Spontaneous Symmetry Breaking in Coupled Bose–Einstein Condensates

Abstract

We study a system of two hardcore bosonic Hubbard models weakly coupled with each other by tunneling. Assuming that the single uncoupled model exhibits off-diagonal long-range order, we prove that the coupled system exhibits spontaneous symmetry breaking (SSB) in the infinite volume limit, in the sense that the two subsystems maintain a definite relative U(1) phase when the tunneling is turned off. Although SSB of the U(1) phase is never observable in a single system, SSB of the relative U(1) phase is physically meaningful and observable by interference experiments. The present theorem is made possible by the rigorous theory of low-lying states and SSB in quantum antiferromagnets developed over the years.

Appendix: Properties of the State \(\left| \Theta ^{L,M}\right\rangle \)

1.1 A.1 Expectation Value of $\hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}$

Let us prove (31). Here we make use of techniques and results from [12]. For $\nu =\mathrm a,b$, let $\hat{o}_\nu ^{+}:=L^{-d}\sum _{x\in \Lambda }\hat{a}^\dagger _{(x,\nu )}$ and $\hat{o}_\nu ^{-}:=L^{-d}\sum _{x\in \Lambda }\hat{a}_{(x,\nu )}$. We here abbreviate $\left| \Theta ^{L,M}\right\rangle$ as $\left| \Theta \right\rangle$. From (40) and (20), we have

$$\begin{aligned} \left| \Theta \right\rangle =&\frac{1}{\sqrt{2M+1}} \left\{ \left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle +\sum _{n=1}^M\frac{(\hat{o}_{\mathrm{a}}^+)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle }{\Vert (\hat{o}_{\mathrm{a}}^+)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert }\frac{(\hat{o}_{\mathrm{b}}^-)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle }{\Vert (\hat{o}_{\mathrm{b}}^-)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \nonumber \right. \\&\left. +\sum _{n=1}^M\frac{(\hat{o}_{\mathrm{a}}^-)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle }{\Vert (\hat{o}_{\mathrm{a}}^-)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert }\frac{(\hat{o}_{\mathrm{b}}^+)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle }{\Vert (\hat{o}_{\mathrm{b}}^+)^n\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \right\} , \end{aligned}$$

(42)

where $\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle$ and $\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle$ are exact copies of the ground state $\left| \Phi _\mathrm {GS}\right\rangle$ of the Hamiltonian (1) for a single system with N particles.

To evaluate the expectation value $\left\langle \Theta \right| \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle$, we first see that

$$\begin{aligned} \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle =\frac{1}{\sqrt{2M+1}}&\left\{ \sum _{n=1}^{M+1} \frac{\hat{a}^\dagger _{(x,\mathrm{a})}(\hat{o}_{\mathrm{a}}^+)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle }{\Vert (\hat{o}_{\mathrm{a}}^+)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert } \frac{\hat{a}_{(x,\mathrm{b})}(\hat{o}_{\mathrm{b}}^-)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle }{\Vert (\hat{o}_{\mathrm{b}}^-)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \nonumber \right. \\&\left. \quad +\sum _{n=0}^{M-1} \frac{\hat{a}^\dagger _{(x,\mathrm{a})}(\hat{o}_{\mathrm{a}}^-)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle }{\Vert (\hat{o}_{\mathrm{a}}^-)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert } \frac{\hat{a}_{(x,\mathrm{b})}(\hat{o}_{\mathrm{b}}^+)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle }{\Vert (\hat{o}_{\mathrm{b}}^+)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \right\} . \end{aligned}$$

(43)

