Multi-observable uncertainty bounds via matrix inequalities

Saptak Bhattacharya

doi:10.4153/S0008439526102197

Multi-observable uncertainty bounds via matrix inequalities

Part of: Basic linear algebra

Published online by Cambridge University Press: 24 June 2026

Saptak Bhattacharya

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Saptak Bhattacharya*: Affiliation:
Indian Statistical Institute, Delhi Centre , India
*: e-mail: saptak21r@isid.ac.in

Article contents

Abstract
Introduction
Main results
Conclusions
Data availability statement
Competing interests
Footnotes
References

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Abstract

We establish new uncertainty bounds for commutators and anti-commutators of observables in both finite- and infinite-dimensional quantum systems. Beginning with a short, elementary proof of Robertson’s standard uncertainty principle, we then derive inequalities whose lower bounds remain nonzero, regardless of the number of observables. The main result is a novel trace inequality comparing the block matrices of commutators and anti-commutators of several observables.

Keywords

Uncertainty principles matrix inequalities commutator bounds

MSC classification

Primary: 15A45: Miscellaneous inequalities involving matrices

Secondary: 81P16: Quantum state spaces, operational and probabilistic concepts

Information

Type: Article
Information: Canadian Mathematical Bulletin , First View , pp. 1 - 19

DOI: https://doi.org/10.4153/S0008439526102197 [Opens in a new window]
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction

Heisenberg’s uncertainty principle [Reference Heisenberg22], one of the cornerstones of quantum mechanics, asserts that the position and momentum of a particle on a line cannot be simultaneously determined with arbitrary accuracy. This places a fundamental limitation on the precision of physical measurements. To make it rigorous, we turn to the Hilbert space formalization of quantum mechanics.

A particle moving on the real line is modeled by the Hilbert space $L^2(\mathbb {R})$ . Each unit vector in this space represents a possible quantum state of the particle. If ${f \in L^2(\mathbb {R})}$ with $||f||_2=1$ denotes the state, then the probability of finding the particle in an interval $(a,b)\subset \mathbb {R}$ is given by the Born rule:

$$\begin{align*}\int_a^b |f(t)|^2\, dt. \end{align*}$$

The position operator $\hat {x}$ is defined on the domain

$$\begin{align*}\mathcal{D}(\hat{x}) = \Big\{ f\in L^2(\mathbb{R}) : \int_{\mathbb{R}} y^2 |f(y)|^2 \, dy < \infty \Big\} \end{align*}$$

$$\begin{align*}\hat{x}(f)(y) = y f(y). \end{align*}$$

Similarly, the momentum operator $\hat {p}$ is defined on

$$\begin{align*}\mathcal{D}(\hat{p}) = \{ f\in L^2(\mathbb{R}) : f' \in L^2(\mathbb{R})\} \end{align*}$$

$$\begin{align*}\hat{p}(f) = -i \hbar f', \end{align*}$$

where $f'$ denotes the distributional derivative of f, and $\hbar $ is Planck’s constant. Both $\hat {x}$ and $\hat {p}$ are unbounded self-adjoint operators, and on the dense subspace of Schwartz functions $\mathcal {S}(\mathbb {R}),$ they satisfy the canonical commutation relation

$$\begin{align*}[\hat{x}, \hat{p}] = i \hbar I. \end{align*}$$

For two self-adjoint operators A and B on a Hilbert space $\mathcal {H}$ , and any unit vector $x \in \mathcal {D}(AB-BA)$ , the Cauchy–Schwarz inequality yields

(1) $$ \begin{align}\begin{aligned}||Ax||^2 ||Bx||^2&\geq|\mathrm{Im} \langle Ax, Bx\rangle|^2\\&=\frac{|\langle ABx, x\rangle-\langle BAx, x\rangle|^2}{4}\\&=\frac{|\langle [A, B]x, x\rangle|^2}{4}.\end{aligned}\end{align} $$

StartLayout 1st Row 1st Column StartAbsoluteValue EndAbsoluteValue upper A x StartAbsoluteValue EndAbsoluteValue squared StartAbsoluteValue EndAbsoluteValue upper B x StartAbsoluteValue EndAbsoluteValue squared 2nd Column greater than or equals StartAbsoluteValue im left angle bracket upper A x comma upper B x right angle bracket EndAbsoluteValue squared 2nd Row 1st Column Blank 2nd Column equals StartFraction StartAbsoluteValue left angle bracket upper A upper B x comma x right angle bracket minus left angle bracket upper B upper A x comma x right angle bracket EndAbsoluteValue squared Over 4 EndFraction 3rd Row 1st Column Blank 2nd Column equals StartFraction StartAbsoluteValue left angle bracket left bracket upper A comma upper B right bracket x comma x right angle bracket EndAbsoluteValue squared Over 4 EndFraction period EndLayout

Note that equality holds in (1) if and only if there exists a real $\lambda $ such that $Ax=i\lambda Bx$ .

Since commutators are unaffected by scalar shifts, one may freely replace A by ${A-aI}$ and B by $B-bI$ in (1) for any $a,b\in \mathbb {R}$ . This observation will lead us to the variance formulation of the uncertainty principle.

Let $f \in \mathcal {S}(\mathbb {R})$ with $\|f\|_2=1$ . Such functions are smooth, rapidly decaying, and dense in $L^2(\mathbb {R})$ . For $a,b\in \mathbb {R}$ , one computes

$$\begin{align*}\|(\hat{x}-aI)f\|_2^2 \;=\; \int_{\mathbb{R}} (y-a)^2 |f(y)|^2 \, dy, \end{align*}$$

the variance of the position observable. On the other hand,

$$\begin{align*}\|(\hat{p}-bI)f\|_2^2 = \int_{\mathbb{R}} | -i \hbar f'(y) - b f(y) |^2 \, dy. \end{align*}$$

Passing to the Fourier transform $\hat f$ and using Parseval’s identity, this expression becomes

$$\begin{align*}\|(\hat{p}-bI)f\|_2^2 = 4\pi^2 \hbar^2 \int_{\mathbb{R}} (\omega-b)^2 |\hat f(\omega)|^2 \, d\omega, \end{align*}$$

which is the variance of the momentum observable.

Substituting into (1) with $A=\hat {x}-aI$ and $B=\hat {p}-bI$ , we obtain the Heisenberg–Weyl uncertainty inequality:

(2) $$ \begin{align} \int_{\mathbb{R}} (y-a)^2 |f(y)|^2 \, dy \; \int_{\mathbb{R}} (\omega-b)^2 |\hat f(\omega)|^2 \, d\omega \;\geq\; \frac{1}{16\pi^2}. \end{align} $$

Choosing $a = \int _{\mathbb {R}} y |f(y)|^2 dy$ and $b = 2\pi \hbar \int _{\mathbb {R}} \omega |\hat f(\omega )|^2 d\omega $ (the expectations of position and momentum, respectively), the terms on the left-hand side become the variances $\sigma _x^2$ and $\sigma _p^2$ . Inequality (2) then reads

$$\begin{align*}\sigma_x^2 \, \sigma_p^2 \;\geq\; \frac{\hbar^2}{4}. \end{align*}$$

Equality holds precisely for the Gaussians.

From the perspective of harmonic analysis, (2) expresses a universal limitation:

$$ \begin{align*} \textit{A function and its Fourier transform cannot both be sharply localized.} \end{align*} $$

This principle underlies many uncertainty relations in analysis (see [Reference Folland and Sitaram17] for an overview). Although we have stated everything for Schwartz functions, the inequality extends to all $f \in \mathcal {D}(\hat {x}) \cap \mathcal {D}(\hat {p})$ , albeit with a more involved proof (see [Reference Hall21] for details).

Remark The essence of the uncertainty principle is the non-commutativity of quantum observables, represented by self-adjoint operators on a Hilbert space. This naturally raises broader questions: what uncertainty relations can be derived for arbitrary self-adjoint operators for more than two observables? Also, can we do it for more general states than those given by the unit vectors? The was first addressed by Robertson in 1934, which later led to a whole family of generalized uncertainty relations. It should be noted here that the framework and terminology we are going to use henceforth will be significantly different from Robertson.

Seemingly, the most natural way to discuss this is to use the operator-algebraic approach to quantum mechanics. Let $\mathcal {A}$ be a unital $C^*$ -algebra of bounded operators modeling a quantum system and let $\phi :\mathcal {A}\to \mathbb {C}$ be a state. Given a self-adjoint ${x\in \mathcal {A}}$ representing an observable, its expectation and variance are given by $\phi (x)$ and ${\phi \big (x-\phi (x))^2}$ , respectively.

