Towards a Resolution of the Spin Alignment Problem

Abstract

Consider minimizing the entropy of a mixture of states by choosing each state subject to constraints. If the spectrum of each state is fixed, we expect that in order to reduce the entropy of the mixture, we should make the states less distinguishable in some sense. Here, we study a class of optimization problems that are inspired by this situation and shed light on the relevant notions of distinguishability. The motivation for our study is the recently introduced spin alignment conjecture. In the original version of the underlying problem, each state in the mixture is constrained to be a freely chosen state on a subset of n qubits tensored with a fixed state Q on each of the qubits in the complement. According to the conjecture, the entropy of the mixture is minimized by choosing the freely chosen state in each term to be a tensor product of projectors onto a fixed maximal eigenvector of Q, which maximally “aligns” the terms in the mixture. We generalize this problem in several ways. First, instead of minimizing entropy, we consider maximizing arbitrary unitarily invariant convex functions such as Fan norms and Schatten norms. To formalize and generalize the conjectured required alignment, we define alignment as a preorder on tuples of self-adjoint operators that is induced by majorization. We prove the generalized conjecture for Schatten norms of integer order, for the case where the freely chosen states are constrained to be classical, and for the case where only two states contribute to the mixture and Q is proportional to a projector. The last case fits into a more general situation where we give explicit conditions for maximal alignment. The spin alignment problem has a natural “dual" formulation, versions of which have further generalizations that we introduce.

Appendices

Appendix A: The overlap Lemma

Lemma A.1

(The overlap lemma) Let $I_{1},\ldots ,I_{\ell }$ be a family of subsets of [n]. For $i \in [\ell ]$, let $Q_{I_{i}^{c}} = \bigotimes _{j \in I_{i}^{c}} Q_{I_{i}^{c},j}$, where for each $j \in I_{i}^{c}$, $Q_{I_{i}^{c},j} \in \mathcal {B}(\mathcal {H}_{j})$ has largest singular value $\alpha _{i,j} > 0$ and is of the form $Q_{I_{i}^{c},j} = \alpha _{i, j} {|{q_1}\rangle }{\langle {q_{1}}|} \oplus W_{i, j}$. For any unitarily invariant norm $||| \cdot |||$, the maximization problem

$$\begin{aligned} \max _{(R_{I_{i}})_{i=1}^{\ell }} ||| \; \;\prod _{i=1}^{\ell } R_{I_{i}} \otimes Q_{I_{i}^{c}} \; |||, \end{aligned}$$

(A.1)

where the variable tuple $(R_{I_{i}})_{i=1}^{\ell }$ is of operators each with trace norm at most $1$, has $({|{q_{1}}\rangle }{\langle {q_{1}}|}^{\otimes I_{i}})_{i=1}^{\ell }$ as a solution.

We argue by induction on the number of sets $\ell$. For that, we need the following statement.

Lemma A.2

Let $\mathcal {K}_1, \mathcal {K}_2,$ and $\mathcal {K}_3$ be given complex inner product spaces. Let $A_{1} \in \mathcal {B}(\mathcal {K}_1)$ and $A_{3} \in \mathcal {B}(\mathcal {K}_3)$ be such that $||A_{1}|| \le 1$ and $||A_{3}|| \le 1$. For all $T_{2,3} \in \mathcal {B}(\mathcal {K}_2 \otimes \mathcal {K}_3)$ and $S_{1,2} \in \mathcal {B}(\mathcal {K}_1 \otimes \mathcal {K}_2)$ of trace norm at most $1$, it holds that $|| (A_{1} \otimes T_{2,3}) (S_{1,2} \otimes A_{3})||_{1} \le 1$.

