1 Introduction

We consider the following Schrödinger operator in \(L^2({\mathbb {R}}^d)\),

$$\begin{aligned} -\Delta -\kappa |x|^{-1} + \Lambda \text { \hspace{0.6cm} with } \kappa> 0, \Lambda > 0, d\ge 2, \end{aligned}$$
(1)

known as Coulomb Hamiltonian. This is a well defined operator and the negative spectrum consists precisely of the eigenvalues

$$\begin{aligned} \Biggl \{\Lambda - \frac{\kappa ^2}{(2k+d-1)^2}: k\in {\mathbb {N}}_0\Biggl \} \text {\hspace{0.3cm} with multiplicity } \mu _{k}=\frac{(d-2+k)!(d-1+2k)}{(d-1)!k!}. \end{aligned}$$

(See for example [1]). Concerning the examination of the spectral properties of the Schrödinger operator family, Cwikel [2], Lieb [3] and Rozenblum [4] found an estimation for the number of negative eigenvalues below zero, usually denoted by \(N(0,-\Delta -V)\), when the dimension is greater or equal to 3.

Theorem 1

(CLR inequality) Let \(d\ge 3\). Then there is a constant \(L_{0,d}<\infty \) such that for all \(V\in L^{1}_{loc}({\mathbb {R}}^d)\) with \(V_{+}\in L^{d/2}({\mathbb {R}}^d)\),

$$\begin{aligned} N(0,-\Delta -V) \le L_{0,d} \int _{{\mathbb {R}}^d} V(x)^{d/2}_{+} dx \end{aligned}$$

Later, Lieb and Thirring [5] generalized the Cwikel–Lieb–Rozenblum inequality to obtain an estimation for the Riesz means of the negative spectrum.

Theorem 2

(Lieb–Thirring inequality) Let \(\gamma \ge 1/2\) if \(d = 1\), \(\gamma > 0\) if \(d = 2\) and \(\gamma \ge 0\) if \(d \ge 3\). Then, there exist a constant \(L_{\gamma ,d}< \infty \) such that, for any V with \(V_{+}\in L^{\gamma + d/2}({\mathbb {R}}^d)\) and \(V_{-} \in L^1_{loc}({\mathbb {R}}^d)\),

$$\begin{aligned} Tr(-\Delta -V)^{\gamma }_{-}\le L_{\gamma ,d}\int _{{\mathbb {R}}^d}V(x)_{+}^{\gamma +d/2} dx. \end{aligned}$$

Although the existence of the constant \(L_{\gamma ,d}\) is known, the sharpest value in some of this inequalities is still an open problem. In order to find the sharpest value, one approach is to study the Lieb–Thirring inequalities in specific cases that can give some information about the constant.

Nevertheless, if we consider the classical Coulomb Hamiltonian,

$$\begin{aligned} -\Delta -\kappa |x|^{-1} \text { \hspace{0.6cm} with } \kappa > 0, d\ge 2, \end{aligned}$$

this operator doesn’t satisfy the hypothesis of Theorem 1. Therefore, it is necessary to add the parameter \(\Lambda > 0\), which acts as a shift, and enables the operator defined in 1 to be an element of \(L^{d/2}({\mathbb {R}}^d)\). Moreover, the integral defined in the right-hand side of Theorem 2 can be explicitely computed and equals

$$\begin{aligned} L_{\gamma ,d}^{cl}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{\gamma +d/2}_{+} dx = 2^{1-d}\kappa ^d\Lambda ^{\gamma -d/2}\frac{\Gamma (\gamma +1) \Gamma \left( \frac{d}{2}-\gamma \right) }{\Gamma (d+1)\Gamma \left( \frac{d}{2}\right) }. \end{aligned}$$

Several months ago, Laptev, Frank and Weidl obtained the following result regarding the 3 dimensional case of the Coulomb Hamiltonian in [1].

Theorem 3

Let \(d=3\) and \(1 \le \gamma < 3/2\). Then for all \(\kappa > 0\) and \(\Lambda > 0\),

$$\begin{aligned} Tr(-\Delta -\kappa |x|^{-1} + \Lambda )^{\gamma }_{-}\le L^{cl}_{\gamma ,3}\int _{{\mathbb {R}}^3}(\kappa |x|^{-1} - \Lambda )^{\gamma +3/2}_{+} dx. \end{aligned}$$

Where

$$\begin{aligned} L_{\gamma ,d}^{cl} = \frac{\Gamma (\gamma +1)}{(4\pi )^{d/2}\Gamma (\gamma + 1+d/2)} \end{aligned}$$

is the classical constant obtained through Weyl asymptotics [6].