Now, let $\{x_j\}_{j=1}^n$ be self-adjoint elements of $\mathcal {A}$ . We normalize them so that $\phi (x_j)=0$ for all $1\leq j\leq n$ . This will be the assumption throughout the article.

Remark We can do this without loss of generality since our lower bounds will depend only on the commutators $[x_i, x_j]$ , and therefore, remain unchanged if the observables are translated by scalars.

The covariance matrix of the n-tuple $\tilde {x}=(x_j)_{j=1}^n$ is the $n\times n$ positive matrix given by

$$\begin{align*}\mathrm{Cov}_{\phi}(\tilde{x})=\big(\mathrm{Re}\hspace{1mm}\phi(x_i x_j)\big).\end{align*}$$

This can also be written as

$$\begin{align*}\mathrm{Cov}_{\phi}(\tilde{x})={M+M^T},\end{align*}$$

where

(3) $$ \begin{align}M=\frac{1}{2}\big(\phi(x_ix_j)\big)\end{align} $$

is a Gram matrix. Note that

$$\begin{align*}M-M^T=\frac{1}{2}\big(\phi[x_i, x_j]\big)\end{align*}$$

and

$$\begin{align*}M+M^T=\mathrm{Cov}_{\phi}(\tilde{x})=\frac{1}{2}\big(\phi\{x_i,x_j\}\big),\end{align*}$$

where $[x_i, x_j]=x_i x_j-x_j x_i$ is the commutator and $\{x_i, x_j\}=x_ix_j+x_jx_i$ is the anti-commutator of $x_i$ and $x_j$ . In this form, Robertson’s standard uncertainty principle (see [Reference Robertson35]) can be stated as

(4) $$ \begin{align}\mathrm{det}|M-M^T|=\mathrm{det}\big|\big(\phi[x_i, x_j]\big)\big|\leq\mathrm{det}\hspace{1mm}\mathrm{Cov}_{\phi}(\tilde{x}).\end{align} $$

Note that for two self-adjoint elements x and y, this reduces to the well-known Schrödinger–Heisenberg uncertainty principle [Reference Schrödinger36]:

$$\begin{align*}\phi(x^2)\phi(y^2)-|\mathrm{Re}\hspace{1mm}\phi(yx)|^2\geq\frac{1}{4}|\phi[x, y]|^2.\end{align*}$$

An immediate consequence is the Heisenberg uncertainty principle, now formulated as

$$\begin{align*}\phi(x^2)\phi(y^2)\geq\frac{1}{4}|\phi[x, y]|^2.\end{align*}$$

If $\mathcal {A}=M_k(\mathbb {C})$ , a state $\phi :M_k(\mathbb {C})\to \mathbb {C}$ can be written as $\phi (X)=\mathrm {tr}\hspace {1mm} AX$ for all ${X\in M_k(\mathbb {C})}$ , for some density matrix A, that is, $A\geq O$ and $\mathrm {tr}\hspace {1mm} A=1$ . In this case, for a pair of Hermitian matrices H and K, we can write the Heisenberg uncertainty principle as

$$\begin{align*}\mathrm{tr}\hspace{1mm} AH^2\hspace{1mm} \mathrm{tr}\hspace{1mm} AK^2\geq\frac{|\mathrm{tr}\hspace{1mm}A[H,K]|^2}{4}.\end{align*}$$

This is very important in the context of quantum computation and quantum information theory, dealing mainly with finite-dimensional quantum systems (see [Reference Wilde40] for more details).

Though inequality (4) generalizes Heisenberg’s uncertainty principle for n observables, it has a downside. Note that $i(M-M^T)$ is a real skew-symmetric matrix, and therefore, if n is odd, its determinant vanishes. Thus, for an odd number of observables, inequality (4) fails to capture any quantum phenomenon arising from non-commutativity. This prompted research on uncertainty relations which do not have this drawback. Andai in [Reference Andai1] and Gibilisco et al. in [Reference Gibilisco, Imparato and Isola19] proved the following.

Theorem Let A be a positive-definite density matrix. Let $\{H_j\}_{j=1}^k$ be Hermitian and let $f:(0,\infty )\to \mathbb {R}$ be a positive operator monotone function satisfying $f(1)=1$ and $xf(\frac {1}{x})=f(x)$ for all $x\in (0,\infty )$ . Then

$$\begin{align*}\mathrm{det}\hspace{1mm}\left(\frac{f(0)}{2}\gamma_f([A, H_i],[A, H_j])\right)\leq\mathrm{det}\hspace{1mm}\big(\mathrm{Re}\hspace{1mm}\mathrm{tr}\hspace{1mm}AH_i H_j\big).\end{align*}$$

Here, $\gamma _f$ denotes the quantum Fisher information metric (see [Reference Petz32, Reference Petz and Ghinea33]) associated with f. This inequality is also known as the dynamical uncertainty principle and was later generalized to arbitrary covariances by Gibilisco et al. in [Reference Gibilisco, Hiai and Petz18].

The question arises, can we have variants of the standard uncertainty principle (4) involving the commutators $[x_i, x_j]$ with a lower bound that does not necessarily vanish depending on the number of observables? Also, can we have a lower bound which is always nonzero whenever any two of the observables do not commute, thus capturing any quantum uncertainty arising from non-commutativity? It is to be noted that multi-observable uncertainty relations involving the sum of variances have been proved by Chen et al. [Reference Chen and Fei13] and Huang [Reference Huang24], but these did not explicitly involve the commutators in the lower bound.

This article is an attempt in that direction. The sum, difference, and geometric mean of positive matrices will all play a key role in the subsequent results.

Lastly, for k Hermitian matrices $\{H_j\}_{j=1}^k$ , we discuss the problem of comparing the block matrices $\big ([H_i, H_j]\big )$ and $\big (\{H_i, H_j\}\big )$ directly without composing a state entrywise like it is done in (4) and prove an interesting trace inequality. This gives a nonzero lower bound whenever there exist $i, j$ such that $H_iH_j \neq H_jH_i$ and generalizes seamlessly to infinite dimensions. It is to be noted that in this case, we cannot write both sides as the sum and difference of two positive matrices, prompting the use of other techniques.

2 Main results

We first give a short proof of Robertson’s standard uncertainty principle (4). For that, we need a lemma.

Lemma 1 Let $A, B$ be two positive matrices. Then

$$\begin{align*}\mathrm{det}(A-B)^2\leq \mathrm{det}(A+B)^2.\end{align*}$$

Proof Suffices to prove for positive definite A and B. If A and B commute, the inequality is obvious. Otherwise, we note that

$$ \begin{align*}\begin{aligned} \textrm{det}(A-B)^2 &=\textrm{det}\big(A^{1/2}(I-A^{-1/2}BA^{-1/2})A^{1/2}\big)^{2}\\ &=\textrm{det}A^2\hspace{1mm}\textrm{det}(I-A^{-1/2}BA^{-1/2})^{2} \\ &\leq \textrm{det} A^2\hspace{1mm}\textrm{det}(I+A^{-1/2}BA^{-1/2})^{2} \\ &=\textrm{det}(A+B)^2. \end{aligned}\end{align*} $$

StartLayout 1st Row 1st Column det left parenthesis upper A minus upper B right parenthesis squared 2nd Column equals det left parenthesis upper A Superscript 1 divided by 2 Baseline left parenthesis upper I minus upper A Superscript negative 1 divided by 2 Baseline upper B upper A Superscript negative 1 divided by 2 Baseline right parenthesis upper A Superscript 1 divided by 2 Baseline right parenthesis squared 2nd Row 1st Column Blank 2nd Column equals det upper A squared det left parenthesis upper I minus upper A Superscript negative 1 divided by 2 Baseline upper B upper A Superscript negative 1 divided by 2 Baseline right parenthesis squared 3rd Row 1st Column Blank 2nd Column less than or equals det upper A squared det left parenthesis upper I plus upper A Superscript negative 1 divided by 2 Baseline upper B upper A Superscript negative 1 divided by 2 Baseline right parenthesis squared 4th Row 1st Column Blank 2nd Column equals det left parenthesis upper A plus upper B right parenthesis squared period EndLayout

Remark Note that the operator inequality $(A-B)^2\leq (A+B)^2$ need not be true for positive A and B unless A and B commute.

As an immediate consequence, we have Robertson’s uncertainty principle.

