Reduction and coherent states

Abstract

We apply a quantum version of dimensional reduction to Gaussian coherent states in Bargmann space to obtain squeezed states on complex projective spaces. This leads to a definition of a family of squeezed spin states (Definition 1.13) with excellent semiclassical properties, governed by a symbol calculus. We prove semiclassical norm estimates and a propagation result.

Appendix A. Propagation of coherent states in Bargmann space

Here, we sketch a derivation of a theorem on the propagation of Gaussian coherent states in Bargmann space. We follow the approach of [6], Chapter 4.

1.1 A.1. Translations

Let $a = (a_1,\ldots , a_N)$ and $a^* = (a_1^*,\ldots , a_N^*)$ be the (vectors of the) creation and annihilation operators. In Bargmann space, these are

$$\begin{aligned} a_j = \hbar \frac{\partial \ }{\partial z_j}\quad \text {and}\quad a_j^* = \text {multiplication by } z_j. \end{aligned}$$

It is clear that $[a_j, a^*_k] = \delta _{jk}\hbar \, I.$ The position and momentum operators are

$$\begin{aligned} {\widehat{Q}} := \frac{1}{\sqrt{2}} (a^* +a) , \qquad {\widehat{P}} := \frac{i}{\sqrt{2}} (a^* -a). \end{aligned}$$

Then, the quantum translation by w (or Weyl operator)

$$\begin{aligned} {\widehat{T}}_w = \exp \left( i\hbar ^{-1}\left[ p\cdot {\widehat{Q}}- q\cdot {\widehat{P}}\right] \right) , \end{aligned}$$

(99)

where $w = \frac{1}{\sqrt{2}}(q-ip)$ and $\hbar = 1/k$, is ${\widehat{T}}_w = e^{\hbar ^{-1}\left( \overline{w}\cdot a^* - w\cdot a\right) }$, which can be seen to be equal to

$$\begin{aligned} {\widehat{T}}_w = e^{-|w|^2/2\hbar } e^{\hbar ^{-1}\overline{w}\cdot a^*}e^{-\hbar ^{-1}w\cdot a}. \end{aligned}$$

(100)

This is equivalent to ${\widehat{T}}_w(f)(z) = e^{-|w|^2/2\hbar } e^{z\overline{w}/\hbar } f(z-w)$, an expression we have used before.

Let $t\mapsto w(t)$ be any smooth curve. Below, it will be necessary to have a formula for $\frac{d\ }{\mathrm{d}t}{\widehat{T}}_{w(t)}$.

Lemma A.1

$$\begin{aligned} \frac{d\ }{\mathrm{d}t}{\widehat{T}}_{w(t)}= \hbar ^{-1}{\widehat{T}}_{w(t)}\left[ \frac{1}{2} \left( w\cdot \dot{\overline{w}}- {\dot{w}}\cdot \overline{w}\right) + \dot{\overline{w}}\cdot a^*- {\dot{w}}\cdot a\right] . \end{aligned}$$

(101)

Proof

We will use (100). By the product rule, we get the sum of three terms, one for each factor. The derivative of the middle factor is

$$\begin{aligned} \frac{\mathrm{d}\ }{\mathrm{d}t} e^{\hbar ^{-1}\overline{w}\cdot a^*} = \hbar ^{-1}e^{\hbar ^{-1}\overline{w}\cdot a^*} \dot{\overline{w}}\cdot a^*. \end{aligned}$$

We want to commute $\dot{\overline{w}}\cdot a^*$ with the third factor. One can show that

$$\begin{aligned} \left[ \dot{\overline{w}}\cdot a^*, e^{-\hbar ^{-1}w\cdot a} \right] =( w\cdot \dot{\overline{w}}) e^{-\hbar ^{-1}w\cdot a}. \end{aligned}$$

(102)

