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Pauli formula for an axis rotation

Statement

For θ∈R and n=(nx,ny,nz)∈R3 with ∥n∥2=1, put Σ=n⋅σ=nxX+nyY+nzZ. Then Rn(θ)=exp⁡ ⁣(−iθ2Σ)=cos⁡(θ/2)I−isin⁡(θ/2)Σ, and Rn(θ) is unitary. For the endpoint θ=0, it equals I.

Facts & Assumptions

Given: A real θ, a real unit vector n∈R3, and Q2=C2 with the ordered computational basis.

[F1]

Qubit spaces and states: Q2 is the finite-dimensional qubit inner-product space.

[F2]

Exact one-qubit gates and rotations: Rn(θ) is the operator-norm series exponential of −iθ2n⋅σ.

[F3]

Pauli matrix algebra and trace identities: the Pauli matrices are Hermitian and satisfy σjσk+σkσj=2δjkI.

[F4]

Sine and cosine defined by their real power series: cos⁡t=∑m≥0(−1)mt2m/(2m)! and sin⁡t=∑m≥0(−1)mt2m+1/(2m+1)! for real t.

[F6]

The exponential series of a bounded operator: for a bounded endomorphism B on a Banach space, ∑Bk/k! converges absolutely in operator norm.

[F8]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary operator on a complex inner-product space is an invertible linear isometry from the space to itself.

[F9]

Every finite-dimensional normed space is Banach: every finite-dimensional normed space, including Q2, is Banach.

[F10]

A linear map from a finite-dimensional normed space is bounded: every linear endomorphism of the finite-dimensional normed space Q2 is bounded.

Proof

1.1givenF3algebra

By [F3], Σ2=∑jnj2σj2+∑j<knjnk(σjσk+σkσj)=(∑jnj2)I=I. Also Σ∗=Σ, since the nj are real and every σj is Hermitian.

2.1givenF1F2F4F6F9F10step 1.1algebra

Let B=−iθΣ/2. By [F1], Q2 is finite-dimensional; [F9] makes it Banach and [F10] makes B bounded, so [F6] makes ∑k≥0Bk/k! absolutely convergent. From [step 1.1], induction gives B2m=(−1)m(θ/2)2mI and B2m+1=−i(−1)m(θ/2)2m+1Σ. Grouping its even and odd terms is valid by absolute convergence; [F4] identifies the two scalar series as cos⁡(θ/2) and sin⁡(θ/2). Thus Rn(θ)=cos⁡(θ/2)I−isin⁡(θ/2)Σ. When θ=0, this is I.

3.1F1F5F7F8step 1.1step 2.1algebra∎

Since Σ∗=Σ, the adjoint of the displayed formula is cos⁡(θ/2)I+isin⁡(θ/2)Σ. Multiplying this by Rn(θ) and using Σ2=I from step 1.1 gives Rn(θ)∗Rn(θ)=(cos⁡2(θ/2)+sin⁡2(θ/2))I=I by [F5]. Hence [F7] makes Rn(θ) unitary in the sense of [F8].

Depends on

Used by

Cited to discharge well-definedness by Exact one-qubit gates and rotations.

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources