The Pauli Problem for Gaussian Quantum States: Geometric Interpretation

Abstract

We solve the Pauli tomography problem for Gaussian signals using the notion of Schur complement. We relate our results and method to a notion from convex geometry, polar duality. In our context polar duality can be seen as a sort of geometric Fourier transform and allows a geometric interpretation of the uncertainty principle and allows to apprehend the Pauli problem in a rather simple way.

1. The Pauli Problem and Quantum Tomography

The problem goes back to Pauli’s question [1]:

The mathematical problem as to whether, for given probability densities$W\left(p\right)$and$W\left(x\right)$, wave function$\psi$(…) is always uniquely determined, has still not been investigated in its generality.

The answer to Pauli’s question is negative [2]; there is a general nonuniqueness of the solution (for a detailed discussion of the Pauli problem and its applications, see [3]). The problem can actually be formulated as from statistical quantum mechanics as follows: can we estimate the density matrix of the said state using repeated measurements on identical quantum systems? After having obtained measurements on these identical systems, can we make a statistical inference about their probability distributions (e.g., [4])? Such a procedure is an instance of quantum state tomography, and is practically implemented using a set of measurements of a so-called quorum of observables. It can be performed using various mathematical techniques, for instance the Radon–Wigner transform that we discussed in [5]; the latter has important applications in medical imaging [6]. For details and explicit constructions, see [7,8,9,10,11,12,13,14], and [15] by Man’ko and Man’ko.

Remark 1. 

Everything in this paper extends mutatis mutandis to time-frequency analysis, replacing the notion of wave function by that of a signal. In this case, one takes $\hslash =1/2\pi$ and replaces phase-space variables $(x,p)$ with time-frequency variables $(x,\omega )$.

2. A Simple Example

Let us discuss the Pauli problem on the simplest possible example, that of a Gaussian wave function in one spatial dimension. Assuming for simplicity, it is centered at the origin and is given by formula

$$\psi \left(x\right)={\left(\frac{1}{2\pi {\sigma }_{xx}}\right)}^{1/4}{e}^{-\frac{x^2}{4{\sigma }_{{}_{xx}}}}{e}^{\frac{i{\sigma }_{xp}}{2\hslash {\sigma }_{{}_{xx}}}{x}^2}$$

(1)

where ${\sigma }_{xx}$ is the variance in the position variable, and ${\sigma }_{xp}$ the covariance in the position and momentum variables. Fourier transform

$$\widehat{\psi }\left(p\right)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }{e}^{-\frac{i}{\hslash }px}\psi \left(x\right)dx$$

of the $\psi$ is explicitly given by

$$\widehat{\psi }\left(p\right)={\left(\frac{1}{2\pi {\sigma }_{pp}}\right)}^{1/4}{e}^{-\frac{p^2}{4{\sigma }_{{}_{pp}}}}{e}^{-\frac{i{\sigma }_{{}_{xp}}}{2\hslash {\sigma }_{{}_{pp}}}{p}^2}$$

(2)

hence, the knowledge of ${\sigma }_{xx}$ and of ${\sigma }_{pp}$, that is, of moduli ${\left|\psi \left(x\right)\right|}^2$ and $|\widehat{\psi }{\left(p\right)|}^2$, determines covariance ${\sigma }_{xp}$ up to a sign because state $\psi$ saturates the Robertson–Schrödinger inequality; so, we have

$${\sigma }_{xx}{\sigma }_{pp}-{\sigma }_{xp}^2=\frac{1}{4}{\hslash }^2$$

(3)

This identity can be solved in ${\sigma }_{xp}$ yielding ${\sigma }_{xp}=\pm {({\sigma }_{xx}{\sigma }_{pp}-\frac{1}{4}{\hslash }^2)}^{1/2}$. The state and its Fourier transform are given by formulas

$${\psi }_{\pm }\left(x\right)={\left(\frac{1}{2\pi {\sigma }_{xx}}\right)}^{1/4}{e}^{-\frac{x^2}{4{\sigma }_{xx}}}{e}^{\pm \frac{i{\sigma }_{xp}}{2\hslash {\sigma }_{xx}}{x}^2}$$