Collecting terms, we get that the left-hand side of (101) is

$$\begin{aligned} \hbar ^{-1}{\widehat{T}}_w \left[ -\frac{1}{2} \left( {\dot{w}}\cdot \overline{w}+ w\cdot \dot{\overline{w}} \right) + w\cdot \dot{\overline{w}} + \dot{\overline{w}}\cdot a^* - {\dot{w}}\cdot a \right] . \end{aligned}$$

$\square$

We will also need:

Lemma A.2

The translation operator acts on the annihilation and creation operators in the following manner:

$$\begin{aligned} {\widehat{T}}_w \, a \, {\widehat{T}}_w^{-1}&= a -\overline{w}I \\ {\widehat{T}}_w \, a^* \, {\widehat{T}}_w^{-1}&= a^* -w I. \end{aligned}$$

The proof follows directly by calculating $\left( {\widehat{T}}_w \, a \, {\widehat{T}}_w^{-1} \right) (f)(z)$ using (100). The formula for the creation operator is found by taking conjugates.

1.2 A.2. Quadratic Hamiltonians and Mp representation

The most general quadratic quantum Hamiltonian in ${\mathbb C}^N$ obtained by Weyl quantization is given by

$$\begin{aligned} {\widehat{{\mathcal Q}}} = a^* R (a^*)^T + a^* S a^T + \hbar \frac{{\text{ Tr }}(S)}{2} + a {\bar{R}} a^T, \end{aligned}$$

(103)

where

$\hbar = 1/k$ and R and S are $N \times N$ complex matrices with $R^T = R$ and ${\bar{S}}^T=S$. This operator acts on $\psi (z) =f(z)e^{-k|z|^2/2}$ by acting on f. The corresponding classical Hamiltonian (the principal symbol of ${\widehat{{\mathcal Q}}}$) is the real quadratic form

$$\begin{aligned} {\mathcal Q}(z) =2\mathfrak {R}( zRz^T) + \overline{z}S z^T. \end{aligned}$$

(104)

Let $A\in {\mathcal D}_N$. We will take R and S to be time dependent (this is needed below). We are interested in solving the initial value problem

$$\begin{aligned} i\hbar \frac{\partial \psi }{\partial t} = {\widehat{{\mathcal Q}}}(t)\psi ,\qquad \psi |_{t=0} = \psi _{A,0}. \end{aligned}$$

(105)

Note that the origin is a fixed point of the Hamilton field of ${\mathcal Q}$.

Proposition A.3

The solution of (105) is

$$\begin{aligned} \psi = \nu (t)\psi _{A(t), 0}, \end{aligned}$$

(106)

where A(t) and $\nu (t)$ solve (89) and (90) with $A(0)=A$ and $\nu (0)=1$.

Proof

We make the ansatz that $\psi$ is of the form (106) and substitute into the equation. After some calculations, we obtain the desired equations for A(t) and $\nu (t)$. $\square$

1.3 A.3. Hamiltonians of degree at most two

Let us now consider an arbitrary Hamiltonian $H:{\mathbb R}^{2N}\rightarrow {\mathbb R}$, $t\mapsto w(t)$ a trajectory of H. For each t, let us write the Taylor approximation of degree at most two centered at w(t), in complex coordinates:

$$\begin{aligned} H_2(z)= & {} H(w(0)) + (z-w(t))\frac{\partial H}{\partial z}(w(t))\nonumber \\&+(\overline{z}-\overline{w}(t))\frac{\partial H}{\partial \overline{z}}(\overline{w}(t)) + {\mathcal Q}(t)(z-w(t), \overline{z}-\overline{w}(t)) \end{aligned}$$

(107)

where ${\mathcal Q}$ is the time-dependent Hamiltonian associated to half the Hessian of H at w(t),

$$\begin{aligned} {\mathcal Q}(t)(\zeta ,\overline{\zeta }) =\frac{1}{2} \left( \zeta H_{zz}\overline{\zeta }^T + \zeta H_{\overline{z}\,\overline{z}}\overline{\zeta }^T + 2\zeta H_{z\overline{z}} \overline{\zeta }^T \right) \end{aligned}$$

(108)

where the partial derivatives are evaluated at w(t).