By using (42) and (43), we get

$$\begin{aligned}&\left\langle \Theta \right| \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle \nonumber \\&\quad =\frac{1}{2M+1} \left\{ \sum _{n=1}^{M+1} \frac{ \bigl \langle (\hat{o}_{\mathrm{a}}^-)^n\hat{a}^\dagger _{(x,\mathrm{a})}(\hat{o}_{\mathrm{a}}^+)^{n-1}\bigr \rangle _{\mathrm{a}}\,\, \bigl \langle (\hat{o}_{\mathrm{b}}^+)^n\hat{a}_{(x,\mathrm{b})}(\hat{o}_{\mathrm{b}}^-)^{n-1}\bigr \rangle _{\mathrm{b}} }{ \Vert (\hat{o}_{\mathrm{a}}^+)^{n}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{a}}^+)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{b}}^-)^{n}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{b}}^-)^{n-1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \nonumber \right. \\&\left. \qquad +\sum _{n=0}^{M-1} \frac{ \bigl \langle (\hat{o}_{\mathrm{a}}^+)^n\hat{a}^\dagger _{(x,\mathrm{a})}(\hat{o}_{\mathrm{a}}^-)^{n+1}\bigr \rangle _{\mathrm{a}}\,\, \bigl \langle (\hat{o}_{\mathrm{b}}^-)^n\hat{a}_{(x,\mathrm{b})}(\hat{o}_{\mathrm{b}}^+)^{n+1}\bigr \rangle _{\mathrm{b}} }{ \Vert (\hat{o}_{\mathrm{a}}^-)^{n}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{a}}^-)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{a}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{b}}^+)^{n}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert \Vert (\hat{o}_{\mathrm{b}}^+)^{n+1}\left| \Phi _{\mathrm{GS}}^{\mathrm{b}}\right\rangle \Vert } \right\} , \end{aligned}$$

(44)

where we wrote $\left\langle \cdots \right\rangle _\nu =\left\langle \Phi ^\nu _{\mathrm{GS}}\right| \cdots \left| \Phi ^\nu _{\mathrm{GS}}\right\rangle$. Note that translation invariance implies, e.g.,

$\left\langle (\hat{o}_{\mathrm{a}}^+)^n\hat{a}^\dagger _{(x,\mathrm{a})}(\hat{o}_{\mathrm{a}}^-)^{n+1}\right\rangle _{\mathrm{a}}=\left\langle (\hat{o}^+)^{n+1}(\hat{o}^-)^{n+1}\right\rangle$, where $\left\langle \cdots \right\rangle =\left\langle \Phi _{\mathrm{GS}}\right| \cdots \left| \Phi _{\mathrm{GS}}\right\rangle$ and $\hat{o}^{\pm }:=\hat{\mathcal{O}}^{\pm }/L^d$ are defined for the single system. We can thus rewrite (44) as

$$\begin{aligned}&\left\langle \Theta \right| \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle \nonumber \\&\quad =\frac{1}{2M+1} \left\{ \sum _{n=1}^{M+1} \frac{ \bigl \langle (\hat{o}^-)^n(\hat{o}^+)^n\bigr \rangle \, \bigl \langle (\hat{o}^+)^n(\hat{o}^-)^n\bigr \rangle }{ \sqrt{\bigl \langle (\hat{o}^-)^n(\hat{o}^+)^n\bigr \rangle \,\bigl \langle (\hat{o}^-)^{n-1}(\hat{o}^+)^{n-1}\bigr \rangle \, \bigl \langle (\hat{o}^+)^n(\hat{o}^-)^n\bigr \rangle \,\bigl \langle (\hat{o}^+)^{n-1}(\hat{o}^-)^{n-1}\bigr \rangle } } \nonumber \right. \\&\left. \qquad +\sum _{n=0}^{M-1} \frac{ \bigl \langle (\hat{o}^+)^{n+1}(\hat{o}^-)^{n+1}\bigr \rangle \, \bigl \langle (\hat{o}^-)^{n+1}(\hat{o}^+)^{n+1}\bigr \rangle }{ \sqrt{ \bigl \langle (\hat{o}^+)^n(\hat{o}^-)^n\bigr \rangle \,\bigl \langle (\hat{o}^+)^{n+1}(\hat{o}^-)^{n+1}\bigr \rangle \, \bigl \langle (\hat{o}^-)^n(\hat{o}^+)^n\bigr \rangle \,\bigl \langle (\hat{o}^-)^{n+1}(\hat{o}^+)^{n+1}\bigr \rangle } } \right\} , \nonumber \\&\quad =\frac{2}{2M+1} \sum _{n=1}^M\sqrt{\frac{ \bigl \langle (\hat{o}^-)^n(\hat{o}^+)^n\bigr \rangle \,\bigl \langle (\hat{o}^+)^n(\hat{o}^-)^n\bigr \rangle }{ \bigl \langle (\hat{o}^-)^{n-1}(\hat{o}^+)^{n-1}\bigr \rangle \,\bigl \langle (\hat{o}^+)^{n-1}(\hat{o}^-)^{n-1}\bigr \rangle }}. \end{aligned}$$