(4)

and

$${\widehat{\psi }}_{\pm }\left(p\right)={\left(\frac{1}{2\pi {\sigma }_{pp}}\right)}^{1/4}{e}^{-\frac{p^2}{4{\sigma }_{pp}}}{e}^{\mp \frac{i{\sigma }_{xp}}{2\hslash {\sigma }_{pp}}{p}^2}.$$

(5)

Both functions ${\psi }_{+}$ and ${\psi }_{-}={\psi }_{+}^{*}$ and their Fourier transforms ${\widehat{\psi }}_{+}$ and ${\widehat{\psi }}_{-}$ satisfy conditions $|{\psi }_{+}{\left(x\right)|}^2={\left|{\psi }_{-}\left(x\right)\right|}^2$ and $|{\widehat{\psi }}_{+}{\left(p\right)|}^2={\left|{\widehat{\psi }}_{-}\left(p\right)\right|}^2$ showing that the Pauli problem does not have a unique solution. In Corbett’s [16] terminology ${\psi }_{+}$ and ${\psi }_{-}$ are “Pauli partners”. Let us now have a look at these things from the perspective of the Wigner transform

$$W\psi (x,p)=\frac{1}{2\pi \hslash }{\int }_{-\infty }^{\infty }{e}^{-\frac{i}{\hslash }py}\psi (x+\frac{1}{2}y){\psi }^{*}(x-\frac{1}{2}y)dy$$

of Gaussian $\psi$. A straightforward calculation involving Gaussian integrals [17] yields, setting $z=(x,p)$, normal distribution

$$W{\psi }_{\pm }\left(z\right)=\frac{1}{2\pi \sqrt{det{\mathsf{\Sigma }}_{\pm }}}{e}^{-\frac{1}{2}{\mathsf{\Sigma }}_{\pm }^{-1}z·z}$$

(6)

where covariance matrix

$${\mathsf{\Sigma }}_{\pm }=\left(\begin{array}{cc}{\sigma }_{xx}& \pm {\sigma }_{xp}\\ \pm {\sigma }_{px}& {\sigma }_{pp}\end{array}\right)$$

has determinant $det{\mathsf{\Sigma }}_{\pm }=\frac{1}{4}{\hslash }^2$ in view of equality (3); hence,

$$W{\psi }_{\pm }\left(z\right)=\frac{1}{\pi \hslash }{e}^{-\frac{1}{2}{\mathsf{\Sigma }}_{\pm }^{-1}z·z}.$$

(7)

Associated covariance matrices are thus

$${\mathsf{\Omega }}_{\pm }=\{z:\frac{1}{2}{\mathsf{\Sigma }}_{\pm }^{-1}z·z\le 1.\}$$

3. Multivariate Case: Asking the Right Questions

We generalize the discussion to the multivariate case where the real variables x and p are replaced with real vectors $x=(x_1,\dots ,x_n)$, $p=(p_1,\dots ,p_n)$.

The Wigner function cannot be directly measured, but its marginal distributions can (they are classical probability densities). In analogy with Formula (6) we determine a (centered) Gaussian, $\psi$ such that

$$W\psi \left(z\right)={\left(\frac{1}{2\pi }\right)}^n\frac{1}{\sqrt{det\mathsf{\Sigma }}}{e}^{-\frac{1}{2}{\mathsf{\Sigma }}^{-1}z·z}$$

(8)

where $z=(x,p)$, and the covariance matrix is

$$\mathsf{\Sigma }=\left(\begin{array}{cc}{\mathsf{\Sigma }}_{XX}& {\mathsf{\Sigma }}_{XP}\\ {\mathsf{\Sigma }}_{PX}& {\mathsf{\Sigma }}_{PP}\end{array}\right),{\mathsf{\Sigma }}_{PX}={\mathsf{\Sigma }}_{XP}^T.$$

(9)

Here, the the n-dimensional Wigner transform $W\psi$ is defined by

$$W\psi (x,p)={\left(\frac{1}{2\pi \hslash }\right)}^n\int e^{-\frac{i}{\hslash }p·y}\psi (x+\frac{1}{2}y){\psi }^{*}(x-\frac{1}{2}y)d^{n}y.$$