Theorem 1 Let $\{x_i\}_{i=1}^n$ be self-adjoint elements in a $C^*$ -algebra $\mathcal {A}$ . Let $\phi :\mathcal {A}\to \mathbb {C}$ be a state. Then,

$$\begin{align*}\mathrm{det}\left(\frac{1}{2}\phi[x_i, x_j]\right)^2\leq\mathrm{det}\hspace{1mm}\mathrm{Cov}_{\phi}(\tilde{x})^2.\end{align*}$$ StartLayout 1st Row det left parenthesis one half phi left bracket x Subscript i Baseline comma x Subscript j Baseline right bracket right parenthesis squared less than or equals det upper C o v Subscript phi Baseline left parenthesis x overtilde right parenthesis squared period EndLayout

Proof Consider the positive matrix M as given in (3) and apply Lemma 1.

Before we state the next lemma, let us have a quick review of the matrix geometric mean, which shall be used subsequently. Given two positive-definite matrices A and B, their geometric mean is given by

$$\begin{align*}A\# B = A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}.\end{align*}$$

This was introduced by Pusz and Woronowicz in [Reference Pusz and Woronowicz34] in the context of positive sesquilinear forms, and later studied in detail from the matrix-theoretic perspective by Ando in [Reference Ando2]. Let $\mathbb {P}(n)$ be the set of all $n\times n$ positive-definite matrices equipped with the Riemannian metric

(5) $$ \begin{align}\langle H, K \rangle_A = \mathrm{tr\hspace{1mm}}(A^{-1}HA^{-1}K)\end{align} $$

for all $A\in \mathbb {P}(n)$ , and Hermitian H and K. It can be shown that for $A, B\in \mathbb {P}(n)$ , $A\# B$ is precisely the geodesic midpoint between A and B with respect to the metric (5). In fact, the geodesic joining A and B is given explicitly by

$$ \begin{align*}t\to A^{1/2}(A^{-1/2}BA^{-1/2})^tA^{1/2},\end{align*} $$

where $t\in [0,1]$ .

This geometric viewpoint was developed independently by Bhatia and Holbrook [Reference Bhatia and Holbrook8] and Moakher [Reference Moakher30] and was used to define the geometric mean of k positive-definite matrices for any $k\in \mathbb {N}$ .

For positive matrices A and B which are not necessarily invertible, we can define

$$\begin{align*}A\#B=\mathop{\lim}\limits_{\varepsilon\to 0+}(A+\varepsilon I)\#(B+\varepsilon I).\end{align*}$$

An important variational characterization is given by

(6) $$ \begin{align}A\#B = \mathrm{max}\{X:X^*=X, \begin{pmatrix}A & X\\X & B\end{pmatrix}\geq O\},\end{align} $$ upper A number sign upper B equals max left brace upper X colon upper X Superscript asterisk Baseline equals upper X comma Start 2 By 2 Matrix 1st Row 1st Column upper A 2nd Column upper X 2nd Row 1st Column upper X 2nd Column upper B EndMatrix greater than or equals upper O right brace comma

as a consequence of which, we get

$$ \begin{align*}A \# B\leq\frac{A+B}{2}.\end{align*} $$

For details, refer to [Reference Ando2, Reference Bhatia4].

The geometric mean is extremely important and finds wide applications in matrix inequalities, quantum information, and even in medical imaging (see [Reference Ando2, Reference Bhatia6, Reference Bhatia, Nielsen and Bhatia7, Reference Congedo, Barachant and Bhatia15, Reference Hiai23, Reference Jenčová25, Reference Lawson and Lim27] for more details).

We now present a couple of lemmas which will be crucial in the later results.

Lemma 2 Let $A, B$ be positive matrices. Then there exists a unitary U such that

$$\begin{align*}|A-B|\leq (A+B)\# U(A+B)U^*.\end{align*}$$

Proof Note that

$$\begin{align*}\begin{pmatrix}A+B & A-B\\A-B & A+B\end{pmatrix}=\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\otimes A +\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix}\otimes B,\end{align*}$$

which is positive. Let

$$\begin{align*}A-B=U|A-B|\end{align*}$$

be the polar decomposition of $A-B$ . Note that U commutes with $A-B$ and hence, $|A-B|$ since $A-B$ is Hermitian. Observe that

$$\begin{align*}\begin{pmatrix}A+B & |A-B|\\|A-B| & U(A+B)U^*\end{pmatrix}=\begin{pmatrix} I & O\\ O & U\end{pmatrix}\begin{pmatrix}A+B & A-B\\A-B & A+B\end{pmatrix}\begin{pmatrix} I & O\\ O & U^*\end{pmatrix}.\end{align*}$$

Hence, the matrix

$$\begin{align*}\begin{pmatrix}A+B & |A-B|\\|A-B| & U(A+B)U^*\end{pmatrix}\end{align*}$$ Start 2 By 2 Matrix 1st Row 1st Column upper A plus upper B 2nd Column StartAbsoluteValue upper A minus upper B EndAbsoluteValue 2nd Row 1st Column StartAbsoluteValue upper A minus upper B EndAbsoluteValue 2nd Column upper U left parenthesis upper A plus upper B right parenthesis upper U Superscript asterisk EndMatrix

is positive and therefore,

$$\begin{align*}|A-B|\leq (A+B)\# U(A+B)U^*\end{align*}$$

by (6).

Lemma 3 Let $A, B\in M_n(\mathbb {C})$ be positive matrices. Then, for any $n\in \mathbb {N}$ ,

$$ \begin{align*}\otimes^n(A \# B)=(\otimes^n \, A) \# (\otimes^n \, B).\end{align*} $$

Proof If A and B are positive definite,

$$\begin{align*}A\# B = A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}.\end{align*}$$

Using multiplicativity under tensor powers,

$$ \begin{align*}\begin{aligned}\otimes^n(A \# B) = (\otimes^n A)^{1/2}[(\otimes^n A)^{-1/2}(\otimes^n B) (\otimes^n A)^{-1/2}]^{1/2}(\otimes^n A)^{1/2}.\end{aligned}\end{align*} $$

When A and B are not positive definite, we proceed in two steps. First, note by (6) that

$$ \begin{align*}\begin{pmatrix} A & A \# B\\ A \# B & B\end{pmatrix}\geq O.\end{align*} $$ Start 2 By 2 Matrix 1st Row 1st Column upper A 2nd Column upper A number sign upper B 2nd Row 1st Column upper A number sign upper B 2nd Column upper B EndMatrix greater than or equals upper O period

By Proposition 1.3.2 of [Reference Bhatia6], a block matrix of the form

$$ \begin{align*}\begin{pmatrix} A & X\\ X^* & B\end{pmatrix}\end{align*} $$ Start 2 By 2 Matrix 1st Row 1st Column upper A 2nd Column upper X 2nd Row 1st Column upper X Superscript asterisk 2nd Column upper B EndMatrix

is positive if and only if there exists a contraction K such that

$$ \begin{align*}X=A^{1/2}K B^{1/2}.\end{align*} $$

Choose $K\in M_n(\mathbb {C})$ such that $||K||\leq 1$ and

$$ \begin{align*}A \# B = A^{1/2}KB^{1/2}.\end{align*} $$

Taking tensor powers,

$$ \begin{align*}\otimes^n (A \# B)=(\otimes^n A)^{1/2}(\otimes^n K)(\otimes^n B)^{1/2}.\end{align*} $$

Since $||\otimes ^n K||=||K||^n\leq 1$ , we get

$$ \begin{align*}\begin{pmatrix}\otimes^n A & \otimes^n (A \# B)\\ \otimes^n (A \# B) & \otimes^n B\end{pmatrix}\geq O.\end{align*} $$

By (6), this implies

$$ \begin{align*}\otimes^n(A \# B)\leq (\otimes^n A) \# (\otimes^n B).\end{align*} $$

The final step is to show the reverse bound

$$ \begin{align*}\otimes^n(A \# B)\geq (\otimes^n A) \# (\otimes^n B).\end{align*} $$

Take $\varepsilon > 0$ . Since $A+\varepsilon I$ and $B+\varepsilon I$ are positive definite,

$$ \begin{align*}\otimes^n[(A+\varepsilon I) \# (B+\varepsilon I)]=[\otimes^n(A+\varepsilon I)] \# [\otimes^n (B+\varepsilon I)].\end{align*} $$

It is easy to see by multilinearity that $\otimes ^n(A+\varepsilon I)\geq \otimes ^n A + \varepsilon ^n\otimes ^n I$ and $\otimes ^n(B+\varepsilon I)\geq \otimes ^n B + \varepsilon ^n\otimes ^n I$ . By monotonicity of the geometric mean (see Theorem 4.1.3 in [Reference Bhatia6]),

$$ \begin{align*}\begin{aligned}\otimes^n[(A+\varepsilon I) \# (B+\varepsilon I)]&=[\otimes^n(A+\varepsilon I)] \# [\otimes^n (B+\varepsilon I)]\\ &\geq (\otimes^n A + \varepsilon^n\otimes^n I) \# (\otimes^n B + \varepsilon^n\otimes^n I).\end{aligned}\end{align*} $$

The desired inequality is now obtained by taking $\varepsilon \to 0+$ .

Remark The comparison argument in the proof was necessary, since the matrix geometric mean is not jointly continuous (see [Reference Bhatia6]). Hence, we cannot directly say that

$$ \begin{align*}\lim_{\varepsilon\to 0+}[\otimes^n(A+\varepsilon I)] \# [\otimes^n (B+\varepsilon I)]=(\otimes^n A) \# (\otimes^n B).\end{align*} $$

Tensor powers are highly non-linear, so we had to bound from below with the more tractable $\otimes ^n A +\varepsilon ^n\otimes ^n I$ and $\otimes ^n B +\varepsilon ^n\otimes ^n I$ .

Before discussing the applications of Lemmas 2 and 3 to uncertainty bounds, let us quickly recap unitarily invariant norms. Let $\{e_1,\dots , e_n\}$ be the standard basis of $\mathbb {C}^n$ . Given any $x,y\in \mathbb {C}^n$ , we denote by $x\otimes y^*$ the rank one operator $v\to \langle v, y\rangle x$ for all $v\in \mathbb {C}^n$ .

Following [Reference Bhatia4], a norm $|||.|||$ on $M_n(\mathbb {C})$ is said to be unitarily invariant if ${|||e_1\otimes e_1^*|||=1}$ and

$$ \begin{align*}|||UAV|||=|||A|||\end{align*} $$

for all $A\in M_n(\mathbb {C})$ and $n\times n$ unitaries U and V. We note, as a natural consequence, that for all unit vectors $x, y$ , $|||x\otimes y^*|||=1$ . A natural class of examples is given by the Schatten p-norms.

Example 1 Given $X\in M_n(\mathbb {C})$ and $p\geq 1$ , define

$$ \begin{align*}||X||_p=(\mathrm{tr}\, |X|^p)^{1/p}.\end{align*} $$

This is known as the Schatten p-norm. If $p=1$ , this gives the trace norm $||X||_1=\mathrm {tr}\, |X|$ . For $p=2$ , it gives the Frobenius norm

$$ \begin{align*}||X||_2=(\mathrm{tr}\, X^*X)^{1/2},\end{align*} $$

while for $p=\infty $ , we get the usual operator norm $||.||$ .

It can be shown that any unitarily invariant norm $|||.|||$ satisfies

$$ \begin{align*}||X||\leq |||X|||\leq ||X||_1\end{align*} $$

for all $X\in M_n(\mathbb {C})$ (see [Reference Bhatia4]). Thus, the trace norm and operator norm are the two extremes of the class of all unitarily invariant norms. The set of all extreme points of the unit ball $\{X\in M_n(\mathbb {C}):||X||_1\leq 1\}$ is precisely $\{x\otimes y^*:||x||=||y||=1\}$ (see [Reference Grone and Marcus20]). While for the operator norm, the extreme points of the unit ball are precisely the unitaries (see [Reference Bhatia6]).

The theory of unitarily invariant norms is very rich. Refer to [Reference Bhatia4] for a thorough discussion, including a complete characterization and applications to interesting matrix inequalities.

Corollary 1 Let $\{x_j\}_{j=1}^n$ be self-adjoint elements in a $C^*$ -algebra $\mathcal {A}$ and let $\phi :\mathcal {A}\to \mathbb {C}$ be a state. Then for any unitarily invariant norm $|||.|||$ on $M_n(\mathbb {C})$ ,

$$\begin{align*}|||\big(\frac{1}{2}\phi[x_i, x_j]\big)|||\leq |||\mathrm{Cov}_{\phi}(\tilde{x})|||.\end{align*}$$

Proof Consider

$$ \begin{align*}M=\frac{1}{2}\big(\phi(x_i x_j)\big)\end{align*} $$

as in (3) and recall that

$$ \begin{align*}M+M^T=\mathrm{Cov}_{\phi}(\tilde{x}),\end{align*} $$

while

$$ \begin{align*}M-M^{T}=\big(\frac{1}{2}\phi[x_i, x_j]\big).\end{align*} $$

By Lemma 2, there exists a unitary U such that

$$ \begin{align*} \begin{aligned} |M-M^T| &\leq (M+M^T)\# U(M+M^T)U^*\\ &\leq\frac{(M+M^T+U(M+M^T)U^*)}{2}. \end{aligned} \end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue upper M minus upper M Superscript upper T Baseline EndAbsoluteValue 2nd Column less than or equals left parenthesis upper M plus upper M Superscript upper T Baseline right parenthesis number sign upper U left parenthesis upper M plus upper M Superscript upper T Baseline right parenthesis upper U Superscript asterisk Baseline 2nd Row 1st Column Blank 2nd Column less than or equals StartFraction left parenthesis upper M plus upper M Superscript upper T Baseline plus upper U left parenthesis upper M plus upper M Superscript upper T Baseline right parenthesis upper U Superscript asterisk Baseline right parenthesis Over 2 EndFraction period EndLayout

Thus,

$$ \begin{align*} \begin{aligned} |||M-M^T||| &=|||\hspace{1mm}|M-M^T|\hspace{1mm}|||\\ &\leq|||M+M^T|||\\ &=|||\mathrm{Cov}_{\phi}(\tilde{x})|||. \end{aligned}\end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue EndAbsoluteValue StartAbsoluteValue upper M minus upper M Superscript upper T Baseline EndAbsoluteValue StartAbsoluteValue EndAbsoluteValue 2nd Column equals StartAbsoluteValue EndAbsoluteValue StartAbsoluteValue EndAbsoluteValue upper M minus upper M Superscript upper T Baseline StartAbsoluteValue EndAbsoluteValue StartAbsoluteValue EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column less than or equals StartAbsoluteValue EndAbsoluteValue StartAbsoluteValue upper M plus upper M Superscript upper T Baseline EndAbsoluteValue StartAbsoluteValue EndAbsoluteValue 3rd Row 1st Column Blank 2nd Column equals StartAbsoluteValue EndAbsoluteValue StartAbsoluteValue upper C o v Subscript phi Baseline left parenthesis x overtilde right parenthesis EndAbsoluteValue StartAbsoluteValue EndAbsoluteValue period EndLayout

Remark Taking the Frobenius norm $||.||_2$ (see Example 1) in Corollary 1, we have

(7) $$ \begin{align}\frac{1}{2}\sum_{i< j}|\phi[x_i, x_j]|^2\leq ||\mathrm{Cov}_{\phi}(\tilde{x})||_2^2\leq\big(\mathrm{tr}\hspace{1mm}\mathrm{Cov}_{\phi}(\tilde{x})\big)^2 = \big[\sum_j\sigma^2(x_j)\big]^2,\end{align} $$

where $\sigma ^2(x_j)=\phi (x_j^2)$ , which is the variance of $x_j$ . This gives a non-vanishing lower bound of the sum of variances in terms of the Lie brackets of the observables, leading to an uncertainty relation.

Given a positive matrix $A=(a_{ij})$ , it is well known that

$$\begin{align*}\mathrm{det}\hspace{1mm}A\leq\prod_j a_{jj}.\end{align*}$$

Using this, we see that

$$\begin{align*}\mathrm{det}|M-M^T|\leq\prod_j\langle |M-M^T|e_j, e_j\rangle,\end{align*}$$

where M is given by (3). The right-hand side is non-vanishing and we would like to have an upper bound for it in terms of $\textrm {Cov}_{\phi }(\tilde {x})$ . We need the following inequality.

