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I was studying Alexander Holevo's book on Quantum Systems, Channels and Information, which is a small but terse introduction to Quantum Information Theory. The theory of Tensor products is used almost everywhere so I decided to study the same topic from different books. There is one property that I find hard to prove and I hope that someone could offer me some assistance in proving it.

N.B: As math stackexchange people may not be comfortable with the physics style notations (a.k.a Dirac's bra and ket notation), I'll use notations from John Conway Functional Analysis. If requested, I'll rephrase the question in Dirac's notation.

Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two finite dimensional Hilbert spaces (over $\mathbb{C}$) with inner products $\langle.\rangle_1$ and $\langle.\rangle_2$ respectively.

Let $\{e_i\}$ and $\{f_j\}$ be bases for $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively.

Let $T \in \mathcal{L}(\mathcal{H}_1 \otimes\mathcal{H}_2)$, where $\otimes$ represents the tensor product. Then for any operator $S \in \mathcal{L}(\mathcal{H}_2)$, show that $$Tr_{\mathcal{H}_2}(T(I_{H_1}\otimes S)) = Tr_{\mathcal{H}_2}((I_{H_1}\otimes S)T)$$

where $Tr_{H_2}(A)$ for any operator $A \in \mathcal{L}(\mathcal{H}_1 \otimes\mathcal{H}_2)$, called the partial trace of $A$ w.r.t $\mathcal{H}_2$, is defined as that operator on $\mathcal{H}_1$ that satisfies the following (this is as per Holevo's book)

$$\langle\phi,Tr_{H_2}(A) \psi\rangle_1 = \sum_{j}\langle\phi \otimes f_j,A (\psi \otimes f_j)\rangle$$ for every $\phi,\psi \in \mathcal{H}_1$. The inner product on RHS is the inner product for $\mathcal{H}_1 \otimes \mathcal{H}_2$ and is given by $$\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle = \langle \phi_1, \psi_1 \rangle_1 \langle \phi_2, \psi_2 \rangle_2$$ for any $\phi_i, \psi_i \in \mathcal{H}_i;$ $i=1,2$.

It is somewhat similar to showing $Tr(AB)=Tr(BA)$ for the usual trace. However I had a lot of trouble showing this. Also I wanted to know if it suffices to prove the result for $\phi = \psi$ as I think I've managed some kind of proof for that part.

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  • $\begingroup$ It would not suffice to prove it for $\phi=\psi$. It would suffice to prove it for $\phi=e_i$ and $\psi=e_j$. $\endgroup$ – luftbahnfahrer Mar 22 '17 at 21:35
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I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $\mathcal{L}(\mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $\mathcal{H}$ is defined as $$ \langle A,B\rangle = \operatorname{Tr}(AB^*) $$ for all $A,B\in\mathcal{L}(\mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^\dagger$ in quantum mechanics notation). Note that $$ A=B \quad\Longleftrightarrow\quad \langle A,X\rangle = \langle B,X\rangle \quad \text{for all }X\in\mathcal{L}(\mathcal{H}) $$ for operators $A,B\in\mathcal{L}(\mathcal{H})$. With this inner product, note that $$ \langle XY,Z\rangle = \langle X,ZY^*\rangle = \langle Y,X^*Z\rangle $$ for any $X,Y,Z\in\mathcal{L}(\mathcal{H})$

Let $A\in\mathcal{L}(\mathcal{H}_1\otimes\mathcal{H}_2)$. The partial trace $\operatorname{Tr}_{\mathcal{H}_2}$ can then be defined as the unique operator such that $$ \langle \operatorname{Tr}_{\mathcal{H}_2}(A), X\rangle = \langle A,X\otimes I_{\mathcal{H}_2}\rangle \quad \text{for all }X\in\mathcal{L}(\mathcal{H_1}). $$ We can then prove the following proposition. It relies on the fact that $X\otimes I_{\mathcal{H}_2}$ commutes with $I_{\mathcal{H_1}}\otimes S^*$.

Proposition. Let $T\in\mathcal{L}(\mathcal{H}_1\otimes\mathcal{H}_2)$ and $S\in\mathcal{L}(\mathcal{H_2})$. Then $\operatorname{Tr}_{\mathcal{H_2}}(T(I_{\mathcal{H}_1}\otimes S)) = \operatorname{Tr}_{\mathcal{H_2}}((I_{\mathcal{H}_1}\otimes S)T)$.

Proof. Let $X\in\mathcal{L}(\mathcal{H_1})$. Then \begin{align*} \langle \operatorname{Tr}_{\mathcal{H_2}}(T(I_{\mathcal{H}_1}\otimes S)) , X\rangle & = \langle T(I_{\mathcal{H}_1}\otimes S), X\otimes I_{\mathcal{H_2}}\rangle\\ & = \langle T, (X\otimes I_{\mathcal{H_2}})(I_{\mathcal{H}_1}\otimes S)^*\rangle\\ %& = \langle T, X\otimes S^*\rangle\\ & = \langle T, (I_{\mathcal{H}_1}\otimes S)^*(X\otimes I_{\mathcal{H_2}})\rangle\\ & = \langle (I_{\mathcal{H}_1}\otimes S)T,X\otimes I_{\mathcal{H_2}}\rangle\\ & = \langle \operatorname{Tr}_{\mathcal{H_2}}((I_{\mathcal{H}_1}\otimes S)T) , X\rangle . \end{align*} Since this holds for all $X$, the result follows.

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  • $\begingroup$ I believe the key statement is "$X \otimes I_{\mathcal{H}_2}$" commutes with... This answer is useful, so let me try again. $\endgroup$ – Gautam Shenoy Mar 23 '17 at 5:22
  • $\begingroup$ Kindly go over my answer. Open for comments. $\endgroup$ – Gautam Shenoy Mar 23 '17 at 6:16
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I think I was able to solve it. This assumes that any operator $T \in \mathcal{L}(\mathcal{H}_1 \otimes \mathcal{H}_2)$ can be written in terms of basis operators

$$T = \sum_{i,j,k,l} \alpha_{i,j,k,l} E_{ij}\otimes F_{kl}$$

where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} \in \mathcal{L}(\mathcal{H}_1)$ and $F_{k,l}\in \mathcal{L}(\mathcal{H}_2)$, such that $$E_{ij}: e_{i} \mapsto e_j$$ and $$F_{kl}: f_{k} \mapsto f_l $$ the other basis vectors being mapped to $0$. We also need to use that (I've proved this before) $$Tr_{\mathcal{H}_2}(A\otimes B) = A. Tr(B)$$ where $A \in \mathcal{L}(\mathcal{H}_1)$ and $B\in \mathcal{L}(\mathcal{H}_2)$.

Now it's easy. Just observe $$Tr_{\mathcal{H}_2}(T(I_{\mathcal{H}_2}\otimes S)) = Tr_{\mathcal{H}_2}(\sum_{i,j,k,l} \alpha_{i,j,k,l} (E_{ij}\otimes F_{kl}) (I_{\mathcal{H}_2}\otimes S)) \\ = \sum_{i,j,k,l} \alpha_{i,j,k,l} Tr_{\mathcal{H}_2}(E_{ij}\otimes F_{kl} S)\\ = \sum_{i,j,k,l} \alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$

Similarly we get $$Tr_{\mathcal{H}_2}((I_{\mathcal{H}_2}\otimes S)T) = \sum_{i,j,k,l} \alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$

But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.

Request everyone to check the steps.

Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|a\otimes b\rangle \langle c \otimes d| = |a\rangle\langle c| \otimes |b\rangle\langle d|$.

Let $e_k$ and $f_j$ be orthonormal basis for $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively. Then we have, \begin{eqnarray} &&\sum_k \langle\phi \otimes f_k~|(I_{H_1} \otimes S)T|~\psi \otimes f_k\rangle \\ &=& \sum_k \langle\phi \otimes f_k~|(I_{H_1} \otimes S)(I_{H_1}\otimes I_{H_2})T|~\psi \otimes f_k\rangle \\ &=& \sum_{k,i,j} \langle\phi \otimes f_k~|(I_{H_1} \otimes S)(|e_i\rangle\langle e_i| \otimes |f_j\rangle\langle f_j|)T|~\psi \otimes f_k\rangle \\ &=& \sum_{k,i,j} \langle\phi \otimes f_k~|(I_{H_1} \otimes S)(|e_i\otimes f_j\rangle \langle e_i \otimes f_j |)T|~\psi \otimes f_k\rangle \\ &=& \sum_{k,i,j} \langle\phi \otimes f_k~||e_i\otimes Sf_j\rangle \langle e_i \otimes f_j |T|~\psi \otimes f_k\rangle \\ &=& \sum_{k,i,j} \langle\phi |e_i\rangle \langle f_k| Sf_j\rangle \langle e_i \otimes f_j| T|~\psi \otimes f_k\rangle \\ &=& \sum_{k,i,j} \langle\phi |e_i\rangle \langle e_i \otimes f_j| T|~\psi \otimes f_k\rangle \langle f_k| Sf_j\rangle\\ &=& \sum_{k,i,j} \langle\phi\otimes f_j |e_i\otimes f_j\rangle \langle e_i \otimes f_j| T|~\psi \otimes f_k\rangle \left(\frac{\langle \psi \otimes f_k||\psi \otimes Sf_j\rangle}{\|\psi\|^2}\right)\\ &=& \sum_{k,i,j} \langle\phi\otimes f_j|(|e_i\rangle\langle e_i| \otimes |f_j\rangle \langle f_j|) T~(|\psi\rangle\langle\psi|\otimes |f_k\rangle\langle f_k|)\left(\frac{|\psi \otimes Sf_j\rangle}{\|\psi\|^2}\right) \\ &=& \sum_{j} \langle\phi\otimes f_j|(I_{H_1} \otimes |f_j\rangle \langle f_j|) T~(|\psi\rangle\langle\psi|\otimes I_{H_2})\left(\frac{(I_{H_1}\otimes S)|\psi \otimes f_j\rangle}{\|\psi\|^2}\right) \\ &\stackrel{(a)}{=}& \sum_{j} \langle\phi\otimes f_j| T (I_{H_1}\otimes S)\left(\frac{(|\psi\rangle\langle\psi|\otimes I_{H_2})|\psi \otimes f_j\rangle}{\|\psi\|^2}\right) \\ &=& \sum_{j} \langle\phi\otimes f_j| T (I_{H_1}\otimes S)|\psi \otimes f_j\rangle \\ \end{eqnarray} where $(a)$ follows from, $$\langle \phi\otimes f_j|(I_{H_1} \otimes |f_j\rangle \langle f_j|) = \langle(I_{H_1} \otimes |f_j\rangle \langle f_j|)^*(\phi\otimes f_j)| = \langle \phi\otimes f_j|$$ where we used the self adjointness of the operator. Also $\langle f_j|f_k \rangle = \delta_{j,k}$. $\blacksquare$

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  • $\begingroup$ Your proof looks good too. $\endgroup$ – luftbahnfahrer Mar 23 '17 at 20:14

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