The most straightforward way to determine this state is to use the properties of the Wigner transform itself. Let us start with the marginal properties [17]:

$$\begin{array}{cc}\hfill \int W\psi (x,p)d^{n}p& ={\left|\psi \left(x\right)\right|}^2\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill \int W\psi (x,p)d^{n}x& =|\widehat{\psi }{\left(p\right)|}^2\hfill \end{array}$$

(11)

where the n-dimensional Fourier transform $\widehat{\psi }$ is given by

$$\widehat{\psi }\left(p\right)={\left(\frac{1}{2\pi \hslash }\right)}^{n/2}\int e^{-\frac{i}{\hslash }px}\psi \left(x\right)d^{n}x.$$

These formulas hold as soon as both $\psi$ and $\widehat{\psi }$ are in $L^1\left({\mathbb{R}}^n\right)\cap L^2\left({\mathbb{R}}^n\right)$ [17]. These quantities allow for determining matrices

$${\mathsf{\Sigma }}_{XX}={\left({\sigma }_{x_j{x}_k}\right)}_{1\le j,k\le n}\mathrm{and}{\mathsf{\Sigma }}_{PP}={\left({\sigma }_{p_j{p}_k}\right)}_{1\le j,k\le n}$$

by usual formulas

$${\sigma }_{x_j{x}_k}=\int x_j{x}_k{\left|\psi \left(x\right)\right|}^2{d}^{n}x,{\sigma }_{p_j{p}_k}=\int p_j{p}_k{\left|\widehat{\psi }\left(p\right)\right|}^2{d}^{n}p$$

and an elementary calculation of Gaussian integrals yields the values

$$\begin{array}{cc}\hfill \left|\psi \right(x\left)\right|& ={\left(\frac{1}{2\pi }\right)}^{n/4}{(det{\mathsf{\Sigma }}_{XX})}^{-1/4}{e}^{-\frac{1}{4}{\mathsf{\Sigma }}_{XX}^{-1}x·x}\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill |\widehat{\psi }\left(p\right)|& ={\left(\frac{1}{2\pi }\right)}^{n/4}{(det{\mathsf{\Sigma }}_{PP})}^{-1/4}{e}^{-\frac{1}{4}{\mathsf{\Sigma }}_{PP}^{-1}p·p}.\hfill \end{array}$$

(13)

Here, we are exactly in the situation discussed by Pauli: $\left|\psi \right(x\left)\right|$ and $|\widehat{\psi }\left(p\right)|$ are what we can measure, so we can determine covariance blocks ${\mathsf{\Sigma }}_{XX}$ and ${\mathsf{\Sigma }}_{PP}$, but not covariance ${\mathsf{\Sigma }}_{XP}$: knowledge of the latter (and hence of ${\mathsf{\Sigma }}_{PX}={\mathsf{\Sigma }}_{XP}^T$) is necessary to entirely determine state $\psi$. In the previous section, the problem was solved: in case $n=1$, blocks ${\mathsf{\Sigma }}_{XX}$, ${\mathsf{\Sigma }}_{PP}$, and ${\mathsf{\Sigma }}_{XP}$ were scalars ${\sigma }_{xx}$, ${\sigma }_{pp}$, and ${\sigma }_{xp}$, and these are related by the uncertainty principle in the form of ${\sigma }_{xx}{\sigma }_{pp}-{\sigma }_{xp}^2=\frac{1}{4}{\hslash }^2$ yielding two possible values ${\sigma }_{xp}=\pm {({\sigma }_{xx}{\sigma }_{pp}-\frac{1}{4}{\hslash }^2)}^{1/2}$, and hence the two states (5). In the multidimensional, case we also have a simple (but not immediately obvious) formula connecting the blocks of the covariance matrix. The way out of this problem consists in using a general formula [17,18,19], which was initially proved by Bastiaans [20] in connection with first-order optics. Let X and Y be real $n\times n$ matrices, such that $X=X^T>0$ and $Y=Y^T$, and set

$${\psi }_{X,Y}\left(x\right)={\left(\frac{1}{\pi \hslash }\right)}^{n/4}{(detX)}^{1/4}{e}^{-\frac{1}{2\hslash }(X+iY)x·x}.$$

(14)