Theorem 4

(Weyl asymptotics) Let \(\gamma \ge 0\) and let V be a continuous function on \({\mathbb {R}}^d\) with compact support. Then

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } \alpha ^{-\gamma -d/2}Tr(0,-\Delta - \alpha V)^{\gamma }_{-} = L_{\gamma ,d}^{cl}\int _{{\mathbb {R}}^d} V(x)_{+}^{\gamma + d/2} dx. \end{aligned}$$

Comparing the Lieb–Thirring inequality to the Weyl asymptotics, it is clear that

$$\begin{aligned} L_{\gamma ,d}^{cl} \le L_{\gamma ,d}. \end{aligned}$$
(2)

A slightly different approach is to investigate the best constant concerning only the first eigenvalue, that is

$$\begin{aligned} |E_1|^{\gamma } \le L_{\gamma ,d}^{(1)}\int _{{\mathbb {R}}^d} V(x)_{+}^{\gamma + d/2} dx. \end{aligned}$$
(3)

In this context, the constant \(L_{\gamma ,d}^{(1)}\) is sometimes referred to as the "one-particle constant". Consequently,

$$\begin{aligned} L_{\gamma ,d}^{(1)} \le L_{\gamma ,d}. \end{aligned}$$
(4)

In 1961, Keller [7] raised the variational problem for \(L_{\gamma ,d}^{(1)}\). Independently, Lieb and Thirring [5] arrived at the same optimization problem and showed that it is intimately related to the problem of finding the sharp constant in Sobolev and Gagliardo-Nirenberg inequalities. Combining 2 and 4, it follows

$$\begin{aligned} L_{\gamma ,d} \ge \max \{L_{\gamma ,d}^{cl}, L_{\gamma ,d}^{(1)}\} \end{aligned}$$
(5)

Lieb and Thirring conjectured in [5] that equality holds in 5. In other words, the optimal constant \(L_{\gamma , d}\) is equal to the maximum of \(L_{\gamma ,d}^{cl}\) and \(L_{\gamma ,d}^{(1)}\). For a further discussion of this conjecture, we refer to [8].

The proof of Theorem 3 resides on the fact that the inequality holds when \(\gamma = 1\), and then they extend the result using the following auxiliary lemma, known as the Aizenman–Lieb principle, presented in [9].

Lemma 1

(Aizenman–Lieb principle) For any \(d\ge 1\), the quotient \(L_{\gamma ,d} / L^{cl}_{\gamma ,d}\) is non increasing in \(\gamma \).

In this paper we will present in which cases the classical constant \(L_{\gamma ,d}^{cl}\) is a valid constant for the CLR inequalities of the Coulomb Hamiltonian when working in dimension \(d\ge 4\).

Moreover, we will extend those results by means of the Aizenman–Lieb principle when possible, obtaining the Lieb–Thirring inequalities. Additionally, we give an starting point \(\gamma _0\) for which the Lieb–Thirring inequalities hold (with the classical constant) in those cases where the CLR inequalities don’t.

2 CLR inequalities

We present now the main two results of this article. The first one shows that the classical constant is valid in the case where the dimension is greater or equal than six.

Theorem 5

Let \(d\ge 6\). Then for all \(\kappa > 0\) and \(\Lambda > 0\),

$$\begin{aligned} N(0,-\Delta -\kappa |x|^{-1} + \Lambda )\le L^{cl}_{0,d}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{\gamma +d/2}_{+} dx. \end{aligned}$$

The next one shows that in dimensions 4 and 5, the classical constant is no longer valid.

Theorem 6

Let \(d = 4\) or \(d=5\). Then, there exists some values of \(\kappa >0\) and \(\Lambda >0\) for which the CLR inequality of the correspondent Coulomb Hamiltonian don’t hold with the classical constant.

Both theorems are a consequence of the following lemma.

