(Quantum Information Theory Series -6)Coherent Protocols

(後面有中文版本!)

Hello, everyone!

Today, let’s delve into Coherent Protocols. In summary, they can implement the “duality” of teleportation and super-dense coding that we learned in the previous QIT-5 under resource reversal.

Duality: If the resources used by one protocol are generated by another protocol, and vice versa, we can say that these two protocols are dual.

But why do we need to make them dual? There are several reasons:

• Understanding and Simplification: Duality can help us understand the interrelationship between two protocols. In this context, redefining resource inequalities so that two protocols become dual under resource reversal may indicate some level of similarity or complementarity between them.
• Capacity Theorems: In quantum information theory, it’s often necessary to prove capacity theorems, the limiting performance of a communication channel. Duality might make it easier to prove such theorems.
• Unified Theory: Through duality, we can unify different protocols or systems, allowing us to understand various aspects of quantum information processing in a more cohesive manner.

Here we present the simplest idea~

Recall from our previous post, the resource inequality for Teleportation is:

The resource inequality for super-dense coding is:

From these two inequalities, it seems there is no “duality” in terms of resources, as mentioned above. However, if we assume that quantum entanglement is a free resource (Here “free” doesn’t mean the kind of generous giveaway that Taiwanese people love 😂, but rather closer to “unaccounted” or “ignored”), then it means we don’t need to include it in the resource calculations.

In this case, the resource inequality for Teleportation becomes:

And the resource inequality for super-dense coding becomes:

This satisfies the resource “duality”: If the resources used by one protocol are the resources generated by another protocol, and vice versa, we can say these two protocols are dual.

Therefore, we can write the resource inequality:

1. Definition of Coherent Communication

A coherent bit channel can be considered as a classical channel, an extension of the classical bit channel. Its behavior is similar to a classical bit channel, but with the distinction that it introduces the element of “Quantum feedback.”

Quantum feedback: involves operations on the quantum system not only during transmission through the channel but also includes some form of feedback or intervention on the information after passing through the channel. This feedback might influence the system’s state after the channel, making it more coherent or maintaining specific quantum properties.

When a classical bit channel undergoes decoherence with a probability of 1/2, it can be represented using a dephasing channel, with the CPTP map given by:

Additionally, its isometric extension can be further expressed as:

After simplification, it can be written as:

Thus, we can represent the mapping of a classical bit channel as:

The coherent bit channel follows the same structure as the above expression. We assume that Alice gains control over the environmental channel in some way.

Its resources can be represented as:

Essentially, this replicates the computational basis while preserving coherent superposition.

The simulated circuit can be illustrated as:

2. Implementations of a Coherent Bit Channel

How do we realize a Coherent Bit Channel? The simplest way is to use a local CNOT gate and a noiseless qubit channel, following these steps:

Step 1:

Alice prepares a qubit

and an auxiliary qubit

Step 2:

Alice applies a CNOT gate to these two qubits, resulting in:

Step 3:

Alice sends the qubit A’ to Bob through a noiseless qubit channel.

This completes a noiseless coherent bit channel.

The resource inequality is:

This demonstrates that quantum communication generates coherent communication. Now, we have the following resource inequality chain:

The capacity of a noiseless coherent bit channel lies between that of a noiseless qubit channel and a noiseless entanglement bit (ebit).

The quantum circuit is:

3.Coherent Super-Dense Coding

Here, we introduce another method to achieve a noiseless coherent bit channel, which can be considered as the coherent version of the super-dense coding protocol. The steps are as follows:

Step 1:

Alice and Bob share an entangled bit (ebit).

Step 2:

Alice prepares two qubits, a1 and a2, and places them in front of the ebit.

Step 3:

Use CNOT and Controlled-Z gates to obtain

Step 4:

Alice gives the A qubit to Bob, naming it B1, and Bob’s original B is named B2.

Step 5: Bob uses CNOT and applies a Hadamard Gate to B1.

It may seem complex, but drawing a circuit can help better understand it.

This protocol implements two coherent bit channels: one from A1 to B1 and the other from A2 to B2.

The resource inequality for this protocol is:

4. Coherent Teleportation

Next up is the coherent version of Teleportation.

As introduced earlier in this post, assuming ∆Z duplicates the eigenstates of Pauli-Z, and ∆X duplicates the eigenstates of Pauli-X.

The protocol unfolds as follows:

Step 1:

Alice possesses a qubit.

Teleports it using ∆Z,

obtaining For convenience, let’s rearrange it for further use:

Step 2:

Passes A through ∆X, resulting in

Step Three:

Bob applies a CNOT to it, transforming the entire state into

The circuit diagram would look something like this:

In the end, both Alice and Bob possess an ebit.