Proof

Because $|| \cdot ||_1$ is homogeneous and convex, it suffices to prove the inequality holds in the cases where $S_{1,2} = {|{s}\rangle }{\langle {\tilde{s}}|}_{1,2}$ and $T_{2,3} = {|{t}\rangle }{\langle {\tilde{t}}|}_{2,3}$ for arbitrary unit vectors ${|{s}\rangle }, {|{\tilde{s}}\rangle } \in \mathcal {K}_1 \otimes \mathcal {K}_2$ and ${|{t}\rangle }, {|{\tilde{t}}\rangle } \in \mathcal {K}_{2}\otimes \mathcal {K}_3$. Since $||A_{1}|| \le 1$ and $||A_{3}|| \le 1$, the norms of ${|{v}\rangle }_{1,2} = A_{1} \otimes \mathbb {1}_2 {|{s}\rangle }_{1,2}$ and ${|{u}\rangle }_{2,3} = \mathbb {1}_2 \otimes A_{3}^{*} {|{\tilde{t}}\rangle }_{2,3}$ are bounded from above by $1$. Hence, it suffices to prove that

$$\begin{aligned} || (\mathbb {1}_1 \otimes {|{t}\rangle }{\langle {u}|}_{2,3}) ({|{v}\rangle }{\langle {\tilde{s}}|}_{1,2} \otimes \mathbb {1}_3||_1 \le 1 \end{aligned}$$

(A.2)

for arbitrary unit vectors ${|{v}\rangle }, {|{\tilde{s}}\rangle } \in \mathcal {K}_1 \otimes \mathcal {K}_2$ and ${|{t}\rangle }, {|{u}\rangle } \in \mathcal {K}_{2}\otimes \mathcal {K}_3$.

Consider the inner factor operator ${\langle {u}|}_{2,3} {|{v}\rangle }_{1,2}: \mathcal {K}_3 \rightarrow \mathcal {K}_1$. To see that its trace norm is at most 1, let ${|{v}\rangle }_{1,2} = \sum _{i} \sqrt{\lambda _i} {|{\alpha _{i}}\rangle }_{1} \otimes {|{\beta _{i}}\rangle }_{2}$ and ${|{u}\rangle }_{2,3} = \sum _{j} \sqrt{\mu _j} {|{\gamma _{j}}\rangle }_{2} \otimes {|{\delta _{j}}\rangle }_{3}$ be Schmidt decompositions. Then,

$$\begin{aligned} {\langle {u}|}_{2,3} {|{v}\rangle }_{1,2} = \sum _{i, j} \sqrt{\mu _j} \sqrt{\lambda _{i}} \langle \gamma _{j}|\beta _{i}\rangle {|{\alpha _{i}}\rangle }_{1}{\langle {\delta _{j}}|}_{3} \end{aligned}$$

(A.3)

and

$$\begin{aligned} |{\langle {u}|}_{2,3} {|{v}\rangle }_{1,2}|^{2} = \sum _{i, j, j'} \sqrt{\mu _{j} \mu _{j'}} \lambda _{i} \langle \gamma _{j}|\beta _{i}\rangle \langle \beta _{i}|\gamma _{j'}\rangle {|{\delta _{j'}}\rangle }{\langle {\delta _{j}}|}_{3}. \end{aligned}$$

(A.4)

Denote $b_j = {\langle {\gamma _j}|} (\sum _{i} \lambda _{i} {|{\beta _i}\rangle }{\langle {\beta _{i}}|}) {|{\gamma _{j}}\rangle }$ and notice that $\sum _{j} b_{j} \le 1$. It follows from the Schur-concavity of $\text {tr}(\sqrt{\cdot })$ and the Schur-Horn theorem that

$$\begin{aligned} || {\langle {u}|}_{2,3} {|{v}\rangle }_{1,2} ||_{1}&= \text {tr}( \sqrt{|{\langle {u}|}_{2,3} {|{v}\rangle }_{1,2}|^{2}}) \end{aligned}$$

(A.5)

$$\begin{aligned}&\le \text {tr}(\sqrt{\text {diag}(|{\langle {u}|}_{2,3} {|{v}\rangle }_{1,2}|^{2})}) \end{aligned}$$

(A.6)

$$\begin{aligned}&= \sum _{j} \sqrt{\mu _j b_j} \le 1. \end{aligned}$$

(A.7)

Finally, $|| {\langle {u}|}_{2,3} {|{v}\rangle }_{1,2} \otimes {|{t}\rangle }_{2,3}{\langle {\tilde{s}}|}_{1,2} ||_1 = || {\langle {u}|}_{2,3} {|{v}\rangle }_{1,2} ||_1 \; ||{|{t}\rangle }_{2,3}{\langle {\tilde{s}}|}_{1,2} ||_1 \le 1.$ $\square$

