Partial trace of action on density matrix












1












$begingroup$


This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: mathcal{N} rightarrow mathcal{N} otimes mathcal{F}$, and we define the superoperator $V cdot V^dagger: rho mapsto Vrho V^dagger$, what is the trace over $mathcal{F}$ of $Vrho V^dagger$? I did a computation which seems to show that $Tr_{mathcal{F}}(Vrho V^dagger) = Tr_mathcal{F}(rho)$, but I don't think this result is right. Where did I go wrong in the following argument?



Let ${phi_i}$ be an orthonormal basis for $mathcal{F}$. Then



begin{equation*}
begin{split}
Tr_{mathcal{F}}(Vrho V^dagger) & = Tr_{mathcal{F}}(V^dagger Vrho) \
& = sum_ilanglephi_i,V^dagger Vrhophi_irangle \
& = sum_ilangle Vphi_i,Vrhophi_irangle \
& = sum_ilanglephi_i,rhophi_irangle \
& = Tr_mathcal{F}(rho)
end{split}
end{equation*}
where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.






functional-analysis quantum-computation quantum-information







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    1












    $begingroup$


    This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: mathcal{N} rightarrow mathcal{N} otimes mathcal{F}$, and we define the superoperator $V cdot V^dagger: rho mapsto Vrho V^dagger$, what is the trace over $mathcal{F}$ of $Vrho V^dagger$? I did a computation which seems to show that $Tr_{mathcal{F}}(Vrho V^dagger) = Tr_mathcal{F}(rho)$, but I don't think this result is right. Where did I go wrong in the following argument?



    Let ${phi_i}$ be an orthonormal basis for $mathcal{F}$. Then



    begin{equation*}
    begin{split}
    Tr_{mathcal{F}}(Vrho V^dagger) & = Tr_{mathcal{F}}(V^dagger Vrho) \
    & = sum_ilanglephi_i,V^dagger Vrhophi_irangle \
    & = sum_ilangle Vphi_i,Vrhophi_irangle \
    & = sum_ilanglephi_i,rhophi_irangle \
    & = Tr_mathcal{F}(rho)
    end{split}
    end{equation*}
    where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.






    functional-analysis quantum-computation quantum-information







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: mathcal{N} rightarrow mathcal{N} otimes mathcal{F}$, and we define the superoperator $V cdot V^dagger: rho mapsto Vrho V^dagger$, what is the trace over $mathcal{F}$ of $Vrho V^dagger$? I did a computation which seems to show that $Tr_{mathcal{F}}(Vrho V^dagger) = Tr_mathcal{F}(rho)$, but I don't think this result is right. Where did I go wrong in the following argument?



      Let ${phi_i}$ be an orthonormal basis for $mathcal{F}$. Then



      begin{equation*}
      begin{split}
      Tr_{mathcal{F}}(Vrho V^dagger) & = Tr_{mathcal{F}}(V^dagger Vrho) \
      & = sum_ilanglephi_i,V^dagger Vrhophi_irangle \
      & = sum_ilangle Vphi_i,Vrhophi_irangle \
      & = sum_ilanglephi_i,rhophi_irangle \
      & = Tr_mathcal{F}(rho)
      end{split}
      end{equation*}
      where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.






      functional-analysis quantum-computation quantum-information







      share|cite|improve this question









      $endgroup$




      This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: mathcal{N} rightarrow mathcal{N} otimes mathcal{F}$, and we define the superoperator $V cdot V^dagger: rho mapsto Vrho V^dagger$, what is the trace over $mathcal{F}$ of $Vrho V^dagger$? I did a computation which seems to show that $Tr_{mathcal{F}}(Vrho V^dagger) = Tr_mathcal{F}(rho)$, but I don't think this result is right. Where did I go wrong in the following argument?



      Let ${phi_i}$ be an orthonormal basis for $mathcal{F}$. Then



      begin{equation*}
      begin{split}
      Tr_{mathcal{F}}(Vrho V^dagger) & = Tr_{mathcal{F}}(V^dagger Vrho) \
      & = sum_ilanglephi_i,V^dagger Vrhophi_irangle \
      & = sum_ilangle Vphi_i,Vrhophi_irangle \
      & = sum_ilanglephi_i,rhophi_irangle \
      & = Tr_mathcal{F}(rho)
      end{split}
      end{equation*}
      where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.






      functional-analysis quantum-computation quantum-information




      functional-analysis quantum-computation quantum-information






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 24 '18 at 16:51









      CameronCameron

      285




      285






















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          $begingroup$

          There are a number of things wrong with your computation.




          1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.


          2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


          3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.


          4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.



          In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
            $endgroup$
            – Cameron
            Nov 26 '18 at 1:23











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          1 Answer
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          $begingroup$

          There are a number of things wrong with your computation.




          1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.


          2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


          3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.


          4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.



          In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
            $endgroup$
            – Cameron
            Nov 26 '18 at 1:23
















          3












          $begingroup$

          There are a number of things wrong with your computation.




          1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.


          2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


          3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.


          4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.



          In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
            $endgroup$
            – Cameron
            Nov 26 '18 at 1:23














          3












          3








          3





          $begingroup$

          There are a number of things wrong with your computation.




          1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.


          2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


          3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.


          4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.



          In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.






          share|cite|improve this answer









          $endgroup$



          There are a number of things wrong with your computation.




          1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.


          2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.


          3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.


          4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.



          In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 14:05









          luftbahnfahrerluftbahnfahrer

          623314




          623314












          • $begingroup$
            Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
            $endgroup$
            – Cameron
            Nov 26 '18 at 1:23


















          • $begingroup$
            Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
            $endgroup$
            – Cameron
            Nov 26 '18 at 1:23
















          $begingroup$
          Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
          $endgroup$
          – Cameron
          Nov 26 '18 at 1:23




          $begingroup$
          Thanks - seems like a lot of the confusion is from a misunderstanding of partial trace. I'll definitely look into that.
          $endgroup$
          – Cameron
          Nov 26 '18 at 1:23


















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