Now let ${\widehat{H}}_2$ denote the Weyl quantization of $H_2$, and let $U_2(t)$ denote its propagator with $U_2(0)= I$. We can express $\widehat{H_2}$ in terms of annihilation and creation operators as:

$$\begin{aligned} {\widehat{H}}_2(t)= & {} H(w(t)) + \left( a^* -w(t) I\right) \cdot \frac{\partial H}{\partial w}(w(t)) \nonumber \\&+ \left( a-\overline{w}_t I\right) \cdot \frac{\partial H}{\partial \overline{w}}(w(t)) + {\widehat{{\mathcal Q}}}(a^* -w(t), a-\overline{w}_t I). \end{aligned}$$

(109)

It turns out one can compute $U_2(t)$, in the following sense:

Proposition A.4

(Proposition 39 in [6]) Let $U_{\mathcal Q}(t)$ be the propagator of ${\widehat{{\mathcal Q}}}$ (a metaplectic operator) satisfying $U_{\mathcal Q}(0)= I$. Then,

$$\begin{aligned} U_2(t) = e^{i\hbar ^{-1}\delta _t}{\widehat{T}}_{w(t)}\circ U_{\mathcal Q}(t)\circ {\widehat{T}}_{w(0)}^{-1} \end{aligned}$$

(110)

where

$$\begin{aligned} \delta _t = - t H(w(0)) + \frac{i}{2} \int _{0}^{t} \left( w(s) \dot{{\bar{w}}}(s) - {\dot{w}}(s) {\bar{w}}(s)\right) . \end{aligned}$$

(111)

Proof

Denote for now the right-hand side of (110) by $U_2$. The proof is to show that

$$\begin{aligned} i\hbar {\dot{U}}_2 = {\widehat{H}}_2 U_2\quad \text {and}\quad U_2(0) = I. \end{aligned}$$

(112)

The second condition is clearly satisfied, so let’s differentiate the right-hand side of (110). We get:

$$\begin{aligned} {\dot{U}}_2 = -i\hbar ^{-1}\dot{\delta _t}\, U_2+ (\mathrm {II}) + \mathrm (III), \end{aligned}$$

where [using (101)]

$$\begin{aligned} \text {(II)}= \hbar ^{-1}e^{i\hbar ^{-1}\delta _t} {\widehat{T}}_{w(t)}\left[ \frac{1}{2} \left( w\cdot \dot{\overline{w}}- {\dot{w}}\cdot \overline{w}\right) + \dot{\overline{w}}\cdot a^*- {\dot{w}}\cdot a\right] U_{\mathcal Q}(t){\widehat{T}}_{w(0)}^{-1} \end{aligned}$$

and

$$\begin{aligned} \text {(III)} = -i\hbar ^{-1}e^{i\hbar ^{-1}\delta _t}{\widehat{T}}_{w(t)}{\widehat{{\mathcal Q}}}(t) U_{\mathcal Q}(t) {\widehat{T}}_{w(0)}^{-1}. \end{aligned}$$

Using again the definition of $U_2$ to solve for $U_{\mathcal Q}{\widehat{T}}_{w(0)}^{-1}$, we can write

$$\begin{aligned} \text {(II)} = \hbar ^{-1}{\widehat{T}}_{w(t)}\left[ \frac{1}{2} \left( w\cdot \dot{\overline{w}}- {\dot{w}}\cdot \overline{w}\right) + \dot{\overline{w}}\cdot a^*- {\dot{w}}\cdot a\right] {\widehat{T}}_{w(t)}^{-1} U_2 \end{aligned}$$

and

$$\begin{aligned} \text {(III)} = -i\hbar ^{-1}{\widehat{T}}_{w(t)}{\widehat{{\mathcal Q}}}(t) {\widehat{T}}_{w(t)}^{-1} U_2. \end{aligned}$$

