2
$\begingroup$

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.

$\endgroup$
2

1 Answer 1

4
$\begingroup$

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 $|e_k\rangle$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.