Traceless Hermitian matrices with simultaneously vanishing Rayleigh quotients

Question

Let $D$ be an integer greater than 1. What is the largest number $N$, such that for all sets of $N$ Hermitian $D\times D$ traceless matrices $M_i$, $i=1,\dots,N$, there exists a non-zero complex vector $v$, such that $v^HM_iv = 0$ for all $i$? ($v^H = (v^*)^T$ is the conjugate transpose).

Example. For $D=2$, $N$ must be less than 3, because for a set $M_i = \sigma_i$, $i=1,2,3$, where $\sigma_i$ are the Pauli matrices, for any vector $v$: $$ \sum_i (v^H M_i v)^2 = \Vert v\Vert^4 \neq 0\,, $$ so $v^H M_i v$ cannot all simultaneously vanish. However, for $N=2$ it is possible: let $v_1,v_2$ be eigenvectors of $M_1$ with eigenvalues $\lambda,-\lambda$. Then for $v = v_1 + \alpha v_2$, with $|\alpha| = 1$, $v^H M_1 v = 0$. Then $$ v^H M_2 v = \alpha (v_1^H M_2 v_2) + \alpha^{-1} (v_2^H M_2 v_1). $$ Then any solution to $\alpha^2 = -(v_2^H M_2 v_1)/(v_1^H M_2 v_2)$ will satisfy $v^H M_2 v = 0$. Thus $N=2$ is the answer.

Conjecture (false). For any $D$, $N$ is upper-bounded by $D^2-2$, because $D^2-1$ is the dimension of space of traceless Hermitian matrices, which for any $v$ contains $P_v - 1_{D\times D}/D$, where $P_v$ is the projector onto $\operatorname{span}\{v\}$, the Rayleigh quotient of which is $1-1/D\neq 0$. My conjecture is that $N = D^2-2$ is the correct answer.

Motivation. In quantum mechanics Rayleigh quotients $(v^HMv) / (v^H v)$ describe expectation values of quantum observables. The question then can be read: for a quantum system in $D$-dimensional Hilbert space, what is the maximum number of independent observables, that can have simultaneously vanishing expectation value.

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New conjecture. As proven in @Nathaniel Johnston's answer and comments thereafter, the answer is bounded by $N < 2D - 1$. The current conjecture (checked numerically with random matrices) is $N = 2D-2$.

Answer 1

The correct value of $N$ is $$ N = \begin{cases} 2 & \text{if $D = 2$},\\ 3 & \text{if $D \geq 3$}. \end{cases} $$ The original question already proved that $N = 2$ when $D = 2$, and Michał's answer here shows that $N \leq 3$ when $D \geq 3$. So all that's left to prove is that $N \geq 3$ when $D \geq 3$.

To this end, for $3$ Hermitian traceless matrices $M_1$, $M_2$, $M_3$, consider the set $$ W = \{(v^HM_1v, v^HM_2v, v^HM_3v) : \|v\| = 1\}. $$ This is called the ``joint numerical range'' of $M_1$, $M_2$, and $M_3$, and it is known to be convex [1] (this depends crucially on both facts that these matrices are at least $3 \times 3$ and that there are 3 of them). In particular, this implies that $$ \left(\int_v v^HM_1v \, dv, \int_v v^HM_2v \, dv, \int_v v^HM_3v \, dv\right) \in W, $$ where the integration is performed over all unit vectors $v$, with respect to uniform spherical measure. Since $M_i$ is traceless for each $i$, we have $\int_v v^HM_iv \, dv = 0$ for each $i$, so $(0,0,0) \in W$. In other words, there exists $v$ with $\|v\| = 1$ such that $v^HM_iv = 0$ for all $i$, and we're done.

[1] Au-Yeung, Yik-Hoi; Poon, Yiu-Tung, A remark on the convexity and positive definiteness concerning Hermitian matrices, Southeast Asian Bull. Math. 3, 85-92 (1979). ZBL0432.15012

Answer 2

I'll prove some bounds that disprove the conjectured $N = D^2 - 2$ when $D \geq 3$.

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Bound 1: $N \leq D(D-1)$.

I'll illustrate this bound when $D = 3$ (so $N \leq 6$). Consider the following 7 of the 8 Gell-Mann matrices: $$ M_1 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix}, $$ $$ M_4 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, M_5 = \begin{pmatrix} 0 & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{pmatrix}, M_6 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, $$ $$ M_7 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix}. $$ It's not too difficult to show that if $v$ has $v^HM_iv = 0$ for $1 \leq i \leq 6$ then $v$ has at most $1$ non-zero entry. But if it has just one non-zero entry then it can't be the case that $v^HM_7v = 0$. So there is no non-zero $v$ for which $v^HM_iv = 0$ for all $1 \leq i \leq 7$.

