High density limit of the Fermi polaron with infinite mass

Ulrich Linden; David Mitrouskas

doi:10.1007/s11005-019-01158-y

Letters in Mathematical Physics

pp 1–21 | Cite as

High density limit of the Fermi polaron with infinite mass

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Ulrich Linden
David Mitrouskas

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First Online: 19 January 2019

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Abstract

We analyze the ground state energy for N identical fermions in a two-dimensional box of volume $L^2$ interacting with an external point scatterer. Since the point scatterer can be considered as an impurity particle of infinite mass, this system is a limit case of the Fermi polaron. We prove that its ground state energy in the limit of high density $N/L^2\gg 1$ is given by the polaron energy. The polaron energy is an energy estimate based on trial states up to first order in particle-hole expansion, which was proposed by Chevy (Phys Rev A 74:063628, 2006) in the physics literature. The relative error in our result is shown to be small uniformly in L. Hence, we do not require a gap of fixed size in the spectrum of the Laplacian on the box. The strategy of our proof relies on a twofold Birman–Schwinger type argument applied to the many-particle Hamiltonian of the system.

Keywords

Point interaction Fermi polaron Energy asymptotics Thermodynamic limit

Mathematics Subject Classification

Primary 81V70 Secondary 35P15

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Notes

Acknowledgements

We thank Marcel Griesemer for extended discussions and helpful remarks. Our work was supported by the Deutsche Forschungsgemeinschaft (DFG) through the Research Training Group 1838: Spectral Theory and Dynamics of Quantum Systems.

Appendix A

In this appendix, we prove (13) by applying Theorem 6.3 and Corollary 6.4 of [9] to the present case.

The one-body operator h is the limit in the norm resolvent sense of

$$\begin{aligned} h_n := -\Delta - g_n \left| \eta _n \right\rangle \left\langle \eta _n \right| \end{aligned}$$

as $n \rightarrow \infty $, where $\eta _n := L^{-1} \cdot \sum _{k^2 \le n} \varphi _k$ and $g_n^{-1} = \left\langle \eta _n,(-\Delta - E_B)^{-1} \eta _n \right\rangle $. Thus, $\exp (ith_n) \rightarrow \exp (ith)$ strongly as $n \rightarrow \infty $ for all $t \in \mathbb {R}$, cf. [20, Theorem VIII.21]. Hence, $(\exp (ith_n))^{\otimes N(\mu )}$ converges strongly to $(\exp (ith))^{\otimes N(\mu )}$ for all $t \in \mathbb {R}$, where the former is the unitary group of

$$\begin{aligned} H_\mu ^{(n)} := d\Gamma (h_n) \! \upharpoonright \!\mathscr {H}_{N(\mu )} = (T - g_n a^*(\eta _n) a(\eta _n))\! \upharpoonright \!\mathscr {H}_{N(\mu )} \end{aligned}$$

and the latter is the unitary group of $H_\mu = d\Gamma (h) \! \upharpoonright \!\mathscr {H}_{N(\mu )}$. This implies that $H_\mu ^{(n)} \rightarrow H_\mu $ as $n \rightarrow \infty $ in the strong resolvent sense.

The resolvent of $H_\mu ^{(n)}$ can be given explicitly. In fact,

$$\begin{aligned} (H_\mu ^{(n)} - z)^{-1} = (T - z)^{-1} + (T - z)^{-1} a^*(\eta _n) \, \phi _\mu ^{(n)}(z)^{-1} \, a(\eta _n) (T - z)^{-1}, \end{aligned}$$

(40)

on $\mathscr {H}_{N(\mu )}$ for all $z \in \rho (T \! \upharpoonright \!\mathscr {H}_{N(\mu )})$ satisfying $0 \in \rho (\phi _\mu ^{(n)}(z))$, where

$$\begin{aligned} \phi _\mu ^{(n)}(z) := (g_n^{-1} - a(\eta _n) (T - z)^{-1} a^*(\eta _n)) \! \upharpoonright \!\mathscr {H}_{N(\mu )- 1}. \end{aligned}$$

For $z < 0$ in normal-ordered form,

$$\begin{aligned} \phi _\mu ^{(n)}(z) = \frac{1}{L^2} \sum _{k^2 \le n} \left( \frac{1}{k^2 - E_B} - \frac{1}{T + k^2 - z} \right) + \frac{1}{L^2} \sum _{k^2,l^2 \le n} a_k^* \, \frac{1}{T + k^2 + l^2 - z} \, a_l \end{aligned}$$

(41)

on $\mathscr {H}_{N(\mu )}$.

