Explicit diagonalization of pair interaction models

Abstract

We provide a general method for constructing bosonic Bogoliubov transformations that diagonalize a general class of quadratic Hamiltonians. These Hamiltonians describe the pair interaction models. Bogoliubov transformations are constructed algebraically, and the resulting Hamiltonians become the second quantizations of explicit one-particle Hamiltonians. Moreover, an explicit formula for the ground state energies is given. Our method systematically diagonalizes various models of quantum field theory, including a model of a harmonic oscillator coupled to a Bose field and the Pauli-Fierz models in the dipole approximation.

Appendices

Inequalities on creation-annihilation operators and second quantizations

Let $(M,\mu )$ be a measure space. Suppose that $L^2(M):=L^2(M,d\mu )$ is separable. The space $\mathop {\otimes }_{{\mathrm {s}}}^nL^2(M)$ can be identified with the set of square integrable symmetric functions.

$$\begin{aligned} L^2_{{\mathrm {sym}}}(M^n) := \big \{\Psi \in L^2(M^n) ~\big |~ \Psi (k_1,\cdots ,k_n)=\Psi (k_{\sigma (1)},\cdots ,k_{\sigma (n)}), \sigma \in S_n\big \}. \end{aligned}$$

Let us consider the Cartesian product space

$$\begin{aligned} {\mathscr {F}}^{\mathsf {x}} := \mathop {{\mathsf {X}}}_{n=0}^\infty L^2_{{\mathrm {sym}}}(M^n), \end{aligned}$$

where ${\mathbb {C}}:=L^2_{{\mathrm {sym}}}(M^0)$. Then the Fock space ${\mathscr {F}}_{{\mathrm {b}}}(L^2(M))$ can be identified with a subset of ${\mathscr {F}}^{\mathsf {x}}$. For $\Psi =(\Psi ^{(n)})_{n=0}^\infty \in {\mathscr {F}}^{\mathsf {x}}$, we define an informal norm by

$$\begin{aligned} \Vert \Psi \Vert ^2 := \sum _{n=0}^\infty \Vert \Psi ^{(n)} \Vert _{L^2(M^n)}^2 \in [0,+\infty ]. \end{aligned}$$

An inner product of $\Psi ,\Phi \in {\mathscr {F}}^{\mathsf {x}}$ is defined by

$$\begin{aligned} \left\langle \Psi ,\Phi \right\rangle := \sum _{n=0}^\infty \left\langle \Psi ^{(n)},\Phi ^{(n)}\right\rangle \end{aligned}$$

if the sum converges. For $\Psi = (\Psi ^{(n)})_{n=0}^\infty \in {\mathscr {F}}^{\mathsf {x}}$ and $k\in M$, we define $A(k)\Psi \in {\mathscr {F}}^{\mathsf {x}}$ by

$$\begin{aligned} (A(k)\Psi )^{(n)}(\cdot ) := \sqrt{n+1}\Psi ^{(n+1)}(k,\cdot ) \in L^2_{{\mathrm {sym}}}(M^n), \qquad n=0,1,2,\cdots . \end{aligned}$$

(A.1)

Note that $A(k)\Psi$ is defined for $\mu$-a.e. $k\in M$. For finite particle state $\Phi \in {\mathscr {F}}_{{\mathrm {b}}0}$, $\left\langle \Phi ,A(k)\Psi \right\rangle$ consists of finite sum and

$$\begin{aligned} \left\langle \Phi ,A(f)\Psi \right\rangle = \int _Md\mu (k)\,\overline{f(k)}\left\langle \Phi ,A(k)\Psi \right\rangle , \qquad \Psi \in \mathop {{\mathrm {dom}}}(A(f)) \end{aligned}$$

holds.

Lemma A.1

Let $Q(k)>0$ be a measurable function. The multiplication operator by Q(k) acting in $L^2(M)$ is also denoted by Q. Then, for all $\Psi \in {\mathscr {F}}^{\mathsf {x}}$, $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{1/2})$ if and only if

$$\begin{aligned} \int _M Q(k)\Vert A(k)\Psi \Vert ^2 \,d\mu (k) < \infty . \end{aligned}$$

In this case, the equality

$$\begin{aligned} \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{1/2}\Psi \Vert ^2 = \int _M Q(k)\Vert A(k)\Psi \Vert ^2 \,d\mu (k) \end{aligned}$$

holds. Moreover, if $f\in \mathop {{\mathrm {dom}}}(Q^{-1/2})$, $\Psi \in \mathop {{\mathrm {dom}}}(A(f))$ and

$$\begin{aligned} \left\langle \Phi ,A(f)\Psi \right\rangle = \int _M \overline{f(k)}\left\langle \Phi ,A(k)\Psi \right\rangle \,d\mu (k) \end{aligned}$$

hold for $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{1/2})$ and $\Phi \in {\mathscr {F}}_{{\mathrm {b}}}(L^2(M))$.

