Entropy inequality of joint von Neumann entropy given some marginal inequalities

Question

Assume that I have a pure, multipartite state $\omega_{ABCD}$ and a unitary $U(\omega_{ABCD})=\tau_{ABCD}$. The effect of $U$ on $\omega$ results in $$ S(\omega_B)<S(\tau_B) $$ and $$ S(\omega_A)>S(\tau_A). $$ $S$ is the von Neumann entropy and the states $\omega_A,\tau_A$ etc are the marginal density matrices (for instance, $\omega_A$= $Tr_{BCD}{\omega_{ABCD}}$). This effect of $U$ is an assumption, I don't say it always holds.

QUESTION: Comparing the joint entropies $S(\omega_{AB})$ and $S(\tau_{AB})$, is this true $$ S(\tau_{AB})-S(\omega_{AB})\leq S(\tau_{B})-S(\omega_{B}) $$ and if yes, why?

EDIT: Another assumption is that $S(\omega_{AB})\geq S(\omega_B)$.

Answer 1

Why should this be true?

Let the dimensions of $A, B, C, D$ all be the same.

Let $\omega_{AB}$ and $\omega_{CD}$ both be pure states, with $$S(\omega_A) = S(\omega_B) = S(\omega_C) = S(\omega_D).$$

Now, let $U$ perform a cyclic permutation on $A, B, C, D$.

We have $S(\tau_{AB}) = 2 S(\tau_B)$, and $S(\omega_{AB}) = 0$, while $S(\tau_B) = S(\omega_B)$.

Now all you have to do is show that you can perturb this construction slightly to get the desired inequalities in your hypothesis.

Answer 2

As suspected by Peter Shor, it is not true.

Almost a counterexample

Let $ω_{ABCD}=Φ_{AC}⊗ψ_{B}⊗ψ_{D}$, with $Φ$ being a maximally entangled state and $ψ$ a pure state. Let $U$ be the unitary $I_A⊗σ_{BC}⊗I_D$ swapping $B$ and $C$, so we have $ω_{ABCD}=Φ_{AB}⊗ψ_{C}⊗ψ_{D}$. All the systems are supposed to be of dimension $d$.

Your assumptions are almost fulfilled, since $$\begin{align} S(ω_B)=0&<S(τ_B)=\log d \\ S(ω_A)=\log d&\ge S(τ_A)=\log d \tag1 \\ S(ω_{AB})=\log d&> S(ω_B)=0 \end{align}$$ but we have $$\begin{align} S(τ_{AB})-S(ω_{AB})=0-\log d&< S(τ_{B})-S(ω_{B})=\log d -0 \end{align}$$.

In this case, the condition (1) is not fulfilled, since you asked for a strict decrease in $A$’s entropy, and kept $S(A)$ constant with a unitary not touching $A$. This can be changed by perturbing $ω$ and $U$, to have a slight decrease in $A$’s entropy.

A real counterexample

A concrete way to do this without a pertubative argument is to add another system $A'$ to $A$, which is entangled to another system $D'$ given to $D$. $$\begin{align} ω_{AA'BCDD'}&=Φ_{AC}⊗Φ_{A'D'}⊗ψ_{B}⊗ψ_{D}\\ U&=σ_{A'D}⊗σ_{BC}⊗I_{D'}\\ τ_{AA'BCDD'}&=ψ_{A'}⊗Φ_{AB}⊗Φ_{DD'}⊗ψ_{C} \end{align}$$

In that case, we have $$\begin{align} S(ω_B)=0&<S(τ_B)=\log d \\ S(ω_{AA'})=2\log d&> S(τ_A)=\log d \\ S(ω_{AA'B})=2\log d&> S(ω_B)=0 \\ S(τ_{AA'B})-S(ω_{AA'B})=0-2\log d&< S(τ_{B})-S(ω_{B})=\log d -0 \end{align}$$ Despite all your assumptions fulfilled by at least a $\log d$ margin, your final inequality is violated by a $3\log d$ margin

The physical intuition begin these counter examples : conditional entropies

The inequality you want to prove and your third assumption are thinly disguised conditional entropies. Moving $S(ω_B)$ to the left-hand side of our third assumption, one obtains $$H(A|B)_{ω}≥0,$$ which is verified by all state which are separable across the $A|B$ split (including my counterexamples.)

Your final condition is equivalent to $$H(A|B)_{τ}\stackrel{?}{≤}H(A|B)_{ω}.$$ Since the right-hand side is positive by assumption, any negative left-hand side is a counterexample. $H(A|B)$ can only be negative for states which are entangled accross the $A|B$ split. In both my counterexamples $H(A|B)_{τ}=-\log d$ because of the entangled state $Φ_{AB}$. The increase in $B$’s entropy is provided by the move of the half EPR pair from $C$ to $B$. In the second counter example, in order to have a decrease of $AA'$’s entropy, I artificially increased the initial entropy of $A$ with the state $Φ_{A'D'}$. This entropy is sent to $D$ by $U$.