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feat: add multi-qubit systems

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Kristofers Solo 2025-06-05 17:31:00 +03:00
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@ -10,6 +10,7 @@
)) ))
#let distance = $space.quad space.quad$ #let distance = $space.quad space.quad$
#let tensor = $lr(times.circle)$
= Bre-Ket Notation = Bre-Ket Notation
== Ket $ket(psi)$ == Ket $ket(psi)$
@ -92,7 +93,7 @@ $
P(-)=abs(braket(-, 1))^2=1/2 P(-)=abs(braket(-, 1))^2=1/2
$ $
=== Example: $ket(psi)=(1+2i)/sqrt(7)ket(0)+(1-i)/sqrt(7)ket(1)$ ==== Example: $ket(psi)=(1+2i)/sqrt(7)ket(0)+(1-i)/sqrt(7)ket(1)$
$ $
P(0) & =abs((1+2i)/sqrt(7))^2=(1^2+2^2)/7=5/7 \ P(0) & =abs((1+2i)/sqrt(7))^2=(1^2+2^2)/7=5/7 \
P(1) & =abs((1-i)/sqrt(7))^2=(1^2+(-1)^2)/7=2/7 P(1) & =abs((1-i)/sqrt(7))^2=(1^2+(-1)^2)/7=2/7
@ -227,3 +228,94 @@ $
U ket(1) = ket(0) U ket(1) = ket(0)
) )
$ $
= Multi-Qubit Systems
== Tensor product
Combines state space.
$
(a ket(0)+b ket(1)) tensor (c ket(0) + d ket(1)) = \ =
a c ket(00)+ a d ket(01) + b c ket(10) + b d ket(11)
$
For $k$ qubits, $2^k$ basis states.
=== Operators
$
(A tensor B)
(ket(psi_A) tensor ket(psi_B))=
(A ket(psi_A)) tensor (B ket(psi_B))
$
== Product States vs. Entangled States
=== Product state
Can be written as $ket(psi_A)tensor ket(psi_B)$.
==== Example
$
1/2(ket(00)+ket(01)+ket(10)+ket(11))= \ =
(1/sqrt(2)(ket(0)+ket(1))) tensor (1/sqrt(2)(ket(0)+ket(1)))
$
=== Entangled States
Cannot be factored
==== Example
$
1/sqrt(2)(ket(00)+ket(11)) ("Bell state" ket(Phi^+))
$
== Multi-Qubit Measurement & Normalization
Measure one qubit from a multi-qubit system.
==== Example
===== State
$
a ket(00)+b ket(01) + c ket(10) + d ket(11)
$
===== Measure 1st qubit
====== Prob of $0$
$
P(q_1=0)=abs(a)^2+abs(b)^2
$
Post-measurement state:
$
(a ket(00)+ b ket(01))/sqrt(abs(a)^2+abs(b)^2)=
ket(0) tensor
(a ket(0) + b ket(1))/(sqrt(abs(a)^2+abs(b)^2))
$
====== Prob of $1$
$
P(q_1=1)=abs(c)^2+abs(d)^2
$
Post-measurement state:
$
(c ket(10)+ d ket(11))/sqrt(abs(c)^2+abs(d)^2)=
ket(1) tensor
(c ket(0) + d ket(1))/(sqrt(abs(c)^2+abs(d)^2))
$
==== Example
===== State
$
2/3 ket(00)+1/3 ket(01)+2/3 ket(10)
$
===== Measure 1st qubit
====== Prob of $0$
$
P(0)=(2/3)^2+(1/3)^2=5/9
$
State of 2nd qubit:
$
(2/3 ket(0) + 1/3 ket(1))/sqrt(5/9)=
1/sqrt(5)(2 ket(0) + ket(1))
$
====== Prob of $1$
$
P(1)=(2/3)^2=4/9
$
State of 2nd qubit:
$
(2/3 ket(0))/sqrt(4/9)=
ket(0)
$