Dynamics of a Tracer Particle Interacting with Excitations of a Bose–Einstein Condensate

Abstract

We consider the quantum dynamics of a large number N of interacting bosons coupled a tracer particle, i.e. a particle of another kind, on a torus. We assume that in the initial state the bosons essentially form a homogeneous Bose–Einstein condensate, with some excitations. With an appropriate mean-field scaling of the interactions, we prove that the effective dynamics for $N\rightarrow \infty$ is generated by the Bogoliubov–Fröhlich Hamiltonian, which couples the tracer particle linearly to the excitation field.

Self-Adjointness of the Hamiltonians

Here, we prove the relevant self-adjointness and domain properties of $H^\mathrm {aux}$ and $H^\mathrm {BF}$, as well as a useful general lemma.

Lemma A.1

Let $(M_{jk})_{j,k\in \mathbb {N}}\in \ell ^2(\mathbb {N}\times \mathbb {N})$. Then, for any $\Phi \in D(\mathcal {N}_+)$

$$\begin{aligned} \Big \Vert \sum _{j,k\in \mathbb {N}} M_{j,k} a_j a_k \Phi \Big \Vert _{\mathcal {F}_+}&\le \Vert M\Vert _{\ell ^2} \Vert \mathcal {N}_+ \Phi \Vert _{\mathcal {F}_+} \\ \Big \Vert \sum _{j,k\in \mathbb {N}} M_{j,k} a_j^* a_k^* \Phi \Big \Vert _{\mathcal {F}_+}&\le \Vert M\Vert _{\ell ^2} \Vert (\mathcal {N}_+ +2) \Phi \Vert _{\mathcal {F}_+}. \end{aligned}$$

Proof

We only prove the first inequality; the second can be proved in a similar way. We have for any $\Psi \in \mathcal {F}_+$ by the Cauchy–Schwarz inequality

$$\begin{aligned} \Big |\left\langle \Psi , \sum _{j,k\in \mathbb {N}} M_{j,k} a_j a_k \Phi \right\rangle \Big | \le \Vert M\Vert _{\ell ^2} \left( \sum _{j,k\in \mathbb {N}} |\langle \Psi , a_j a_k \Phi \rangle |^2\right) ^{1/2}. \end{aligned}$$

(64)

With (43), we get the claim. $\square$

Lemma A.2

Let $V, W\in L^2(T^d, \mathbb {R})$ satisfy Assumption 2.1. The operator $H_N^\mathrm {aux}$ defined by the expression (27) is self-adjoint on $U_N D(H_N)$ and essentially self-adjoint on $D(H_0)$.

Proof

For the second claim, we prove that $H_N^\mathrm {aux}$ is a perturbation of $-\tfrac{1}{2m}\Delta _x + d\Gamma (-\Delta )^{\le N}$, the projection of $H_0$ to $\mathcal {H}_+^{\le N}$, by a bounded operator.

For the quadratic terms in (27), this follows from the fact that $H_N^\mathrm {aux}$ acts non-trivially only on $\mathcal {H}_+^{\le N}$ and Lemma A.1 together with Parseval’s identity, which yields

$$\begin{aligned} \Vert V_{jk00}\Vert _{\ell ^2(\mathbb {N}\times \mathbb {N})}^2&= \Vert V_{00jk}\Vert _{\ell ^2(\mathbb {N}\times \mathbb {N})}^2 \nonumber \\&\le \sum _{j,k\in \mathbb {N}_0} \Big | \int _{T^d } \int _{T^d } \bar{\varphi }_j(y) \bar{\varphi }_k(y') V(y-y') \mathrm{d}y \mathrm{d}y' \Big |^2 \nonumber \\&= \Vert V(y-y')\Vert _{L^2(T^d \times T^d)}^2 = \Vert V\Vert _{L^2(T^d) }^2, \nonumber \\ \Vert V_{j0k0}\Vert _{\ell ^2(\mathbb {N}\times \mathbb {N})}^2&\le \sum _{j,k\in \mathbb {N}_0} \big |\langle \varphi _j \otimes \varphi _0, V(y_1-y_2) \varphi _0 \otimes \varphi _k \rangle \big |^2 = \Vert V\Vert _{L^2(T^d) }^2.\nonumber \\ \end{aligned}$$

(65)

For the linear term in (27), this follows from the bound $\Vert a(W_x)\Psi \Vert _{\mathcal {H}_+}\le \Vert W\Vert _{L^2} \Vert \sqrt{\mathcal {N}_+} \Psi \Vert _{\mathcal {H}_+}$ by the same reasoning. Hence, $H^\mathrm {aux}_N$ is self-adjoint on the domain of $-\tfrac{1}{2m}\Delta _x + d\Gamma (-\Delta )^{\le N}$ and essentially self-adjoint on $D(H_0)$ by the Kato-Rellich theorem.