Proof of Lemma A.1

Since $||| \cdot |||$ is homogeneous, it may be assumed without loss of generality for each $i \in [\ell ]$ that $||Q_{I_{i}^{c}, j} || = 1$ for all $j \in I_{i}^{c}$. We proceed via induction on $\ell$. If $\ell = 1$, the statement of the lemma follows from the facts that $|| {|{q_1}\rangle }{\langle {q_1}|}^{\otimes I_1}||_1 = 1$ and that $||| \cdot |||$ is convex and unitarily invariant. Suppose that the statement holds in cases where the family of sets has $\ell > 1$ elements. Consider $\prod _{i=1}^{\ell +1} R_{I_{i}} \otimes Q_{I_{i}^{c}}$ and notice that

$$\begin{aligned} R_{I_{1}} \otimes Q_{I_{1}^{c}} R_{I_{2}} \otimes Q_{I_{2}^{c}} = \underbrace{Q_{I_{1}^{c} \cap I_2} R_{I_{1}} R_{I_{2}} Q_{I_{2}^{c} \cap I_{1}} }_{ \tilde{R}_{ I_{1} \cup I_{2} }} \otimes \; \underbrace{Q_{I_{1}^{c} \setminus I_2} Q_{I_{2}^{c} \setminus I_1}}_{\tilde{Q}_{ (I_{1} \cup I_{2})^{c}}}. \end{aligned}$$

(A.8)

The operator $\tilde{Q}_{ (I_{1} \cup I_{2})^{c}}$ is completely factorizable on $(I_{1} \cup I_{2})^c$ and has maximal singular value $1$. Moreover, each of its factors can be written as ${|{q_{1}}\rangle }{\langle {q_1}|} \oplus W$ for some operator $W$. By Lemma A.2, $|| \tilde{R}_{ I_{1} \cup I_{2}} ||_{1} \le 1$ and so we may estimate

$$\begin{aligned} ||| \prod _{i=1}^{\ell +1} R_{I_{i}} \otimes Q_{I_{i}^{c}} |||&= ||| \tilde{R}_{ I_{1} \cup I_{2} } \otimes \; \tilde{Q}_{ (I_{1} \cup I_{2})^{c}} \prod _{i=3}^{\ell + 1} R_{I_{i}} \otimes Q_{I_{i}^{c}}||| \end{aligned}$$

(A.9)

$$\begin{aligned}&\le ||| {|{q_{1}}\rangle }{\langle {q_1}|}^{\otimes I_{1} \cup I_{2}} \otimes \; \tilde{Q}_{ (I_{1} \cup I_{2})^{c}} \prod _{i=3}^{\ell + 1} {|{q_1}\rangle }{\langle {q_1}|}^{\otimes I_{i}} \otimes Q_{I_{i}^{c}} ||| \end{aligned}$$

(A.10)

$$\begin{aligned}&= ||| \prod _{i=1}^{\ell +1} {|{q_{1}}\rangle }{\langle {q_{1}}|}^{\otimes I_{i}} \otimes Q_{I_{i}^{c}} |||, \end{aligned}$$

(A.11)

where the inequality is by the induction hypothesis. $\square$

Appendix B: On the Relationship Between the Sum of Two Projectors and Their Product

In this appendix, we prove lemmas necessary to elucidate the relationship between a non-negative linear combination $s_{1} P_{1} + s_{2} P_{2}$ of two projectors $P_{1}, P_{2} \in \mathcal {S}(\mathcal {K})$ and their product $P_{1} P_{2}$. By [25], $\mathcal {K}$ may be decomposed into a direct sum of subspaces, each of dimension at most 2, that are invariant under the action of both $P_{1}$ and $P_{2}$. Moreover, when restricted to each invariant subspace, the two projectors have rank at most 1. So, we may write $\mathcal {K}$ as an orthogonal direct sum of one- and two-dimensional minimal invariant subspaces

$$\begin{aligned} \mathcal {K} = \bigoplus _{i_{1} = 1}^{m_{1}} \mathcal {K}_{i_{1}}^{(1)} \oplus \bigoplus _{i_{2} = 1}^{m_{2}} \mathcal {K}_{i_{2}}^{(2)}, \end{aligned}$$

