Does increasing “momentum excess” change the position width?

No. σ_x is controlled independently to stress that σ_x σ_p can exceed ħ/2 when the state is no longer a single minimum-uncertainty Gaussian in both spaces simultaneously (here modeled as a wider momentum Gaussian paired with the same position Gaussian — a pedagogical illustration rather than a single Schrödinger eigenstate of free space).

Heisenberg Product σ_x σ_p

Gaussian position and momentum widths with ℏ = 1; momentum “excess” over the minimum-uncertainty floor.

About this model

Gaussian wavefunctions minimize the Robertson uncertainty product for position and momentum: σ_x σ_p = ħ/2 for a minimum-uncertainty packet. With ħ set to 1, the simulator fixes σ_x and introduces a multiplicative “momentum excess” over the minimal σ_p = 1/(2σ_x). The product σ_x σ_p grows linearly with that factor. Two bar charts show |ψ(x)|² and |φ(p)|² with matched widths, reinforcing that narrowing one distribution widens the other unless excess is added.

Who it's for: Introductory quantum classes linking Fourier width intuition to the Heisenberg relation.

Key terms

Heisenberg Uncertainty
Gaussian Wavepacket
Standard Deviation
Momentum Space

How it works

Position and momentum widths for minimum-uncertainty Gaussians (ℏ = 1). Slide σ_x and an extra momentum spread to see the Heisenberg product σ_x σ_p stay at or above ℏ/2 with matched |ψ(x)|² and |φ(p)|² plots.

Frequently asked questions

Does increasing “momentum excess” change the position width?: No. σ_x is controlled independently to stress that σ_x σ_p can exceed ħ/2 when the state is no longer a single minimum-uncertainty Gaussian in both spaces simultaneously (here modeled as a wider momentum Gaussian paired with the same position Gaussian — a pedagogical illustration rather than a single Schrödinger eigenstate of free space).

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About this model

How it works

Frequently asked questions

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Frequently asked questions