Bij Quantum Engineering & Applications (TN2306) wordt ingegaan op thema's uit de categorieën Quantum bits and entanglement met daarbijbehorende quantum circuits, Quantum communication en Quantum computing over zowel algoritmen als hardware. Het vak wordt gegeven in het vierde octaal van de Minor M...
I've already run into several errors with minus signs in the first 6 pages, so the effect is actually not perfect, but otherwise everything is very neat.
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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
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.
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
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