To generalize this to higher $D$, replace $M_1, \ldots, M_6$ by the $D(D-1)$ matrices $E_{j,k} + E_{k,j}$ and $iE_{j,k} - iE_{k,j}$ for $1 \leq j < k \leq D$, where $E_{j,k}$ is the matrix with $1$ in the $(j,k)$-entry and $0$ elsewhere, and replace $M_7$ by any traceless diagonal matrix with all diagonal entries non-zero.

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The next bound is strictly better than Bound 1 above, but we have to work a bit harder to prove it.

Bound 2: When $D \geq 3$, we have $N \leq D(D-1)/2 + D - 2$.

Again, I'll illustrate this bound first when $D = 3$. This time, take the following 5 matrices:

$$ M_1 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} $$ $$ M_4 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, M_5 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}. $$ If $v$ has $v^HM_iv = 0$ for $1 \leq i \leq 3$, then the off-diagonal entries of $vv^H$ are all purely imaginary (i.e., of the form $bi$, where $b$ is a real number (maybe equal to $0$)). However, $vv^H$ is also a positive semidefinite rank-1 matrix, and the only positive semidefinite rank-1 matrices whose off-diagonal entries are purely imaginary are equal to zero outside of a single $2 \times 2$ principal submatrix (proof: $vv^H + (vv^H)^*$ is diagonal with rank at most $2$, so it has at most $2$ non-zero diagonal entries, so $vv^H$ also hat at most $2$ non-zero diagonal entries).

But then $v^HM_iv = 0$ for $4 \leq i \leq 5$ forces the diagonal entries of $vv^H$ to all equal each other, which (when $D \geq 3$) can only happen if $v$ is the zero vector.

Generalizing this to higher $D$ makes use of the $D(D-1)/2$ matrices of the form $E_{j,k} + E_{k,j}$ for $1 \leq j < k \leq D$ and the $D-1$ matrices of the form $E_{j,j} - E_{j+1,j+1}$ for $1 \leq j < D$.

Answer 3

The new conjecture is false for $D=4$ and $N=2D-2=6$. Consider the Pauli matrices $$\sigma_1=\left[\begin{matrix} & 1 \\ 1 & \end{matrix}\right],\ \sigma_2=\left[\begin{matrix} & -i \\ i & \end{matrix}\right],\ \sigma_3=\left[\begin{matrix} 1 & \\ & -1 \end{matrix}\right]$$ and the following sextuple of $4\times4$ trace-zero hermitian matrices: $$\left[\begin{matrix} \sigma_1 & \\ & 0 \end{matrix}\right],\ \left[\begin{matrix} \sigma_2 & \\ & 0 \end{matrix}\right],\ \left[\begin{matrix} 0 & \\ & \sigma_1 \end{matrix}\right],\ \left[\begin{matrix} 0 & \\ & \sigma_2 \end{matrix}\right],\ \left[\begin{matrix} \sigma_3 & \\ & \sigma_3 \end{matrix}\right],\ \left[\begin{matrix} \sigma_3 & \\ & -\sigma_3 \end{matrix}\right].$$ Suppose that $\mathbf{x}=(x_1,\ldots,x_4)^T$ satisfies $\mathbf{x}^H A\mathbf{x}=0$ for each $A$ in the above. The first two implies $x_1\overline{x_2}=0$. The middle two implies $x_3\overline{x_4}=0$. And the last two implies $|x_1|^2+|x_3|^2=|x_2|^2+|x_4|^2$ and $|x_1|^2+|x_4|^2=|x_2|^2+|x_3|^2$, or equivalently $|x_1|=|x_2|$ and $|x_3|=|x_4|$. These altogether imply $\mathbf{x}=0$.

ADDED: Let's consider the smallest $N=N(D)$ that such that violates the conjecture for $D$. We claim $N$ has at most logarithmic growth in $D$.

There are $D\times D$ matrices $A_1,\ldots,A_{N(D)}$ with the following property: if $y\in\mathbb{C}^D$ satisfies $y^HA_iy=0$ for all $i$, then $y=0$. Then, the diagonal sums $1_D\oplus-1_D, A_1\oplus0_D,\ldots,A_{N(D)}\oplus0_D$ witness $N(2D) \le 1+N(D)$.

Answer 4

I believe this proves that $N<4$ for all $D>2$.

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Take 4 matrices with dimension $D>2$: \begin{gather} M_\alpha = \begin{pmatrix} \sigma_\alpha \\ & 0_{D-2} \end{pmatrix}\,,\qquad \alpha=1,2,3\,,\\ M_4 = \begin{pmatrix} D-1 \\ & -I_{D-1} \end{pmatrix} \end{gather} where $I_n$ and $0_D$ are $n\times n$ identity and zero matrices respectively. $v$ satisfies $v^H M_i v = 0$ for $i=1,\dots 3$ iff $v_1 = v_2 = 0$. But this implies $v^H M_4 v = -\lVert v \rVert^2 \neq 0$.