In section 8.2.3 of Nielsen and Chuang, they discuss how unitary dynamics of a system and environment arise from quantum operations (i.e. Kraus operators $E_k$ such that $\sum_k E_k^*E_k=I$). Importantly, they mention that

"we are trying to find a model environment giving rise to a dynamics described by the operation elements $\{E_k\}$."

In what follows they let $|e_k\rangle$ be an orthonormal basis for the environment. They then go on to say

"Define an operator $U$ which has the following action on states of the form $|\psi\rangle|e_0\rangle$, $$U|\psi\rangle|e_0\rangle=\sum_k E_k|\psi\rangle|e_k\rangle,\quad\quad\quad\quad\quad (8.37)$$ where $|e_0\rangle$ is just some standard state of the model environment."

Let $\mathcal{H}$ and $\mathcal{K}$ be the Hilbert spaces of the system and environment respectively and suppose we fix the environment $\mathcal{K}$.

My question is whether every unitary on $\mathcal{H}\otimes \mathcal{K}$ has the form of $U$ in equation (8.37)?

I see how defining the action of $U$ in (8.37) leads to these unitary dynamics. This is just the same thing as saying that any isometry on $\mathcal{H}$ extends to a unitary on $\mathcal{H}\otimes \mathcal{K}$. I guess my question is whether the converse is true, does every unitary on the joint system decompose into something of the form in (8.37)? Is there some counterexample, or argument to see that any such unitary has this decomposition?


1 Answer 1


You are asking whether any unitary operator $U:\mathcal H\otimes\mathcal K\to \mathcal H\otimes\mathcal K$ can be written, given $|\psi\rangle\in\mathcal H$ and $|e_0\rangle\in\mathcal K$, as $$U |\psi\rangle|e_0\rangle = \sum_k E_k |\psi\rangle|e_k\rangle,$$ for some $E_k$ and $|e_k\rangle$.

The answer is yes, because this essentially just amounts to a way of writing $U$ with respect to a choice of basis for $|e_k\rangle$. More precisely, given any orthonormal basis $\{|e_k\rangle\}_{k\ge0}\subset\mathcal K$, we can write $$U|\psi\rangle|e_0\rangle = \left(I\otimes \sum_k |e_k\rangle\!\langle e_k|\right)U|\psi\rangle|e_0\rangle = \sum_k \left(I\otimes |e_k\rangle\right) \left(I\otimes\langle e_k|\right)U|\psi\rangle|e_0\rangle \\ = \sum_k \color{red}{\left(I\otimes\langle e_k|\right)U(I\otimes |e_0\rangle)} (|\psi\rangle \otimes |e_k\rangle) \equiv \sum_k (\color{red}{E_k} |\psi\rangle)\otimes|e_k\rangle.$$ In other words, the Kraus operators $E_k$ are nothing but the "matrix elements" of $U$ with respect to a choice of basis for $\mathcal K$. Because we are not really choosing a basis for the full space to represent $U$, but rather a "partial basis", we get operators instead of scalars as "components".

In fact, $U$ doesn't need to be unitary for this to be the case. The above argument works for any operator in $\mathcal H\otimes\mathcal K$. The fact that $U$ is unitary only affects the normalisation of the "coefficients" in the expansion: $\sum_k E_k^\dagger E_k=I$.

See also

  1. How does $\mathcal E(\rho)=\mathrm{Tr}_{env}[U(\rho\otimes\rho_{env})U^\dagger]$ turn into $P_0\rho P_0+P_1\rho P_1$?
  2. Equivalent statement of the unitary freedom of Kraus operator?

and links therein.

  • $\begingroup$ thank you, this is precisely what I was wondering. $\endgroup$
    – Condo
    Commented Nov 1, 2021 at 16:17
  • 1
    $\begingroup$ Ah if only this question and answer were also on QCSE! physics.stackexchange.com/questions/27657/… $\endgroup$
    – Condo
    Commented Nov 1, 2021 at 16:50

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