Theorem 2 Let $A, B\in M_n(\mathbb {C})$ be positive matrices. Then

$$\begin{align*}\big[\prod_j\langle |A-B|e_j, e_j\rangle\big]^{2/n}\leq \frac{\mathrm{tr}\hspace{1mm}A+\mathrm{tr}\hspace{1mm} B}{n}\big[\prod_j\langle(A+B)e_j, e_j\rangle\big]^{1/n}.\end{align*}$$

Proof By lLemma 2, there exists a unitary U such that

$$\begin{align*}|A-B|\leq(A+B)\# U(A+B)U^*.\end{align*}$$

Taking tensor powers and using Lemma 3,

$$\begin{align*}\otimes^n |A-B|\leq[\otimes^n(A+B)]\#[\otimes^n U(A+B)U^*].\end{align*}$$

Let $\{e_j\}_{j=1}^n$ be the standard basis vectors of $\mathbb {C}^n$ . Consider the state $\xi : \otimes ^n M_n(\mathbb {C})\to \mathbb {C}$ given by

$$\begin{align*}\xi(X)=\langle X(\otimes_{j=1}^n e_j), \otimes_{j=1}^ne_j\rangle.\end{align*}$$

Now,

$$ \begin{align*} \Pi_j\langle|A-B|e_j, e_j\rangle^2 &=\xi\big(\otimes^n|A-B|\big)^2\\&\leq \xi\big(\otimes^n (A+B)\big)\hspace{1mm}\xi\big(\otimes^n U(A+B)U^*\big)\\&\leq\prod_j\langle(A+B)U^*e_j, U^*e_j\rangle\hspace{1mm}\prod_j\langle(A+B)e_j, e_j\rangle\\&\leq\left[\frac{\mathrm{tr}\hspace{1mm}(A+B)}{n}\right]^n\hspace{1mm} \prod_j\langle(A+B)e_j, e_j\rangle .\end{align*} $$

StartLayout 1st Row 1st Column normal upper Pi Subscript j Baseline left angle bracket StartAbsoluteValue upper A minus upper B EndAbsoluteValue e Subscript j Baseline comma e Subscript j Baseline right angle bracket squared 2nd Column equals xi left parenthesis circled times Superscript n Baseline StartAbsoluteValue upper A minus upper B EndAbsoluteValue right parenthesis squared 2nd Row 1st Column Blank 2nd Column less than or equals xi left parenthesis circled times Superscript n Baseline left parenthesis upper A plus upper B right parenthesis right parenthesis xi left parenthesis circled times Superscript n Baseline upper U left parenthesis upper A plus upper B right parenthesis upper U Superscript asterisk Baseline right parenthesis 3rd Row 1st Column Blank 2nd Column less than or equals product Underscript j Endscripts left angle bracket left parenthesis upper A plus upper B right parenthesis upper U Superscript asterisk Baseline e Subscript j Baseline comma upper U Superscript asterisk Baseline e Subscript j Baseline right angle bracket product Underscript j Endscripts left angle bracket left parenthesis upper A plus upper B right parenthesis e Subscript j Baseline comma e Subscript j Baseline right angle bracket 4th Row 1st Column Blank 2nd Column less than or equals left bracket StartFraction trace left parenthesis upper A plus upper B right parenthesis Over n EndFraction right bracket Superscript n Baseline product Underscript j Endscripts left angle bracket left parenthesis upper A plus upper B right parenthesis e Subscript j Baseline comma e Subscript j Baseline right angle bracket period EndLayout

Taking powers of $1/n$ , we are done.

Remark The inequality

$$\begin{align*}\xi\big(\otimes^n|A-B|\big)^2\leq \xi\big(\otimes^n (A+B)\big)\hspace{1mm}\xi\big(\otimes^n U(A+B)U^*\big)\end{align*}$$

holds because for any state $\phi :M_n(\mathbb {C})\to \mathbb {C}$ , we have $\phi (A\#B)\leq [\phi (A)\phi (B)]^{1/2}$ for any two positive matrices A and B. For details, see [Reference Congedo, Barachant and Bhatia15].

Applying Theorem 2 to the positive matrices M and $M^T$ , we have the following.

Corollary 2 Let $\{x_j\}_{j=1}^n$ be self-adjoint elements in a $C^*$ -algebra $\mathcal {A}$ and let $\phi :\mathcal {A}\to \mathbb {C}$ be a state. Let $\sigma (x_j)=\sqrt {\phi (x_j^2)}$ be the standard deviation of $x_j$ . Then

(8) $$ \begin{align}\frac{1}{4}\prod_{k=1}^n\langle |(\phi[x_i, x_j])|e_k,e_k\rangle^{2/n}\leq \left[\frac{\sum\sigma^2(x_k)}{n}\right] \big[\prod_k\sigma^2(x_k)\big]^{1/n}. \end{align} $$

From inequality (8), we also get

(9) $$ \begin{align}\frac{1}{2}\prod\langle |(\phi[x_i, x_j])|e_k,e_k\rangle^{1/n}\leq \frac{\sum\sigma^2(x_k)}{n}.\end{align} $$

This gives a non-vanishing lower bound for the arithmetic mean of the variances and hence, an uncertainty relation. Note that for two self-adjoint elements $x_1$ and $x_2$ , the matrix

$$ \begin{align*}\big|\big(\phi[x_i, x_j]\big)\big|\end{align*} $$

is just

$$ \begin{align*}\begin{pmatrix} |\phi[x_1, x_2]| & 0\\ 0 & |\phi[x_1, x_2]|\end{pmatrix}\end{align*} $$ StartLayout 1st Row Start 2 By 2 Matrix 1st Row 1st Column StartAbsoluteValue phi left bracket x 1 comma x 2 right bracket EndAbsoluteValue 2nd Column 0 2nd Row 1st Column 0 2nd Column StartAbsoluteValue phi left bracket x 1 comma x 2 right bracket EndAbsoluteValue EndMatrix EndLayout

and hence, inequality (9) becomes

$$ \begin{align*}|\phi[x_1, x_2]|\leq \sigma^2(x_1)+\sigma^2(x_2).\end{align*} $$

Note that this is weaker than the Heisenberg uncertainty principle. The importance of inequality (9) lies in the fact that for several self-adjoint elements $\{x_j\}_{j=1}^n$ , the matrix

$$ \begin{align*}|(\phi[x_i, x_j])|\end{align*} $$

is no longer scalar. Hence, unlike the case for two observables, the left-hand side of inequality (9) is bigger than

$$ \begin{align*}\frac{1}{2}[\mathrm{det}\,|(\phi[x_1,x_j])|]^{1/n}.\end{align*} $$

Compare this to Robertson’s bound in (4) and notice the improvement on the left-hand side. Weakening the right-hand side of (4) with the arithmetic mean of variances is the price being paid for the non-vanishing improvement on the left.

The next inequality gives another nonzero lower bound in terms of the entries $\phi [x_i, x_j]$ .

Theorem 3 Let $\mathcal {A}$ be a $C^{*}$ -algebra and let $\{x_j\}_{j=1}^n$ be self-adjoint elements of $\mathcal {A}$ . Let $\phi :\mathcal {A}\to \mathbb {C}$ be a state. Then

$$\begin{align*}\frac{1}{4}\prod_k\left[\sum_{i\neq k}|\phi[x_i, x_k]|^2\right]^{1/n}\leq||\mathrm{Cov}_{\phi}(\tilde{x})||\hspace{1mm}\left[\frac{\sum \sigma^2(x_k)}{n}\right]^{1/2}\big[\prod_k\sigma(x_k)\big]^{1/n}. \end{align*}$$

Proof For each k,

$$ \begin{align*} \frac{1}{4}\sum_{i\neq k}|\phi[x_i, x_k]|^2 &=\langle (M-M^T)^2e_k e_k\rangle\\&\leq ||M-M^T||\hspace{1mm}\langle |M-M^T|e_k, e_k\rangle\\&\leq||\mathrm{Cov}_{\phi}(\tilde{x})||\hspace{1mm}\langle |M-M^T|e_k, e_k\rangle,\end{align*} $$

StartLayout 1st Row 1st Column one fourth sigma summation Underscript i not equals k Endscripts StartAbsoluteValue phi left bracket x Subscript i Baseline comma x Subscript k Baseline right bracket EndAbsoluteValue squared 2nd Column equals left angle bracket left parenthesis upper M minus upper M Superscript upper T Baseline right parenthesis squared e Subscript k Baseline e Subscript k Baseline right angle bracket 2nd Row 1st Column Blank 2nd Column less than or equals StartAbsoluteValue EndAbsoluteValue upper M minus upper M Superscript upper T Baseline StartAbsoluteValue EndAbsoluteValue left angle bracket StartAbsoluteValue upper M minus upper M Superscript upper T Baseline EndAbsoluteValue e Subscript k Baseline comma e Subscript k Baseline right angle bracket 3rd Row 1st Column Blank 2nd Column less than or equals StartAbsoluteValue EndAbsoluteValue upper C o v Subscript phi Baseline left parenthesis x overtilde right parenthesis StartAbsoluteValue EndAbsoluteValue left angle bracket StartAbsoluteValue upper M minus upper M Superscript upper T Baseline EndAbsoluteValue e Subscript k Baseline comma e Subscript k Baseline right angle bracket comma EndLayout

where the last inequality follows from Corollary 1. Taking geometric mean on both sides and using Corollary 2, we are done.

An important question that arises now is, can we get an uncertainty relation for which the lower bound is always nonzero whenever any two of the observables do not commute? To answer this, we need to compare the block matrices $\big ([x_i, x_j]\big )$ and $\big (\{x_i,x_j\}\big )$ without composing a state entry-wise.