Lemma 2

Let \(d\ge 4\). Then, there exists \(k_0 > 0\) such that for all \(\kappa > 0\) and \(\Lambda > 0\) satisfying \(\frac{\kappa }{\sqrt{\Lambda }} \ge k_{0}\),

$$\begin{aligned} N(0,-\Delta -\kappa |x|^{-1} + \Lambda )\le L^{cl}_{0,d}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{\gamma +d/2}_{+} dx. \end{aligned}$$

Proof of Lemma 2

In order to compute \(N(0,-\Delta -\kappa |x|^{-1} + \Lambda )\), the first goal is to determine when the quantity

$$\begin{aligned} \frac{\kappa ^2}{(2k+d-1)^2} - \Lambda \end{aligned}$$
(6)

is positive. Making the computations, we obtain that (6) is positive for those \(k \in {\mathbb {N}}\) such that

$$\begin{aligned} k \le \frac{\frac{\kappa }{\sqrt{\Lambda }}-(d-1)}{2}. \end{aligned}$$
(7)

In the following, we will denote \(M:= \lfloor \frac{\frac{\kappa }{\sqrt{\Lambda }}-(d-1)}{2}\rfloor \) the greatest integer less than or equal to \(\frac{\frac{\kappa }{\sqrt{\Lambda }}-(d-1)}{2}\). Hence,

$$\begin{aligned} N(0,-\Delta -\kappa |x|^{-1} + \Lambda )= & {} \sum _{k=0}^{\infty } \mu _{k} \left( \frac{\kappa ^2}{(2k+d-1)^2} - \Lambda \right) ^{0}_{+} \nonumber \\= & {} \sum _{k=0}^{M}\frac{(d-2+k)!(d-1+2k)}{(d-1)!k!}. \end{aligned}$$
(8)

Similar to the case of the Laplace–Beltrami operator, since our potential only depends on |x|, we can reduce the study on \({\mathbb {R}}^d\) to problems on the real line using the spherical harmonics. If \(\nu (l)\) denotes the dimension of the space of spherical harmonics of degree l, then the multiplicity \(\mu _k\) associated to 6 can be written as:

$$\begin{aligned} \mu _k = \sum _{l=0}^{k} \nu (l) = \frac{(d-2+k)!(d-1+2k)}{(d-1)!k!} \end{aligned}$$

The sum in 8 can be explicitly computed and equals

$$\begin{aligned} \sum _{k=0}^{M}\frac{(d-2+k)!(d-1+2k)}{(d-1)!k!} = \frac{(d+2M)\Gamma (d+M)}{\Gamma (d+1)\Gamma (M+1)}. \end{aligned}$$

Simplifying the expression we get,

$$\begin{aligned} N(0,-\Delta -\kappa |x|^{-1} + \Lambda ) \le \frac{1}{2}(d-1)\left( \frac{\kappa }{\sqrt{\Lambda }} + 1\right) ^2\frac{1}{\Gamma (d+1)}. \end{aligned}$$

On the other hand,

$$\begin{aligned} L_{0,d}^{cl}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{0+d/2}_{+} dx = 2^{1-d}\kappa ^d\Lambda ^{-d/2}\frac{1}{\Gamma (d+1)}. \end{aligned}$$

Comparing both expressions, it is easy to see that CLR inequality hold if and only if

$$\begin{aligned} (d-1)\left( 1+\frac{1}{\left( \frac{\kappa }{\sqrt{\Lambda }}\right) }\right) ^2 \le \left( \frac{\kappa }{2\sqrt{\Lambda }}\right) ^{d-2}. \end{aligned}$$
(9)

We now recall two well known facts. Firstly, for any non zero \(x \in {\mathbb {R}}, 1 +x < e^x\). Secondly, by the definition of exponentiation,

$$\begin{aligned} \left( \frac{\kappa }{2\sqrt{\Lambda }}\right) ^{d-2} = e^{(d-2)ln\left( \frac{\kappa }{2\sqrt{\Lambda }}\right) }. \end{aligned}$$

Combining both of them with the strictly increasing condition of the exponential function we get,

$$\begin{aligned} d-1 = 1+(d-2) < e^{d-2} \le e^{(d-2)ln\left( \frac{\kappa }{2\sqrt{\Lambda }}\right) } \,\text { if } \,\frac{\kappa }{\sqrt{\Lambda }} \ge 2e. \end{aligned}$$

To finish our argument, we conclude considering the following limits:

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\left( 1+\frac{1}{\alpha }\right) ^2 = 1 \text { and } \lim _{\alpha \rightarrow \infty } ln\left( \frac{\alpha }{2}\right) = \infty . \end{aligned}$$

