Q1L5e.txt

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Any graph state can be constructed
using a simple unitary circuit with Hadamard gates
and controlled Z gates, as we now show.
Given a graph g, specified by its binary symplectic matrix
representation with an adjacency matrix,
what quantum circuit can be used to construct the state?
Well, the stabilizer has X entries and Z entries,
and this simple pattern leads to a simple construction circuit.
First, Hadamard operations are used on every qubit
to start in the state plus.
That is Hadamard applied to the zero state.
These are the vertices of the graph.
Second, a controlled Z gate is applied on every edge.
That is, every connection between two qubits as
indicated by the adjacency matrix.
For example, consider this graph with three qubits
linked by two edges.
Label the qubits zero, one, and two.
I claim that a circuit creating this graph state
starts with plus states and then applies two controlled Z
gates, one between qubits two and one,
and one between qubits one and zero.
Note that the order of the controlled Z gates
does not matter, because they commute with each other.
The output state is the desired graph state
as can be shown in this proof for just the example.
But the idea generalizes to any case.
We analyze this by considering the three stabilizers,
describing the input, the state after the first controlled Z,
and the state after the second controlled Z gate.
By the way, a controlled Z gate is symmetric,
and that is why it is noted as two dots with no direction
of control and target.
The initial stabilizer state S zero
is the graph state stabilizer shown here.
Label the qubits two, one, and zero.
Recall that an X gate applied before a controlled NOT
gate is equivalent to 2x gates applied after the controlled
NOT gate.
That is, the controlled NOT gate conjugates a Pauli X operator
to become two Pauli X operators if the X is on the control
gate.
This Heisenberg evolution picture of Pauli operators
is useful to consider systematically.
The controlled NOT gate conjugates XI to XX, as we saw,
IX to IX, ZI to ZI, and IZ to ZZ.
Similarly, the controlled Z gate conjugates XI to XZ, IX to ZX,
and leaves the ZI and IZ unchanged.
The special cases to remember are highlighted in red here.
Let us use this transformation picture to understand
what the output of the first controlled NOT gate is,
that is the stabilizer S1.
We see that XI is conjugated to XZ by the SZ gate,
and therefore, the first stabilizer changes
to XZI, the second one to ZXI, and the third one
is left alone.
The second controlled Z gate changes
the second and third stabilizer states,
so that XZ is the last two on the second stabilizer
and the third stabilizer becomes ZX in the last two qubits.
This resulting state is exactly the graph state
which we desire.
That is, three qubits connected to each other.
The same idea behind this proof generalizes to any graph state
construction circuit.
And in fact, it also generalizes to stabilizer states simply
by adding the local Clifford operation, which
is used to transform the graph state to the stabilizer state desired.