Saturation of the Mandelstam–Tamm integral bound by fixed-axis two-level pulses #
For a time-dependent Hamiltonian the integral form of the Mandelstam–Tamm bound
(Mandelstam–Tamm 1945) reads arccos |⟨ψ(0), ψ(T)⟩| ≤ ∫₀ᵀ ΔE(t) dt (ℏ = 1).
Here we show that it is an equality for every fixed-axis two-level pulse
H(t) = f(t) σₓ / 2 with f ≥ 0, as long as the total rotation angle θ(T) = ∫₀ᵀ f is at most π.
The state is ψ = exp(−i θ σₓ / 2) |0⟩ = cos(θ/2) |0⟩ − i sin(θ/2) |1⟩; it depends on the pulse only
through the accumulated angle θ(t) = ∫₀ᵗ f. Then ⟨σₓ⟩ = 0, so ΔE(t) = f(t)/2,
∫₀ᵀ ΔE = θ(T)/2 and |⟨ψ(0), ψ(T)⟩| = |cos(θ(T)/2)|.
Scope: the trajectory ψ(t) = qubitState (∫₀ᵗ f) is used as the evolution; for continuous f we
prove (hasDerivAt_trajectory) that it solves the Schrödinger equation ψ′ = −i (f σₓ / 2) ψ
with ψ(0) = |0⟩ (uniqueness is not proved). The statements without the Continuous f
hypothesis are about the trajectory only.
Main results #
QSL.FixedAxis.energyStd_qubitState:ΔE = a/2forH = (a/2) σₓon everyqubitState θQSL.FixedAxis.arccos_overlap_qubitState:arccos ‖⟪ψ(0), ψ(θ)⟫‖ = θ/2for0 ≤ θ ≤ πQSL.FixedAxis.mandelstam_tamm_integral_eq: forf ≥ 0with∫₀ᵀ f ≤ π,∫₀ᵀ ΔE(t) dt = arccos ‖⟪ψ(0), ψ(T)⟫‖QSL.FixedAxis.pi_pulse_quantum_speed_limit_ratio: for∫₀ᵀ f = πthe final state is orthogonal to the initial one and∫₀ᵀ ΔE = π/2: the Mandelstam–Tamm ratio is exactly1, whateverfQSL.FixedAxis.arccos_overlap_lt_of_pi_lt: strict inequality forπ < θ < 3π
References #
- L. Mandelstam, I. Tamm, The uncertainty relation between energy and time in non-relativistic quantum mechanics, J. Phys. USSR 9 (1945) 249–254.
Qubit vectors.
Equations
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Single-qubit operators.
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The Hamiltonian (a/2) σₓ.
Equations
- QSL.FixedAxis.hamX a = ↑(a / 2) • QSL.FixedAxis.sigmaX
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Action of a matrix on a vector.
Equations
- QSL.FixedAxis.act M v = WithLp.toLp 2 (Matrix.mulVec M v.ofLp)
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The vector c |0⟩ − i s |1⟩ for real c, s.
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The state exp (−i θ σₓ / 2) |0⟩ = cos (θ/2) |0⟩ − i sin (θ/2) |1⟩.
Equations
- QSL.FixedAxis.qubitState θ = QSL.FixedAxis.qubitVec (Real.cos (θ / 2)) (Real.sin (θ / 2))
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Expectation value ⟨v, M v⟩.
Equations
- QSL.FixedAxis.expect M v = inner ℂ v (QSL.FixedAxis.act M v)
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Energy standard deviation ΔE = √(⟨H²⟩ − ⟨H⟩²) of the state v for the Hamiltonian H.
Equations
- QSL.FixedAxis.energyStd H v = √((QSL.FixedAxis.expect (H * H) v).re - (QSL.FixedAxis.expect H v).re ^ 2)
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Pointwise: for H = (a/2) σₓ the energy standard deviation of every state
exp (−i θ σₓ / 2) |0⟩ equals |a|/2 (= a/2 for a ≥ 0).
The overlap ⟨ψ(0), ψ(θ)⟩ = cos (θ/2).
Saturation, trajectory form: for 0 ≤ θ ≤ π, arccos ‖⟪ψ(0), ψ(θ)⟫‖ = θ/2.
Time-dependent form. For a pulse f ≥ 0 with θ(T) = ∫₀ᵀ f ≤ π, the trajectory
ψ(t) = qubitState (∫₀ᵗ f) of H(t) = f(t) σₓ / 2 satisfies the Mandelstam–Tamm integral bound
with equality: ∫₀ᵀ ΔE(t) dt = arccos ‖⟪ψ(0), ψ(T)⟫‖.
The trajectory solves the Schrödinger equation. For continuous f, the curve
ψ(t) = qubitState (∫₀ᵗ f) satisfies ψ′(t) = −i H(t) ψ(t) with H(t) = f(t) σₓ / 2
(ℏ = 1).
Mandelstam–Tamm saturation for the Schrödinger solution. For continuous f ≥ 0 on [0,T]
with ∫₀ᵀ f ≤ π, the curve ψ(t) = qubitState (∫₀ᵗ f) is a solution of
ψ′ = −i (f σₓ/2) ψ with ψ(0) = |0⟩ (qubitState 0), and along it the Mandelstam–Tamm integral
bound is an equality, ∫₀ᵀ ΔE(t) dt = arccos ‖⟪ψ(0), ψ(T)⟫‖. (Uniqueness of the solution is
not proved; it is standard for linear ODEs.)
Corollary (shape independence). If f ≥ 0 and ∫₀ᵀ f = π, the final state of the pulse
is orthogonal to the initial state and ∫₀ᵀ ΔE = π/2: the Mandelstam–Tamm ratio
(∫ΔE)/arccos|⟨ψ₀,ψ_T⟩| is exactly 1, for every shape f.