Trial states give upper bounds with a precise scope
Use the Rayleigh quotient, optimize a Gaussian, and distinguish an energy bound from an error certificate.
On this page
The question for today
What you’ll understand
- Prove the ground-energy upper bound for admissible trial states.
- Optimize a normalized Gaussian trial family.
- State what additional conditions are needed for excited-state bounds.
When a Hamiltonian cannot be diagonalized exactly, a sensible trial wavefunction can still give a rigorous upper bound on its lowest energy. The variational method uses an expectation value rather than an expansion in a weak coupling. Its strength is the direction of the bound; its limitation is that a low trial energy does not automatically tell us how close the trial state is to the true ground state. We will make both points explicit.
The bound follows from spectral weights
For a self-adjoint Hamiltonian bounded below, expand an admissible normalized trial state in energy eigenstates when a discrete expansion is available. Its expectation is a weighted average of the eigenvalues, with nonnegative weights summing to one. Every contribution lies at or above the ground energy, so the average does too. More generally the spectral measure gives the same lower-spectral-edge statement when continuum components are present. The trial must belong to the appropriate operator or quadratic-form domain and satisfy the physical boundary conditions.
Normalization can be handled through the Rayleigh quotient, dividing the energy matrix element by the norm squared of any nonzero admissible trial state. Multiplying that state by a constant then leaves the quotient unchanged. Equality requires support entirely in the ground eigenspace when the ground energy is an attained eigenvalue. This proof establishes an upper bound on the true ground energy; reversing the inequality would turn a useful variational calculation into an incorrect lower bound.
Optimization searches only the chosen family
Choose a parameterized trial family that respects the expected symmetry and spatial decay. Minimizing its Rayleigh quotient gives the best upper bound available within that family, not necessarily the global ground state. Enlarging a nested family cannot raise its exact variational minimum, because every earlier trial remains available. Numerical integration error or an incorrectly normalized expectation can spoil that monotonicity in a computed result; such behavior should trigger a check rather than a new physical interpretation.
For a concrete one-dimensional example, use units in which ħ=m=1 and take H=p²/2+λx⁴ with positive λ. A normalized Gaussian with width parameter a has moments ⟨x²⟩=1/(2a), ⟨x⁴⟩=3/(4a²), and kinetic expectation a/4. The energy trades localization against potential cost. A very narrow Gaussian lowers its quartic potential expectation but raises its kinetic energy, while a broad Gaussian does the reverse. Minimizing this competition is a physical balance rather than an arbitrary fit.
Energy quality and state quality are different questions
If the first excitation gap g is known and a normalized trial has energy excess δE above the exact ground energy, its weight outside the ground eigenspace is at most δE/g. Without a known gap and a controlled ground-energy reference, the upper bound alone does not supply that overlap estimate. A small energy error can coexist with substantial mixing among nearly degenerate low states. The method also does not bound an arbitrary excited energy merely because a trial has nodes.
Excited-state bounds require orthogonality to the exact lower eigenspaces or a valid min–max construction in a trial subspace. Orthogonality to approximate lower trial states is not automatically the same condition. The quantum lab illustrates the spectral-weight average in a specified two-level model: Reset and set both controls to θ=60° and φ=0°, predicting P(1)=1/4; then Reset and set θ=0° and φ=0°, predicting zero. If the assigned basis energies are 2 and 10, compute mean energies 4 and 2 yourself. The lab displays populations, not a variational optimizer or energy estimator.
Work it through
Optimize a quartic Gaussian bound
In scaled units ħ=m=1, let H=p²/2+(4/3)x⁴. Using the normalized Gaussian family above, find the minimizing width parameter and the resulting upper bound on the ground energy.
Insert λ=4/3 into the trial energy to obtain E(a)=a/4+1/a². The kinetic and potential terms are both positive and diverge at opposite ends of the allowed width range.
Differentiate: E′(a)=1/4−2/a³. Setting it to zero gives a³=8 and hence a=2. The second derivative is 6/a⁴>0, so this stationary point is the unique minimum.
Evaluate E(2)=2/4+1/4=3/4. The variational principle therefore gives E₀≤0.75 in these scaled energy units; no equality claim follows because the quartic ground state need not be Gaussian.
The optimum is a=2 and the Gaussian calculation establishes the upper bound E₀≤0.75.
The living notebook
Explore the model
Interactive lab · quantum
Energy as a population-weighted average
Reset and set BOTH controls to θ=60° and φ=0°, predicting P(1)=1/4. Reset and set θ=0° and φ=0°, predicting P(1)=0. For assigned energies 2 and 10, infer expectations 4 and 2 manually; no energy readout is provided.
- Probability |0⟩
- 71.1%
- Probability |1⟩
- 28.9%
- Bloch (x, y, z)
- 0.78, 0.45, 0.42
A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forTrial populations determine an expectation for a specified diagonal Hamiltonian, while the lab performs no minimization.
Pause & reason
Check your understanding
Your turn
Try it yourself
A Hamiltonian has orthonormal eigenstates of energies 2 and 10. A normalized trial has amplitudes √3/2 and 1/2 on those states. Find its energy, the energy excess above the ground state, and its excited-state weight. Check the gap-based inequality.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Square the amplitudes before weighting energies. The spectral gap is eight.
Compare with the solution
The weights are 3/4 and 1/4, giving energy (3/4)×2+(1/4)×10=4. The excess is 4−2=2 and the excited weight is 1/4. The gap bound δE/g=2/8=1/4 is saturated because all the nonground weight lies exactly at the first excitation energy.
Keep the idea
Take it with you
- An admissible trial expectation is an upper bound on the ground energy.
- Optimization improves a chosen family without proving that family contains the exact state.
- Excited-state bounds and overlap estimates need additional spectral constraints.
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