Proof

The lemma directly follows from the definitions of second quantization operator and A(k). $\square$

Lemma A.2

Let T be an injective non-negative self-adjoint operator acting in a separable Hilbert space and $f_1,\cdots ,f_n\in \mathop {{\mathrm {dom}}}(T^{-1/2})$. Then

$$\begin{aligned} \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(T)^{n/2})\subset \mathop {{\mathrm {dom}}}(A(f_1)\cdots A(f_n)), \end{aligned}$$

and the bound

$$\begin{aligned} \Vert A(f_1)\cdots A(f_n)\Psi \Vert \le \Vert T^{-1/2}f_1 \Vert \cdots \Vert T^{-1/2}f_n \Vert \cdot \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(T)^{n/2}\Psi \Vert \end{aligned}$$

(A.2)

holds for $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(T)^{n/2})$. In the case of $n=2$, the bound

$$\begin{aligned} \Vert A(f_1) A(f_2)\Psi \Vert \le \Vert T^{-1/2}f_1 \Vert \cdot \Vert T^{-1/2}f_2 \Vert \big ( \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(T)\Psi \Vert ^2 - \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(T^2)^{1/2}\Psi \Vert ^2 \big )^{1/2} \end{aligned}$$

(A.3)

holds.

Proof

Since any self-adjoint operator T is unitarily equivalent to a multiplication operator on an $L^2$-space, it is enough to prove the lemma in the case of ${\mathscr {H}}=L^2(M)$, $T=Q$. By Lemma A.1, for $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(T)^{n/2})$ and $\Phi \in {\mathscr {F}}_{{\mathrm {b}}0}$, we have

$$\begin{aligned} \left\langle A^*(f_1)\cdots A^*(f_n)\Phi ,\Psi \right\rangle&= \int _{M^n} \overline{f_1(k_1)}\cdots \overline{f_n(k_n)} \left\langle \Phi ,A(k_1)\cdots A(k_n)\Psi \right\rangle \,d\mu (k_1)\cdots d\mu (k_n). \end{aligned}$$

We set $\Vert \Phi \Vert =1$, $f_j:=f_j(k_j)$, $Q_j^{1/2}:=Q(k_j)^{1/2}$ and $d\mu :=d\mu (k_1)\cdots d\mu (k_n)$. Then we have

$$\begin{aligned}&|\left\langle A^*(f_1)\cdots A^*(f_n)\Phi ,\Psi \right\rangle | \\&\quad \le \int _{M^n} |f_1 \cdots f_n| \cdot \Vert A(k_1)\cdots A(k_n)\Psi \Vert \,d\mu \\&\quad \le \bigg ( \int _{M^n} \prod _{j=1}^n |Q_j^{-1/2}f_j|^2\,d\mu \bigg )^{1/2} \bigg ( \int _{M^n} Q_1\cdots Q_n \Vert A(k_1)\cdots A(k_n)\Psi \Vert ^2\,d\mu \bigg )^{1/2} \\&\quad \le \Vert Q^{-1/2}f_1 \Vert \cdots \Vert Q^{-1/2}f_n \Vert \bigg ( \int _{M^n} Q_1\cdots Q_n \Vert A(k_1)\cdots A(k_n)\Psi \Vert ^2\,d\mu \bigg )^{1/2}. \end{aligned}$$

