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This is a question about an exercise, but since it about a generalization of a definition, I think it can be useful for someone else. The problem is an exercise from Nielsen's and Chuang's Quantum Computation and Quantum Information book, the exercise 8.3. I could get most of it answered but there is a problem in the end, as I will show. Also, this is my first question here, so if there is something I could have written in a better way, please tell me. Any help is appreciated!

The section of the book starts by defining the operator-sum representation for a quantum operation of a system in the state $\rho$ that gets in contact with an environment in the state $\lvert e_0 \rangle$. The whole system evolves accordingly to an operator $U$ and then we discard the environment, through a partial trace over the environment space. Then, if $\lvert e_k\rangle$ is a basis for the environment space, the system of interest ends in the state

$$\varepsilon(\rho) = \sum_k \langle e_k\rvert U [\rho \otimes \lvert e_0 \rangle \langle e_0 \rvert]U^\dagger \lvert e_k \rangle = \sum_k E_k \rho E_k^\dagger$$

Then the problem is to show that this holds also for the more general process when the system of interest is a composite system AB, in the state $\rho$, and it gets in contact with an environment that is also composite CD, initially in the state $\lvert 0\rangle$. The whole system evolves accordingly to $U$ and then we discard A and D, through a partial trace in A and D, so that

$$ \varepsilon(\rho) =\rho'= Tr_{AD}(U [\rho \lvert 0 \rangle \langle 0 \rvert]U^\dagger) = \sum_k E_k \rho E_k^\dagger $$ Where $E_k$ is an operator from AB to BC and $\sum E_k^\dagger E_k=I$. This completeness relation for $E_k$ is my problem, I can't show that it is true. Here is what I did:

  1. Throughout the calculations $\lvert x_Z \rangle$ is always an element $\lvert x \rangle$ of the basis of the space Z. I expanded the partial trace as $$ \sum_{a,b} \langle a_A \rvert \langle b_D \rvert (U \rho \lvert 0 \rangle \langle 0 \rvert U^\dagger) \lvert b_D \rangle \lvert a_A \rangle $$

  2. Keeping in mind that objects in different spaces (e.g. one in A and another in B) commute, rearranged it to $$ \sum_{a,b} \langle a_A \rvert \langle b_D \rvert U \lvert 0 \rangle \rho \langle 0 \rvert U^\dagger \lvert b_D \rangle \lvert a_A \rangle $$

  3. I expanded $\rho$ and $\lvert 0 \rangle$ as $$ \rho = \sum_{ijkl} \alpha_{ij} \alpha_{kl}^* \lvert i_A \rangle \lvert j_B \rangle \langle k_A \rvert \langle l_B \rvert $$ $$ \lvert 0 \rangle = \sum_{mn} \beta_{mn} \lvert m_C \rangle \lvert n_D \rangle $$

  4. And substitute these $\rho$ and $\lvert 0 \rangle$ into $\varepsilon(\rho)$ obtaining $$ \sum_{a,b} \sum_{mnop} \sum_{ijkl} \alpha_{ij} \alpha_{kl}^* \beta_{mn} \beta_{op}^* \langle a_A \rvert \langle b_D \rvert U \lvert m_C \rangle \lvert n_C \rangle \lvert i_A \rangle \lvert j_B \rangle \langle k_A \rvert \langle l_B \rvert \langle o_C \rvert \langle p_D \rvert U^\dagger \lvert a_A \rangle \lvert b_D \rangle $$

  5. Remembering who commutes with who, rearranged it to

$$ \sum_{a,b}\bigg( \underbrace{\sum_{mn} \beta_{mn} \langle a_A \rvert \langle b_D \rvert U \lvert m_C \rangle \lvert n_D \rangle}_{=E_{a,b}} \bigg) \bigg( \underbrace{\sum_{ijkl} \alpha_{ij} \alpha_{kl}^* \lvert i_A \rangle \lvert j_B \rangle \langle k_A \rvert \langle l_B \rvert }_{=\rho} \bigg) \bigg( \underbrace{\sum_{op} \beta_{op}^* \langle o_C \rvert \langle p_D \rvert U^\dagger \lvert a_A \rangle \lvert b_D \rangle}_{=E_{a,b}^\dagger} \bigg) $$

  1. Then I simply argued that for each pair (a,b) we can associate a number k and write $\varepsilon(\rho) = \sum_k E_k \rho E_k^\dagger$

OK, with these type of reasoning, I could see the operator-sum representation and also I can see $E_k$ is an operator from AB to BC by expanding U in a similiar way to everything that was done before.

$$ U = \sum_{cdef\\ghij} \omega_{cdef\\ghij} \lvert c_A \rangle \lvert d_B \rangle \lvert e_C \rangle \lvert f_D \rangle \langle g_A \rvert \langle h_B \rvert \langle i_C \rvert \langle j_D \rvert $$

Inserting this into the expression for $E_{a,b}$ that we obtained before, we see clearly that $E_{a,b}$ is a sum of operators from AB to BC:

$$ E_{a,b} = \sum_{mn}\sum_{degh} \beta_{mn} \omega_{adeb\\ghmn} \lvert d_B \rangle \lvert_C \rangle \langle g_A \rvert \langle h_B \rvert $$

So these calculations were useful until here, but when I try to do analogously to show that $\sum_k E_k E_k^\dagger = I$ I get to nowhere.

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  • $\begingroup$ Shouldn't the first step be $ \rho \otimes |0\rangle \langle0| $ ? And not $ \rho $ applied on $ |0\rangle \langle 0| $ ? $\endgroup$ – Mahathi Vempati Jan 17 at 14:20
  • $\begingroup$ And the second step, why should tensor products in different spaces commute? $\endgroup$ – Mahathi Vempati Jan 17 at 14:24
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$\newcommand{\ket}[1]{\left|#1\right>}$ $\newcommand{\bra}[1]{\left<#1\right|}$ $\newcommand{\abs}[1]{\left|#1\right|}$ $\newcommand{\I}{\mathrm{I}}$

Using your notation $$ E_k=E_{a,b}=\sum_{m,n} \beta_{mn}\bra{a_A b_D}U\ket{m_C n_D},$$

So, $$ \sum_{k} E_k^\dagger E_k = \sum_{m,n} \sum_{m',n'} \beta^*_{m'n'} \beta_{m n} \bra{m_C' n_D'}U^\dagger \underbrace{\left(\sum_{a,b} \ket{a_A b_D}\bra{a_A b_D}\right)}_{\I_{AD}} U \ket{m_C n_D} \\ =\sum_{m,n} \sum_{m',n'} \beta^*_{m'n'} \beta_{m n} \bra{m_C' n_D'}U^\dagger \I_{AD}U \ket{m_C n_D}=\sum_{m,n} \sum_{m',n'} \beta^*_{m'n'} \beta_{m n} \bra{m_C' n_D'} \I\ket{m_C n_D}\\ =\I \sum_{m,n} \abs{\beta_{mn}}^2 = \I. $$

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  • $\begingroup$ It might be more useful, going forwards, to add some details to what you're doing, rather than posting just the maths. $\endgroup$ – Kyle Kanos Apr 16 '18 at 9:56

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