This function is normalized to unity: $\left|\right|{\psi }_{X,Y}{\left|\right|}_{L^2}=1$, and its Wigner transform is given by

$$W{\psi }_{X,Y}\left(z\right)={\left(\frac{1}{\pi \hslash }\right)}^n{e}^{-\frac{1}{\hslash }Gz·z}$$

(15)

where G is the symmetric matrix

$$G=\left(\begin{array}{cc}X+Y{X}^{-1}Y& Y{X}^{-1}\\ X^{-1}Y& X^{-1}\end{array}\right).$$

(16)

A fundamental fact, which is related to the uncertainty principle, is that G is a symplectic matrix, i.e., it belongs to symplectic group $Sp\left(n\right)$. Equivalently, since $G=G^T$,

$$G^{T}JG=GJG=J$$

where

$$J=\left(\begin{array}{cc}{0}_{n\times n}& I_{n\times n}\\ -I_{n\times n}& 0_{n\times n}\end{array}\right)$$

the standard symplectic matrix. We have $G=S^{T}S$, where

$$S=\left(\begin{array}{cc}{X}^{1/2}& 0\\ X^{-1/2}Y& X^{-1/2}\end{array}\right)$$

(17)

is clearly symplectic. Assuming that function $\psi$ for which we are looking is a Gaussian, comparing Formulas (8) and (15) leads to identification

$$\mathsf{\Sigma }=\frac{\hslash }{2}{G}^{-1}.$$

Since $GJG=J$ the inverse $G^{-1}$ is $-JGJ$, explicit formula

$$G^{-1}=\left(\begin{array}{cc}{X}^{-1}& -X^{-1}Y\\ -Y{X}^{-1}& X+Y{X}^{-1}Y,\end{array}\right)$$

so that there remains to solve matrix equation

$$\left(\begin{array}{cc}{\mathsf{\Sigma }}_{XX}& {\mathsf{\Sigma }}_{XP}\\ {\mathsf{\Sigma }}_{PX}& {\mathsf{\Sigma }}_{PP}\end{array}\right)=\frac{\hslash }{2}\left(\begin{array}{cc}{X}^{-1}& -X^{-1}Y\\ -Y{X}^{-1}& X+Y{X}^{-1}Y\end{array}\right).$$

(18)

It immediately follows that we have $X=\frac{\hslash }{2}{\mathsf{\Sigma }}_{XX}^{-1}$ and $Y=-\frac{2}{\hslash }{\mathsf{\Sigma }}_{XP}{\mathsf{\Sigma }}_{XX}^{-1}$, so the unknown Gaussian for which we were looking is

$$\psi \left(x\right)={\left(\frac{1}{2\pi }\right)}^{n/4}{(det{\mathsf{\Sigma }}_{XX})}^{-1/4}exp\left[-\left(\frac{1}{4}{\mathsf{\Sigma }}_{XX}^{-1}+\frac{i}{2\hslash }{\mathsf{\Sigma }}_{XP}{\mathsf{\Sigma }}_{XX}^{-1}\right)x·x\right],$$

(19)

which is the n-dimensional variant of (1), replacing ${\sigma }_{xx}$ with ${\mathsf{\Sigma }}_{XX}$ and ${\sigma }_{xp}$ with ${\mathsf{\Sigma }}_{XP}$. This does not solve completely our problem, however, because we do not know matrix ${\mathsf{\Sigma }}_{XP}$. The crucial step is to notice that, as a bonus, we obtained from (18) the matrix form of the saturated Robertson–Schrödinger equality, namely,

$${\mathsf{\Sigma }}_{PP}{\mathsf{\Sigma }}_{XX}-{\mathsf{\Sigma }}_{XP}^2=\frac{1}{4}{\hslash }^2{I}_{n\times n}.$$

(20)

From this formula we can deduce ${\mathsf{\Sigma }}_{XP}^2$, and one finds two Pauli partners

$${\psi }_{\pm }\left(x\right)={\left(\frac{1}{2\pi }\right)}^{n/4}{(det{\mathsf{\Sigma }}_{XX})}^{-1/4}exp\left[-\left(\frac{1}{4}{\mathsf{\Sigma }}_{XX}^{-1}\pm \frac{i}{2\hslash }{\mathsf{\Sigma }}_{XP}{\mathsf{\Sigma }}_{XX}^{-1}\right)x·x\right]$$

