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The Golden–Thompson trace inequality, which states that Tr  e H+ K ≤ Tr  e H e K , has proved to be very useful in quantum statistical mechanics. Golden used it to show that the classical free energy is less than the quantum one. Here, we make this G–T inequality more explicit by proving that for some operators, notably the operators of interest in quantum mechanics, H = Δ or [Formula: see text] and K = potential, Tr  e H+(1− u) K e uK is a monotone increasing function of the parameter u for 0 ≤ u ≤ 1. Our proof utilizes an inequality of Ando, Hiai, and Okubo (AHO): Tr  X s Y t X 1− s Y 1− t ≤ Tr  XY for positive operators X, Y and for [Formula: see text], and [Formula: see text]. The obvious conjecture that this inequality should hold up to s + t ≤ 1 was proved false by Plevnik [Indian J. Pure Appl. Math. 47, 491–500 (2016)]. We give a different proof of AHO and also give more counterexamples in the [Formula: see text] range. More importantly, we show that the inequality conjectured in AHO does indeed hold in the full range if X, Y have a certain positivity property—one that does hold for quantum mechanical operators, thus enabling us to prove our G–T monotonicity theorem. 

Abstract We consider the inequality $$f \geqslant f\star f$$ for real functions in $$L^1({\mathbb{R}}^d)$$ where $$f\star f$$ denotes the convolution of $$f$$ with itself. We show that all such functions $$f$$ are nonnegative, which is not the case for the same inequality in $L^p$ for any $$1 < p \leqslant 2$$, for which the convolution is defined. We also show that all solutions in $$L^1({\mathbb{R}}^d)$$ satisfy $$\int _{{\mathbb{R}}^{\textrm{d}}}f(x)\ \textrm{d}x \leqslant \tfrac 12$$. Moreover, if $$\int _{{\mathbb{R}}^{\textrm{d}}}f(x)\ \textrm{d}x = \tfrac 12$$, then $$f$$ must decay fairly slowly: $$\int _{{\mathbb{R}}^{\textrm{d}}}|x| f(x)\ \textrm{d}x = \infty $$, and this is sharp since for all $r< 1$, there are solutions with $$\int _{{\mathbb{R}}^{\textrm{d}}}f(x)\ \textrm{d}x = \tfrac 12$$ and $$\int _{{\mathbb{R}}^{\textrm{d}}}|x|^r f(x)\ \textrm{d}x <\infty $$. However, if $$\int _{{\mathbb{R}}^{\textrm{d}}}f(x)\ \textrm{d}x =: a < \tfrac 12$$, the decay at infinity can be much more rapid: we show that for all $$a<\tfrac 12$$, there are solutions such that for some $$\varepsilon>0$$, $$\int _{{\mathbb{R}}^{\textrm{d}}}e^{\varepsilon |x|}f(x)\ \textrm{d}x < \infty $$.