(45)

Following [12], let $\hat{p}=(\hat{o}^+\hat{o}^-+\hat{o}^-\hat{o}^+)/2$. Since $[\hat{o}^-,\hat{o}^+]=L^{-d}$, one finds that $(\hat{o}^-)^n(\hat{o}^+)^n=\hat{p}^n+O(L^{-d})$ and $(\hat{o}^+)^n(\hat{o}^-)^n=\hat{p}^n+O(L^{-d})$. See Lemma 4.1 of [12]. We therefore get

$$\begin{aligned} \left\langle \Theta \right| \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle = \frac{2}{2M+1} \sum _{n=1}^M\frac{\left\langle \hat{p}^n\right\rangle }{\left\langle \hat{p}^{n-1}\right\rangle }+O(L^{-d}). \end{aligned}$$

(46)

It was proved in Lemma 4.2 of [12] that

$$\begin{aligned} \lim _{n\uparrow \infty }\liminf _{L\uparrow \infty }\frac{\left\langle \hat{p}^n\right\rangle }{\left\langle \hat{p}^{n-1}\right\rangle }=(m^*)^2. \end{aligned}$$

(47)

This, with (46), implies

$$\begin{aligned} \lim _{M\uparrow \infty }\liminf _{L\uparrow \infty }\left\langle \Theta \right| \hat{a}^\dagger _{(x,\mathrm{a})}\hat{a}_{(x,\mathrm{b})}\left| \Theta \right\rangle =(m^*)^2, \end{aligned}$$

(48)

which is the desired (31).

1.2 A.2 Energy Expectation Value

We shall prove (32). Let $E_{\mathrm{GS}}$ be the ground state energy of the Hamiltonian (1) for a single system with N particles. From (21), which shows that $\left| \Gamma _M\right\rangle$ has very low excitation energy, one readily finds that

$$\begin{aligned} \lim _{L\uparrow \infty }\frac{1}{L^d}\left\{ \left\langle \Theta ^{L,M}\right| \hat{H}^{\mathrm{tot}}_0\left| \Theta ^{L,M}\right\rangle -2E_{\mathrm{GS}}\right\} =0 \end{aligned}$$

(49)

for any M. This is almost the desired (32) since it is very likely that $E^{\mathrm{tot}}_{\mathrm{GS},0}=2E_{\mathrm{GS}}$. But this equality cannot be proved in general and we need some work.

Define the ground state energy density (for the single system) by

$$\begin{aligned} {\tilde{\epsilon }}(\rho ):=\lim _{L\uparrow \infty }\frac{E_{\mathrm{GS}}}{L^d}, \end{aligned}$$

(50)

where L and N always satisfy $\rho =N/L^d$. It is standard that the limit exists, and ${\tilde{\epsilon }}(\rho )$ extends to a convex function of $\rho \in [0,1]$. See, e.g., [31, 32].

We then note that

$$\begin{aligned} \lim _{L\uparrow \infty }\frac{E^{\mathrm{tot}}_{\mathrm{GS},0}}{L^d}=\min _{\delta }\{{\tilde{\epsilon }}(\rho +\delta )+{\tilde{\epsilon }}(\rho -\delta )\}, \end{aligned}$$

(51)

where $E^{\mathrm{tot}}_{\mathrm{GS},0}$ is the ground state energy of $\hat{H}^{\mathrm{tot}}_0=\hat{H}_{\mathrm{a}}+\hat{H}_{\mathrm{b}}$ with total particle number 2N, and we again fix $\rho =N/L^d$. But the right-hand side is equal to $2{\tilde{\epsilon }}(\rho )$ because of the convexity. We have thus proved that

$$\begin{aligned} \lim _{L\uparrow \infty }\frac{1}{L^d}\bigl \{2E_{\mathrm{GS}}-E^{\mathrm{tot}}_{\mathrm{GS},0}\bigr \}=0. \end{aligned}$$

(52)

Then (49) and (52) imply the desired (32).