(21)

once a value of ${\mathsf{\Sigma }}_{XP}$ is determined (even if ${\mathsf{\Sigma }}_{XP}^2=0$, we can have ${\mathsf{\Sigma }}_{XP}\ne 0$). Here, we solved a so-called “phase retrieval problem” (see Klibanov et al. [21] for a good review of the topic): in view of Formula (12), we know that

$$\psi \left(x\right)=e^{i\mathsf{\Phi }\left(x\right)}{\left(\frac{1}{2\pi }\right)}^{n/4}{(det{\mathsf{\Sigma }}_{XX})}^{-1/4}{e}^{-\frac{1}{4}{\mathsf{\Sigma }}_{XX}^{-1}x·x}$$

(22)

where $\mathsf{\Phi }$ is an unknown real function of the position variable. We identified this phase here as being function

$$\mathsf{\Phi }\left(x\right)=-\left(\frac{1}{2\hslash }{\mathsf{\Sigma }}_{XP}{\mathsf{\Sigma }}_{XX}^{-1}\right)x·x.$$

4. Geometric Interlude

We introduce the notion of ℏ-polarity and duality; we see in the next section that this notion from convex geometry is quite unexpectedly related to the Pauli problem, of which it gives a limpid geometric interpretation. For a very detailed study of polarity, see Charalambos and Aliprantis [22]. In both sources, alternative competing definitions are also described; the one we use here is the most common and the best fitted to our needs.

Let X be a nonempty subset of n-dimensional configuration space ${\mathbb{R}}_x^n$; this may be, for instance, a set of position measurements performed on some physical system with n degrees of freedom. One defines the polar set of X as the set $X^o$ of all points $p=(p_1,\dots ,p_n)$ in the momentum space ${\mathbb{R}}_p^n$, such that

$$px=p_1{x}_1+\cdots +p_n{x}_n\le 1$$

for all points $x=(x_1,\dots ,x_n)$ in X. Similarly, if P is a subset of ${\mathbb{R}}_p^n$, one defines its polar $P^o$ as the set of all x in ${\mathbb{R}}_x^n$, such that $px\le 1$ for all p in P. We use a rescaled variant of the notion of polarity here, which we call ℏ polarity. By definition, the ℏ-polar $X^{\hslash }$ of X is the set of all p, such that

$$px=p_1{x}_1+\cdots +p_n{x}_n\le \hslash$$

for all points x in X. We have $X^{\hslash }=\hslash X^o$ and $P^{\hslash }=\hslash P^o$ likewise.

From now on, we assume for simplicity that X and P are convex bodies, i.e., they are convex, compact, and with a nonempty interior; we also assume that they are symmetric (i.e., $X=-X$), which implies, by convexity, that they contain 0 in their interior. Simple examples of such sets are balls and ellipsoids centered at the origin. Polar duals have the following remarkable properties:

• Biduality: ${\left(X^{\hslash }\right)}^{\hslash }=X$
• Antimonotonicity: $X\subset Y⟹Y^{\hslash }\subset X^{\hslash }$
• Scaling property: $L\in GL(n,\mathbb{R})⟹{\left(LX\right)}^{\hslash }={\left(L^T\right)}^{-1}{X}^{\hslash }$.

Let ${\mathcal{B}}_X^n\left(R\right)$ (resp. ${\mathcal{B}}_P^n\left(R\right)$) be the ball $\{x:|x|\le R\}$ in ${\mathbb{R}}_x^n$ (resp. $\{p:|p|\le R\}$ in ${\mathbb{R}}_p^n$). We have

$${\mathcal{B}}_X^n{\left(\sqrt{\hslash }\right)}^{\hslash }={\mathcal{B}}_P^n\left(\sqrt{\hslash }\right)$$

