Arbitrary Single Qubit Gate using L(4) Cluster State

Perform three equatorial measurements to a linear cluster state to obtain a general one qubit transformation

A general one qubit transformation



can also be represented by three Euler angles indicating rotations about the Cartesian axes. One possible sequence of rotations that is equivalent to a general transformation is the following:

U(α,β,γ)

(γ)

(β)

(α)

. It's possible to achieve the rotations doing three measurements to a linear cluster state. Each of the measurements is performed in an equatorial basis with the required angles.

Install the Quantum Framework paclet:

In[1]:=

PacletInstall["Wolfram/QuantumFramework"]

Load the paclet:

In[2]:=

Needs["Wolfram`QuantumFramework`"]

Represent the graph and the measurements in the equatorial basis:

In[3]:=

HighlightGraphg=GridGraph{4,1},

,{4,3,2},





Out[3]=

Define the equatorial basis using

QuantumBasis

with the angle as parameter option:

In[4]:=

eb=QuantumBasis"

"1

1,

Iθ

,"

"1

1,-

Iθ

,"Parameters"θ;

Create the cluster state with 4 qubits:

In[5]:=

L4cluster=QuantumState["Graph"[g]];

Assume the results of the 3 measurements were the

〉

state to avoid Pauli corrections (see notes). Get their respective bra from the "BasisStates":

In[6]:=

projectors=eb[#]["BasisStates"]〚1〛["Dagger"]&/@{γ,β,α}

Out[6]=

{

〈

}

Simulate the

(

)

measurement projecting the cluster state sequentially, and use the "DensityMatrix" property to have a state to compare without global phases. Simplify symbolically (the 8 is for normalization):

In[7]:=

projectedState=8FullSimplify[Composition[Sequence@@projectors][L4cluster]["DensityMatrix"]//Normal,AssumptionsElement[{α,β,γ},Reals]]

Out[7]=



(1+Cos[α]Cos[γ]-Cos[β]Sin[α]Sin[γ]),

(Sin[α](Cos[β]Cos[γ]+Sin[β])+Cos[α]Sin[γ]),-

(Cos[β]Cos[γ]Sin[α]+Sin[α]Sin[β]+Cos[α]Sin[γ]),

(1-Cos[α]Cos[γ]+Cos[β]Sin[α]Sin[γ])

Consider the following circuit:

In[8]:=

q=QuantumCircuitOperator[{"RZ"[α],"RX"[β],"RZ"[γ],"H"}];q["Diagram","WireLabels"{Style["|+〉",14]}]

Out[8]=

Calculate the density matrix of applying that circuit to the |+〉 state.

ComplexExpand

and

FullSimplify

are used to get a symbolic simplification:

In[9]:=

FullSimplify[ComplexExpand[q[QuantumState["Plus"]]["DensityMatrix"]//Normal],AssumptionsElement[{α,β,γ},Reals]]

Out[9]=



(1+Cos[α]Cos[γ]-Cos[β]Sin[α]Sin[γ]),

(Cos[β]Cos[γ]Sin[α]+Sin[α]Sin[β]+Cos[α]Sin[γ]),-

(Cos[β]Cos[γ]Sin[α]+Sin[α]Sin[β]+Cos[α]Sin[γ]),

(1-Cos[α]Cos[γ]+Cos[β]Sin[α]Sin[γ])

Verify the matrices are the same:

In[10]:=

Expand[%]Expand[projectedState]

Out[10]=

True

The extra Hadamard gate is not a problem because if the initial state is |0〉 the sequences of gates are H RZ(α) RX(β) RZ(γ) H which are equivalent to RX(α) RZ(β) RX(γ). That also is a valid way to describe an arbitrary transformation:

In[11]:=

FullSimplify[ComplexExpand[QuantumCircuitOperator[{"RX"[α],"RZ"[β],"RX"[γ]}][]["DensityMatrix"]//Normal],AssumptionsElement[{α,β,γ},Reals]]

Out[11]=



(1+Cos[α]Cos[γ]-Cos[β]Sin[α]Sin[γ]),

(Cos[β]Cos[γ]Sin[α]+Sin[α]Sin[β]+Cos[α]Sin[γ]),-

(Cos[β]Cos[γ]Sin[α]+Sin[α]Sin[β]+Cos[α]Sin[γ]),

(1-Cos[α]Cos[γ]+Cos[β]Sin[α]Sin[γ])

External Links

Wolfram/QuantumFramework

Source Metadata

Citation:
- WALTHER, Philip, et al. Experimental one-way quantum computing. Nature, 2005

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Contributed by: Wolfram Research, Quantum Computation Framework Team

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Arbitrary Single Qubit Gate using L(4) Cluster State

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Arbitrary Single Qubit Gate using L(4) Cluster State

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