I have a few confusions regarding quantum communication. Please bear with me as I ask what might appear to be basic or naive questions:

In the case of classical communication the capacity of wireless channel is given by: \begin{align} C_c= B \log_2\left(1+\dfrac{P h}{\sigma^2}\right) \end{align}
where $B$ is the bandwidth, $P$ is the transmission power, $h$ is the channel gain and $\sigma^2$ is the variance of AWGN. So we can increase the per second capacity by increasing the transmission power and/or the bandwidth.

Now in the case of amplitude dampening quantum channel the capacity is given by: \begin{align} Q=\max_{\tau}[H(\tau(1-\gamma))-H(\tau(\gamma))] \end{align} where $\tau\in [0,1]$, $H$ is the Shannon entropy function and $\gamma$ is the damping parameter. I think this is also referred to one-shot capacity of the channel at some places.

I have the following questions/confusions:

  1. It appears that the capacity of the quantum channel is inherently limited to $\leq$ 1. Unlike classical communication where we can enhance capacity by increasing transmission power or bandwidth, I wonder if a quantum channel's bandwidth has a similar impact on the capacity. Is there an analogous method in the realm of quantum communication that could potentially boost its capacity?

  2. If what we have is one-shot capacity (capacity per channel use). Does that mean we can transmit on the channel multiple times in a second? As the amplitude dampening channel is degradable, the capacity will be N*Q qbits/second? Where N is the number of transmission per second. Is there an upper bound to the value of N (maximum number of transmissions per second)?

  3. What if we need to transmit hundreds of qbits per-second? Can we assume multiple quantum channels between the sender and the receiver over the same optical fiber? Or is there some other resource that can help in increasing the capacity?

I understand that these questions might appear elementary, and I have attempted to search for answers online; however, I am facing some challenges in finding the information I need. If you have any familiarity with the field, I would greatly appreciate your insights, even if you cannot provide a definitive answer. Additionally, if you could suggest any references or resources that I could explore to deepen my understanding, that would be immensely helpful. Thank you for your support!

  • $\begingroup$ I think part of the confusion is that you're comparing a continuous classical channel to a ``discrete'' (finite dimensional) quantum channel. This is why you see the quantum channel has a capacity bounded by $1$, note this bound on the capacity would be the same if you considered a classical channel with a 1-bit output. To get a fairer comparison you probably want to look at infinite-dimensional quantum systems. $\endgroup$
    – Rammus
    Commented Jul 26, 2023 at 9:46

2 Answers 2


We define the rate to be $R = \frac{\log_k |\mathcal{M}(n)|}{n}$ where $\mathcal{M}(n)$ is the set of messages you can successfully transmit using $n$ i.i.d. copies of the channel and $k$ is some constant. The capacity is just the maximum possible rate in the asymptotic limit of $n\rightarrow\infty$. As a matter of convention, if we are discussing finite dimensional output alphabets, we choose $k$ such that a noiseless channel gives us a capacity of $1$ (see below).

Question 1

The AWGN channel has real numbers as outputs and therefore you can send arbitrarily many messages over it if the noise is $0$. Hence, it's capacity is infinite when there is no noise, per the definition.

The quantum setting you are looking at considers finite dimensional quantum states over a channel. This is not special to quantum though. If you set $k=2$ and consider a classical channel whose output is a bit, it also cannot have a capacity exceeding $1$ because $|\mathcal{M}| = 2$ per copy of the channel is the best you can hope for.

Question 2

The capacity is given as the maximum rate per use of the channel in the asymptotic limit. We take $n$ i.i.d. copies of the channel, compute the rate as a function of $n$ and then take the limit $n\rightarrow\infty$. There is no upper bound on number of channel copies you can use. This is the same classically and quantumly.

Question 3

As above, yes, you can assume arbitrarily many i.i.d. copies of the channel between the sender and receiver. You are not necessarily increasing the rate because the definition of rate divides by the number of channel copies you used.

There is a quantum-specific phenomenon that boosts quantum communication rates in a way that is classically impossible. This is called superadditivity and a good resource to understand it is to start with Section 13.1 of Mark Wilde's book for classical channels before moving onto later chapters for what changes when you look at quantum channels.


Your question is a bit tricky to me.

You are trying to compare a theoretical quantum channel capacity with the capacity of a real technology for classical communication. I'd like to help here but first please find a single format to your questions.

As regards 1st question, I published an experimental paper, exploiting indefinite causal orders to achieve communication by means of 0-capacity channels. In brief, within the document you will find a detailed example of quantum advantage in communication (by superposing channels).


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