ 2 0 ⎟⎠ ⎝ i 0 ⎟⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎡⎣σ y , σ z ⎤⎦ = ⎜ 0 −i ⎟ ⎜ 1 0 ⎟ − ⎜ 1 0 ⎟ ⎜ 0 −i ⎟ = ⎜ 0 i ⎟ − ⎜ 0 −i ⎟ = ⎜ 0 2i ⎟ = 2i ⎜ 0 1 ⎟ = 2iσ x
⎝ i 0 ⎠ ⎝ 0 −1 ⎠ ⎝ 0 −1 ⎠ ⎝ i 0 ⎠ ⎝ i 0 ⎠ ⎝ −i 0 ⎠ ⎝ 2i 0 ⎠ ⎝ 1 0 ⎠

Problem 2
Consider the basis states
⎛ 1 ⎞ ⎛ 0 ⎞
0 →⎜ , 1 →⎜
⎝ 0 ⎟⎠ ⎝ 1 ⎟⎠
Using the bra-ket notation, we can write σ z in the following form: σ z = 0 0 − 1 1
a) Write σ x and σ y in bra-ket notation.
b) Use bra-ket notation to calculate the commutators between the three matrices.
c) The operator σ + and σ − take 0 to 1 and 1 to 0 , respectively.
Write σ + and σ − in bra-ket notation and as matrices.

A.
⎛ 0 1 ⎞
σx = ⎜ ⎟
⎝ 1 0 ⎠
⎛ 1 ⎞
= 0 1 + 1 0 =⎜
⎝ 0 ⎟⎠
( 0 1 ) + ⎛⎜⎝ 0 ⎞
1 ⎟⎠
( 1 0 ) = ⎛⎜⎝ 0 1 ⎞ ⎛ 0 0 ⎞ ⎛ 0 1 ⎞
+ =
0 0 ⎟⎠ ⎜⎝ 1 0 ⎟⎠ ⎜⎝ 1 0 ⎟⎠
⎛ 0 −i ⎞
σy = ⎜
⎝ i 0 ⎠⎟
⎛ 0 ⎞
= i 1 0 − i 0 1 = i⎜
⎝ 1 ⎟⎠
1 0 ( ) − i ⎛⎜⎝ 1 ⎞
0 ⎟⎠
( 0 1 ) = i ⎛⎜⎝ 0 0 ⎞ ⎛ 0 1 ⎞ ⎛ 0 −i ⎞
−i =
1 0 ⎟⎠ ⎜⎝ 0 0 ⎟⎠ ⎜⎝ i 0 ⎟⎠

B.
⎡⎣σ x , σ y ⎤⎦ = ( 0 1 + 1 0 ) ( i 1 0 − i 0 1 ) − ( i 1 0 − i 0 1 ) ( 0 1 + 1 0 )
= i 0 1 |1 0 − i 0 1 | 0 1 + i 1 0 |1 0 − i 1 0 | 0 1 − i 1 0 | 0 1 − i 1 0 |1 0 + i 0 1 | 0 1 + i 0 1 |1 0
= i 0 •1• 0 − i 0 • 0 • 1 + i 1 • 0 • 0 − i 1 •1• 1 − i 1 •1• 1 − i 1 • 0 • 0 + i 0 • 0 • 1 + i 0 •1• 0
= i 0 0 − i 1 1 − i 1 1 + i 0 0 = 2i 0 0 − 2i 1 1 = 2i ( 0 0 − 1 1 ) = 2iσ z
⎡⎣σ x , σ z ⎤⎦ = ( 0 1 + 1 0 ) ( 0 0 − 1 1 ) − ( 0 0 − 1 1 ) ( 0 1 + 1 0 )
= 0 1 | 0 0 − 0 1 |1 1 + 1 0 | 0 0 − 1 0 |1 1 − 0 0 | 0 1 − 0 0 |1 0 + 1 1 | 0 1 + 1 1 |1 0
= 0 • 0 • 0 − 0 •1• 1 + 1 •1• 0 − 1 • 0 • 1 − 0 •1• 1 − 0 • 0 • 0 + 1 • 0 • 1 + 1 •1• 0
= − 0 1 + 1 0 − 0 1 + 1 0 = −2 0 1 + 2 1 0 = 2 ( 1 0 − 0 1 ) = −2i ( i 1 0 − i 0 1 ) = −2iσ y
⎡⎣σ y , σ z ⎤⎦ = ( i 1 0 − i 0 1 ) ( 0 0 − 1 1 ) − ( 0 0 − 1 1 ) ( i 1 0 − i 0 1 )
= i 1 0 | 0 0 − i 1 0 |1 1 − i 0 1 | 0 0 + i 0 1 |1 1 − i 0 0 |1 0 + i 0 0 | 0 1 + i 1 1 |1 0 − i 1 1 | 0 1
= i 1 •1• 0 − i 1 • 0 • 1 − i 0 • 0 • 0 + i 0 •1• 1 − i 0 • 0 • 0 + i 0 •1• 1 + i 1 •1• 0 − i 1 • 0 • 1
=i 1 0 +i 0 1 +i 0 1 +i 1 0
= 2i 1 0 + 2i 0 1 = 2i ( 0 1 + 1 0 ) = 2iσ x
Pagina 4 van 33