We do this by obtaining a trace inequality, which as remarked later, holds for tracial $C^*$ -algebras as well but for the sake of better clarity, we stick to $M_n(\mathbb {C})$ , our observables being Hermitian matrices $\{H_j\}_{j=1}^k$ .

Note that, in this case, the block matrix $\big (H_j H_i\big )$ is not necessarily positive even though $\big (H_i H_j\big )$ always is. This is demonstrated by the following example.

Example 2. Let $\theta \in \mathbb {R}$ and let

$$\begin{align*}H_1=\begin{pmatrix}\frac{\sin\hspace{1mm} \theta}{2} & \cos\hspace{1mm} \theta\\ \cos\hspace{1mm} \theta & \sin\hspace{1mm} \theta\end{pmatrix}\end{align*}$$ upper H 1 equals Start 2 By 2 Matrix 1st Row 1st Column StartFraction sine theta Over 2 EndFraction 2nd Column cosine theta 2nd Row 1st Column cosine theta 2nd Column sine theta EndMatrix

and

$$\begin{align*}H_2=\begin{pmatrix}\cos\hspace{1mm} \theta & \sin\hspace{1mm} \theta \\ \sin\hspace{1mm} \theta & \frac{\cos\hspace{1mm} \theta}{2}\end{pmatrix}.\end{align*}$$ upper H 2 equals Start 2 By 2 Matrix 1st Row 1st Column cosine theta 2nd Column sine theta 2nd Row 1st Column sine theta 2nd Column StartFraction cosine theta Over 2 EndFraction EndMatrix period

Let $e_1, e_2$ be the standard basis vectors of $\mathbb {C}^2$ . Then the matrix

$$\begin{align*}\big(\langle H_i e_j, H_j e_i\rangle\big)\end{align*}$$

has negative determinant and hence is not positive.

Hence, unlike the previous results, we cannot use the idea of writing both sides as the sum and difference of two positive matrices. However, we can still obtain an inequality with the trace norm (see Example 1) such that the left-hand side vanishes if and only if all the observables commute. For two Hermitian matrices $H_1$ and $H_2$ , this takes the form

$$\begin{align*}\left|\left|\begin{pmatrix}O & [H_1, H_2] \\ [H_2, H_1] & O\end{pmatrix}\right|\right|_1\leq2(\mathrm{tr}\hspace{1mm}H_1^2+\mathrm{tr}\hspace{1mm}H_2^2). \end{align*}$$

This is easy to prove, since,

$$ \begin{align*}\left|\left|\begin{pmatrix}O & [H_1, H_2] \\ [H_2, H_1] & O\end{pmatrix}\right|\right|_1&=2\hspace{1mm}\mathrm{tr}\hspace{1mm}\big|[H_1,H_2]\big|\\&\leq 2\big(||H_1 H_2||_1+||H_2 H_1||_1\big)\\&\leq 4\sqrt{\mathrm{tr}\hspace{1mm}H_1^2\hspace{1mm} \mathrm{tr}\hspace{1mm}H_2^2}\\&\leq2(\mathrm{tr}\hspace{1mm}H_1^2+\mathrm{tr}\hspace{1mm}H_2^2).\end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue StartAbsoluteValue Start 2 By 2 Matrix 1st Row 1st Column upper O 2nd Column left bracket upper H 1 comma upper H 2 right bracket 2nd Row 1st Column left bracket upper H 2 comma upper H 1 right bracket 2nd Column upper O EndMatrix EndAbsoluteValue EndAbsoluteValue Subscript 1 2nd Column equals 2 trace StartAbsoluteValue left bracket upper H 1 comma upper H 2 right bracket EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column less than or equals 2 left parenthesis StartAbsoluteValue EndAbsoluteValue upper H 1 upper H 2 StartAbsoluteValue EndAbsoluteValue Subscript 1 Baseline plus StartAbsoluteValue EndAbsoluteValue upper H 2 upper H 1 StartAbsoluteValue EndAbsoluteValue Subscript 1 Baseline right parenthesis 3rd Row 1st Column Blank 2nd Column less than or equals 4 StartRoot trace upper H 1 squared trace upper H 2 squared EndRoot 4th Row 1st Column Blank 2nd Column less than or equals 2 left parenthesis trace upper H 1 squared plus trace upper H 2 squared right parenthesis period EndLayout

The next theorem generalizes this to k Hermitian matrices $\{H_j\}_{j=1}^k$ .

Theorem 4 Let $\{H_j\}_{j=1}^k$ be $n\times n$ Hermitian matrices. Then

$$\begin{align*}\big|\big|\big([H_i, H_j]\big)\big|\big|_1\leq (k-1)\mathrm{tr}\hspace{1mm}\big(\{H_i, H_j\}\big).\end{align*}$$

Proof Let

$$ \begin{align*}A=\big([H_i, H_j]\big).\end{align*} $$

Note that A is Hermitian and

$$\begin{align*}||A||_1=\mathrm{tr}\hspace{1mm}(A^2)^{1/2}.\end{align*}$$

Let $\varphi :M_k(M_n(\mathbb {C}))\to M_k(M_n(\mathbb {C}))$ be given by

$$ \begin{align*}\Phi \begin{pmatrix} A_{11} &\dots & A_{1k}\\ \vdots & & \vdots\\ A_{k1} &\dots & A_{kk}\end{pmatrix}=\begin{pmatrix}A_{11} & \\ & \ddots &\\ & & A_{kk}\end{pmatrix}\end{align*} $$

normal upper Phi Start 3 By 3 Matrix 1st Row 1st Column upper A 11 2nd Column ellipsis 3rd Column upper A Subscript 1 k Baseline 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column vertical ellipsis 3rd Row 1st Column upper A Subscript k Baseline 1 Baseline 2nd Column ellipsis 3rd Column upper A Subscript k k Baseline EndMatrix equals Start 3 By 2 Matrix 1st Row 1st Column upper A 11 2nd Column Blank 2nd Row 1st Column Blank 2nd Column down right diagonal ellipsis 3rd Column Blank 3rd Row 1st Column Blank 2nd Column Blank 3rd Column upper A Subscript k k EndMatrix

for all block matrices $(A_{ij})\in M_k(M_n(\mathbb {C}))$ . This operation is known as the pinching and is a central tool in matrix analysis, with numerous applications to matrix inequalities (see [Reference Bhatia4, Reference Bhatia5]). It also constitutes an important example of a quantum channel known as a conditional expectation (see [Reference Carlen12] for details) and plays a significant role in the derivation of quantum entropy inequalities.

Proceeding with our proof, consider the matrix

$$\begin{align*}D=\Phi(A^2)=\begin{pmatrix}-\sum_{j\neq 1}[H_1, H_j]^2 & &\\ & \ddots\\& & -\sum_{j\neq n}[H_n, H_j]^2\end{pmatrix}.\end{align*}$$

upper D equals normal upper Phi left parenthesis upper A squared right parenthesis equals Start 3 By 3 Matrix 1st Row 1st Column minus sigma summation Underscript j not equals 1 Endscripts left bracket upper H 1 comma upper H Subscript j Baseline right bracket squared 2nd Column Blank 3rd Column Blank 2nd Row 1st Column Blank 2nd Column down right diagonal ellipsis 3rd Row 1st Column Blank 2nd Column Blank 3rd Column minus sigma summation Underscript j not equals n Endscripts left bracket upper H Subscript n Baseline comma upper H Subscript j Baseline right bracket squared EndMatrix period

Note that

$$\begin{align*}D=\frac{1}{2\pi}\int_{0}^{2\pi}U(\theta)A^2U(\theta)^* d\theta,\end{align*}$$

where

$$\begin{align*}U(\theta)=\begin{pmatrix}I & & &\\ & e^{i\theta}I & &\\& & \ddots &\\ & & & e^{i(k-1)\theta}I\end{pmatrix}.\end{align*}$$

upper U left parenthesis theta right parenthesis equals Start 4 By 4 Matrix 1st Row 1st Column upper I 2nd Column Blank 3rd Column Blank 4th Column Blank 2nd Row 1st Column Blank 2nd Column e Superscript i theta Baseline upper I 3rd Column Blank 4th Column Blank 3rd Row 1st Column Blank 2nd Column Blank 3rd Column down right diagonal ellipsis 4th Column Blank 4th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column e Superscript i left parenthesis k minus 1 right parenthesis theta Baseline upper I EndMatrix period

By operator concavity of the square root,

$$\begin{align*}D^{1/2}\geq \frac{1}{2\pi}\int_{0}^{2\pi}U(\theta)|A|U(\theta)^* d\theta.\end{align*}$$

