# Proof of subadditivity of quantum entropy

1. Aug 15, 2013

### 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(ρAB||ρA$\otimes$ρB) = Tr(ρABlogρAB) - Tr(ρABlogρA$\otimes$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.

2. Aug 15, 2013

### wle

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 $\rho_{\mathrm{B}}$.

3. Aug 16, 2013

### 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 ρAB. I do not want to make the assumption that ρAB = ρA$\otimes$ρB. However, this assumption is made in the literature without comment at times, while one is often adviced to recall that ρAB $\neq$ ρA$\otimes$ρB in general. How should the combined density operator be interpreted? I would accept the logic of operator algebra in this case.

4. Aug 16, 2013

### Narvalen

So I get

Tr(ρABlog(ρA$\otimes$ρB)) = Tr(ρAB(logρA$\otimes$IB + IA$\otimes$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?

5. Aug 17, 2013

### wle

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 $S(\rho_{\mathrm{AB}} \,\Vert\, \rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}) = 0$.

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 $\mathcal{H}_{\mathrm{B}}$.

Last edited: Aug 17, 2013
6. Aug 17, 2013

### kith

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.

7. Aug 19, 2013

### 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:

ρABlog(ρA)$\otimes$$\mathbb{I}$B

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

ρAlogρA$\|\left\{a_{i}\right\}\rangle$ ρB$\mathbb{I}_{B}\|\left\{b_{i}\right\}\rangle$

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

Many thanks. :)