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Let $\Phi_S$ be an operator acting on a space $\mathcal H_S$. If we introduce an ancilla $A$, the total space becomes $\mathcal H_S\otimes \mathcal H_A$ and I can naturally extend the operator $\Phi_S$ to act on the whole space by defining $\Phi_{SA}=\Phi_S\otimes \mathbb{I}_A$, where $\mathbb{I}_A$ is the identity on $A$.

What is the correct way of extending a state or more generally a density operator $\rho_S$ to the space $\mathcal H_S\otimes \mathcal H_A$? Since $\rho_S$ is also an operator, I would write again $\rho_{SA}=\rho_S\otimes \mathbb I_A$. Is this correct?

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The way that you propose does not preserve the trace of the state -- if you compute the trace you'll find it equal to the dimension of $\mathcal{A}$.

There are many ways to extend your state $\rho_S$ to the joint system such that the marginal state on $S$ stays the same. For example, if the systems $S$ and $A$ are assumed to be independent then you can pick any state $\rho_A$ on system $A$ and define $$ \rho_{SA} = \rho_S \otimes \rho_A. $$

Another possibility would be to take $A$ to be a purifying system of $S$ and define $\rho_{SA}$ to be the purification of $\rho_S$ (assuming the dimension of the system $A$ is large enough.

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    $\begingroup$ Notably, in the spirit of the question being posed, one can choose $\rho_A=\mathbb{I}/\mathrm{Tr}(\mathbb{I})$. $\endgroup$ Commented Jul 14, 2021 at 19:36
  • $\begingroup$ @QuantumMechanic This is not at all in the spirit of the question. $\endgroup$ Commented Jul 23, 2021 at 17:04
  • $\begingroup$ @NorbertSchuch you don't think? The question asked about using $\rho_{SA}=\rho_S\otimes\mathbb{I}$... ahh, your answer is much more in the spirit of the question, this is explicitly just the letter $\endgroup$ Commented Jul 23, 2021 at 19:20
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This is not a meaningful question -- in the sense that there is no "well-defined" way to extend a state to a larger system, without further assumptions.

In analogy, it is like asking what is written on the second page of a book after having read the first page. (Maybe more accurately, what's written in the second chapter of a collection of essays after having read the first.)

On the contrary, extending operations is well-defined: Just as it is clear how to extend the operation "strike out the first page" to a whole book: You still just strike out the first page, and leave the rest of the book untouched.

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  • $\begingroup$ This analogy is quite useful $\endgroup$ Commented Jul 23, 2021 at 19:23
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There is no more natural way of doing the extension. Yet, two are most often used:

  1. You do not want to make hypotheses about the state of A besides its dimension so that the state of SA will be $\rho_S \otimes I / N$ where $I$ is the identity matrix, and $N$ is the dimension of A.

  2. You want to purify your system $S$ so that the global state of SA is now pure. This is useful when you want to argue that the state of the system --that you are about to purify-- could be viewed as a resulting from a unitary transformation acting on SA. In such case, a purification can be obtained by diagonalizing $\rho_S = \sum_i \lambda_i | e_i\rangle\langle e_i|$ and writing $\rho_{SA} = \sum_i \sqrt{\lambda_i} |e_i\rangle\otimes|f_i\rangle$ where {$|f_i\rangle$} is any orthonormal set of state vectors on $A$.

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