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Suppose there is some $n$-qubit state $\rho$. It is well known fact that, given some orthonormal basis $U = \{|u_i\rangle\}$, if $p_i = \langle u_i| \rho |u_i \rangle$ (that is, measuring $\rho$ with $U$ produce the result $u_i$ with probability $p_i$) and $P = (p_1,...)$, then $H(P) \geq S(\rho)$.

Although that statement is written in papers and even in Wikipedia, I haven't found any proper proof nor I have been able to prove it myself. I would be glad for some help.

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3 Answers 3

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This can be seen through "twirling" with a bunch of unitaries.

Call your density operator $\rho$. Let $U_i$ be a unitary with $\pm 1$ on the diagonal, and zeros everywhere else when expressed in your basis. Consider all $2^d$ such unitaries where $d$ is the dimension of your density matrix. I leave it to you to show that $\rho_D = \frac{1}{2^d}\sum\limits_i U_i\rho U^\dagger_i$, where $\rho_D$ is the diagonal matrix with entries $p_i$.

Using the concavity of entropy and the unitary invariance of entropy, we have that $$H(P) = S(\rho_D) = S\left(\frac{1}{2^d}\sum_i U_i\rho U^\dagger_i\right) \geq \frac{1}{2^d}\sum\limits_i S(U_i\rho U^\dagger_i) = S(\rho).$$

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  • $\begingroup$ Thanks! thats seems correct (i fixed the direction of the inequality, it was exactly the opposite). Do you have any intuition for the matrices $U_i$? $\endgroup$
    – Woka
    Sep 6, 2020 at 10:56
  • $\begingroup$ Yes, sorry about the typo and thanks for the fix. The construction of the $U_i$ is just to acheive the same result as the measurement but I'm not sure I can offer any deeper insight there. Rammus' comment is also a very clean way to do it if you're happy with using data processing. $\endgroup$
    – rnva
    Sep 6, 2020 at 16:59
  • $\begingroup$ After a few thoughts... Are you sure the entropy in invariant for unitary multiplication? Look at the proof which actually does it much easier. Define $\Pi_t = |u_t \rangle \langle u_t|$. Then $S(\sum_t \Pi_t \rho \Pi_t) = S(\sum_t \Pi_t p_t) = S(P) = H(P)$. And on the other hand, due to invariant of unitary + concaviity, $S(\sum_t \Pi_t \rho \Pi_t) \geq \sum_t S(\Pi_t \rho \Pi_t) = 2^n S(\rho) \geq S(\rho)$ which concludes the proof. $\endgroup$
    – Woka
    Sep 9, 2020 at 13:32
  • $\begingroup$ The von Neumann entropy is a function of the eigenvalues of the state. This cannot change after a unitary transformation. However, the projector $\Pi_t$ is not a unitary. Also there are $n$, not $2^n$ possibilities for the index $t$, if $n$ is meant to be the dimension of the state. $\endgroup$
    – rnva
    Sep 10, 2020 at 14:16
  • $\begingroup$ You are right, missed that they aren't unitary. Thanks! $\endgroup$
    – Woka
    Sep 11, 2020 at 13:23
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I will expand on my comment as an answer because it is not as immediate as I initially thought it was. Let $D(\rho \| \sigma ) := \mathrm{Tr}[\rho( \log \rho - \log \sigma)]$ be the relative entropy where $\rho$ is a state and $\sigma$ is a positive semidefinite operator. We can write the von Neumann entropy of a state $\rho$ in terms of the relative entropy, $$ S(\rho) = -D(\rho \| \mathbb{1}). $$

Now the relative entropy satisfies something known as the data processing inequality) DPI which states that for any CPTP map $\mathcal{N}$ we have $$ D(\rho \| \sigma) \geq D(\mathcal{N}(\rho) \| \mathcal{N}(\sigma)). $$

Let us take the CPTP map $\mathcal{M}(\rho) = \sum_i |i \rangle \langle i | \rho | i \rangle \langle i |$ which is defined by the measurement in your question. This map when applied to $\rho$ prepares the state $ \sum_i p(i) | i \rangle \langle i |$ where $p(i)$ is the probability of obtaining outcome $i$ when measuring the state $\rho$. Now by the above we have $$ \begin{aligned} S(\rho) &= -D(\rho \| \mathbb{1}) \\ &\leq -D(\mathcal{M}(\rho) \| \mathcal{M}( \mathbb{1})) \\ &= -D(\mathcal{M}(\rho) \| \mathbb{1}) \\ &= S(\mathcal{M}(\rho)) \\ &= H(p). \end{aligned} $$

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  • $\begingroup$ nice approach (+1). Do you have reference to read about this form of the DPI? $\endgroup$
    – glS
    Sep 7, 2020 at 16:16
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    $\begingroup$ @glS My go to reference on all things quantum divergences is Marco Tomamichel's Quantum Information Processing with Finite Resources $\endgroup$
    – Rammus
    Sep 7, 2020 at 17:42
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Let $\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\braket}[2]{\langle #1 | #2\rangle}\{\ket{u_k}\}_k$ be some orthonormal basis for the space, and define $p_k\equiv \langle u_k|\rho| u_k\rangle$. Let $\newcommand{\bs}[1]{\boldsymbol{#1}}\bs p\in\mathbb R^n$ denote the corresponding vector. Write the eigendecomposition of $\rho$ as $$\rho = \sum_\ell \lambda_\ell |\lambda_\ell\rangle\!\langle\lambda_\ell\rvert, \qquad \lambda_\ell\ge0, \quad \sum_\ell \lambda_\ell=1.$$ From these, we see that $$p_k = \sum_\ell \lambda_\ell \lvert \braket{u_k}{\lambda_\ell}\rvert^2 = \sum_\ell M_{k,\ell}\lambda_\ell \equiv (M\bs\lambda)_k,$$ where $M_{k,\ell}\equiv \lvert \braket{u_k}{\lambda_\ell}\rvert^2$ is easily seen to be a bistochastic matrix, due to the completeness of both the bases $\{\ket{u_k}\}_k$ and $\{\ket{\lambda_\ell}\}_\ell$. See also Schur-Horn's lemma.

We therefore proved that $\bs p=M\bs\lambda$ for some bistochastic matrix $M$. This is equivalent to $\bs p\preceq \bs\lambda$, where $\preceq$ here denotes the majorization preorder. This, in turn, implies that $H(\bs p)\ge H(\bs \lambda)$ (see e.g. this related answer on math.SE). Because by definition of the von Neumann entropy $H(\bs\lambda)=S(\rho)$, we reach the conclusion.

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