Therefore, if the coefficient \(\frac{\kappa }{\sqrt{\Lambda }}\) is big enough, the last inequality holds. We denote by \(k_{0}\) the number such that if \(\frac{\kappa }{\sqrt{\Lambda }} \ge k_{0}\) then the inequality holds for all \(d\ge 4\). Although we can not give an explicit computation of that value for the moment, numerical analysis show us that the inequality is valid when considering a really low value \(k_{0}\). Namely,

$$\begin{aligned} 4.27451< k_{0} < 4.27452. \end{aligned}$$

However, we can proof analytically that the inequality 9 holds when \(\frac{\kappa }{\sqrt{\Lambda }} \ge 4.5\). Although it is not the optimal value for \(k_0\), will be sufficient to justify the argument in the following proofs. If \(\frac{\kappa }{\sqrt{\Lambda }} \ge 4.5\), then

$$\begin{aligned} \left( 1+\frac{1}{\left( \frac{\kappa }{\sqrt{\Lambda }}\right) }\right) ^2 \le \left( 1+\frac{1}{(4.5)}\right) ^2 \text { and } \left( \frac{4.5}{2}\right) ^{d-2} \le \left( \frac{\kappa }{2\sqrt{\Lambda }}\right) ^{d-2}\text { for any } d\ge 4. \end{aligned}$$

Hence, inequality 9 will hold if the function \(h(d) = (d-1)(1+\frac{1}{(4.5)})^2 - (\frac{4.5}{2})^{d-2} \le 0\) when \(d\ge 4\). We note that \(h(4) < 0\) and computing the derivative we see that h(d) is strictly decreasing in \([4.5,\infty )\), completing the proof. \(\square \)

Proof of Theorem 5

We move now to study the case where \(\frac{\kappa }{\sqrt{\Lambda }} < k_0\). We first observe that there exist some cases for which the CLR inequalities hold trivially. In particular, in those cases where we do not have eigenvalues.

$$\begin{aligned} \frac{\frac{\kappa }{\sqrt{\Lambda }}-(d-1)}{2}< 0 \iff \left\{ \begin{array}{lcc} d=4 \text { or } d=5: \frac{\kappa }{\sqrt{\Lambda }}< d-1. \\ \\ d \ge 6: \frac{\frac{\kappa }{\sqrt{\Lambda }}-(d-1)}{2}< \frac{k_0-(d-1)}{2} < 0.\\ \end{array} \right. \end{aligned}$$

Therefore, if \(d\ge 6\) the CLR inequalities hold trivially when \(\frac{\kappa }{\sqrt{\Lambda }} < k_0\). This completes the proof of Theorem 5. \(\square \)

Proof of Theorem 6

For the two remaining cases, \(d=4\) and \(d=5\), we have already proved that if

$$\begin{aligned} \frac{\kappa }{\sqrt{\Lambda }} \in (0,d-1) \cup [k_0,\infty ), \end{aligned}$$

then the CLR inequalities for the classical constant are valid. If we move to the remaining case, due to the value of \(k_0\), the condition 7 holds if and only if \(k=0\). Therefore, we consider the CLR inequalities associated with one eigenvalue, \(-\frac{\kappa ^2}{(d-1)^2}\).

$$\begin{aligned} 1 \le 2^{1-d}\kappa ^d\Lambda ^{-d/2}\frac{1}{\Gamma (d+1)} = 2^{1-d}\frac{1}{\Gamma (d+1)}\left( \frac{\kappa }{\sqrt{\Lambda }}\right) ^d \end{aligned}$$

In those cases, the CLR inequalities hold if the quotient \(\frac{\kappa }{\sqrt{\Lambda }}\) satisfies the following condition:

$$\begin{aligned} 2^{(1-1/d)}\Gamma (d+1)^{1/d} \le \frac{\kappa }{\sqrt{\Lambda }} \end{aligned}$$

Nevertheless, the interval \([d-1, 2^{(1-1/d)}\Gamma (d+1)^{1/d})\) is non empty in both these cases. In particular, if we take \(\Lambda = 1\) and \(\kappa = d-1\), we get an example of the Coulomb Hamiltonian for which the CLR inequalities with the classical constant don’t hold in dimension 4 and dimension 5. Concluding the proof of Theorem 6. \(\square \)

3 Lieb–Thirring inequalities

As we advanced in the introduction, we extend Theorem 5 by means of the Aizenman–Lieb principle (Lemma 1), obtaining

Corollary 1

Let \(d\ge 6\) and \(0 \le \gamma < d/2\). Then for all \(\kappa > 0\) and \(\Lambda > 0\),

$$\begin{aligned} Tr(-\Delta -\kappa |x|^{-1} + \Lambda )^{\gamma }_{-}\le L^{cl}_{\gamma ,d}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{\gamma +d/2}_{+} dx. \end{aligned}$$

In the cases where \(d=4\) or \(d=5\), we obtain the following corolary.

