Is there an expression for the partial trace of a vectorized density matrix?

Question

Is there an expression for the partial trace of vectorized density matrix? I did some literature review but didn't find not much relevant information.

Answer 1

Here's a simple way to compute it. We know the partial trace over system $B$ can be defined as $$ \mathrm{Tr}_B[\rho_{AB}] = \sum_k (I_A \otimes \langle k|) \rho_{AB} (I_A \otimes |k\rangle) = \sum_k T_k \rho_{AB} T_k^{\dagger} $$ where $\{|k\rangle\}_k$ is an orthonormal basis of $B$ and for the second equality we define the matrices $T_k = I_A \otimes \langle k |$.

Now we also have a well-known identity for vectorization (see Watrous - Eq. 1.132): $$ \mathrm{vec}(XYZ) = (X \otimes Z^T) \mathrm{vec}(Y)\,. $$ Which implies in our case that $$ \mathrm{vec}(\mathrm{Tr}_B[\rho_{AB}]) = \sum_k (T_k \otimes \overline{T_k})\mathrm{vec}(\rho_{AB}) $$ where $\overline{T}$ is the elementwise complex conjugate of $T$.

Answer 2

If you want to know what the superoperator/representation matrix of the partial trace ${\rm tr}_{\mathbb C^\ell}:\mathbb C^{n\times n}\otimes\mathbb C^{\ell\times\ell}\to\mathbb C^{n\times n}$ with respect to (column) vectorization is—that is, what the matrix $\widehat{{\rm tr}_{\mathbb C^\ell}}\in\mathbb C^{n^2\times n^2\ell^2}$ that satisfies $${\rm vec}({\rm tr}_{\mathbb C^\ell}(X))=\widehat{{\rm tr}_{\mathbb C^\ell}}{\rm vec}(X)$$ for all $X\in\mathbb C^{n\times n}\otimes\mathbb C^{\ell\times\ell}$ is—the answer is:

Result. For all $\ell,n\in\mathbb N$ it holds that $$ \widehat{{\rm tr}_{\mathbb C^\ell}}=\big({\bf1}_{n}\otimes{\bf1}_n\otimes({\rm vec}({\bf1}_\ell)^\dagger\big)( {\bf1}_{n}\otimes\mathbb F_{\ell n}\otimes {\bf1}_{\ell})\,, $$ where $F_{\ell n}:\mathbb C^\ell\otimes\mathbb C^n\to\mathbb C^n\otimes\mathbb C^\ell$, $x\otimes y\mapsto y\otimes x$ is the usual flip operator.

While this is, implicitly, shown here already, I would like to present an indepdent proof.:

By linearity it suffices to verify this on product vectors. For all $a,j=1,\ldots,n$ and all $b,k=1,\ldots,\ell$, using that ${\rm vec}(|a\rangle\langle b|)=|b\otimes a\rangle$ we find \begin{align*} \widehat{{\rm tr}_{\mathbb C^\ell}}|j\otimes k\otimes a\otimes b\rangle&=\widehat{{\rm tr}_{\mathbb C^\ell}}{\rm vec}(|a\otimes b\rangle\langle j\otimes k|)\\ &\overset{\hphantom{\eqref{eq:vecG}}}={\rm vec}\big({\rm tr}_{\mathbb C^\ell}(|a\rangle\langle j|\otimes |b\rangle\langle k|)\big)\\ &\overset{\hphantom{\eqref{eq:vecG}}}={\rm vec}(|a\rangle\langle j|\langle k|b\rangle) =\langle k|b\rangle|j\otimes a\rangle\,. \end{align*} On the other hand \begin{align*} \big({\bf1}_{n}\otimes{\bf1}_n\otimes({\rm vec}({\bf1}_\ell)^\dagger\big)( {\bf1}_{n}\otimes\mathbb F_{\ell n}\otimes &{\bf1}_{\ell})|j\otimes k\otimes a\otimes b\rangle\\&=\big({\bf1}_{n}\otimes{\bf1}_n\otimes({\rm vec}({\bf1}_\ell)^\dagger\big)|j\otimes a\otimes k\otimes b\rangle\\ &=|j\otimes a\rangle\langle{\rm vec}({\bf1}_\ell)|k \otimes b\rangle\\ &=|j\otimes a\rangle\langle{\rm vec}({\bf1}_\ell)|{\rm vec}(|b\rangle\langle k|)\rangle\\ &=|j\otimes a\rangle{\rm tr}( |b\rangle\langle k| )=\langle k|b\rangle|j\otimes a\rangle\,. \end{align*} Because these two expressions coincide, we are done. $\square$

Answer 3

Let $\rho$ be a bipartite linear operator (it doesn't really matter here whether it's Hermitian, positive, or anything else). Denote with $\operatorname{vec}(\rho)$ its vectorisation.

The partial trace of $\rho$ equals, in components: $$\operatorname{Tr}_2(\rho) = \sum_{ij} |i\rangle\!\langle j| \, \sum_k \rho_{ik,jk}.$$ In terms of the vectorisation, you could write this as $(I\otimes \langle m|)\operatorname{vec}(\rho)$ with $|m\rangle\equiv \sum_k |k,k\rangle$. You can verify directly that this gives the correct expression for $\operatorname{vec}(\operatorname{Tr}_2(\rho))$.