(23)

and one can show that ${\mathcal{B}}_X^n\left(\sqrt{\hslash }\right)$ is the only self ℏ-dual set in ${\mathbb{R}}_x^n$. Let us extend this to the case of ellipsoids. An ellipsoid in ${\mathbb{R}}_x^n$ centered at the origin (which is just an ordinary plane ellipse when $n=1$) can always be viewed as the image of ball ${\mathcal{B}}_X^n\left(\sqrt{\hslash }\right)$ by some invertible linear transformation L, in which case, it is given by inequality

$$L^{-1}x·L^{-1}x={\left(L{L}^T\right)}^{-1}x·x\le \hslash .$$

Conversely, if A is a positive definite symmetric matrix, inequality $Ax·x\le \hslash$ always defines an ellipsoid, since it is equivalent to the above inequality, taking for L inverse square root $A^{-1/2}$ of A. It immediately follows from the scaling property that the ℏ-polar of the ellipsoid is obtained by inverting the matrix of the ellipsoid:

$$X:A{x}^2\le \hslash ⟺X^{\hslash }:A^{-1}p·p\le \hslash$$

(24)

(that we have an equivalence follows from biduality property ${\left(X^{\hslash }\right)}^{\hslash }=X$).

5. The Pauli Problem and Polar Duality

Let us return to the Wigner transform of Gaussian states; using Formula (15), we can explicitly calculate $W{\psi }_{\pm }$, and one finds

$$W{\psi }_{\pm }\left(z\right)={(\pi \hslash )}^{-n}{e}^{-\frac{1}{2}{\mathsf{\Sigma }}_{\pm }^{-1}z·z}$$

where covariance matrices ${\mathsf{\Sigma }}_{\pm }$ are given by

$${\mathsf{\Sigma }}_{\pm }=\left(\begin{array}{cc}{\mathsf{\Sigma }}_{XX}& \pm {\mathsf{\Sigma }}_{XP}\\ \pm {\mathsf{\Sigma }}_{PX}& {\mathsf{\Sigma }}_{PP}\end{array}\right),$$

with ${\mathsf{\Sigma }}_{PX}={\mathsf{\Sigma }}_{XP}^T$. Two ellipsoids ${\mathsf{\Omega }}_{\pm }$ centered at the origin correspond to ${\mathsf{\Sigma }}_{\pm }$. Let us determine orthogonal projections ${\mathsf{\Omega }}_{X,\pm }$ and ${\mathsf{\Omega }}_{P\pm }$ of ${\mathsf{\Omega }}_{\pm }$ on the position and momentum spaces ${\mathbb{R}}_x^n$ and ${\mathbb{R}}_p^n$.

5.1. Case $n=1$

We begin with case $n=1$, and projections are line segments. Here, ${\mathsf{\Sigma }}_{XX}={\sigma }_{xx}$, ${\mathsf{\Sigma }}_{PP}={\sigma }_{pp}$, and ${\mathsf{\Sigma }}_{XP}={\mathsf{\Sigma }}_{PX}={\sigma }_{xp}$ and covariance ellipses ${\mathsf{\Omega }}_{\pm }$ are defined by

$${\displaystyle \frac{{\sigma }_{pp}}{2D}}{x}^2\mp \frac{{\sigma }_{xp}}{D}px+{\displaystyle \frac{{\sigma }_{xx}}{2D}}{p}^2\le 1$$

(25)

where $D={\sigma }_{xx}{\sigma }_{pp}-{\sigma }_{xp}^2=\frac{1}{4}{\hslash }^2$ (cf. Formula (3)). Orthogonal projections ${\mathsf{\Omega }}_{X,\pm }$ and ${\mathsf{\Omega }}_{P\pm }$ of ${\mathsf{\Omega }}_{\pm }$ on the x and p axes are the same:

$${\mathsf{\Omega }}_X=[-\sqrt{2{\sigma }_{xx}},\sqrt{2{\sigma }_{xx}}],{\mathsf{\Omega }}_P=[-\sqrt{2{\sigma }_{pp}},\sqrt{2{\sigma }_{pp}}].$$

(26)

Let ${\mathsf{\Omega }}_X^{\hslash }$ be the polar dual of ${\mathsf{\Omega }}_X$: it is the set of all numbers p, such that $px\le \hslash$ for $-\sqrt{2{\sigma }_{xx}}\le x\le \sqrt{2{\sigma }_{xx}}$ and is thus the interval

$${\mathsf{\Omega }}_X^{\hslash }=[-\hslash /\sqrt{2{\sigma }_{xx}},\hslash /\sqrt{2{\sigma }_{xx}}].$$

