16 Partial Traces in Orthonormal Coordinates

We record the coordinate and integrability facts used to estimate physical output operators in [ Ope26a , proof of Theorem 5.2 ] , 04-compression.tex, lines 383–549. The matrices need not be positive or self-adjoint. All norms denoted by \(\| \cdot \| _1\) are rectangular trace norms.

Theorem 16.1 Isometric coordinates and the reduced trace norm

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Let \(X,Y,D,E\) be finite index sets and let \(U:\mathbb {C}^X\to \mathbb {C}^Y\) and \(V:\mathbb {C}^D\to \mathbb {C}^E\) satisfy \(U^{\dagger }U=I\) and \(V^{\dagger }V=I\). Every matrix \(M\) on \(\mathbb {C}^X\otimes \mathbb {C}^D\) satisfies

\begin{align} \bigl\| \operatorname {tr}_{E} ((U\otimes V)M(U\otimes V)^{\dagger })\bigr\| _1 & =\| \operatorname {tr}_{D}M\| _1. \label{eq:physical_trace_isometries} \end{align}

Neither isometry is required to be onto.

Proof ▶

The partial trace on the left equals \(U(\operatorname {tr}_{D}M)U^{\dagger }\) because \(V^{\dagger }V=I\). Left and right multiplication by isometries preserve the nonzero singular values, giving (1).

Theorem 16.2 Independence of product orthonormal coordinates

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Let \(\mathcal H,\mathcal K\) be complex inner-product spaces with finite orthonormal bases \(b_X,b_Y\) and \(b_D,b_E\), respectively. Write \(C_{YX}\) and \(C_{ED}\) for the two changes of coordinates. The change from \(b_X\otimes b_D\) to \(b_Y\otimes b_E\) is \(C_{YX}\otimes C_{ED}\). If \([z]_{XD}\) denotes product coordinates, then for arbitrary \(z,t\in \mathcal H\otimes \mathcal K\),

\begin{align} (C_{YX}\otimes C_{ED})[z]_{XD}[t]_{XD}^{\dagger } (C_{YX}\otimes C_{ED})^{\dagger } & =[z]_{YE}[t]_{YE}^{\dagger }. \label{eq:physical_trace_rank_one_coordinates} \end{align}

Consequently, for every finite family \(z_i,t_i\) and complex coefficients \(c_i\),

\begin{align} \left\| \operatorname {tr}_{E} \sum _i c_i[z_i]_{YE}[t_i]_{YE}^{\dagger }\right\| _1 & =\left\| \operatorname {tr}_{D} \sum _i c_i[z_i]_{XD}[t_i]_{XD}^{\dagger }\right\| _1. \end{align}

Proof ▶

A product-basis coordinate is the product of the two endpoint coordinates, which gives the Kronecker formula for the change of basis. Multiplication by that matrix sends \([z]_{XD}\) to \([z]_{YE}\), proving (2). Distribute the coordinate changes through the finite sum and apply Theorem 16.1.

Theorem 16.3 Reconstruction from rectangular blocks

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Let \(M\) have rows indexed by \(X\times I\) and columns indexed by \(Y\times J\), where \(X,Y\) are finite. Put \((M_{xy})_{ij}=M_{(x,i),(y,j)}\). Then

\begin{align} M=\sum _{x\in X}\sum _{y\in Y}|x\rangle \langle y|\otimes M_{xy}. \label{eq:physical_trace_block_reconstruction} \end{align}

If \(I,J\) are also finite, then

\begin{align} \| M\| _1\leq \sum _{x\in X}\sum _{y\in Y}\| M_{xy}\| _1. \label{eq:physical_trace_block_bound} \end{align}

No factor depending on the dimensions of the blocks appears.

Proof ▶

At entry \(((x,i),(y,j))\), all matrix-unit summands except \((x,y)\) vanish. For the inequality, apply the triangle inequality to (4) and use \(\| |x\rangle \langle y|\otimes N\| _1=\| N\| _1\).

Theorem 16.4 Blocks commute with the discarded partial trace

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Let \(X,Y\) be arbitrary index sets, let \(D\) be finite, and let \(M\) be a matrix with rows and columns indexed by \((X\times Y)\times D\). Fix \(x,x'\in X\). Taking the \((x,x')\) block after tracing out \(D\) gives the same matrix as first taking that block and then tracing out \(D\):

\begin{align} (\operatorname {tr}_D M)_{xx'} & =\operatorname {tr}_D(M_{xx'}). \end{align}

Proof ▶

Both sides have entry \(\sum _{d\in D}M_{((x,y),d),((x',y'),d)}\) at \((y,y')\).

Theorem 16.5 An averaged block bound

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Let \((\Omega ,\mu )\) be a measure space and let \(M(\omega )\) be a finite rectangular block matrix as in Theorem 16.3. Suppose every function \(\omega \mapsto \| M_{xy}(\omega )\| _1\) is integrable. Then

\begin{align} \int \| M(\omega )\| _1\, d\mu (\omega ) & \leq \sum _{x,y}\int \| M_{xy}(\omega )\| _1\, d\mu (\omega ). \label{eq:physical_trace_integral_blocks} \end{align}

Here the Bochner integral is assigned the value zero when its integrand is not integrable. The assertion does not itself imply integrability of the full trace norm.

Proof ▶

The sum of the block norms is integrable and dominates the nonnegative full trace norm by (5). If the latter is integrable, integrate this inequality and exchange the finite sum with the integral. Otherwise the left side is zero by the stated convention, whereas the right side is nonnegative.

Theorem 16.6 Integrability of finite matrix combinations

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Let \(\mu \) be a finite measure, let \(J\) be finite, and fix finite rectangular complex matrices \(A_j\). Suppose the scalar functions \(z_j:\Omega \to \mathbb {C}\) satisfy \(z_i\overline{z_j}\in L^1(\mu )\) for all \(i,j\in J\). Then \(\omega \mapsto \| \sum _jz_j(\omega )A_j\| _1\) is integrable. In particular, it is enough that each \(z_j\) belongs to \(L^2(\mu )\). No independence or prescribed covariance is assumed.

Proof ▶

The diagonal product hypothesis gives \(|z_j|^2\in L^1\). Taking its square root gives measurability of \(|z_j|\), and finiteness of the measure then gives \(|z_j|\in L^1\). The Gram matrix of \(\sum _jz_jA_j\) depends only on the measurable pair products. Its positive square root therefore shows that the trace norm is measurable. The pointwise bound

\begin{align} \left\| \sum _j z_jA_j\right\| _1\leq \sum _j|z_j|\, \| A_j\| _1 \end{align}

gives integrability by domination. Finally, if all coefficients are square-integrable, complex conjugation preserves that property and Cauchy–Schwarz gives the required pairwise integrability.