Proof of subadditivity of quantum entropy

[Narvalen]

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(ρ_AB||ρ_A\otimesρ_B) = Tr(ρ_ABlogρ_AB) - Tr(ρ_ABlogρ_A\otimesρ_B) = Tr_AB(ρ_ABlogρ_AB) - Tr(ρ_Alogρ_A) - Tr(ρ_Blogρ_B)

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

Please help!

 

[wle]

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)?

You use that \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}}), where \mathbb{I} is the identity operator. You can easily verify this by diagonalising \rho_{\mathrm{A}} and \rho_{\mathrm{B}}.

 

[Narvalen]

Thank you for your response, it was very helpful. Back to basics of tensor product relations. My next problem is with the interpretation of ρ_AB. I do not want to make the assumption that ρ_AB = ρ_A\otimesρ_B. However, this assumption is made in the literature without comment at times, while one is often adviced to recall that ρ_AB \neq ρ_A\otimesρ_B in general. How should the combined density operator be interpreted? I would accept the logic of operator algebra in this case.

 

[Narvalen]

So I get

Tr(ρ_ABlog(ρ_A\otimesρ_B)) = Tr(ρ_AB(logρ_A\otimesI_B + I_A\otimeslogρ_B))

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?

 

[wle]

Thank you for your response, it was very helpful. Back to basics of tensor product relations. My next problem is with the interpretation of ρ_AB. I do not want to make the assumption that ρ_AB = ρ_A\otimesρ_B.

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

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

So I get

Tr(ρ_ABlog(ρ_A\otimesρ_B)) = Tr(ρ_AB(logρ_A\otimesI_B + I_A\otimeslogρ_B))

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?

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

 

[kith]

How should the combined density operator be interpreted?

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.

 

[Narvalen]

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:

ρ_ABlog(ρ_A)\otimes\mathbb{I}_B

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

ρ_Alogρ_A\|\left\{a_{i}\right\}\rangle ρ_B\mathbb{I}_{B}\|\left\{b_{i}\right\}\rangle

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

Many thanks. :)