Rotations on the Bloch Sphere

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Ian Glendinning

May 20, 2010

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Outline

• The Bloch Sphere • The Density Operator • Rotation Operators • Rotation about thez ˆ Axis • Rotation about an Arbitrary Axis • Arbitrary Unitary Operator • Future Topics

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The Bloch Sphere

An arbitrary single qubit state can be written: µ ¶ θ θ |ψi = eiγ cos |0i + eiφ sin |1i 2 2 where θ, φ and γ are real numbers. The numbers 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π deﬁne a point on a unit three-dimensional sphere. This is the Bloch sphere. Qubit states with arbitrary values of γ are all represented by the same point on the Bloch sphere because the factor of eiγ has no observable eﬀects, and we can therefore choose to write: θ θ |ψi = cos |0i + eiφ sin |1i 2 2

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The Bloch Sphere

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The Density Operator

In order to relate unitary operations on a qubit state |ψi to rotations on the Bloch sphere it turns out to be convenient to use the corresponding density operator ρ , deﬁned as:

ρ = |ψihψ| = |ψi ⊗ hψ|

where hψ| = |ψi†

so   cos θ ³ ´ ρ = |ψi ⊗ hψ| =  2  ⊗ cos θ e−iφ sin θ iφ θ 2 2 e sin 2

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The Density Operator

By the deﬁnition of the outer product   cos2 θ e−iφ cos θ sin θ ρ =  2 2 2  iφ θ θ 2 θ e cos 2 sin 2 sin 2 and using standard trigonometric identities   1 1 + cos θ cos φ sin θ − i sin φ sin θ ρ =   2 cos φ sin θ + i sin φ sin θ 1 − cos θ

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then grouping terms in the basis {I,X,Y,Z}, where       0 1 0 −i 1 0 X =   ,Y =   ,Z =   1 0 i 0 0 −1

are the Pauli matrices, we have 1 ρ = (I + X cos φ sin θ + Y sin φ sin θ + Z cos θ) 2 1 = (I + ~r · ~σ) 2 ρ where I is the identity matrix, ~σ is the 3-element ‘vector’ of Pauli

Matrices (X,Y,Z), and ~rρ is the unit Bloch vector

~rρ = (rx, ry, rz) = (cos φ sin θ, sin φ sin θ, cos θ) & % Ian Glendinning 7 May 20, 2010 ' $

Unitary Evolution of the Density Operator

Quantum circuits consist of combinations of quantum gates, each corresponding to a unitary operation on a qubit state. That is:

|ψi 7→ U|ψi

where U is a unitary operator (matrix), i.e. U †U = UU † = I, so, recalling that the density operator |ψihψ| = |ψi ⊗ (|ψi)†, it evolves as

|ψihψ| 7→ (U|ψi) ⊗ (U|ψi)† 7→ (U|ψi) ⊗ (hψ|U †) 7→ U|ψihψ|U †

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Rotation Operators

The Pauli X, Y and Z matrices are so-called because when they are exponentiated, they give rise to the rotation operators, which rotate

the Bloch vector ~rρ about thex ˆ,y ˆ andz ˆ axes, by a given angle θ:

−i θ X Rx(θ) ≡ e 2 −i θ Y Ry(θ) ≡ e 2 −i θ Z Rz(θ) ≡ e 2

Now, if operator A satisﬁes A2 = I, it can be shown that

eiθA = cos(θ)I + i sin(θ)A

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Rotation Operators

And since the Pauli matrices satisfy X2 = Y 2 = Z2 = I, the rotation operators can be expanded as:   θ θ −i θ X θ θ cos 2 −i sin 2 Rx(θ) ≡ e 2 = cos I − i sin X =   2 2 θ θ −i sin 2 cos 2   θ θ −i θ Y θ θ cos 2 − sin 2 Ry(θ) ≡ e 2 = cos I − i sin Y =   2 2 θ θ sin 2 cos 2   −iθ/2 −i θ Z θ θ e 0 Rz(θ) ≡ e 2 = cos I − i sin Z =   2 2 0 eiθ/2