With this at hand, one shows the following facts about the terms in (40).

(i)

$a(\eta _n) (T-z)^{-1}$ converges in $\mathscr {L}(\mathscr {H}_{N(\mu )},\mathscr {H}_{N(\mu )-1})$ to an operator $R_z$. The existence of this limit is easily established with the help of (17).
(ii)

Let $D \subseteq \mathscr {H}_{N(\mu )-1}$ denote the space of finite linear combinations of states of the form $\varphi _{k_1} \wedge ... \wedge \varphi _{k_{N(\mu )-1}}$. Then for $\psi \in D$ and $z < 0$,
$$\begin{aligned} \phi _\mu ^{(n)}(z) \psi \rightarrow \phi _\mu (z) \psi \qquad (n \rightarrow \infty ) \end{aligned}$$
(cf. (14), (41)), and $\phi _\mu (z)$ is essentially self-adjoint on D. For the latter fact note that the last term in (14) is a bounded operator.
(iii)

Since the last term in (41) is a positive operator (cf. (19)) and $T \! \upharpoonright \!\mathscr {H}_{N(\mu )-1} \ge E_0(\mu ) - \mu $, there is a $c_z > 0$ for each $z < E_0(\mu ) - \mu + E_B$ such that $\phi _\mu ^{(n)}(z) \ge c_z$ for all $n \in \mathbb {N}$.

These facts imply that we are in the situation of Theorem 4.2 in [9]. Hence, by this theorem

$$\begin{aligned} (H_\mu - z)^{-1} = (T - z)^{-1} + R_{\overline{z}}^* \phi _\mu (z)^{-1} R_z \end{aligned}$$

(42)

on $\mathscr {H}_{N(\mu )}$ for $z < E_0(\mu ) - \mu + E_B$.

Since $(T - z)^{-1} \! \upharpoonright \!\mathscr {H}_{N(\mu )}$ is compact and $\mathrm {Ker}R_z^* = \{ 0 \}$, we can apply the results of [9, Section 6] including Theorem 6.3 and Corollary 6.4, which yields (13).

Appendix B

Lemma B.1

(a) Let $f:[0,\infty )\rightarrow [0,\infty )$ be monotonically decreasing. Then,

$$\begin{aligned} \left| \frac{1}{L^2} \sum _{k}f(k^2) - \frac{1}{2\pi }\int \limits _0^{\infty } f(t^2)t\, {\hbox {d}}t \right| \le \frac{2}{\pi L}\int \limits _0^{\infty }f(t^2)\,{\hbox {d}}t + \frac{3 f(0)}{L^2}. \end{aligned}$$

(43)

(b) Let $m>0$ and $f:[m,\infty )\rightarrow [0,\infty )$ be monotonically decreasing. Then,

$$\begin{aligned} \left| \frac{1}{L^2} \sum _{k^2>m}f(k^2) - \frac{1}{2\pi }\int \limits _{\sqrt{m}}^{\infty } f(t^2)t\, {\hbox {d}}t \right| \le \frac{2}{\pi L} \int \limits _{\sqrt{m}}^{\infty } f(t^2)\, {\hbox {d}}t + \left( \frac{4 \sqrt{m}}{\pi L} + \frac{6}{L^2} \right) f(m). \end{aligned}$$

(44)

Proof

(a) By the symmetry of the function $k\mapsto f(k^2)$,

$$\begin{aligned} \sum _{k\in \frac{2\pi }{L}\mathbb {Z}^2}f(k^2) = 4 \sum _{k\in \frac{2\pi }{L}\mathbb {Z}_{+}^2}f(k^2) + 4 \sum _{k\in \frac{2\pi }{L}(\mathbb {Z}_{+}\times \{0\})}f(k^2) + f(0), \end{aligned}$$

which, by the monotonicity of f is bounded from above by

$$\begin{aligned} \frac{L^2}{2\pi }\int \limits _0^{\infty } f(t^2)t\, {\hbox {d}}t + \frac{2L}{\pi }\int \limits _0^{\infty }f(t^2)\,{\hbox {d}}t + f(0). \end{aligned}$$

