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Proof of subadditivity of quantum entropy

  1. Aug 15, 2013 #1
    I'm having problems understanding the trace of tensor products when the density matrix is expressed in its reduced density operators. The proof of subadditivity is quite simple.

    S(ρAB||ρA[itex]\otimes[/itex]ρB) = Tr(ρABlogρAB) - Tr(ρABlogρA[itex]\otimes[/itex]ρB) = TrABABlogρAB) - Tr(ρAlogρA) - Tr(ρBlogρB)

    This carries on to finalize the proof. But this last step is the step where I'm at a loss. How is the trace over AB (the second term in the first step) expanded into the respective partial traces over A and B (the second and third term in the last step)?

    Here, ρAB is a density operator acting on the Hilbert space of the bipartite system and the rest should be self-explanatory.

    Please help!
     
  2. jcsd
  3. Aug 15, 2013 #2

    wle

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    You use that [itex]\log(\rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}) = \log(\rho_{\mathrm{A}}) \otimes \mathbb{I}_{\mathrm{B}} + \mathbb{I}_{\mathrm{A}} \otimes \log(\rho_{\mathrm{B}})[/itex], where [itex]\mathbb{I}[/itex] is the identity operator. You can easily verify this by diagonalising [itex]\rho_{\mathrm{A}}[/itex] and [itex]\rho_{\mathrm{B}}[/itex].
     
  4. Aug 16, 2013 #3
    Thank you for your response, it was very helpful. Back to basics of tensor product relations. My next problem is with the interpretation of ρAB. I do not want to make the assumption that ρAB = ρA[itex]\otimes[/itex]ρB. However, this assumption is made in the literature without comment at times, while one is often adviced to recall that ρAB [itex]\neq[/itex] ρA[itex]\otimes[/itex]ρB in general. How should the combined density operator be interpreted? I would accept the logic of operator algebra in this case.
     
  5. Aug 16, 2013 #4
    So I get

    Tr(ρABlog(ρA[itex]\otimes[/itex]ρB)) = Tr(ρAB(logρA[itex]\otimes[/itex]IB + IA[itex]\otimes[/itex]logρB))

    This is obviously wrong if you exploit the trace of a tensor product and the trace of an identity matrix which is just the number of dimensionality of each vector space. Where am I loosing it?
     
  6. Aug 17, 2013 #5

    wle

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    It is not being assumed that [itex]\rho_{\mathrm{AB}} = \rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}[/itex]. [itex]\rho_{\mathrm{A}}[/itex] and [itex]\rho_{\mathrm{B}}[/itex] are simply the reduced density operators defined by, e.g., [itex]\rho_{\mathrm{A}} = \mathrm{Tr}_{\mathrm{B}}[\rho_{\mathrm{AB}}][/itex].

    You can think of the relative entropy [itex]S(\rho_{\mathrm{AB}} \,\Vert\, \rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}})[/itex] as a measure of how different [itex]\rho_{\mathrm{AB}}[/itex] is from the product state [itex]\rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}[/itex]. If [itex]\rho_{\mathrm{AB}}[/itex] is a product state, then [itex]S(\rho_{\mathrm{AB}} \,\Vert\, \rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}) = 0[/itex].


    I don't follow. The second part of the equation you write can be explicitly split into the sum of two terms:
    [tex]\mathrm{Tr} [ \rho_{\mathrm{AB}} ( \log(\rho_{\mathrm{A}}) \otimes \mathbb{I}_{\mathrm{B}} ) ] + \mathrm{Tr} [ \rho_{\mathrm{AB}} ( \mathbb{I}_{\mathrm{A}} \otimes \log(\rho_{\mathrm{B}}) ) ] \,.[/tex]
    In general, you can always write the trace on some Hilbert space [itex]\mathcal{H}_{\mathrm{A}} \otimes \mathcal{H}_{\mathrm{B}}[/itex] as [itex]\mathrm{Tr}[\,\cdot\,] = \mathrm{Tr}_{\mathrm{A}} \{ \mathrm{Tr}_{\mathrm{B}} [ \,\cdot\, ] \}[/itex]. Working on the first term for instance, you should find
    [tex]\mathrm{Tr}_{\mathrm{B}}[ \rho_{\mathrm{AB}} ( \log(\rho_{\mathrm{A}}) \otimes \mathbb{I}_{\mathrm{B}} ) ] = \rho_{\mathrm{A}} \log(\rho_{\mathrm{A}})[/tex]
    with [itex]\rho_{\mathrm{A}} = \mathrm{Tr}_{\mathrm{B}}[\rho_{\mathrm{AB}}][/itex]. If this isn't obvious to you, you can always see it just by writing [itex]\mathrm{Tr}_{\mathrm{B}}[/itex] explicitly in terms of a sum over any orthonormal basis [itex]\{ \lvert n \rangle_{\mathrm{B}} \}[/itex] of [itex]\mathcal{H}_{\mathrm{B}}[/itex].
     
    Last edited: Aug 17, 2013
  7. Aug 17, 2013 #6

    kith

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    It is the density operator of the combined physical system. If it isn't seperable, you have entanglement. Then, performing a measurement on subsystem A may influence subsystem B.
     
  8. Aug 19, 2013 #7
    Thank you both for your response. I managed to clear the logic after some time. It's really annoying to get stuck like that. I started thinking in terms of operators and then, progress.

    I had a really hard time interpreting what this really means in terms of vector spaces:

    ρABlog(ρA)[itex]\otimes[/itex][itex]\mathbb{I}[/itex]B

    I decided to think of ρAB as expressed in terms of the A and B bases. Sort of like this:

    ρAlogρA[itex]\|\left\{a_{i}\right\}\rangle[/itex] ρB[itex]\mathbb{I}_{B}\|\left\{b_{i}\right\}\rangle[/itex]

    Not sure if I brained it correctly (please correct me if I'm wrong) but at least I can proceed.

    Many thanks. :)
     
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