Taking traces, we get the inequality

(10) $$ \begin{align}||A||_1\leq\sum_i\mathrm{tr}\hspace{1mm}\big(\sum_{j} \big|[H_i, H_j]\big|^2)^{1/2}\big).\end{align} $$

Now, for each i, consider the block matrix

$$\begin{align*}\tilde{H}_i=\begin{pmatrix}{}[H_i, H_1] &O &\cdots &O\\ \vdots & & &\\{}[H_i, H_k] & O &\cdots &O \\\end{pmatrix}.\end{align*}$$

upper H overtilde Subscript i Baseline equals Start 3 By 4 Matrix 1st Row 1st Column left bracket upper H Subscript i Baseline comma upper H 1 right bracket 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column upper O 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 3rd Row 1st Column left bracket upper H Subscript i Baseline comma upper H Subscript k Baseline right bracket 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column upper O EndMatrix period

Note that inequality (10) now becomes

$$\begin{align*}||A||_1\leq \sum_i ||\tilde{H}_i||_1.\end{align*}$$

Let

$$\begin{align*}Y_i = \begin{pmatrix}H_i H_1 & O &\cdots &\cdots & O\\\vdots & & & & &\\ H_i H_{i-1 } & & & &\\\\ O & & & &\\\\ H_i H_{i+1} & & & &\\\vdots & & & &\\ H_i H_k & O &\cdots &\cdots & O\\\end{pmatrix}\end{align*}$$

upper Y Subscript i Baseline equals Start 9 By 5 Matrix 1st Row 1st Column upper H Subscript i Baseline upper H 1 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 3rd Row 1st Column upper H Subscript i Baseline upper H Subscript i minus 1 Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 4th Row 1st Column Blank 5th Row 1st Column upper O 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Row 1st Column Blank 7th Row 1st Column upper H Subscript i Baseline upper H Subscript i plus 1 Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 8th Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 9th Row 1st Column upper H Subscript i Baseline upper H Subscript k Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O EndMatrix

and

$$\begin{align*}Z_i = \begin{pmatrix}H_1 H_i & O &\cdots &\cdots & O\\\vdots & & & & &\\ H_{i-1} H_{i} & & & &\\\\ O & & & &\\\\ H_{i+1} H_{i} & & & &\\\vdots & & & &\\ H_k H_i & O &\cdots &\cdots & O\\\end{pmatrix},\end{align*}$$

upper Z Subscript i Baseline equals Start 9 By 5 Matrix 1st Row 1st Column upper H 1 upper H Subscript i Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 3rd Row 1st Column upper H Subscript i minus 1 Baseline upper H Subscript i Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 4th Row 1st Column Blank 5th Row 1st Column upper O 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Row 1st Column Blank 7th Row 1st Column upper H Subscript i plus 1 Baseline upper H Subscript i Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 8th Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 9th Row 1st Column upper H Subscript k Baseline upper H Subscript i Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O EndMatrix comma

where the O in the first column is at the $i^{\mathrm {th}}$ position. Note that $\tilde {H}_i = Y_i - Z_i$ . Inequality (10) now gives

(11) $$ \begin{align} \begin{aligned} ||A||_1 &\leq\sum_i ||Y_i-Z_i||_1\\ &\leq \sum_i ||Y_i||_1+||Z_i||_1. \end{aligned}\end{align} $$

StartLayout 1st Row 1st Column StartAbsoluteValue EndAbsoluteValue upper A StartAbsoluteValue EndAbsoluteValue Subscript 1 2nd Column less than or equals sigma summation Underscript i Endscripts StartAbsoluteValue EndAbsoluteValue upper Y Subscript i Baseline minus upper Z Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 1 Baseline 2nd Row 1st Column Blank 2nd Column less than or equals sigma summation Underscript i Endscripts StartAbsoluteValue EndAbsoluteValue upper Y Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 1 Baseline plus StartAbsoluteValue EndAbsoluteValue upper Z Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 1 Baseline period EndLayout

Consider the $k\times k$ block matrices

$$\begin{align*}\widehat{H}_i=\begin{pmatrix}H_1 & O &\cdots &\cdots & O &\\\vdots & & & & \\ H_{i-1} & & & &\\ \\ O & & & &\\ \\ H_{i+1} & & & &\\\vdots & & & &\\ H_k & O &\cdots &\cdots & O &\\\end{pmatrix}, \hspace{2mm}\tilde{Y}_i=\begin{pmatrix}H_i & & & & & &\\ &\ddots & & & & &\\ & & H_i & & & &\\ & & & O & & &\\ & & & & H_i & &\\ & & & & &\ddots &\\ & & & & & & H_i\end{pmatrix}\end{align*}$$

ModifyingAbove upper H With caret Subscript i Baseline equals Start 9 By 6 Matrix 1st Row 1st Column upper H 1 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 6th Column Blank 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 3rd Row 1st Column upper H Subscript i minus 1 Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 4th Row 1st Column Blank 5th Row 1st Column upper O 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Row 1st Column Blank 7th Row 1st Column upper H Subscript i plus 1 Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 8th Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 9th Row 1st Column upper H Subscript k Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 6th Column Blank EndMatrix comma upper Y overtilde Subscript i Baseline equals Start 7 By 7 Matrix 1st Row 1st Column upper H Subscript i Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 2nd Row 1st Column Blank 2nd Column down right diagonal ellipsis 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 3rd Row 1st Column Blank 2nd Column Blank 3rd Column upper H Subscript i Baseline 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 4th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column upper O 5th Column Blank 6th Column Blank 7th Column Blank 5th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column upper H Subscript i Baseline 6th Column Blank 7th Column Blank 6th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column down right diagonal ellipsis 7th Column Blank 7th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column upper H Subscript i Baseline EndMatrix

and

$$\begin{align*}\tilde{K}_i=\begin{pmatrix}H_i & O &\cdots &\cdots & O &\\\vdots & & & & \\ H_{i} & & & &\\ \\ O & & & &\\ \\ H_{i} & & & &\\\vdots & & & &\\ H_i & O &\cdots &\cdots & O &\\\end{pmatrix}, \hspace{2mm}\widehat{Z}_i=\begin{pmatrix}H_1 & & & & & &\\ &\ddots & & & & &\\ & & H_{i-1} & & & &\\ & & & O & & &\\ & & & & H_{i+1} & &\\ & & & & &\ddots &\\ & & & & & & H_k\end{pmatrix}.\end{align*}$$

upper K overtilde Subscript i Baseline equals Start 9 By 6 Matrix 1st Row 1st Column upper H Subscript i Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 6th Column Blank 2nd Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 3rd Row 1st Column upper H Subscript i Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 4th Row 1st Column Blank 5th Row 1st Column upper O 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Row 1st Column Blank 7th Row 1st Column upper H Subscript i Baseline 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 8th Row 1st Column vertical ellipsis 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 9th Row 1st Column upper H Subscript i Baseline 2nd Column upper O 3rd Column midline horizontal ellipsis 4th Column midline horizontal ellipsis 5th Column upper O 6th Column Blank EndMatrix comma ModifyingAbove upper Z With caret Subscript i Baseline equals Start 7 By 7 Matrix 1st Row 1st Column upper H 1 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 2nd Row 1st Column Blank 2nd Column down right diagonal ellipsis 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 3rd Row 1st Column Blank 2nd Column Blank 3rd Column upper H Subscript i minus 1 Baseline 4th Column Blank 5th Column Blank 6th Column Blank 7th Column Blank 4th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column upper O 5th Column Blank 6th Column Blank 7th Column Blank 5th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column upper H Subscript i plus 1 Baseline 6th Column Blank 7th Column Blank 6th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column down right diagonal ellipsis 7th Column Blank 7th Row 1st Column Blank 2nd Column Blank 3rd Column Blank 4th Column Blank 5th Column Blank 6th Column Blank 7th Column upper H Subscript k Baseline EndMatrix period

Here, O is the $i^{\mathrm {th}}$ entry in the first column of $\widehat {H}_i$ and $\tilde {K}_i$ , and the $i^{\mathrm {th}}$ diagonal entry of $\tilde {Y}_i$ and $\widehat {Z}_i$ . Note that $Y_i = \widehat {H_i}\tilde {Y}_i$ and $Z_i = \tilde {K}_i\widehat {Z_i}$ . By Cauchy–Schwarz and A.M–G.M,

(12) $$ \begin{align} \begin{aligned} ||Y_i||_1 &\leq||\widehat{H}_i||_2 ||\tilde{Y}_i||_2\\ &\leq\frac{||\widehat{H}_i||_2^{2}+||\tilde{Y}_i||_2^{2}}{2}\\ &\leq\frac{(k-1)\hspace{1mm}\mathrm{tr}\hspace{1mm}H_i^2+\sum_{j\neq i}\mathrm{tr}\hspace{1mm}H_j^2}{2}. \end{aligned}\end{align} $$

