1 Erratum to: J Stat Phys (2014) 154:866–876 DOI 10.1007/s10955-014-0915-0

A sign misprint in the statement of Proposition 3.1 in [1] has a consequence in the statement of the main Theorem 1.1 in [1]. We write a short note in which we present the correct statements.

The correct version of Proposition 3.1 in [1] is the following one.

Proposition 0.1

Let \(A:{\mathbb R}^3\rightarrow {\mathbb R}^3\). For any \(\phi =\phi (x)\in \mathcal C^{\infty }_0({\mathbb R}^3;{\mathbb C}^2)\), the following identity holds

$$\begin{aligned} \int _{{\mathbb R}^3} r\left| \sigma \cdot \nabla _A\phi \right| ^2&= \int _{{\mathbb R}^3} r\left| \partial _r^A\phi \right| ^2 \,dx\\ \nonumber&\qquad +\int _{{\mathbb R}^3} r\left| \frac{1}{r}\left( \sigma \cdot L_A+1\right) \phi \right| ^2 \,dx -\int _{{\mathbb R}^3} \frac{|\phi |^2}{r} \,dx\\ \nonumber&\qquad +\int _{{\mathbb R}^3}\langle \sigma \cdot [\partial _r(x\wedge A)]\phi ,\phi \rangle \,dx -\int _{{\mathbb R}^3}\left\langle \sigma \cdot \left( x\wedge \nabla A_r\right) \phi ,\phi \right\rangle \,dx \end{aligned}$$
(0.1)

where \(r:=|x|\), \(\partial _r^A:=\frac{x}{r}\cdot \nabla _A\), and \(A_r:=A\cdot \frac{x}{r}\).

The main difference between Proposition 3.1 in [1] and the proposition above is the negative sign in from of the last term in (0.1).

We also underline that the identity (0.1) follows by the identities (3.7) and (3.11) in [1], whose proof given in [1], is correct.

In order to present a correct version of Theorem 1.1 in [1] we first need the following version of Proposition 2.1 in [1].

Proposition 0.2

Let \(B:{\mathbb R}^3\setminus \{0\}\rightarrow {\mathbb R}^3\), and assume that there exists a vector potential \(A:{\mathbb R}^3\rightarrow {\mathbb R}^3\) such that the following holds

$$\begin{aligned} A\cdot x=0, \qquad A(\lambda x)=\lambda ^{-1}A(x), \qquad \text {curl}\,A(x)=B(x), \end{aligned}$$
(0.2)

for all \(x\in {\mathbb R}^3\setminus \{0\}\), \(\lambda \ne 0\). Then we have

$$\begin{aligned} \partial _r\left( x\wedge A\right) -x\wedge \nabla \left( A\cdot \frac{x}{r}\right) =0 \end{aligned}$$
(0.3)

for \(\in {\mathbb R}^3\setminus \{0\}\).

Proof

The proof is quite immediate. Due to the first condition in (0.2), one has \(x\wedge \nabla (A\cdot x/r)=0\). Moreover, since \(A\) is homogeneous of degree \(-1\), then \(x\wedge A\) is homogeneous of degree \(0\), hence \(\partial _r(x\wedge A)\equiv 0\). \(\square \)

We can now state the new and correct version of Theorem 1.1 in [1].

Theorem 0.1

Let \(B:{\mathbb R}^3\setminus \{0\}\rightarrow {\mathbb R}^3\), and assume that there exists a vector potential \(A:{\mathbb R}^3\rightarrow {\mathbb R}^3\) such that the following holds

$$\begin{aligned} A\cdot x=0, \qquad A(\lambda x)=\lambda ^{-1}A(x), \qquad \text {curl}\,A(x)=B(x), \end{aligned}$$
(0.4)

for all \(x\in {\mathbb R}^3\setminus \{0\}\), \(\lambda \ne 0\). Then, for any \(\phi =\phi (x)\in \mathcal C^{\infty }_0({\mathbb R}^3;{\mathbb C}^2)\), the following inequality holds

$$\begin{aligned} \mu _1\int _{{\mathbb R}^3}\frac{|\phi |^2}{|x|}\,dx \le \int _{{\mathbb R}^3} \frac{1}{|x|}\left| \left( \sigma \cdot L_A+1\right) \phi \right| ^2 \,dx \le \int _{{\mathbb R}^3}|x|\left| \sigma \cdot \nabla _A\phi \right| ^2\,dx, \end{aligned}$$
(0.5)

where \(\mu _1=\mu _1(A)=\inf \left\{ \text {spec} (\sigma \cdot {L}_{A}+1) \cap [0, \infty )\right\} \).

Proof

The proof is an immediate corollary of Propositions 0.1 and 0.2, and the usual radial Hardy inequality. \(\square \)