By the definition of A(k), one has

$$\begin{aligned}&\Vert A(k_1)\cdots A(k_n)\Psi \Vert ^2 \\&\quad = \sum _{N=0}^\infty (N+1)\Vert (A(k_2)\cdots A(k_n)\Psi )^{(N+1)}(k_1,\cdot ) \Vert ^2 \\&\quad = \sum _{N=0}^\infty (N+1)(N+2)\Vert (A(k_3)\cdots A(k_n)\Psi )^{(N+2)}(k_1,k_2,\cdot ) \Vert ^2 \\&\quad = \sum _{N=0}^\infty (N+1)\cdots (N+n) \Vert \Psi ^{(N+n)}(k_1,\cdots ,k_n,\cdot ) \Vert ^2 \\&\quad = \sum _{N=0}^\infty (N+1)\cdots (N+n) \int _{M^N} d\mu (k_{n+1})\cdots d\mu (k_{n+N})\,|\Psi ^{(N+n)}(k_1,\cdots ,k_{N+n})|^2. \end{aligned}$$

Therefore, we have

$$\begin{aligned}&\int _{M^n} Q_1\cdots Q_n \Vert A(k_1)\cdots A(k_n)\Psi \Vert ^2\,d\mu \nonumber \\&\quad = \sum _{N=0}^\infty \int _{M^{N+n}} d\mu (k_1)\cdots d\mu (k_{n+N})\,\frac{(N+n)!}{N!} Q_1\cdots Q_n| \Psi ^{(N+n)}(k_1,\cdots ,k_{N+n})|^2 \nonumber \\&\quad = \sum _{N=n}^\infty \int _{M^N} d\mu (k_1)\cdots d\mu (k_N)\,\frac{N!}{(N-n)!} Q_1\cdots Q_n| \Psi ^{(N)}(k_1,\cdots ,k_N)|^2 \nonumber \\&\quad = \sum _{N=n}^\infty \int _{M^N} d\mu (k_1)\cdots d\mu (k_N) \mathop {\sum _{j_1,\cdots ,j_n=1}^N}_{\sharp \{j_1,\cdots ,j_n\}=n} Q_{j_1}\cdots Q_{j_n}| \Psi ^{(N)}(k_1,\cdots ,k_N)|^2, \end{aligned}$$

(A.4)

where, in the last step, we used the symmetry of $\Psi ^{(N)}$. In the case $n=2$, we have

$$\begin{aligned} (A.4)&= \sum _{N=2}^\infty \int _{M^N} d\mu (k_1)\cdots d\mu (k_N)\,\left\{ \left( \sum _{j=1}^N Q_j \right) ^2 -\sum _{j=1}^N Q_j^2 \right\} |\Psi ^{(N)}(k_1,\cdots ,k_N)|^2 \nonumber \\&= \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q)\Psi \Vert ^2 - \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q^2)^{1/2}\Psi \Vert ^2, \end{aligned}$$

(A.5)

and, for $n\ge 2$, we have

$$\begin{aligned} (A.4)&\le \sum _{N=n}^\infty \int _{M^N} d\mu (k_1)\cdots d\mu (k_N)\,\Big ( \sum _{j=1}^N Q(k_j)\Big )^n | \Psi ^{(N)}(k_1,\cdots ,k_N)|^2 \\&\le \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{n/2}\Psi \Vert ^2 < \infty . \end{aligned}$$

Hence, for all $\Phi \in {\mathscr {F}}_{{\mathrm {b}}0}$ and $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{n/2})$, it holds that

$$\begin{aligned} |\left\langle A^*(f_1)\cdots A^*(f_j)\Phi ,\Psi \right\rangle | \le \prod _{\ell =1}^j \Vert Q^{-1/2}f_\ell \Vert \cdot \Vert \Phi \Vert \cdot \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{j/2}\Psi \Vert ,\qquad j=1,2,\cdots ,n. \end{aligned}$$

Since ${\mathscr {F}}_{{\mathrm {b}}0}$ is a core for A(f), by setting $j=1$ in the above inequality, we have $\Psi \in \mathop {{\mathrm {dom}}}(A(f_1))$. Next, by setting $j=2$, one has $A(f_1)\Psi \in \mathop {{\mathrm {dom}}}(A(f_2))$. Therefore $\Psi \in \mathop {{\mathrm {dom}}}(A(f_n)\cdots A(f_1))$ follows by induction. We also have the bound

$$\begin{aligned} \Vert A(f_n)\cdots A(f_1)\Psi \Vert&= \mathop {\sup _{\Phi \in {\mathscr {F}}_{{\mathrm {b}}0}}}_{{\Vert \Phi \Vert =1}} |\left\langle A^*(f_1)\cdots A^*(f_n)\Phi ,\Psi \right\rangle | \\&\le \prod _{\ell =1}^n \Vert Q^{-1/2}f_\ell \Vert \cdot \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(Q)^{n/2}\Psi \Vert . \end{aligned}$$

