Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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


    User Avatar

    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


    User Avatar
    Science Advisor

    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:


    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. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook