The Variational Principle

The variational theorem says that for any admissible trial state ψ the Rayleigh quotient ⟨ψ|H|ψ⟩/⟨ψ|ψ⟩ is at least the ground-state energy E₀, with equality only when ψ is itself a ground state. Use the supplied topic model Y_out = (cosd(th) + sin(n*pi/2)*sind(th))^2, where Y_out is CENTRE DENSITY ÷ EXACT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. th is Mixing angle θ, measured in °. Keep n (Admixed level n (odd, same symmetry)) = 3 fixed. For this exercise use only 47.3625 ° ≤ th ≤ 80.4375 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.29383796, with a bounded reading uncertainty ±0.043566. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0033075 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

A one-parameter Gaussian trial is the normalised function (2b/π)^(1/4) e^(−bx²) whose width parameter b is varied to minimise ⟨H⟩; the minimum is an upper bound on the ground-state energy, and comparing it with an exact answer measures how well the Gaussian shape fits. Use the supplied topic model Y_out = 100*(b/2+1/(8*b)-0.5)/0.5, where Y_out is ½x²: E ABOVE EXACT (% of |E₀|). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Gaussian width parameter b, dimensionless. Keep no other adjustable input fixed. For this exercise use only 0.261562 ≤ b ≤ 0.482062 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 2.7273039 % of |E₀|, with a bounded reading uncertainty ±0.39908 % of |E₀|. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00002205, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

If a trial state has exactly zero overlap with the true ground state, its ⟨H⟩ is an upper bound on the first excited energy E₁; a symmetry of H is the usual way to guarantee that zero without knowing the ground state. Use the supplied topic model Y_out = ((nu-1)*(1.5*beta+0.375/beta)+(2-nu)*(3*beta+0.797885/sqrt(beta))), where Y_out is ODD TRIAL ⟨H⟩ (ε). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. beta is Width parameter β = bℓ², dimensionless. Keep w (Even-trial probability w) = 0; nu (Power ν: 1 for F|x|, 2 for ½mω²x²) = 1 fixed. For this exercise use only 0.3035 ≤ beta ≤ 1.08312 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 3.2411789 ε, with a bounded reading uncertainty ±0.11623 ε. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown beta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000779625, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

First-order helium replaces the electron–electron repulsion by its average over the state in which both electrons occupy the unscreened hydrogenic 1s orbital of the full nuclear charge, giving E ≈ E⁽⁰⁾ + ⟨V_ee⟩ = −(Z² − 5Z/8) e²/(4πε₀a₀). Use the supplied topic model Y_out = -27.211*z^2 + 17.007*z, where Y_out is E⁽⁰⁾ + E⁽¹⁾, UPPER BOUND (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Nuclear charge Z, dimensionless. Keep r1 (Radius of electron 1, r₁) = 0.5 a₀ fixed. For this exercise use only 1 ≤ z ≤ 3, at the specified test settings separated by 1. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -74.83 eV, with a bounded reading uncertainty ±14.694 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of z consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.