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The question:

In this paper (DOI: 10.1142/S012918311750005X) Maziero cites the following definition for the partial trace from this book ("The Theory of Quantum Information" by John Watrous from 2018):

$$\text{Tr}_b(O)=\sum_{j=1}^{d_b}(\mathbb{I}_a\otimes\langle b_j|)O(\mathbb{I}_a\otimes|b_j\rangle)\tag{1} $$

where $d_b$ is the dimension of the considered hilbert-space $\mathcal{H}=\mathcal{H}_a\otimes\mathcal{H}_b$.

Where exactly is it taken from and how does one arrive mathematically (e.g. using matrix representation of operators) at (1)?

More context:

Since no page number is given, I assume it is taken from page 22, although I'm not sure. I paraphrase what is being defined there in the following quote:

There the Trace function is defined as a linear mapping of the form $$\text{Tr}:\mathcal{X}\rightarrow\mathbb{C}$$ with $\mathcal{X}$ being a euclidean space.

The partial trace then is defined as the unique map $\text{Tr}\otimes\mathbb{1}_{L(\mathcal{Y})}\in T(\mathcal{X}\otimes\mathcal{Y}),\mathcal{Y})$ that fulfills $$\text{Tr}\otimes\mathbb{1}_{L(\mathcal{Y})}(X\otimes Y)=\text{Tr}(X)Y$$ for any arguments $X$ and $Y$. In this definition, $L(\mathcal{Y})$ is the collection of linear mappings from $\mathcal{Y}$ to $\mathcal{Y}$ and $T(\mathcal{X}\otimes\mathcal{Y}),\mathcal{Y})$ the set of all linear maps from $L(\mathcal{X}\otimes\mathcal{Y})$ to $L(\mathcal{Y})$.

If that's the case, I fail to see how one could arrive mathematically (e.g. using matrix representation of operators) at (1) from this.

I'd appreciate anyone trying to scan the book (downloadable at the link given above) for the phrase "partial" or "partial trace" to find relevant pages. But it seems that page 22 is what Maziero cites.

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  • $\begingroup$ What exactly is the question: Where you can find $(1)$ in the book or how one 'arrives' at $(1)$? After all, it is a definition and it certainly fulfills the definition of Watrous... $\endgroup$ Jun 22 at 10:06
  • $\begingroup$ Well, both actually - what Maziero uses from Watrous and how he arrives with this at (1). I just want to understand the thoughts of the author, that clearly used Watrous to end up at (1). The reason is that I am looking for the definition (1) that I can cite in a thesis. Since Maziero cited it, I cannot cite him but I also cannot cite Watrous, since I don't know how one gets to (1) using Watrous. $\endgroup$
    – manuel459
    Jun 22 at 10:18
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    $\begingroup$ I see. There are a lot of sources using the definition $(1)$, tho. $\endgroup$ Jun 22 at 10:23
  • $\begingroup$ If you could suggest a book for me, my question would be answered (or rather dispensable). Thank you. $\endgroup$
    – manuel459
    Jun 22 at 10:28
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    $\begingroup$ Among several lecture notes (one of which I mentioned in an answer to another question by you), you can find $(1)$ in e.g. definition 4.3.4 of From Classical to Quantum Shannon Theory $\endgroup$ Jun 22 at 10:32

2 Answers 2

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I'm not sure exactly what you're looking for, but I can try to provide some intuition for the formula you're citing:

Recall that the standard trace takes a linear operator $O$ and spits out a number $\mathrm{Tr}(O)$, by doing an appropriate sum over some basis, i.e. $\mathrm{Tr}(O) = \sum_i \langle i | O|i \rangle$. Now, let's think about a Hilbert space that is a product: $H = H_a \otimes H_b$. What would we want out of a partial trace over the $H_b$ degrees of freedom?

One reasonable desire would be that if we had an operator $O = O_a\otimes O_b$, then our partial trace over the $b$ degrees gives $\mathrm{Tr}_b(O) = O_a \mathrm{Tr}(O_b)$. That is, our partial trace essentially ignores the degrees of freedom pertaining to $H_a$, and does the usual trace over the degrees of freedom pertaining to $H_b$. Note from this expression that the partial trace returns an operator, not on $H_a\otimes H_b$, but on $H_a$.

In any case, this desire that our partial trace (1) does nothing to the $H_a$ degrees of freedom, and (2) does the usual trace on the $H_b$ degrees of freedom is exactly what is encoded in your formula! You can see (1) from the identity operators $\mathbb{1}_a$ on the $H_a$ degrees of freedom, and (2) from the sum over the $H_b$ degrees of freedom.