Thus we get (A.2). The bound (A.3) follows from (A.5). $\square$

Lemma A.3

Let T be an injective non-negative self-adjoint operator and $g\in \mathop {{\mathrm {dom}}}(T^{-1/2})$. Then $\mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(T))\subset \mathop {{\mathrm {dom}}}(\Phi _{{\mathrm {S}}}(g)^2)$ and for all $\Psi \in \mathop {{\mathrm {dom}}}({\mathrm {d}}\Gamma _{\mathrm{b}}(T))$,

$$\begin{aligned} \frac{1}{2} \Vert \Phi _{{\mathrm {S}}}(g)^2\Psi \Vert \le \Vert T^{-1/2}g \Vert ^2 \Vert {\mathrm {d}}\Gamma _{\mathrm{b}}(T)\Psi \Vert + \Vert g \Vert ^2\Vert \Psi \Vert \end{aligned}$$

(A.6)

holds.

Proof

In this proof, we write

$$\begin{aligned} a:=A(g), ~~ a^*:=A^*(g), ~~ c:=\Vert g \Vert ^2, ~~ d:=\Vert T^{-1/2}g \Vert ^2, ~~ H_0 := {\mathrm {d}}\Gamma _{\mathrm{b}}(T) \end{aligned}$$

for short. We first assume $\Psi \in {\mathscr {F}}_{\mathrm {b,fin}}(\mathop {{\mathrm {dom}}}(T))$. By the triangle inequality,

$$\begin{aligned} \Vert \Phi _{{\mathrm {S}}}(g)^2\Psi \Vert ^2&= \frac{1}{4} \Vert (a^2 + a^*a + aa^* + a^{*2})\Psi \Vert ^2 \\&\le \Vert a^2\Psi \Vert ^2 + \Vert a^*a\Psi \Vert ^2 + \Vert aa^*\Psi \Vert ^2 + \Vert a^{*2}\Psi \Vert ^2. \end{aligned}$$

By the CCRs and Lemma A.2, we have

$$\begin{aligned} \Vert a^2\Psi \Vert ^2&\le d^2 \Vert H_0\Psi \Vert ^2, \\ \Vert a^*a\Psi \Vert ^2&= \Vert a^2\Psi \Vert ^2 + c\Vert a\Psi \Vert ^2 \le d^2\Vert H_0\Psi \Vert ^2 + cd\Vert H_0^{1/2}\Psi \Vert ^2, \\ \Vert aa^*\Psi \Vert ^2&= \Vert (a^*a+c)\Psi \Vert ^2 = \Vert a^*a\Psi \Vert ^2 + 2c\Vert a\Psi \Vert ^2 + c^2\Vert \Psi \Vert ^2 \\&\le d^2\Vert H_0\Psi \Vert ^2+3cd\Vert H_0^{1/2}\Psi \Vert ^2 + c^2\Vert \Psi \Vert ^2, \\ \Vert (a^*)^2\Psi \Vert ^2&= \Vert aa^*\Psi \Vert ^2 + c\Vert a^*\Psi \Vert ^2 \\&\le d^2\Vert H_0\Psi \Vert ^2+3cd\Vert H_0^{1/2}\Psi \Vert ^2 + c^2\Vert \Psi \Vert ^2 + c(d\Vert H_0^{1/2}\Psi \Vert ^2+c\Vert \Psi \Vert ^2) \\&= d^2\Vert H_0\Psi \Vert ^2+4cd\Vert H_0^{1/2}\Psi \Vert ^2 + 2c^2\Vert \Psi \Vert ^2. \end{aligned}$$

Thus, we have

$$\begin{aligned} \Vert \Phi _{{\mathrm {S}}}(g)^2\Psi \Vert ^2&\le 4d^2\Vert H_0\Psi \Vert ^2 + 8cd\Vert H_0^{1/2}\Psi \Vert ^2 + 3c^2\Vert \Psi \Vert ^2 \\&\le \Vert (2dH_0+2c)\Psi \Vert ^2, \end{aligned}$$

and the bound (A.6) holds for all $\Psi \in {\mathscr {F}}_{\mathrm {b,fin}}(\mathop {{\mathrm {dom}}}(T))$. By a limiting argument, the lemma follows. $\square$

On the domains of $T^{-3/2}$ and $S^{-3/2}$

It is shown that $\mathop {{\mathrm {dom}}}(T^{p})=\mathop {{\mathrm {dom}}}(S^{p})$ for all $|p|\le 1$ and $p=2$ in Lemma 3.1. Here, we show the equality for $p=-3/2$ under the infrared regularity condition.

