structural:reformatting, information:classification, linguistic:translation, quality:corruption
Phase space

In the phase space formulation of quantum mechanics, the Robertson–Schrödinger relation follows from a positivity condition on a real star-square function. Given a Wigner function W (x, p) {\displaystyle W(x,p)} with star product ★ and a function f, the following is generally true: ⟨ f ∗ ⋆ f ⟩ = ∫ (f ∗ ⋆ f) W (x, p) d x d p ≥ 0. {\displaystyle \langle f^{*}\star f\rangle =\int (f^{*}\star f)\,W(x,p)\,dx\,dp\geq 0~.}

Choosing f = a + b x + c p {\displaystyle f=a+bx+cp}, we arrive at ⟨ f ∗ ⋆ f ⟩ = ≥ 0. {\displaystyle \langle f^{*}\star f\rangle ={\begin{bmatrix}a^{*}&b^{*}&c^{*}\end{bmatrix}}{\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x\star x\rangle &\langle x\star p\rangle \\\langle p\rangle &\langle p\star x\rangle &\langle p\star p\rangle \end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}\geq 0~.}

Since this positivity condition is true for all a, b, and c, it follows that all the eigenvalues of the matrix are non-negative.

The non-negative eigenvalues then imply a corresponding non-negativity condition on the determinant, det = det ≥ 0 , {\displaystyle \det {\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x\star x\rangle &\langle x\star p\rangle \\\langle p\rangle &\langle p\star x\rangle &\langle p\star p\rangle \end{bmatrix}}=\det {\begin{bmatrix}1&\langle x\rangle &\langle p\rangle \\\langle x\rangle &\langle x^{2}\rangle &\left\langle xp+{\frac {i\hbar }{2}}\right\rangle \\\langle p\rangle &\left\langle xp-{\frac {i\hbar }{2}}\right\rangle &\langle p^{2}\rangle \end{bmatrix}}\geq 0~,} or, explicitly, after algebraic manipulation, σ x 2 σ p 2 = (⟨ x 2 ⟩ − ⟨ x ⟩ 2) (⟨ p 2 ⟩ − ⟨ p ⟩ 2) ≥ (⟨ x p ⟩ − ⟨ x ⟩ ⟨ p ⟩) 2 + ℏ 2 4. {\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}=\left(\langle x^{2}\rangle -\langle x\rangle ^{2}\right)\left(\langle p^{2}\rangle -\langle p\rangle ^{2}\right)\geq \left(\langle xp\rangle -\langle x\rangle \langle p\rangle \right)^{2}+{\frac {\hbar ^{2}}{4}}~.}