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A first excited oscillator state has negative Wigner values

Statement refuted

The assertion to be refuted is: for every $0 < ϵ \leq 1$ and every normalized $u \in S (R)$ , $W_{ϵ} [u] (x, ξ) \geq 0$ for all $(x, ξ) \in R^{2}$ .

Assume Countable Choice and $0 < ϵ \leq 1$ . In the one-dimensional oscillator with $m = ω = 1$ , $ℏ = ϵ$ , the normalized first excited state is $ψ_{1}^{ϵ} (x) = \sqrt{\frac{2}{ϵ}} (π ϵ)^{- 1 / 4} x e^{- x^{2} / (2 ϵ)} .$ Its Wigner function is $W_{ϵ} [ψ_{1}^{ϵ}] (x, ξ) = \frac{1}{π ϵ} (\frac{2 (x^{2} + ξ^{2})}{ϵ} - 1) e^{- (x^{2} + ξ^{2}) / ϵ} .$ It is strictly negative on the disk $x^{2} + ξ^{2} < ϵ / 2$ . Consequently the assertion that every normalized Schwartz vector has a nonnegative Wigner function is false.

Statement references

Wigner pairing and marginals

Oscillator scaling and exact realization

Facts & Assumptions

Given: Countable Choice, $0 < ϵ \leq 1$ , and the one-dimensional oscillator conventions of the Statement.

[A1]

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

$H_{1} (y) = 2 y$ , $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2}$ , and $g_{1} = H_{1} g_{0} / \sqrt{2} = \sqrt{2} y g_{0}$ ; scaling by $ℓ = \sqrt{ϵ}$ gives the displayed normalized oscillator state (Hermite recurrence, Oscillator scaling and exact realization).

[F2]

The Wigner integral uses the normalization and Fourier sign in Schwartz Weyl quantization and Wigner distributions; for $u \in S$ it is real and integrable (Wigner pairing and marginals).

[F3]

$G_{ϵ} (ξ) : = \int e^{- y^{2} / (4 ϵ)} e^{- i y ξ / ϵ} d y = 2 \sqrt{π ϵ} e^{- ξ^{2} / ϵ}$ (Euclidean Gaussian transform with the 2π normalization).

[F4]

Differentiation under this Gaussian integral is valid because the first two differentiated integrands are dominated by integrable polynomial multiples of $e^{- y^{2} / (4 ϵ)}$ ; dominated convergence justifies the limit (Dominated convergence).

Counterexample

Given: The data of the Statement and [A1], [F1]–[F4].

1.1A1F1algebra

By [F1] and the oscillator scaling $ℓ = \sqrt{ϵ}$ , the first excited state is $ℓ^{- 1 / 2} g_{1} (x / ℓ) = ϵ^{- 1 / 4} \sqrt{2} \frac{x}{\sqrt{ϵ}} π^{- 1 / 4} e^{- x^{2} / (2 ϵ)} = \sqrt{\frac{2}{ϵ}} (π ϵ)^{- 1 / 4} x e^{- x^{2} / (2 ϵ)} .$ The oscillator realization makes it a normalized Schwartz vector.

1.2F1F2F3F4algebra

Write $A^{2} = 2 ϵ^{- 1} (π ϵ)^{- 1 / 2}$ . The product in the Wigner integrand is $ψ_{1}^{ϵ} (x + y / 2) \overline{ψ_{1}^{ϵ} (x - y / 2)} = A^{2} (x^{2} - \frac{y^{2}}{4}) e^{- x^{2} / ϵ - y^{2} / (4 ϵ)} .$ By [F4], $G_{ϵ}^{''} (ξ) = \int (- y^{2} / ϵ^{2}) e^{- y^{2} / (4 ϵ)} e^{- i y ξ / ϵ} d y$ . Differentiating the explicit Gaussian in [F3] gives $\int y^{2} e^{- y^{2} / (4 ϵ)} e^{- i y ξ / ϵ} d y = - ϵ^{2} G_{ϵ}^{''} (ξ) = (2 ϵ - 4 ξ^{2}) G_{ϵ} (ξ) .$ Substitution in the Wigner definition [F2] therefore yields $W_{ϵ} [ψ_{1}^{ϵ}] (x, ξ) = (2 π ϵ)^{- 1} A^{2} e^{- x^{2} / ϵ} (x^{2} G_{ϵ} (ξ) - \frac{1}{4} (2 ϵ - 4 ξ^{2}) G_{ϵ} (ξ)) = \frac{1}{π ϵ} (\frac{2 (x^{2} + ξ^{2})}{ϵ} - 1) e^{- (x^{2} + ξ^{2}) / ϵ} .$ Every factor and sign follows from the displayed negative-sign Fourier convention.

2.1step 1.1step 1.2algebra∎

The exponential and the prefactor $1 / (π ϵ)$ are strictly positive. Thus $W_{ϵ} [ψ_{1}^{ϵ}] (x, ξ) < 0$ exactly when $2 (x^{2} + ξ^{2}) / ϵ - 1 < 0$ , which is the disk $x^{2} + ξ^{2} < ϵ / 2$ . This normalized Schwartz vector refutes universal Wigner nonnegativity.

A first excited oscillator state has negative Wigner values

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Depends on

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