Lemma B.1

Suppose (B1)–(B5). We further suppose that $g_n\in \mathop {{\mathrm {dom}}}(T^{-1})$ for all $n\in {\mathbb {N}}$, and that

$$\begin{aligned} \sum _{n=1}^{\infty }|\lambda _n|\cdot \Vert g_n\Vert \cdot \Vert T^{-1}g_n\Vert <\infty . \end{aligned}$$

Then $\mathop {{\mathrm {dom}}}(T^{-3/2})=\mathop {{\mathrm {dom}}}(S^{-3/2})$ holds, where S is defined in (5.2).

Proof

Let $A:=\sum _{n=1}^\infty \lambda _n|g_n\rangle \langle T^{-1}g_n|$. By assumption, A is of trace class. We first show that $1+A$ is bijective. Let $u\in \ker (1+A)$ be arbitrary. Since the range of A is contained in $\mathop {{\mathrm {dom}}}(T^{-1/2})$, the equality $(1+A)u=0$ implies that $u\in \mathop {{\mathrm {dom}}}(T^{-1/2})$. Hence it follows that

$$\begin{aligned} 0&= T^{-1/2}(1+A)u = T^{-1/2}u + \sum _{n=1}^\infty \lambda _n\langle T^{-1/2}g_n,T^{-1/2}u\rangle T^{-1/2}g_n\\&=\left( 1+\sum _{n=1}^\infty \lambda _n|T^{-1/2}g_n\rangle \langle T^{-1/2}g_n|\right) T^{-1/2}u. \end{aligned}$$

This together with the condition (B5) implies that $u=0$, and thus $1+A$ is injective. The Fredholm alternative now tells us that $1+A$ is bijective.

Set

$$\begin{aligned} {\mathscr {D}}:=\mathop {{\mathrm {dom}}}(T^2)\cap \mathop {{\mathrm {dom}}}(T^{-1/2})=\mathop {{\mathrm {dom}}}(S^2)\cap \mathop {{\mathrm {dom}}}(S^{-1/2}), \end{aligned}$$

and take an arbitrary $u\in {\mathscr {D}}$. Then we have

$$\begin{aligned} S^{3/2}u&=S^{-1/2}S^2u=S^{-1/2}\left( T^2+\sum _{n=1}^\infty \lambda _n|T^{1/2}g_n\rangle \langle T^{1/2}g_n|\right) u\\&=S^{-1/2}T^{1/2}\left( 1+\sum _{n=1}^\infty \lambda _n|g_n\rangle \langle T^{-1}g_n|\right) T^{3/2}u =\overline{S^{-1/2}T^{1/2}}(1+A)T^{3/2}u. \end{aligned}$$

Note that $\overline{T^{1/2}S^{-1/2}}$ is bijective with inverse $\overline{S^{1/2}T^{-1/2}}$ by Lemma 3.2. Since ${\mathscr {D}}$ is a core of both $S^{3/2}$ and $T^{3/2}$, we get the operator equality

$$\begin{aligned} S^{3/2} = \overline{S^{-1/2}T^{1/2}}(1+A)T^{3/2}. \end{aligned}$$

By taking the conjugation of both sides, we obtain the operator equality

$$\begin{aligned} S^{3/2} = T^{3/2}(1+A^*)\overline{T^{1/2}S^{-1/2}}, \end{aligned}$$

(B.1)

which in particular implies that the range of $S^{3/2}$ is contained in the range of $T^{3/2}$, and thus $\mathop {{\mathrm {dom}}}(S^{-3/2})\subset \mathop {{\mathrm {dom}}}(T^{-3/2})$ holds. On the other hand, it follows from (B.1) that we have the operator equality

$$\begin{aligned} S^{3/2}\overline{S^{1/2}T^{-1/2}}(1+A^*)^{-1} = T^{3/2}, \end{aligned}$$

whence $\mathop {{\mathrm {dom}}}(S^{-3/2})\supset \mathop {{\mathrm {dom}}}(T^{-3/2})$ follows. This completes the proof. $\square$