<p>Why should this be true?</p>

<p>Let the dimensions of $A, B, C, D$ all be the same. </p>

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

<p>Now, let $U$ perform a cyclic permutation on $A, B, C, D$. </p>

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

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

CC BY-SA 3.0
2016-01-05T12:55:46.330Z
It does not need to be true, you are right. I forgot an assumption, though. Please see the edit.
RebeccaM
CC BY-SA 3.0
2016-01-05T17:13:10.950Z
Are $A, B, C, D$ all the same dimension?
Peter Shor 
CC BY-SA 3.0
2016-01-05T17:25:38.440Z
No. Would it help?
RebeccaM
CC BY-SA 3.0
2016-01-05T04:11:50.887Z
2016-01-05T04:11:50.887Z
Peter Shor 
<p>As suspected by Peter Shor, it is not true.</p>

<h2>Almost a counterexample</h2>

<p>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 $d$.</p>

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

<p>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.</p>

<h1>A real counterexample</h1>

<p>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 $D$. 
$$\begin{align}
  ω_{AA'BCDD'}&amp;=Φ_{AC}⊗Φ_{A'D'}⊗ψ_{B}⊗ψ_{D}\\
  U&amp;=σ_{A'D}⊗σ_{BC}⊗I_{D'}\\
  τ_{AA'BCDD'}&amp;=ψ_{A'}⊗Φ_{AB}⊗Φ_{DD'}⊗ψ_{C}
\end{align}$$  </p>

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

<h1>The physical intuition begin these counter examples : conditional entropies</h1>

<p>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.) </p>

<p>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 $U$.</p>

CC BY-SA 3.0
2016-02-04T18:36:48.327Z
2016-02-04T19:56:44.613Z
2016-02-04T19:56:44.613Z
Frédéric Grosshans
<p>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
$$
S(\omega_B)&lt;S(\tau_B) 
$$ 
and
$$
S(\omega_A)&gt;S(\tau_A).
$$
$S$ 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.</p>

<p>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?</p>

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

<quantum-information>
Entropy inequality of joint von Neumann entropy given some marginal inequalities
RebeccaM
# Entropy inequality of joint von Neumann entropy given some marginal inequalities

(Asked by: username_0 on 2016-01-03)

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
$$
S(\omega\_B)S(\tau\_A).
$$
$S$ 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 $S(\omega\_{AB})\geq S(\omega\_B)$.

**Tags:** quantum-information


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## 2 answers

### Answer by username_1 on 2016-01-05. Score: 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.

> **Comments:**
> **username_0**: It does not need to be true, you are right. I forgot an assumption, though. Please see the edit.  
> **username_1**: Are $A, B, C, D$ all the same dimension?  
> **username_0**: No. Would it help?  

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### Answer by username_2 on 2016-02-04. Score: 1

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
$$\begin{align}
S(ω\_B)=0& S(ω\_B)=0
\end{align}$$
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
$$\begin{align}
S(ω\_B)=0& S(τ\_A)=\log d \\
S(ω\_{AA'B})=2\log d&> S(ω\_B)=0 \\
S(τ\_{AA'B})-S(ω\_{AA'B})=0-2\log d&< S(τ\_{B})-S(ω\_{B})=\log d -0
\end{align}$$
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 $U$.


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