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Rotation about the zˆ Axis

We can check that these operators do what they’re supposed to by

considering their action on the state ρ. For example, Rz(θ) evolves ρ to:

0 † ρ = Rz(θ)ρRz(θ) 1 = R (θ) (I + ~r · ~σ) R (θ)† z 2 ρ z 1 = R (θ) (I + r X + r Y + r Z) R (θ)† z 2 x y z z † It is easily veriﬁed that Rz(θ) is unitary, so Rz(θ)Rz(θ) = I, and 1 ρ0 = (I + r R (θ)XR (θ)† + r R (θ)YR (θ)† + r R (θ)ZR (θ)†) 2 x z z y z z z z z

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Rotation about the zˆ Axis

Now expand µ ¶ µ ¶ θ θ θ θ R (θ)XR (θ)† = cos I − i sin Z X cos I + i sin Z z z 2 2 2 2 θ θ θ θ θ = cos2 X + i sin cos XZ − i sin cos ZX 2 2 2 2 2 θ + sin2 ZXZ 2 To evaluate this expression we use the algebra of the Pauli matrices.

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Algebra of the Pauli Matrices

The algebra of the Pauli matrices can be summarised by the equation:

X2 = Y 2 = Z2 = −iXY Z = I

All the products of pairs of Pauli matrices can be calculated from the above equation.

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For example:

I = −iXY Z (I)Z = (−iXY Z)Z Z = −iXY (Z)Y = (−iXY )Y ZY = −iX (ZY )X = (−iX)X ZYX = −iI Z(ZYX) = Z(−iI) YX = −iZ & % Ian Glendinning 14 May 20, 2010 ' $

Algebra of the Pauli Matrices

The products of pairs of Pauli matrices are:

XY = −YX = iZ YZ = −ZY = iX ZX = −XZ = iY

which can be summarised as X σiσj = δijI + i ²ijkσk k

where σ1 = X, σ2 = Y and σ3 = Z. Notice that non-identical Pauli

matrices anticommute, i.e. σiσj = −σjσi if i 6= j.

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Rotation about the zˆ Axis

We can now write θ θ θ θ θ R (θ)XR (θ)† = cos2 X + i sin cos XZ − i sin cos ZX z z 2 2 2 2 2 θ + sin2 ZXZ 2 θ θ θ θ θ θ = cos2 X + sin cos Y + sin cos Y − sin2 X 2 2 2 2 2 2 µ ¶ θ θ θ θ = cos2 − sin2 X + 2 sin cos Y 2 2 2 2 = cos θ X + sin θ Y

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Rotation about the zˆ Axis

Similarly we can show that

† Rz(θ)YRz(θ) = cos θ Y − sin θ X † Rz(θ)ZRz(θ) = Z

So we can now rewrite 1 ρ0 = (I + r R (θ)XR (θ)† + r R (θ)YR (θ)† + r R (θ)ZR (θ)†) 2 x z z y z z z z z 1 = (I + r (cos θ X + sin θ Y ) + r (cos θ Y − sin θ X) + r Z) 2 x y z 1 = (I + (r cos θ − r sin θ)X + (r sin θ + r cos θ)Y + r Z) 2 x y x y z

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Rotation about the zˆ Axis

and recalling that we can write

0 1 ρ = (I + ~r 0 · ~σ) 2 ρ 1 = (I + r0 X + r0 Y + r0 Z) 2 x y z we can see that

0 rx = rx cos θ − ry sin θ 0 ry = rx sin θ + ry cos θ 0 rz = rz

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Rotation about the zˆ Axis

so the Bloch vector of the new state is   cos θ − sin θ 0     ~rρ0 =  sin θ cos θ 0 ~rρ 0 0 1

and the matrix is the usual 3D-rotation matrix for a rotation about thez ˆ axis by and angle of θ, as required. We can similarly show that

Rx(θ) and Ry(θ) perform rotations of the Bloch vector about the x and y axes by an angle θ.

