Imagine using the following quantum communication scheme between Alice and Bob:

  • Alice has a set of mixed states $\{\hat{\rho}_1, ..., \hat{\rho}_N\}$ and she draws them according to a classical variable $X$ distributed according to the discrete probability distribution $P = \{p_1, ..., p_N\}$, then she sends these mixed states to Bob.
  • Bob performs a measurements on these states and registers the classical variable $Y$.

According to Holevo's theorem, the amount of accessible information $I(Y;X)$ that Bob can extract from $X$ is upper bounded by the Holevo capacity, namely

$$ \chi(P, \hat{\rho}) = S\Big(\sum_i p_i \hat{\rho}_i\Big)-\sum_ip_i S(\hat{\rho}_i) $$

where $S(\hat{\rho}) = Tr\big[\hat{\rho}log(\hat{\rho})\big]$ is the von Neumann entropy. This is also proved to be attainable, therefore a capacity (HWS theorem).

My question

I am basically trying to gain an intuitive idea of this quantity. The relative wikipedia page says that

In essence, the Holevo bound proves that given n qubits, although they can "carry" a larger amount of (classical) information (thanks to quantum superposition), the amount of classical information that can be retrieved, i.e. accessed, can be only up to n classical (non-quantum encoded) bits.

Why can we say that the amount of information that can be retrieved can only be up to n bits? My intuition says that the quantity $S\Big(\sum_i p_i \hat{\rho}_i\Big)$ gives us the amount of information carried by $\hat{\rho}$, but most of it is encoded in the "quantum nature" of the state, which is however lost when performing a measurement, this part is quantified by $\sum_ip_i S(\hat{\rho}_i)$. Then by subtracting them we are only considering the classical part of information encoded in the quantum scheme, does this make any sense? Any help would be very much appreciated.


1 Answer 1


Let me start by saying that it doesn't sound like what you are asking is actually the Holevo capacity. What you're referring to is more commonly referred to as the Holevo $\chi$. The Holevo capacity is the (least) upper bound for the classical capacity of a classical channel. These are tightly related, but if you fix a choice of input probabilities $p_x$, then you might as well talk about the mutual information of the associated joint probability distribution. The capacity of the channel is the mutual information maximised over input probability distributions.

That aside, the setting where we talk about Holevo bounds is that Alice encodes classical information into quantum states, and Bob seeks to retrieve said classical information from the outcome probabilities resulting from measuring the state he received.

The encoding is thus in the form of a mapping $x\mapsto \rho_x$. In the simplest case imagine $x\in\{0,1\}$, so that the encoding amounts to choosing of a pair of states to send over. Alice sends $\rho_0$ if she wants to communicate "$0$", or $\rho_1$ if she wants to communicate "$1$". Let $p_x$ be the probability distribution with which different inputs are chosen.

Bob chooses a measurement setting, say, some POVM $\mu$, and correspondingly obtains a probability distribution. If the state he received was $\rho_x$, the outcome probabilities will be $p_\mu(y|x)=\operatorname{Tr}(\mu_y \rho_x)$, where I'm denoting with $b$ the labels for the possible outcomes (again, imagine $y\in\{0,1\}$ for measurements with binary outcomes). Note that this probability distribution depends on the measurement choice $\mu$.

The accessible correlations between Alice and Bob is then the mutual information of the joint probability distribution maximised over the possible measurement choices $\mu$. This mutual information can also be written as the quantum mutual information of a "classical-classical" state, maximised over measurements: $$I_{\rm acc}(X:Y) \equiv \max_\mu I\left(\sum_{x,y}p_x p_\mu(y|x) \, (\mathbb{P}_x\otimes\mathbb{P}_y)\right), \qquad \mathbb{P}_x\equiv |x\rangle\!\langle x|.$$ Note that this has a fairly intuitive interpretation: it's the amount of correlations between inputs and outputs that can actually be observed ("accessed" you might say) from measurement outcomes. The problem is that performing this maximisation isn't trivial in general. So a trick that is often employed is to instead use an upper bound for it. One can indeed see that $$I_{\rm acc}(X:Y) \le I\left(\sum_x p_x \mathbb{P}_x\otimes\rho_x\right) = \chi(\{p_x,\rho_x\}),$$ where $\chi(\{p_x,\rho_x\})$ is the Holevo chi of the ensemble, and it's a relatively simple exercise to show that this equals the (quantum) mutual information of the "classical-quantum state" $\sum_x p_x \mathbb{P}_x\otimes\rho_x$.

The difference between the actual value of the accessible information and the Holevo $\chi$ is the so-called discord of the associated classical-quantum state. This is a quantity that measures how much of the quantum mutual information corresponds to actually observable correlations. See also this other answer for more details on this point. The gist of the matter is that the quantum mutual information is easy (or at least easier) to compute compared to the accessible one, which does not involve maximisations, but also doesn't simply quantify accessible correlations as one might expect.

In summary, the amount of "information that can be retrieved i.e. accessed" is quantified by the mutual information between measurement outcomes, that is, the accessible mutual information of the corresponding quantum states. This can be upper bounded by the Holevo $\chi$ of the ensemble $\{(p_x,\rho_x)\}_x$, which equals the quantum mutual information of the classical-quantum state $\sum_x p_x \mathbb{P}_x\otimes\rho_x$. Because it's easy to see that this quantity is smaller than the entropy of the source, $H(p)$, this immediately tells you that the accessible mutual information is also upper bounded by $H(p)$, hence you cannot exploit quantum features to send more information than you would classically.


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