This question/answer might be helpful to you as well.

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  • $\begingroup$ I see my question was unclear. I thank you for your answer, which does help on a "more formal" way. I think what I was trying to ask is how one arrives at (1) mathematically, especially when you consider the Definition of the Trace in matrix representation (so to speak a multiplication with a vector and its transpose conjugate). $\endgroup$
    – manuel459
    Jun 22 at 9:51
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Let $A:\mathcal{X}\to\mathcal{Y}$ and $B:\mathcal Z\to\mathcal W$ be two linear operators (note that input and output spaces are not required to be equal here). Then $A\otimes B:\mathcal X\otimes\mathcal Z\to\mathcal Y\otimes \mathcal W$ is the linear operator characterised by its action on simple tensors: $$(A\otimes B)(v\otimes w) = (Av)\otimes (Bw),\qquad \forall v\in \mathcal X,\, w\in\mathcal Z.$$

Consider now the trace operator. This is the linear operator $\operatorname{Tr}:\operatorname{Lin}(\mathcal X)\to\mathbb{C}$, where $\mathcal X$ is an arbitrary finite-dimensional complex vector space, and $\mathbb{C}$ is understood here as the trivial one-dimensional complex vector space. Note that $\operatorname{Lin}(\mathcal X)$ stands for the vector space of linear operators acting on $\mathcal X$. Also, strictly speaking, one could argue that $\operatorname{Tr}$ is an operator $\operatorname{Tr}:\operatorname{Lin}(\mathcal X)\to\operatorname{Lin}(\mathbb{C})$. But $\operatorname{Lin}(\mathbb{C})\simeq\mathbb{C}$, so the difference is immaterial. You could call such $\operatorname{Tr}$ as a "superoperator", to stress its acting on spaces of operators, but really this is just another type of linear operator as the ones used at the beginning of the post.

This operator is defined as the one sending any linear operator to its trace, which can also be written as $\operatorname{Tr}(X)\equiv\sum_i \langle v_i, X v_i\rangle$, for any orthonormal basis $\{v_i\}_i$ for $\mathcal X$.

Now the operators $\operatorname{Tr}\otimes \,I$ are necessarily defined on extensions of the space. In other words, for any finite-dimensional space $\mathcal Y$, we have $\operatorname{Tr}\otimes\, I_{\cal Y}:\operatorname{Lin}(\mathcal{X}\otimes\mathcal Y)\to \operatorname{Lin}(\mathcal{Y})$. Here a word of caution is warranted: strictly speaking, this should be an operator of the form $$\operatorname{Tr}\otimes\, I_{\cal Y}:\operatorname{Lin}(\mathcal{X})\otimes\operatorname{Lin}(\mathcal Y)\to \mathbb{C}\otimes \operatorname{Lin}(\mathcal{Y}).$$ However, we have natural linear isomorphisms $\operatorname{Lin}(\mathcal X)\otimes\operatorname{Lin}(\mathcal{Y})\simeq \operatorname{Lin}(\mathcal{X}\otimes\mathcal{Y})$, and $\mathbb{C}\otimes\operatorname{Lin}(\mathcal{Y})\simeq \operatorname{Lin}(\mathcal{Y})$.

In conclusion, knowing that the action of $\operatorname{Tr}$ on operators can be written as a sum of expectation values over a basis, we conclude that $\operatorname{Tr}\otimes\, I_{\cal Z}$, for any finite-dimensional $\mathcal Z$, is characterised by its action on simple tensors, which reads $$(\operatorname{Tr}\otimes\, I_{\cal Z})(A\otimes B) = \operatorname{Tr}(A)\otimes B\simeq \operatorname{Tr}(A) B.$$ Using the previously discussed characterisation, this can also be written as $$(\operatorname{Tr}\otimes\, I_{\cal Z})(A\otimes B) = \left(\sum_i \langle v_i,Av_i\rangle\right) B.$$ The action of a linear operator on simple products then immediately determines its action on general vectors, because any vector $v\in\mathcal X\otimes\mathcal Y$ can be decomposed as $v=\sum_i x_i\otimes y_i$ for some $x_i\in\mathcal X$ and $y_i\in\mathcal Y$, and because our operator $\operatorname{Tr}\otimes\, I$ oughts to be linear, we have $$(\operatorname{Tr}\otimes \,I_{\cal Z}) C =\sum_i (\operatorname{Tr}\otimes \,I_{\cal Z})(X_i\otimes Y_i) = \sum_i \left(\sum_j \langle v_j, X_i v_j\rangle\right)Y_i = \sum_j (\langle v_j|\otimes I)C(|v_j\rangle\otimes I).$$

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