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Can someone help me with the following question?

Let $M$ be a general operator on the composite system $\mathcal{H}_A\otimes \mathcal{H}_B$ and let $O_A$ be an operator on $\mathcal{H}_A$. Using the definition of a partial trace that was given in class, together with the properties of the full trace, prove that $Tr_B(O_AM) = O_ATr_B(M)$.

I got stuck here:

$$M = \sum\limits_{\alpha}M^{(A)}_{\alpha}\otimes M^{(B)}_{\alpha}; O_AM = (O\otimes\mathbb{1}_B)M = \sum\limits_{\alpha}OM_\alpha^{(A)}\otimes M_\alpha^{(B)}\rightarrow$$ $$Tr_B(O_AM) = Tr_B\left(\sum\limits_{\alpha} OM_\alpha^{(A)}\otimes M_\alpha^{(B)}\right)=\sum\limits_\alpha Tr_B\left(OM_\alpha^{(A)}\otimes M_\alpha^{(B)}\right)$$$$=\sum\limits_\alpha OM_\alpha^{(A)}Tr\left(M_\alpha^{(B)}\right)$$


Editor's note: the $\mathbb{1}_B$ actually looked more like $\mathbb{R}$ except with a 1; however, I couldn't get the font to work properly in mathjax. Apologies.

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    $\begingroup$ Hi, welcome to QCSE. This clearly sounds like homework, which, while not necessarily disallowed at this site, is generally disfavored. I recommend you revise your question. Start by rewriting it in Latex, and then indicating clearly what you have done, and where you are stuck. $\endgroup$ – Mark S Apr 24 '20 at 21:31
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Firstly, just because an operator $M$ is defined on a composite space, it does not mean that the operator itself has a tensor product structure. You need $$ M=\sum_{i,j,k,l}M_{ij,kl}|ij\rangle\langle kl| $$ Now let $$ O=\sum_{i,k}O_{i,k}|i\rangle\langle k|. $$ With these in place, you just calculate the two sides of the equation and see if they're the same. (In a strict mathematical sense, this is not the best way to present the answer, but it's the best way to start to understand it before working out a sequence so that you can start from the left-hand side and finally arrive at the right-hand side.)

So, I can work out the product \begin{align*} \Phi=(O\otimes I)M&=\left(\sum_{i,k}O_{i,k}|i\rangle\langle k|\otimes\sum_j|j\rangle\langle j|\right)\left(\sum_{j,k,l,m}M_{kj,ml}|kj\rangle\langle ml|\right) \\ &=\left(\sum_{i,j,k}O_{i,k}|ij\rangle\langle kj|\right)\left(\sum_{j,k,l,m}M_{kj,ml}|kj\rangle\langle ml|\right) \\ &=\sum_{i,j,k,l,m}O_{i,k}M_{kj,ml}|ij\rangle\langle ml|. \end{align*} Taking the partial trace of this, I get $$ \text{Tr}_B\Phi=\sum_{i,j,k,m}O_{i,k}M_{kj,mj}|i\rangle\langle m| $$ (basically summing over indices $j=l$).

Now, I need to compare this to $$ O\cdot\text{Tr}_B(M)=\left(\sum_{i,k}O_{i,k}|i\rangle\langle k|\right)\left(\sum_{j,k,m}M_{kj,mj}|k\rangle\langle m|\right)=\sum_{i,j,k,m}O_{i,k}M_{kj,mj}|i\rangle\langle m|, $$ so the two are the same.

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  • $\begingroup$ First of all thanks. Second i only have problem with :O⋅TrB(M) because you ignored the fact that it operates only on the A system and then it should be :(O⊗I) and that is where i am stuck $\endgroup$ – Vladimir kozlov Apr 25 '20 at 13:12
  • $\begingroup$ @Vladimirkozlov Made an edit to more explicitly show taking identity into account. $\endgroup$ – DaftWullie Apr 27 '20 at 8:40
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A way that's perhaps more compact (if less terse), we can write

$$ \mathrm{Tr}_B(O_AM)=\mathrm{Tr}_B((O_A\otimes \mathbb 1_B)M)=\sum_i (\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B)M(\mathbb 1_A\otimes |i\rangle_B)=\\=\sum_i O_A(\mathbb 1_A\otimes \langle i|_B) M(\mathbb 1_A\otimes |i\rangle_B)=O_A\mathrm{Tr}_B(M)$$

The only nontrivial thing I used is $(\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B)=O_A(\mathbb 1_A\otimes \langle i|_B)$, it's pretty clearly true, but to see it explicitely just write

$$(\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B)= (\mathbb 1_A\otimes \langle i|_B) O_A\otimes\sum_j |j\rangle\langle j|=O_A\otimes\langle i|_B=O_A(\mathbb 1_A\otimes\langle i|_B) $$

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  • $\begingroup$ It's even simpler: $(\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B) = O_A\otimes \langle i|_B$ due to mixed-product rule. Also I wouldn't use the same symbol for $O_A$ and $O_A \otimes \mathbb 1_B$, it's confusing $\endgroup$ – Danylo Y Apr 28 '20 at 10:40

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