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Rotation About an Arbitrary Axis

We can use the rotation operators about the y and z axes to construct the operator for rotation by and angle α about an arbitrary axisn ˆ, since we can decompose it as:

Rnˆ (α) = Rz(φ)Ry(θ)Rz(α)Ry(−θ)Rz(−φ) † † = Rz(φ)Ry(θ)Rz(α)Ry(θ) Rz(φ)

Taking the rotations one by one, Rz(−φ) ﬁrst rotatesn ˆ into the x-z

plane, Ry(−θ) then rotates it into the z axis, Rz(α) performs the

desired rotation aboutn ˆ, and Ry(θ) and Rz(φ) rotaten ˆ back to its original orientation.

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Rotation About an Arbitrary Axis

Working from the inside out, we can rewrite

† † Rnˆ (α) = Rz(φ)Ry(θ)Rz(α)Ry(θ) Rz(φ) h α α i = R (φ)R (θ) cos I − i sin Z R (θ)†R (φ)† z y 2 2 y z And using the Pauli matrix algebra as earlier, we can show that

† Ry(θ)ZRy(θ) = cos θ Z + sin θ X

† and Ry(θ)Ry(θ) = I, so h α α i R (α) = R (φ) cos I − i sin (cos θ Z + sin θ X) R (φ)† nˆ z 2 2 z

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Rotation About an Arbitrary Axis

h α α i R (α) = R (φ) cos I − i sin (cos θ Z + sin θ X) R (φ)† nˆ z 2 2 z † † and using Rz(θ)ZRz(θ) = Z and Rz(θ)XRz(θ) = cos θ X + sin θ Y we obtain α α R (α) = cos I − i sin (cos θ Z + sin θ [cos φ X + sin φ Y ]) nˆ 2 2 α α = cos I − i sin (sin θ cos φ X + sin θ sin φ Y + cos θ Z) 2 2 α α = cos I − i sin (n X + n Y + n Z) 2 2 x y z α α = cos I − i sin nˆ · ~σ 2 2

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We can rewrite this result as an operator exponential, because

2 (ˆn · ~σ) = (nxX + nyY + nzZ)(nxX + nyY + nzZ) 2 2 = nxX + nxnyXY + nxnzXZ + 2 2 nynxYX + nyY + nynzYZ + 2 2 nznxZX + nznyZY + nzZ 2 2 2 = (nx + ny + nz) I = I

sincen ˆ is a unit vector, so we can use A =n ˆ · ~σ in the identity

eiθA = cos(θ)I + i sin(θ)A

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Rotation About an Arbitrary Axis

and we can write α α R (α) = cos I − i sin nˆ · ~σ nˆ 2 2 α = exp(−i nˆ · ~σ) 2 which is our ﬁnal result for the operator to transform a state ρ such

that its Bloch vector ~rρ is rotated about then ˆ axis by an angle α, i.e.

0 † ρ = Rnˆ (α)ρRnˆ (α)

or equivalently 0 |ψ i = Rnˆ (α)|ψi

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Since Rnˆ (α) can rotate a Bloch vector into any other Bloch vector, which includes all possible qubit states up to a global phase factor, an arbitrary single qubit unitary operator can be written in the form:

U = exp(iγ)Rnˆ (α) for some real numbers γ and α and a real three-dimensional unit vectorn ˆ. For example, consider γ = π/2, α = π, andn ˆ = ( √1 , 0, √1 ) 2 2 · ¸ ³π ´ ³π ´ 1 U = exp(iπ/2) cos I − i sin √ (X + Z) 2 2 2   1 1 1 = √   2 1 −1

which is the Hadamard gate H. & % Ian Glendinning 25 May 20, 2010 ' $

Future Topics

• Generalisation of the Bloch sphere to mixed states • Generalizations to more qubits

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