Since ${\sigma }_{xx}{\sigma }_{pp}\ge \frac{1}{2}\hslash$, we have inclusion

$${\mathsf{\Omega }}_X^{\hslash }\subset {\mathsf{\Omega }}_P$$

(27)

and this inclusion reduces to equality ${\mathsf{\Omega }}_X^{\hslash }={\mathsf{\Omega }}_P$ if and only if the Heisenberg inequality is saturated, i.e., ${\sigma }_{xx}{\sigma }_{pp}=\frac{1}{4}{\hslash }^2$, which is equivalent to ${\sigma }_{xp}=0$.

5.2. General Case

We have similar properties in arbitrary dimension n. To study this case, we first must find the orthogonal projections of covariance ellipsoid $\mathsf{\Omega }$ on the position and momentum spaces. Ellipsoid $\mathsf{\Omega }$ is given by equation $Mz·z\le \hslash$ where $M=\frac{\hslash }{2}{\mathsf{\Sigma }}^{-1}$ is symmetric and positive definite ($M>0$). Writing M in block form

$$M=\left(\begin{array}{cc}{M}_{XX}& M_{XP}\\ M_{PX}& M_{PP}\end{array}\right)$$

where $M_{XX}=M_{XX}^T$, $M_{PP}=M_{PP}^T$, and $M_{XP}=M_{XP}^T$ are $n\times n$ matrices; since $M>0$, we also have $M_{XX}>0$ and $M_{PP}>0$. Then, the projections of $\mathsf{\Omega }$ on ${\mathbb{R}}_x^n$ and ${\mathbb{R}}_p^n$ are ellipsoids given by, respectively [23],

$${\mathsf{\Omega }}_X:(M/M_{PP})x·x\le \hslash \},{\mathsf{\Omega }}_P=(M/M_{XX})p·p\le \hslash$$

(28)

where symmetric matrices

$$\begin{array}{cc}\hfill M/M_{PP}& =M_{XX}-M_{XP}{M}_{PP}^{-1}{M}_{PX}\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill M/M_{XX}& =M_{PP}-M_{PX}{M}_{XX}^{-1}{M}_{XP}\hfill \end{array}$$

(30)

are Schur complements in M of $M_{PP}$ and $M_{XX}$; we have $M/M_{PP}>0$ and $M/M_{XX}>0$ so that ${\mathsf{\Omega }}_X$ and ${\mathsf{\Omega }}_P$ are nondegenerate (see Zhang’s treatise [24] for a detailed study of the Schur complement). To prove that inclusion ${\mathsf{\Omega }}_X^{\hslash }\subset {\mathsf{\Omega }}_P$ holds, we must show that cf. implication (24)) that

$$(M/M_{PP})(M/M_{XX})\le I_{n\times n},$$

(31)

that is, that the eigenvalues of $(M/M_{PP})(M/M_{XX})$ must be smaller than 1. To prove this, we use the following essential remark: we showed above that matrix $M=\frac{\hslash }{2}{\mathsf{\Sigma }}^{-1}$ is symplectic; therefore, its entries obey some constraints. Considering that M is also symmetric, these constraints are

$$\begin{array}{c}{M}_{XX}{M}_{PP}-M_{XP}^2=I_{n\times n}\end{array}$$

(32)

$$\begin{array}{c}{M}_{XX}{M}_{PX}=M_{XP}{M}_{XX}\end{array}$$

(33)

$$\begin{array}{c}{M}_{PX}{M}_{PP}=M_{PP}{M}_{XP}.\end{array}$$

(34)

Using Identities (33) and (34), it follows that Schur complements (29) and (30) can be rewritten as

$$\begin{array}{cc}\hfill M/M_{PP}& =M_{XX}-M_{PP}^{-1}{M}_{PX}^2\hfill \\ \hfill & =M_{PP}^{-1}(M_{PP}{M}_{XX}-M_{PX}^2)\hfill \\ \hfill & =M_{PP}^{-1}\hfill \end{array}$$

the last equality by using the transpose of Identity (32). Similarly,

$$M/M_{XX}=M_{PP}-M_{XX}^{-1}{M}_{XP}^2=M_{XX}^{-1}$$

So, summarizing, Schur complements are given by

$$M/M_{PP}=M_{PP}^{-1},M/M_{XX}=M_{XX}^{-1}.$$

(35)