(B.1)

where the $\mathcal {K}_{i}^{(1)}$ are one-dimensional and the $\mathcal {K}_{i}^{(2)}$ are two dimensional. By minimal, we mean that the subspaces contain no proper nonzero invariant subspace. For each $i_{2} \in [m_{2}]$, we notate

$$\begin{aligned} P_{1}|_{\mathcal {K}_{i_{2}}^{(2)}} =: {|{\alpha _{i_{2}}}\rangle }{\langle {\alpha _{i_{2}}}|} \; , \; P_{2}|_{\mathcal {K}_{i_{2}}^{(2)}} =: {|{\beta _{i_{2}}}\rangle }{\langle {\beta _{i_{2}}}|}. \end{aligned}$$

(B.2)

$P_{1}$ and $P_{2}$ commute if and only if $m_{2} = 0$. The difficulty in reasoning about the eigenvalues of a linear combination of two projectors lies in these subspaces where they do not commute.

When restricted to $\bigoplus _{i_{2} = 1}^{m_{2}} \mathcal {K}_{i_{2}}^{(2)}$, the nonzero singular values of $P_{1} P_{2}$ are $(| \langle \alpha _{i_{2}}|\beta _{i_{2}}\rangle |)_{i_{2} = 1}^{m_{2}}$. The eigenvalues of the restriction $(s_{1} P_{1} + s_{2} P_{2}) |_{\mathcal {K}_{i_{2}}^{(2)}}$ may be computed as

$$\begin{aligned} \frac{1}{2} ( (s_{1} + s_{2}) \pm \sqrt{(s_{1} - s_{2})^{2} + 4 s_{1} s_{2} | \langle \alpha _{i_{2}}|\beta _{i_{2}}\rangle |^{2}}). \end{aligned}$$

(B.3)

Hence, the eigenvalues of $s_{1} P_{1} + s_{2} P_{2}$ are a function of the singular values of the product $P_{1} P_2$. We show next that it is in fact a strictly isotone function (see page 41 of Ref. [10]). A strictly isotone function G is one that preserves the majorization ordering in the sense that if $v\succeq w$, then $G(v)\succeq G(w)$. That is, the less dispersed the singular values of $P_{1} P_{2}$, the less dispersed the eigenvalues of $s_{1} P_{1} + s_{2} P_{2}$. This is a consequence of the fact that for $a, b \ge 0$, the map $x \mapsto \sqrt{a + b x^{2}}$ is convex on $\mathbb {R}_{\ge 0}$.

Lemma B.1

Let $A \subseteq \mathbb {R}$ be convex and $g: A \rightarrow \mathbb {R}_{\ge 0}$ be a convex function. For $t \in \mathbb {R}$, define the mapping $G: A^{m} \rightarrow \mathbb {R}^{2 m}$ with action

$$\begin{aligned} (v_{1},\ldots , v_{m}) \mapsto (t + g(v_{1}), \ldots , t + g(v_{m}) ) \oplus (t - g(v_{1}), \ldots , t - g(v_{m}) ). \end{aligned}$$

(B.4)

Then, G is strictly isotone.

Proof

Let $v, w \in A^{m}$ be such that $v \succeq w$. Observe that for $k \in [m]$,

$$\begin{aligned} \sum _{j=1}^{k} G(v)^{\downarrow }_{j} = k t + \sum _{j=1}^{k} g(v)_{j}^{\downarrow } \ge k t + \sum _{j=1}^{k} g(w)_{j}^{\downarrow } = \sum _{j=1}^{k} G(w)^{\downarrow }_{j}, \end{aligned}$$

(B.5)

where the inequality follows from the convexity of g and the doubly-stochastic characterization of majorization (see, for example, Theorem II.3.3 on page 41 of Ref. [10]). If $k > m$, then

$$\begin{aligned} \sum _{j=1}^{k} G(v)^{\downarrow }_{j}&= k t + \sum _{j=1}^{m} g(v)_{j}^{\downarrow } - \sum _{j=1}^{k - m} g(v)_{j}^{\uparrow } = k t + \sum _{j=1}^{2m - k} g(v)_{j}^{\downarrow } \end{aligned}$$

(B.6)

$$\begin{aligned}&\ge k t + \sum _{j=1}^{2m - k} g(w)_{j}^{\downarrow } = \sum _{j=1}^{k} G(w)^{\downarrow }_{j}. \end{aligned}$$

(B.7)

Since $\sum _{j=1}^{2\,m} G(\cdot )_{j} = 2\,m t$, $G(v) \succeq G(w)$. $\square$

The following lemma exhibits the maximal elements in the majorization ordering in a superset of the possible tuples of nonzero singular values of $P_{1} P_{2}|_{\bigoplus _{i_{2} = 1}^{m_{2}} \mathcal {K}_{i_{2}}^{(2)}}$.