StartLayout 1st Row 1st Column StartAbsoluteValue EndAbsoluteValue upper Y Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 1 2nd Column less than or equals StartAbsoluteValue EndAbsoluteValue ModifyingAbove upper H With caret Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 2 Baseline StartAbsoluteValue EndAbsoluteValue upper Y overtilde Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 2 Baseline 2nd Row 1st Column Blank 2nd Column less than or equals StartFraction StartAbsoluteValue EndAbsoluteValue ModifyingAbove upper H With caret Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 2 Superscript 2 Baseline plus StartAbsoluteValue EndAbsoluteValue upper Y overtilde Subscript i Baseline StartAbsoluteValue EndAbsoluteValue Subscript 2 Superscript 2 Baseline Over 2 EndFraction 3rd Row 1st Column Blank 2nd Column less than or equals StartFraction left parenthesis k minus 1 right parenthesis trace upper H Subscript i Superscript 2 Baseline plus sigma summation Underscript j not equals i Endscripts trace upper H Subscript j Superscript 2 Baseline Over 2 EndFraction period EndLayout

Similarly,

(13) $$ \begin{align} \begin{aligned} ||Z_i||_1 &\leq\frac{(k-1)\hspace{1mm}\mathrm{tr}\hspace{1mm}H_i^2+\sum_{j\neq i}\mathrm{tr}\hspace{1mm}H_j^2}{2}. \end{aligned}\end{align} $$

By (11)–(13),

$$ \begin{align*} ||A||_1 &\leq(k-1)\sum_i\mathrm{tr}\hspace{1mm}H_i^2+ \sum_i \big[\sum_{j}\mathrm{tr}\hspace{1mm}H_j^2 -\mathrm{tr}\hspace{1mm}H_i^2\big]\\&=2(k-1)\sum_i \mathrm{tr}\hspace{1mm}H_i^2\\&=(k-1)\hspace{1mm}\mathrm{tr}\hspace{1mm}\big(\{H_i, H_j\}\big).\end{align*} $$

It is to be noted that even though we have stuck to Hermitian matrices keeping up with the spirit of the article, a slight variant of Theorem 4 can be obtained for a k-tuple $\{X_j\}_{j=1}^k$ of matrices which are not necessarily Hermitian. In fact, the same proof, with very minor modifications, shows that

(14) $$ \begin{align}\big|\big|\big([X_i, X_j]\big)\big|\big|_1\leq 2(k-1)\sum_{i=1}^k ||X_i||_2^2.\end{align} $$

This gives a trace norm bound for the block matrix formed by the commutators $[X_i, X_j]$ . Commutator bounds for two matrices are well-studied in the literature. The Böttcher–Wenzel inequality (see [Reference Audenaert3, Reference Bóttcher and Wenzel10, Reference Bóttcher and Wenzel11, Reference Cheng, Jin and Vong14, Reference Lu29, Reference Vong and Jin37, Reference Wenzel38]) states that for two matrices X and Y,

(15) $$ \begin{align}|| [X,Y] ||_2 \leq \sqrt{2}||X||_2||Y||_2.\end{align} $$

Later, Wenzel and Audenaert [Reference Wenzel and Audenaert39] used complex interpolation theory to generalize inequality (15) for arbitrary Schatten-p norms. Bhatia and Kittaneh [Reference Bhatia and Kittaneh9] gave unitarily invariant norm bounds on the commutator under the assumption of normality of one of the matrices and showed connections to spectral variation.

Inequality (14), to the best of our knowledge, is the first trace norm bound for the block matrix of commutators $\big ([X_i, X_j]\big )$ . Note that upper bounds for the trace norm are generally very useful in matrix theory, since it is the largest unitarily invariant norm (see the discussion following Example 1).

Remarks

1. Note that Theorem 4 gives a refinement of Corollary 1 when $\phi $ is the normalized trace on $M_n(\mathbb {C})$ and $|||.|||$ is the trace norm. In this case, the left-hand side of the inequality in Corollary 1 vanishes, but Theorem 4 gives a non-vanishing lower bound.
2. Theorem 4 always gives a nonzero lower bound whenever any two of the observables do not commute.
3. Let $\mathcal {A}$ be a $C^*$ -algebra with a faithful finite trace $\tau $ (see [Reference Kadison and Ringrose26, Reference Li28] for details). Given $x\in \mathcal {A}$ , its trace norm is given by
$$\begin{align*}||x||_1=\tau(|x|)\end{align*}$$
for all $x\in \mathcal {M}$ . We can lift $\tau $ to $\tau _k : M_k(\mathcal {A})\to \mathbb {C}$ via
$$\begin{align*}\tau_k (x_{ij})=\sum_{i}\tau(x_{ii}).\end{align*}$$

Note that $\tau _k$ is positive, faithful, and tracial. Let $\{x_j\}_{j=1}^k$ be elements of $\mathcal {A}$ . In this context, inequality (14) reads
(16) $$ \begin{align} \big|\big|\big([x_i, x_j]\big)\big|\big|_1\leq 2(k-1)\sum_{i=1}^k\tau(x_i^*x_i),\end{align} $$
where $||.||_1$ is the trace norm induced by $\tau _k$ on $M_n(\mathcal {A})$ . Indeed, the same proof holds, since the only properties of the trace being used there are invariance under unitary conjugation, triangle inequality, and Cauchy–Schwarz inequality, all of which hold in the $C^*$ -algebra setting (see [Reference da Silva16, Reference Nelson31]).

3 Conclusions

In this article, we first obtained several multivariate uncertainty relations for observables in a unital $C^*$ -algebra equipped with a state. The starting point was a determinant inequality for positive matrices, using which we gave a simple proof of Robertson’s uncertainty principle [Reference Robertson35]. Using the matrix geometric mean, we then obtained some interesting inequalities for a pair of positive matrices A and B. These matrix inequalities were then applied to covariance matrices associated with observables, yielding lower bounds involving their commutators.

The resulting uncertainty relations (see (9), Corollary 1, and Theorem 4) provide nontrivial lower bounds for sums and arithmetic means of variances, regardless of the number of observables. This is an improvement on Robertson’s determinant inequality, where the lower bound is zero when the number of observables is odd.

Of particular importance is the trace norm inequality in Theorem 4 comparing the block matrix of commutators with the corresponding block matrix of anti-commutators. As a consequence, one obtains a state-independent lower bound which is nonzero whenever any two observables fail to commute.

One advantage of Theorem 4 is that it extends to tracial $C^*$ -algebras as shown in (16). The standard Böttcher–Wenzel inequality, however, does not. To see this, consider the normalized trace $\tau _n:M_n(\mathbb {C})\to \mathbb {C}$ given by $\tau _n(A)=\frac {\mathrm {tr}\, A}{n}$ for all ${A\in M_n(\mathbb {C})}$ . This is a tracial state on $M_n(\mathbb {C})$ . We can then modify the Böttcher–Wenzel inequality in (15) as

$$ \begin{align*}\tau_n |[X,Y]|^2\leq 2n \, \tau_n(X^*X)\tau_n(Y^*Y).\end{align*} $$

The sharpness of the Böttcher–Wenzel bound (see [Reference Bóttcher and Wenzel11]) and the blow-up of the factor $2n$ imply that we cannot have a bound of the form

$$ \begin{align*}\tau|[x,y]|^2\leq C\tau(x^*x)\tau(y^*y)\end{align*} $$

on a tracial $C^*$ -algebra $(\mathcal {A},\tau )$ for any universal $C > 0$ . Our inequality is novel in this sense, since it solves the more general problem of giving a state-independent bound with the block matrix of commutators

$$ \begin{align*}([x_i,x_j])\end{align*} $$

with a factor depending only on the number of observables, and not on the algebra or the ambient trace.

Data availability statement

No available data have been used.

Acknowledgements

I thank my Ph.D. supervisor Prof. Tanvi Jain for her helpful comments during the preparation of this article. I also thank the anonymous referee for reading the article thoroughly and giving detailed, insightful suggestions to improve the overall quality of the article.

Competing interests

I declare no competing interests.

Footnotes

I thank the Indian Statistical Institute for the financial support during my ongoing Ph.D. through their Institute Fellowship Programme.

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Article contents

Multi-observable uncertainty bounds via matrix inequalities

Abstract

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MSC classification

Information

1 Introduction

2 Main results

3 Conclusions

Data availability statement

Acknowledgements

Competing interests

Footnotes

References

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