Document 13587309

```Lecture 14: Cluster States
Scribed by: Ilia Mirkin
Department of Mathematics, MIT
October 21, 2003
A cluster state is a highly entangled rectangular array of qubits. We measure qubits
one at a time. The wiring diagram tells us which basis to measure each qubit in, and which
order to measure them in. A wiring diagram is represented by connecting up the dots (which
represent qubits) with lines. A junction of two separate lines represents a gate. These gates
do not have to be unitary, but if done right, are.
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Figure 1: Unconnected cluster states
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Figure 2: Wiring diagram for two gates
Measurement on a wiring diagram is done by ﬁrst measuring all the qubits that are not
in the wiring diagram (i.e. unconnected) in the σz basis. Once those qubits are measured,
we measure the qubits in the circuit from left to right in the speciﬁed basis.
1
P. Shor – 18.435/2.111 Quantum Computation – Lecture 14
2
Cluster state given by eigenvalue equations. The neighborhood of a qubit are the up,
down, left, and right qubits.
K (a) = σx(a) ⊗
�
σz(b) |b ∈ neighborhood(a)
(1)
b
The claim is that K (a) commutes with K (b) when a �= b. To show that this is true, we
can look at the following cases:
neighborhood(a) ∩ neighborhood(b) = ∅
(2)
When this is true, then there is absolutely no overlap between K (a) and K (b) and thus
the two commute.
neighborhood(a) �� b
(3)
This means that neighborhoods overlap, but that the qubit b is not in the neighborhood
of a in an arrangement such as
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ap u p
ap l ap
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ap r bp
ap d p
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Figure 3: Neighborhoods overlap
K (a) = σx(a) ⊗ σz(ar ) ⊗ &middot; &middot; &middot;
K (b) = σx(b) ⊗ σz(ar ) ⊗ &middot; &middot; &middot;
K (a) K (b) = σx(a) ⊗ σz(ar ) ⊗ &middot; &middot; &middot; ⊗ σx(b) ⊗ σz(ar ) ⊗ &middot; &middot; &middot;
And in the third case, a and b are adjacent to each other.
p
ap u p
ap l ap
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bp
ap d p
p
bp r
p
Figure 4: a and b are adjacent
(4)
(5)
(6)
P. Shor – 18.435/2.111 Quantum Computation – Lecture 14
3
K (a) K (b) = σx(a) ⊗ σz(b) ⊗ &middot; &middot; &middot; ⊗ σx(a) σz(b) &middot; &middot;&middot;
(7)
In all three cases K (a) and K (b) both commute, so the claim holds. This means that K (a)
are all simultaneously diagonalizable. Any simultaneous eigenvector of K (a) , a ∈ C (cluster)
is a cluster state. Each K (a) has eigenvalue &plusmn;1, making for 2n vectors of eigenvalues
{Ka }�.
�
�
�
�φ{κa } is a cluster state with eigenvalue κa on qubit a, {κa } = {&plusmn;1}. Thus φ{κa } |φ �
=
{κa }
C
C
�
0 if {κ
a } =
� {κa }. For example,
κb = +1
κ = −1
�
�
�
�
�a
�
φ{κa } |φ{κ�a }
= − φ{κa } |Kb |φ{κ�a }
= − φ{κa } |φ{κ�a }
=0
C
�
If {κa } = {κa } except for κb = −κb , then
(b)
σz
(10)
C
C
�
(8)
(9)
�
�
�
�
�
φ{κa } = ��
φ �
{κa }
C
C
�
�
�
�
Ka σz(b) �
φ{κa } C = (−1)δab σz(b) Ka �φ{κa } C
�
�
= (−1)δab σz(b) κa �φ{κa } C
�
�
�
= κa σz(b) �φ{κa } C
Cluster state for ∀a κa = 1, start in state |ψ�C =
then apply Sab to all neighbors a, b.
�
a
|+�a , where |+� =
(11)
(12)
(13)
√1 (|0� + |1�).
2
⎛
Sab
⎞
1 0 0 0
⎜ 0 1 0 0
⎟
⎟
=
⎜
⎝
0 0 1 0 ⎠
0 0 0 −1
�
�
1
=
I + σz(a) + σz(b) − σz(a) ⊗ σz(b)
2
We
Here are a few examples of gates that can be made using wiring diagrams:
Transmission Line
CNOT Gate σ
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σp x σp x σp x σp x
σp x p σy p
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σp x σp y σp y σp y σp x σp x
σ σ σ
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σp x σp x σp y p σx p
y
p
p
y
Rotation
σp x σp x &plusmn;θ
σp x σp x
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Figure 5: CNOT Gate, Transmission Line, Hadamard, Rotation
(14)
(15)
P. Shor – 18.435/2.111 Quantum Computation – Lecture 14
4
|ψ�
p |+�
p |+�
p
Sab
Figure 6: Transmission line
We also know that Sab commutes with Sa� b� . In the |0�, |1� basis, Sab�
can be represented
by a diagonal matrix, which means that they have to commute. K (a) a,b Sab |+�⊗n is an
eigenvector of K (a) .
Demonstration of a transmission line eﬀect:
1
Sab |+� |+� = Sab (|00� + |01� + |10� + |11�)
2
1
=
(|00� + |01� + |10� − |11�)
2
1
= √ (|+� |0� + |−� |1�)
2
(16)
(17)
(18)
With this, we apply Sab and measure both a and b in the |+�, |−� basis. This is equivalent
to measuring in the �++| Sab , �+−| Sab , �−+| Sab , �−−| Sab basis, which is also equivalent to
measuring in the √12 (�0+| + �1−|) basis. In this way we get the teleportation eﬀect on the
original |ψ�.
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