Lemma B.2

Given $m \in \mathbb {N}, e \ge 0$, define the set

$$\begin{aligned} S_{e} := \{ x \in \mathbb {R}^{m} \; | \; \forall i \in [m], x_{i} \in [0,1], \sum _{i=1}^{m} x_{i} = e\}. \end{aligned}$$

(B.8)

The element $(\underbrace{1, \ldots , 1}_{\lfloor e \rfloor \; \text {times}}, e - \lfloor e \rfloor , 0, \ldots ,0)$ is a majorant of $S_{e}$.

Proof

Let $x \in S_{e}$ be arbitrary. For $k \in [m]$, $k \le \lfloor e \rfloor$, observe that $\sum _{i = 1}^{k} x^{\downarrow }_{i} \le \sum _{i=1}^{k} 1 = k$. And for $k > \lfloor e \rfloor$, $\sum _{i = 1}^{k} x^{\downarrow }_{i} \le \sum _{i = 1}^{m} x^{\downarrow }_{i} = e$. $\square$

Appendix C: Proof of the Conjecture in Example 1.2

Let $\sum _{i = 1}^{d-1} \lambda _i(\tau ) {|{\tau _i}\rangle }{\langle {\tau _i}|}$ be a spectral decomposition of $\tau$. Let arbitrary $p \in [0, 1]$ be given and consider the convex mixture $\tau _{\gamma }:= p \tau + (1-p) {|{v_{\gamma }}\rangle }{\langle {v_{\gamma }}|}$. We wish to prove that $\tau _\gamma \preceq \tau _1$. Define

$$\begin{aligned} w := (\sqrt{p(1-p)\lambda _1 (\tau )} \langle \tau _1 | \alpha \rangle ,\ldots , \sqrt{p(1-p)\lambda _{d - 1} (\tau ) } \langle \tau _{d-1} | \alpha \rangle )^T. \end{aligned}$$

(C.1)

The Gram matrix for the $d$ vectors $\sqrt{p\lambda _{1}}{|{\tau _{1}}\rangle }, \ldots \sqrt{p\lambda _{d-1}}{|{\tau _{d-1}}\rangle }, \sqrt{1-p}{|{v_{\gamma }}\rangle }$ is

(C.2)

It is not difficult to show that (see [29] for example), up to zeros, the spectrum of $M_{\gamma }$ is equal to the spectrum of $\tau _{\gamma }$. Hence, it suffices to show that $M_{\gamma } \preceq M_{1}$ for all $\gamma \in [0,1]$. This statement is proven in the next paragraph.

Let ${|{e_1}\rangle }, \ldots , {|{e_d}\rangle }$ denote the orthonormal basis (ordered in the obvious way) used to write down the Gram matrices in Eq. C.2. Consider the unitary

$$\begin{aligned} V := (\sum _{i = 1}^{d-1} {|{e_i}\rangle }{\langle {e_i}|}) - {|{e_d}\rangle }{\langle {e_d}|}. \end{aligned}$$

(C.3)

Supposing it occurs with a probability $q \in [0,1]$, we have

$$\begin{aligned} (1-q) M_1 + q V M_1 V^* = \begin{pmatrix} \begin{matrix} p \lambda _{1} (\tau ) &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} p \lambda _{d - 1} (\tau ) \end{matrix} \; \vline&\begin{matrix} (1-2q) w \end{matrix} \\ \hline \begin{matrix} \quad \quad (1-2q) w^* \end{matrix} \vline&(1 - p) \end{pmatrix} \end{aligned}$$

(C.4)

If $q = \frac{1 - \sqrt{\gamma }}{2}$, then $(1-q) M_1 + q V M_1 V^* = M_{\gamma }$. Hence, for all $\gamma \in [0,1]$, there exists a mixed-unitary channel that takes $M_1$ to $M_\gamma$.