In Nielsen and Chuang, 10th Anniversary Edition, there is a definition of the operator sum representation of a quantum operation: $\mathcal{E}(\rho)=\sum_{k}\langle e_k|U[\rho\otimes|e_0\rangle\langle e_0|]U^\dagger|e_k\rangle$ and then the operation element is defined as $E_k\equiv \langle e_k|U|e_0 \rangle$. In both cases the dimensions of $U$ on the one hand, and $\langle e_k|$, and $|e_0\rangle$ are different ($U$ is an operator acting on composite system, whereas $|e\rangle$ is a state of the component system).

How can the inner product be calculated in this case? Is this some sort of notational convention or is there a different explanation? It's a trivial question, but I can't find an explicit answer anywhere. It comes down to the identity $I\otimes |e_k\rangle = |e_k\rangle$ the meaning of which I'm struggling to understand.

  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/28290/55 $\endgroup$
    – glS
    Feb 22, 2023 at 10:12
  • 1
    $\begingroup$ There is an implicit $I\otimes \langle e_k|$ so that nothing happens on the system and the environment is traced out. $\endgroup$
    – DaftWullie
    Feb 22, 2023 at 10:53

1 Answer 1


TL;DR: We can understand the object $E_k=\langle e_k|U|e_0\rangle$ rigorously in two steps. First, think of $\langle e_k|$ and $|e_0\rangle$ as linear functions. Next, treat the implicit operation in $\langle e_k|U|e_0\rangle$ as function composition.

$\langle e_k|$ and $|e_0\rangle$ are functions

Usually, we think of elements of a set $A$ and functions to $A$ as different types of objects. However, we can view elements of $A$ as just a special type of function to $A$: namely, one whose domain is some singleton set $\{\star\}$. Then we identify $a\in A$ with $a:\{\star\}\to A$ defined by $a(\star)=a$ where I allowed myself a slight abuse of notation by denoting both objects as $a$.

Similarly, we typically think of an object like $|e_0\rangle$ as an element of some complex Hilbert space $\mathcal{H}_E$. However, we can also think of $|e_0\rangle$ as the linear function $|e_0\rangle:\mathbb{C}\to\mathcal{H}_E$ defined by $|e_0\rangle(z):=z|e_0\rangle$ for all $z\in\mathbb{C}$. In fact, this function is just the linear extension of $|e_0\rangle:\{\star\}\to\mathcal{H}_E$ defined by $|e_0\rangle(\star)=|e_0\rangle$ which we saw earlier. Physically, the function $|e_0\rangle$ can be understood as state preparation.

Next, an object such as $\langle e_k|$ is a dual of $|e_k\rangle$. In other words, $\langle e_k|$ is a linear function $\langle e_k|:\mathcal{H}_E\to\mathbb{C}$ defined by $\langle e_k|(|e_j\rangle)=\langle e_k|e_j\rangle=\delta^j_k$ and extended by linearity to all of $\mathcal{H}_E$. Physically, the function $\langle e_k|$ can be thought of as an "effect" that is a possible outcome in a measurement. It evaluates to the amplitude between the input and the fixed state $|e_k\rangle$ it is the dual of.

Function composition

Thus, the object $E_k=\langle e_k|U|e_0\rangle$ is built from three functions: $$ \begin{align} |e_0\rangle&:\mathbb{C}\to\mathcal{H}_E\\ U&:\mathcal{H}_S\otimes\mathcal{H}_E\to\mathcal{H}_S\otimes\mathcal{H}_E\\ \langle e_k|&:\mathcal{H}_E\to\mathbb{C}. \end{align}\tag1 $$ We would like to compose them together, but unfortunately their domains and codomains disagree. We can fix this by extending any function that doesn't explicitly act on $\mathcal{H}_S$ to all of $\mathcal{H}_S\otimes\mathcal{H}_E$ by assuming that it acts on $\mathcal{H}_S$ as identity. Then $(1)$ becomes $$ \begin{align} |e_0\rangle&:\mathcal{H}_S\otimes\mathbb{C}\to\mathcal{H}_S\otimes\mathcal{H}_E\\ U&:\mathcal{H}_S\otimes\mathcal{H}_E\to\mathcal{H}_S\otimes\mathcal{H}_E\\ \langle e_k|&:\mathcal{H}_S\otimes\mathcal{H}_E\to\mathcal{H}_S\otimes\mathbb{C} \end{align}\tag2 $$ where I continue to abuse notation by not giving new names to new objects made of old ones. We can further simplify $(2)$ by noting that $\mathcal{H}\otimes\mathbb{C}=\mathcal{H}$ $$ \begin{align} |e_0\rangle&:\mathcal{H}_S\to\mathcal{H}_S\otimes\mathcal{H}_E\\ U&:\mathcal{H}_S\otimes\mathcal{H}_E\to\mathcal{H}_S\otimes\mathcal{H}_E\\ \langle e_k|&:\mathcal{H}_S\otimes\mathcal{H}_E\to\mathcal{H}_S. \end{align}\tag3 $$ Finally, we can compose the three functions to obtain $$ E_k=\langle e_k|U|e_0\rangle:\mathcal{H}_S\to\mathcal{H}_S.\tag4 $$

Physical meaning

Mathematically, this is just the composition of the three functions. Physically, we can understand the action of $E_k=\langle e_k|U|e_0\rangle$ as consisting of three steps. First, $E_k$ sends any $|\psi\rangle\in\mathcal{H}_S$ from the smaller Hilbert space of system $S$ to $|\psi\rangle\otimes|e_0\rangle$ in the larger Hilbert space of both the system $S$ and the environment $E$, then it applies unitary $U$ on the larger Hilbert space, and finally it measures the environment obtaining whatever (non-degenerate) outcome is associated with $|e_k\rangle$.


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