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Niesen and Chuang, 2nd edition, page 107, Box 2.6, in its motivation for partial trace, says that if M is an observable on system A and $\tilde{M}$ is the corresponding observable on system AB, then it is physically reasonable that the average measurement outcomes computed for the two operators are equal. That is,

$$\mathrm{tr}(M\rho^A)=\mathrm{tr}(\tilde{M}\rho^{AB})=\mathrm{tr}((M\otimes I_B)\rho^{AB})$$

I have no problem with these statements, but they go on to say that this is "obviously satisfied" if $\rho^A=\mathrm{tr}_B(\rho^{AB})$.

This loses me.

The partial trace is defined by $\mathrm{tr}_B(A\otimes B)\equiv A \mathrm{tr}(B)$, I believe.

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It's easy to see that it's true for $\rho^{AB} = A \otimes B$ for any matrices $A,B$ (even when $\rho^{AB}$ is not a state, but just a matrix).

Any matrix (not only states) is a linear combination of such products, that is $\rho^{AB} = \sum_i A_i \otimes B_i$, where $A_i,B_i$ are some matrices.

Thus $tr(M\rho^A)=tr((M\otimes I_B)\rho^{AB})$ since both sides are linear over $\rho^{AB}$.

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  • $\begingroup$ As far as seeing it is true for $\rho^{AB}=A\otimes B$, I get as far as $tr(M\rho_A)=tr(MA)tr(B)$, but not to $tr((M\otimes I_B)A\otimes B)=tr(MA\otimes B)$ $\endgroup$
    – Anna Naden
    Jan 14 at 9:35
  • $\begingroup$ The trace of a tensor product is the product of traces $\endgroup$
    – Danylo Y
    Jan 14 at 9:44
  • $\begingroup$ I looked in the Wikipedia article on Kronecker products and I don't believe it mentioned this property, although I was able to verify it for a pair of 2 by 2 matrices when the second one is the identity $\endgroup$
    – Anna Naden
    Jan 14 at 9:49
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    $\begingroup$ The English version of the wiki article states it under the section "Abstract properties. 1. Spectrum". Though, it can be deduced straight from the definition. The Kronecker product has blocks $a_{ii}B$ on the diagonal. Thus, the sum of all diagonal elements is the sum of $tr(a_{ii}B)$. Which equals to $tr((\sum_i a_{ii}) B) = tr(tr(A)B) = tr(A)tr(B)$. $\endgroup$
    – Danylo Y
    Jan 14 at 10:01
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The equality follows by hitting both sides of

$$ X_A\mathrm{tr}_B(Y_{AB})=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB})\tag1 $$

with the trace and setting $X_A=M$ and $Y_{AB}=\rho^{AB}$.


We can establish $(1)$ for $Y_{AB}$ of the form $Y_{AB}=Y_A\otimes Y_B$ using

$$ \mathrm{tr}_B(A\otimes B)=A\mathrm{tr}(B)\tag2 $$

as follows

$$ \begin{align} X_A\mathrm{tr}_B(Y_{AB})&=X_A\mathrm{tr}_B(Y_A\otimes Y_B)\\ &\stackrel{(2)}=X_AY_A\mathrm{tr}(Y_B)\\ &\stackrel{(2)}=\mathrm{tr}_B((X_AY_A)\otimes Y_B)\\ &=\mathrm{tr}_B((X_A\otimes I_B)(Y_A\otimes Y_B))\\ &=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB}). \end{align}\tag3 $$

We generalize to arbitrary $Y_{AB}$ by linearity of the partial trace and the fact that there exists a basis consisting of operators of the form $Y_A\otimes Y_B$.

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  • $\begingroup$ This gives $tr(\rho^A)=tr(tr_B((M\otimes I_B)\rho^{AB}))$. How does that lead to $tr(\rho^A)=tr((M\otimes I_B)\rho^{AB}$? $\endgroup$
    – Anna Naden
    Jan 14 at 9:25
  • $\begingroup$ I guess it follows from @DaftWullie, below $\endgroup$
    – Anna Naden
    Jan 14 at 9:29
  • $\begingroup$ You're missing $M$ on LHS. The substitution into $(1)$ gives $M\mathrm{tr}_B(\rho^{AB})=\mathrm{tr}_B((M\otimes I_B)\rho^{AB})$ which becomes $\mathrm{tr}(M\mathrm{tr}_B(\rho^{AB}))=\mathrm{tr}((M\otimes I_B)\rho^{AB})$ once you hit both sides with the trace and use $\mathrm{tr}(.)=\mathrm{tr}_A(\mathrm{tr}_B(.))=\mathrm{tr}(\mathrm{tr}_B(.))$ which follows from expansion of trace in a product basis and $\sum_{ij}...=\sum_i\sum_j\dots$. $\endgroup$ Jan 14 at 17:35
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If you're tracing over two systems, $A$ and $B$, you can split this into two steps $$ \text{Tr}(Q_{AB})=\text{Tr}\left(\text{Tr}_B(Q_{AB})\right) $$ (To see this, let the basis you use for taking the first trace be the standard basis, $|ij\rangle$. All I'm doing here is separating out the sums over $i$ and $j$.)

So, if you let $Q_{AB}=\tilde M\rho^{AB}$, then $\text{Tr}_B(\tilde M\rho^{AB})=M\rho_A$ (because the only thing your tracing over is $\rho_{AB}$), and it follows very easily.

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I will give an utter matrix method, to be more clear, the notation $\cdot$ will stand for matrix multiplication.

Mind that $$Tr\left( \left( M\otimes I \right) \rho ^{AB} \right) =\sum_{ij}{\langle ij|\left( M\otimes I \right) \rho ^{AB}|ij\rangle} \\ =\sum_{ij}{\left( \langle i| \right) \cdot \left( I\otimes \langle j| \right) \cdot \left( M\otimes I \right) \cdot \rho ^{AB}\cdot \left( I\otimes |j\rangle \right) \cdot \left( |i\rangle \right)}$$ The second equation can be easy to understand if you think $\langle i|$ is a $1\times n$ matrix and $I$ a $n\times n$ matrix. Then we can have $$ \sum_{ij}{\left( \langle i| \right) \cdot \left( I\otimes \langle j| \right) \cdot \left( M\otimes I \right) \cdot \rho ^{AB}\cdot \left( I\otimes |j\rangle \right) \cdot \left( |i\rangle \right)}\\ =\sum_{ij}{\left( \langle i| \right) \cdot \left( M\otimes \langle j| \right) \cdot \rho ^{AB}\cdot \left( I\otimes |j\rangle \right) \cdot \left( |i\rangle \right)}\\ =\sum_{ij}{\left( \langle i| \right) \cdot \left( M \right) \cdot \left( I\otimes \langle j| \right) \cdot \rho ^{AB}\cdot \left( I\otimes |j\rangle \right) \cdot \left( |i\rangle \right)} \\ =\sum_i{\left( \langle i| \right) \left( M \right)}\rho ^A\left( |i\rangle \right) \\ =Tr\left( M\rho ^A \right).$$

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