To obtain the first claim, it is now sufficient to prove that the difference of $H_N^\mathrm {aux}$ and $U_N H_N U_N^*$ is bounded relative to $d\Gamma (-\Delta )^{\le N}$, with relative bound zero. This difference consists of cubic (49) and quartic (50) terms involving V, and a quadratic term (48) with W. The cubic terms are bounded by an argument analogous to (40). The relative bound for $N^{-1/2}d\Gamma (QW_xQ)^{\le N}$ (i.e. (48)) is a consequence of the bound for $\Psi \in D(d\Gamma (-\Delta ))$, $\Phi \in \mathcal {H}_+^{\le N}$, and $\lambda \ge 0$ (cf. [32, Eqs.2.9, 2.10])

$$\begin{aligned} |\langle \Phi , N^{-1/2} d\Gamma (QW_xQ)^{\le N} \Psi \rangle |&\le N^{-1/2}\Vert \mathcal {N}_+^{1/2} \Phi \Vert _{\mathcal {H}_+^{\le N}} \Vert d\Gamma ((QW_xQ)^2)^{1/2}\Psi \Vert _{\mathcal {H}_+^{\le N}} \nonumber \\&\le \varepsilon \Vert \Phi \Vert _{\mathcal {H}_+^{\le N}} \Vert d\Gamma ((\lambda -\Delta )^2)^{1/2}\Psi \Vert _{\mathcal {H}_+^{\le N}} \nonumber \\&\le \varepsilon \Vert \Phi \Vert _{\mathcal {H}_+^{\le N}}\Vert d\Gamma (\lambda -\Delta )\Psi \Vert _{\mathcal {H}_+^{\le N}}, \end{aligned}$$

(66)

with $\varepsilon =\Vert (\lambda -\Delta )^{-1}W_x\Vert _{L^2 \rightarrow L^2}=\Vert W_x(\lambda -\Delta )^{-1}\Vert _{L^2 \rightarrow L^2}$. Since $\varepsilon$ goes to zero as $\lambda \rightarrow \infty$, because W is infinitesimally $-\Delta$-bounded, this gives an infinitesimal bound. The quartic term is $d\Gamma (-\Delta )^{\le N}$-bounded by the same reasoning and the estimates that give (55).

Proposition A.3

Let $V, W\in L^2(T^d)$, then $H^\mathrm {BF}-H_0$ is $\mathcal {N}_+$-bounded and $H^\mathrm {BF}$ is essentially self-adjoint on $D(H_0)$.

Proof

The first statement follows from Lemma A.1 as above. Consequently, there exists a constant c such that $H^\mathrm {BF}+ c\mathcal {N}_+$ is self-adjoint and positive on $D(H_0)=D(H_0+c\mathcal {N}_+)$ by the Kato–Rellich theorem, since $H^\mathrm {BF}-H_0$ is $H_0 + c\mathcal {N}_+$-bounded with bound less than one.

Essential self-adjointness of $H^\mathrm {BF}$ can now be obtained by applying the commutator theorem [45, Thm.X.36], with $H^\mathrm {BF}+ c\mathcal {N}_+$ as a comparison operator. For this, it is sufficient to prove that

$$\begin{aligned} |\langle \Phi , [\mathcal {N}_+, H^\mathrm {BF}] \Phi \rangle | = |\langle \Phi , {[}\mathcal {N}_+, H^\mathrm {BF}-H_0 ] \Phi \rangle | \le C \langle \Phi , \mathcal {N}_+ \Phi \rangle , \end{aligned}$$

for some constant $C>0$ and all $\Phi \in D(H_0^{1/2})$. Since the commutator of $\mathcal {N}_+$ with $H^\mathrm {Bog}$ is again a quadratic operator, composed of the same terms up to signs, this follows from Lemma A.1 as above. $\square$