It follows that

$$(M/M_{PP})(M/M_{XX})=M_{PP}^{-1}{M}_{XX}^{-1}={\left(M_{XX}{M}_{PP}\right)}^{-1}.$$

We show that $(M/M_{PP})(M/M_{XX})\le I_{n\times n}$; equivalently, $M_{XX}{M}_{PP}\ge I_{n\times n}$. Now, since $M=\frac{\hslash }{2}{\mathsf{\Sigma }}^{-1}$ is symplectic, so is matrix

$$M^{-1}=\frac{2}{\hslash }\mathsf{\Sigma }=\left(\begin{array}{cc}\frac{2}{\hslash }{\mathsf{\Sigma }}_{XX}& \frac{2}{\hslash }{\mathsf{\Sigma }}_{XP}\\ \frac{2}{\hslash }{\mathsf{\Sigma }}_{PX}& \frac{2}{\hslash }{\mathsf{\Sigma }}_{PP};\end{array}\right)$$

hence, reinverting,

$$M=\left(\begin{array}{cc}\frac{2}{\hslash }{\mathsf{\Sigma }}_{PP}& -\frac{2}{\hslash }{\mathsf{\Sigma }}_{PX}\\ -\frac{2}{\hslash }{\mathsf{\Sigma }}_{XP}& \frac{2}{\hslash }{\mathsf{\Sigma }}_{XX}\end{array}\right)$$

(36)

so that $M_{XX}{M}_{PP}=\frac{4}{{\hslash }^2}{\mathsf{\Sigma }}_{PP}{\mathsf{\Sigma }}_{XX}$. In view of the generalized RSUP (20), we have

$${\mathsf{\Sigma }}_{PP}{\mathsf{\Sigma }}_{XX}-{\mathsf{\Sigma }}_{XP}^2=\frac{1}{4}{\hslash }^2{I}_{n\times n}$$

(37)

hence

$$M_{XX}{M}_{PP}=I_{n\times n}+\frac{4}{{\hslash }^2}{\mathsf{\Sigma }}_{XP}^2$$

(38)

and we are finished, provided that we can prove that ${\mathsf{\Sigma }}_{XP}^2\ge 0$ (which is obvious if $n=1$), or, which amounts to the same $M_{XP}^2\ge 0$. For this, since $M_{XX}{M}_{PX}=M_{XP}{M}_{XX}$ (Formula (33)), we have

$$M_{XP}=M_{XX}{M}_{PX}{M}_{XX}^{-1};$$

(39)

hence, $M_{XP}$ and $M_{PX}$ have the same eigenvalues; since $M_{PX}=M_{XP}^T$, these eigenvalues must be real, and those of $M_{XP}^2$ must be $\ge 0$.

For completeness, we still need to discuss what happens when ${\mathsf{\Omega }}_X^{\hslash }={\mathsf{\Omega }}_P$. In view of Formulas (28) and Equivalence (24), this means that (31) reduces to equality

$$(M/M_{PP})(M/M_{XX})=I_{n\times n}$$

that is, by (35), $M_{XX}{M}_{PP}=I_{n\times n}$. Taking (38) into account, we must thus have $M_{XP}^2=0$, which does not imply that $M_{XP}=0$. We are in the presence of states (21) in this case, saturating the Heisenberg inequalities.

6. Discussion and Outlook

Our discussion of polar duality suggests that a quantum system localized in the position representation in a set X cannot be localized in the momentum representation in a set smaller than that of its polar dual $X^{\hslash }$. The notion of polar duality thus appears informally as a generalization of the uncertainty principle of quantum mechanics, as expressed in terms of variances and covariances (see [23]). The idea of such generalizations is not new, and can already be found in the work of Uffink and Hilgevoord [25,26]; see Butterfield’s discussion in [27]. It would certainly be interesting to explore the connection between convex geometry and quantum mechanics, but very little work has been conducted so far.