Moreover, by the symmetry of the function $k\mapsto f(k^2)$,

$$\begin{aligned} \sum _{k\in \frac{2\pi }{L}\mathbb {Z}^2}f(k^2) = 4 \sum _{k\in \frac{2\pi }{L}\mathbb {Z}_{\ge 0}^2}f(k^2) - 4 \sum _{k\in \frac{2\pi }{L}(\mathbb {Z}_{+}\times \{0\})}f(k^2) - 3f(0), \end{aligned}$$

which, by the monotonicity of f is bounded from below by

$$\begin{aligned} \frac{L^2}{2\pi }\int \limits _0^{\infty } f(t^2)t\, {\hbox {d}}t - \frac{2L}{\pi }\int \limits _0^{\infty }f(t^2)\,{\hbox {d}}t - 3 f(0). \end{aligned}$$

The combination of upper and lower bound yields the statement of the lemma.

(b) We write

$$\begin{aligned} \frac{1}{L^2}\sum _{k^2>m} f(k^2) = \frac{1}{L^2}\sum _{k} g_m(k^2) - \frac{1}{L^2}\sum _{k} h_m(k^2), \end{aligned}$$

with the nonnegative, monotonically decreasing functions

$$\begin{aligned} g_m(k^2)&:= \chi _{(m,\infty )}(k^2) f(k^2) + \chi _{[0,m]}(k^2) f(m), h_m(k^2) := \chi _{[0,m]}(k^2) f(m). \end{aligned}$$

The left-hand side of (44) can thus be estimated in terms of

$$\begin{aligned} \left| \frac{1}{L^2} \sum _{k} g_m(k^2) - \frac{1}{2\pi } \int \limits _{0}^\infty g_m (t^2) t\; {\hbox {d}}t \right| + \left| \frac{1}{L^2} \sum _{k} h_m(k^2) - \frac{1}{2\pi } \int \limits _{0}^\infty h_m (t^2) t\; {\hbox {d}}t \right| . \end{aligned}$$

To the latter expression, we apply (43) leading to the stated bound in (b). $\square $

Appendix C

Lemma C.1

Let A and B be two self-adjoint operators on a Hilbert space $\mathscr {H}$ such that $A - B$ is of rank one. Then,

$$\begin{aligned} \mu _{i+1}(A) \ge \mu _i(B) \end{aligned}$$

for all $i \in \mathbb {N}$. Here, $\mu _i(A)$ and $\mu _i(B)$ denote the min–max values of A and B, respectively.

Proof

Let $i \in \mathbb {N}$. By definition of the min–max values,

$$\begin{aligned} \mu _{i+1}(A) = \sup _{\psi _1,...,\psi _i} \, \inf _{\varphi \perp \psi _1, ... ,\psi _i} \frac{\left\langle \varphi ,A \varphi \right\rangle }{\left\| \varphi \right\| ^2}. \end{aligned}$$

Restricting the supremum in this expression by fixing $\psi _i$ to a nonzero $v \in \mathrm {Ran}(A-B)$ yields a lower bound for $\mu _{i+1}(A)$. Since $A - B$ is of rank one, $0 \ne \varphi \perp v$ implies $v \in \mathrm {Ran}(A-B)^\perp = \mathrm {Ker}(A-B)$ and $\left\langle \varphi ,A \varphi \right\rangle = \left\langle \varphi ,B \varphi \right\rangle $. Hence,

$$\begin{aligned} \mu _{i+1}(A) \ge \sup _{\psi _1,...,\psi _{i-1}} \, \inf _{\varphi \perp \psi _1, ... ,\psi _{i-1},v} \frac{\left\langle \varphi ,B \varphi \right\rangle }{\left\| \varphi \right\| ^2}. \end{aligned}$$

Lowering the right-hand side by extending the infimum to all $\varphi $ perpendicular to $\psi _1, ..., \psi _{i-1}$ only, yields the statement of the lemma. $\square $

Proof of (8)

We apply Lemma C.1 to $A := -(h-z)^{-1}$ and $B := -(-\Delta -z)^{-1}$ for $z< E_B$. Since the spectra of these operators are purely discrete and bounded from below, the min–max values of A and B coincide with the respective eigenvalues. Then, $\lambda _i = z - 1/\mu _i(A)$ and $\lambda _i^0 = z - 1/{\lambda _i(B)}$ for all $i \in \mathbb {N}$, and (8) is a direct consequence of Lemma C.1. $\square $

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Ulrich Linden
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David Mitrouskas
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1.Fachbereich MathematikUniversität StuttgartStuttgartGermany

High density limit of the Fermi polaron with infinite mass

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