Corollary 2

Let \(d=4\) or \(d=5\). Then, there exist \(0<\gamma _{0,d}<\frac{d}{2}\) as in (14) such that for all \(\gamma _{0,d} \le \gamma < d/2\), \(\kappa > 0\) and \(\Lambda > 0\),

$$\begin{aligned} Tr(-\Delta -\kappa |x|^{-1} + \Lambda )^{\gamma }_{-}\le L^{cl}_{\gamma ,d}\int _{{\mathbb {R}}^d}(\kappa |x|^{-1} - \Lambda )^{\gamma +d/2}_{+} dx. \end{aligned}$$

Proof of Corolary 2

From Lemma 2, the study of trivial cases in the proof of Theorem 5 and the study of the CLR inequalities associated with one eigenvalue in the proof of Theorem 6, we know that if

$$\begin{aligned} \frac{\kappa }{\sqrt{\Lambda }} \in (0,d-1) \cup [2^{(1-1/d)}\Gamma (d+1)^{1/d},\infty ), \end{aligned}$$

then we can use the Aizenman–Lieb principle to extend the validity of the classical constant. To solve the remaining case, \(\frac{\kappa }{\sqrt{\Lambda }}\in [d-1, 2^{(1-1/d)}\Gamma (d+1)^{1/d})\), we consider the following variational problem associated to one eigenvalue,

$$\begin{aligned} |E_1|^{\gamma } \le L_{\gamma ,d}^{cl}\int _{{\mathbb {R}}^d} V(x)_{+}^{\gamma + d/2} dx. \end{aligned}$$

In our case, it reads as

$$\begin{aligned} \left( \frac{\kappa }{d-1}\right) ^{2\gamma } \le 2^{1-d}\kappa ^d\Lambda ^{\gamma - d/2}\frac{\Gamma {(\gamma + 1)}\Gamma {\left( \frac{d}{2}-\gamma \right) }}{\Gamma {(d + 1)}\Gamma {\left( \frac{d}{2}\right) }}. \end{aligned}$$

When \(d=4\), can be rewritten as

$$\begin{aligned} \left( \frac{\kappa }{\sqrt{\Lambda }}\right) ^{2(\gamma -2)} \le \frac{3^{2\gamma -1}}{2^6}\Gamma (\gamma +1)\Gamma (2-\gamma ). \end{aligned}$$
(10)

And when \(d=5\), we obtain

$$\begin{aligned} \left( \frac{\kappa }{\sqrt{\Lambda }}\right) ^{2\gamma -5} \le \frac{2^{4\gamma -5}}{5\cdot 3^2\cdot \sqrt{\pi }}\Gamma (\gamma +1)\Gamma \left( \frac{5}{2}-\gamma \right) . \end{aligned}$$
(11)

In order to solve 10 and 11 with respect to the parameter \(\gamma \), we note that for any \(0< \gamma < \frac{d}{2}\) the left hand side of both expressions is always less or equal to it’s value at the point \(d-1\). If we substitute the value and simplify, we can rewrite 10 and 11 as

$$\begin{aligned}{} & {} \frac{2^6}{3^3} \le \Gamma (\gamma +1)\Gamma (2-\gamma ). \end{aligned}$$
(12)
$$\begin{aligned}{} & {} \frac{5\cdot 3^2\cdot \sqrt{\pi }}{2^5} \le \Gamma (\gamma +1)\Gamma \left( \frac{5}{2}-\gamma \right) . \end{aligned}$$
(13)

Using the properties of the Gamma function, in particular using the fact that the right hand side of both equations tends to infinity when \(\gamma \) tends to \(\frac{d}{2}\), we know both equations have solution. We will denote it by \(\gamma _{0,d}\). Solving numerically,

$$\begin{aligned} 1,5< \gamma _{0,4}<1,51 \text { and } 1,86< \gamma _{0,5} <1,87. \end{aligned}$$
(14)

Therefore, we can extend the result by the Aizenman–Lieb principle.\(\square \)