We analyze (II) further, the key step being

$$\begin{aligned} {\widehat{T}}_{w(t)}\left[ \dot{\overline{w}}\cdot a^*- {\dot{w}}\cdot a\right] {\widehat{T}}_{w(t)}^{-1}= & {} {\widehat{T}}_{w(t)}(\dot{\overline{w}}\cdot a^*) {\widehat{T}}_{w(t)}^{-1} - {\widehat{T}}_{w(t)}( {\dot{w}}\cdot a) {\widehat{T}}_{w(t)}^{-1}\\= & {} \dot{\overline{w}} \cdot (a^* - wI) - {\dot{w}} \cdot (a-\overline{w}I) . \end{aligned}$$

After some calculations, one finds that

$$\begin{aligned} i\hbar {\dot{U}}_2\,U_2^{-1} = -\dot{\delta }_t +\frac{i}{2} \left( w\cdot \dot{\overline{w}}- {\dot{w}}\cdot \overline{w}\right) - H(w(t)) + {\widehat{H}}_2(t), \end{aligned}$$

so $\dot{\delta _t} = -H(w(0)) + \frac{i}{2} \left( w\cdot \dot{\overline{w}}- {\dot{w}}\cdot \overline{w}\right)$ using $H(w(t)) = H(w(0))$. Integrating gives (111). $\square$

Corollary A.5

$$\begin{aligned} U_2(t)\left( \psi _{A,w(0)}\right) = \nu (t) e^{ i k \delta _t}\psi _{A(t),w(t)} \end{aligned}$$

where $\nu (t)$ and A(t) satisfy (89) and (90).

1.4 A.4. Propagation

First, we need a preliminary estimate which we state without proof:

Proposition A.6

Let ${\widehat{H}}, {\widehat{H}}_2$ be semiclassical pseudodifferential operators acting on the Bargmann space of ${\mathbb C}^N$, with principal symbols H and $H_2$. Let $w\in {\mathbb C}^N$ and assume that $H-H_2$ vanishes at w, together with its first and second derivatives. Then, for any $A\in {\mathcal D}_N$

$$\begin{aligned} \Vert ({\widehat{H}}-{\widehat{H}}_2)\psi _{A,w}\Vert = \Vert \psi _{A,w}\Vert \cdot O(\hbar ^{3/2}). \end{aligned}$$

(113)

To finish the proof of (88), we follow the argument of Chapter 4 in [6]. By Duhamel’s principle

$$\begin{aligned} U(t) - U_2(t) = \frac{1}{i\hbar }\int _0^t U(t,s)\left( {\widehat{H}}-{\widehat{H}}_2(s) \right) U_2(t,s)\, \mathrm{d}s \end{aligned}$$

(114)

where U(t, s) is the propagator for ${\widehat{H}}$ such that $U(t,t)= I$ (and similarly for $U_2(t,s)$), it follows that

$$\begin{aligned} \Vert U(t)(\psi _{A,w(0)}) - U_2(t)(\psi _{A,w(0)})\Vert \le \hbar ^{-1}\int _0^t \Vert \left( {\widehat{H}}-{\widehat{H}}_2(s) \right) \phi _{t,s}\Vert \,\mathrm{d}s \end{aligned}$$

where $\phi _{t,s} = U_2(t,s)(\psi _{A,w(0)})$. This, combined with (113), yields

$$\begin{aligned} \Vert U(t)(\psi _{A,w(0)}) - U_2(t)(\psi _{A,w(0)})\Vert = \Vert \psi _{A,w(0)}\Vert \cdot O(\hbar ^{1/2}). \end{aligned}$$

(115)

Using Corollary A.5, we obtain Theorem 5.5.

On behalf of all authors, the corresponding author states that there is no conflict of interest.