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How do I prove that $$tr_B(\rho^{AB})=\sum_{i}(I_A\otimes \langle i|)\rho^{AB}(I_A\otimes |i\rangle)\\\quad\quad\quad\text{ where }\quad (tr_B(\rho^{AB}))_{aa'}=\sum_{b}\rho^{AB}_{ab,a'b}=\rho^{A}_{aa'}$$

The trace of $\rho^A\in\mathcal{L}(\mathbb{C}^d)$ is defined as $tr(\rho^A)=\sum_i\langle i|\rho^A|i\rangle=\sum_j\langle j|\rho^A|j\rangle$ were $\{i\}$ and $\{j\}$ are any set of orthonormal basis for $\mathbb{C}^d$.

We define the partial trace by extending this idea, except that we only apply it to one subsystem.i.e., for $\rho^{AB}\in\mathcal{L}(\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2})$ we can define

$tr_B(\rho^{AB})=\sum_{i=1}^{d_2}\sum_{i}(I_A\otimes \langle i|)\rho^{AB}(I_A\otimes |i\rangle)=\sum_{j=1}^{d_2}\sum_{j}(I_A\otimes \langle j|)\rho^{AB}(I_A\otimes |j\rangle)$ were $\{i\}$ and $\{j\}$ are any set of orthonormal basis for $\mathbb{C}^{d_2}$.

Note: Refer to Understanding the Expression $tr\Big(\rho(X\otimes I)\Big)=\sum_{a,b,a',b'} \rho_{ab,a'b'}X_{a,a'}\delta_{b,b'}$ for the definition of $\rho_{aba'b'}^{AB}$

Quoting my Reference

Refer to Page 18 part2-pic

In my reference it is defined that $\rho^{A}_{aa'}=(tr_B(\rho^{AB}))_{aa'}=\sum_{b}\rho^{AB}_{ab,a'b}$ but how do I fit it into the original definition $tr_B(\rho^{AB})=\sum_{i=1}^{d_2}\sum_{i}(I_A\otimes \langle i|)\rho^{AB}(I_A\otimes |i\rangle)$ ?

Note : $\rho^{A},\rho^{AB}$ are positive definite.

SOORAJ SOMAN
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1 Answers1

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Write $$ \rho^{AB} = \sum_{a,a',b,b'} \rho^{AB}_{a,a',b,b'} |a\rangle \langle a'| \otimes |b\rangle \langle b'| $$ where we've chosen some orthonormal basis to express the state in. Now we take the partial trace of the $B$ system in the orthonormal basis $\{|b\rangle\}$ that we chose above. Then we have $$ \begin{aligned} \sum_{i}(I_A\otimes \langle i|)\rho^{AB}(I_A\otimes |i\rangle) &=\sum_{i}(I_A\otimes \langle i|)\left(\sum_{a,a',b,b'} \rho^{AB}_{a,a',b,b'} |a\rangle \langle a'| \otimes |b\rangle \langle b'|\right)(I_A\otimes |i\rangle) \\ &= \sum_{a,a',b,b', i} \rho^{AB}_{a,a',b,b'} |a\rangle \langle a'| \otimes \langle i|b\rangle \langle b'|i\rangle \\ &= \sum_{a,a',b,b',i} \rho^{AB}_{a,a',b,b'} |a\rangle \langle a'| \delta_{ib}\delta_{b'i} \\ &= \sum_{a,a',b} \rho^{AB}_{a,a',b,b} |a\rangle \langle a'| \end{aligned} $$ where $\delta_{ij}$ is the Kronecker delta. This is the same as what you get up to how I defined $\rho^{AB}_{a,a',b,b}$ in the definition of the state, you didn't provide a definition in the question so I picked one (it appears to be slightly different from the one your reference used but the logic is exactly the same).

Rammus
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  • I have chosen your definition $\rho^{AB}=\sum_{a,a',b,b'}\rho^{AB}{a,a',b,b'} |a\rangle\langle a'|\otimes |b\rangle\langle b'|$ to prove equation (35) in my reference, which gives $tr(\rho(X\otimes I))=\sum{a,a',b,b'}\rho^{AB}{a,a',b,b'} tr(|a\rangle\langle a'|X)tr(|b\rangle\langle b'|)=\sum{a,a',b,b'}\rho^{AB}{a,a',b,b'} tr(|a\rangle\langle a'|X)\delta{b',b})=\sum_{a,a',b,b'}\rho^{AB}{a,a',b} \langle a'|X|a\rangle$. Can we say that $\langle a'|X|a\rangle=X{a',a}$ for any orthonormal basis ${|a\rangle}$ ? – SOORAJ SOMAN Sep 17 '21 at 12:11
  • If $X = \sum_{a,a'} X_{a',a} |a'\rangle \langle a|$ then yes. When specifying $X_{a,a'}$ you are specifying a coefficient in a matrix once you have fixed some basis. – Rammus Sep 17 '21 at 12:32
  • We are free to choose the standard orthonormal basis as ${|a\rangle}$, right ? Or can there be other possibilities of orthonormal basis which satisfies $X=\sum_{a,a'} X_{a',a}|a'\rangle\langle a|$ ? – SOORAJ SOMAN Sep 17 '21 at 13:02
  • Even in you answer, since $tr_B(\rho^{AB})=\sum_{a,a',b}\rho_{a,a',b,b}|a\rangle\langle a'|$, inorder for $[tr_B(\rho^{AB})]{a,a'}=\sum{a,a',b}\rho_{a,a',b,b}$, we have to choose ${|a\rangle}$ as the standard basis, right ? – SOORAJ SOMAN Sep 17 '21 at 13:39
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    You choose whatever orthonormal basis you want, my comment is just to say that these coefficients (the elements of your matrix) depend on what basis you choose. In my answer ${|a\rangle}$ and ${|b\rangle}$ is whatever basis $\rho_{a,a',b,b'}$ are defined with respect to. Sure, you can think of them as the standard basis if you want to. – Rammus Sep 17 '21 at 13:51