The resource inequality is:

At this point, combining the resource inequalities obtained in sections 3 and 4,

we find And this is the coveted duality (X) we’ve been yearning for!

That concludes QIT Series, Part 6. It’s the last post for February 2024. I hope everyone has had a joyful start to the year!

(中文開始~)

鴕鳥來囉~

今天要來討論一下Coherent Protocols，總結來說，他可以實現我們在上一篇QIT-5時學到的teleportation和super-dense coding使他們在資源反轉下的”對偶”。

對偶(dual): 如果一個協議所使用的資源是另一個協議生成的資源，反之亦然，我們就可以說這兩個協議是對偶的。

那為甚麼會需要使他們對偶呢? 有以下幾個原因

• 理解和簡化： 對偶性可以幫助我們理解兩個協議之間的相互關係。在這個情境下，重新定義資源不等式使得兩個協議在資源反轉下變得對偶，可能代表著它們在某種程度上的相似性或互補性。
• 容量定理： 在量子信息理論中，經常需要證明容量定理，即一個通信通道的極限性能。對偶性可能使得這樣的定理更容易證明。
• 統一理論：通過對偶性，我們可以將不同的協議或系統統一起來，使得我們能夠以更統一的方式理解量子信息處理的各個方面。

我們這邊舉一個最簡單的想法~

回想在上一篇我們學到的，Teleportation的資源不等式為:

super-dense coding的資源不等式為:

從這兩個不等式來看，並沒有符合我們上面提到的資源上的”對偶”，但是如果我們假設量子糾纏是一個免費的資源(這邊的免費不是指台灣人很喜歡的大放送XD，而是更接近”unaccounted” 或 “ignored”的意思)，此時代表著我們不需要將其納入資源計算裡面，此時Teleportation的資源不等式會變為:

super-dense coding的資源不等式變為:

這樣就會滿足資源上的”對偶”: 如果一個協議所使用的資源是另一個協議生成的資源，反之亦然，我們就可以說這兩個協議是對偶的。

因此我們可以寫出資源不等式:

1. Definition of Coherent Communication

Coherent bit channel可以視為是一種經典通道，可以想成為Classical bit channel的擴展。他的行為與Classical bit channel相似，但是不一樣的地方是他引入了”Quantum feedback”的元素在內。

Quantum feedback: 通道的操作不僅僅涉及對量子系統的傳輸，還包括對通道後的信息進行某種形式的回饋或干預。這種反饋可能影響通道後的系統狀態，使其更具相干性或保持一些特定的量子特性。

當Classical bit channel以1/2的機率進行去相干化，Classical bit channel可以使用dephasing通道表示，其CPTP映射可以寫成:

並且其isometric extension可以將其進一步寫成:

經過整理可以寫成

因此我們可以將classical bit channe的映射寫成:

而coherent bit channel就與上式雷同，我們假設Alice以某種方式獲得對環境通道的控制

其資源可以表示成

相當於它複製了基態，同時保持了相干的Superposition

模擬Circuit可以畫成:

2. Implementations of a Coherent Bit Channel

我們要怎麼實現Coherent Bit Channel呢? 最簡單的方式就是使用一個local CNOT gate和一個 noiseless qubit channel步驟如下:

步驟一:

Alice準備一個qubit

以及一個輔助qubit

步驟二:

Alice對這兩個qubit使用CNOT，得到

步驟三:

Alice將A’的qubit透過noiseless qubit channel傳給Bob

這樣就完成了一個noiseless coherent bit channel。

資源不等式為:

這證明了量子通信生成相干通信。現在我們有以下的資源不等式鏈：

noiseless coherent bit channel能力介於noiseless qubit channel和noiseless ebit 之間。

Quantum circuit為:

3.Coherent Super-Dense Coding

這邊我們要介紹另一個實現noiseless coherent bit channel的方法，這也可以說是super-dense coding protocol的coherent版本。步驟如下:

步驟一:

Alice和Bob有一個共享ebit

步驟二:

Alice準備兩個qubit，a1,a2。並將其放在ebit前面。

步驟三:

使用CNOT和Controlled-Z gate得到

步驟四:

Alice將A qubit給Bob並命名為B1，Bob原來的B命名為B2

步驟五:

Bob使用CNOT，並對B1做Hadamard Gate

看著很複雜，畫出Circuit可以比較好去理解它

這個Protocol實現了兩個coherent bit channels，一個為A1->B1，另一個為A2->B2。

這個Protocol的資源不等式為:

4. Coherent Teleportation

接下來是Teleportation的Coherent版本。

如同這篇前面介紹的，假設∆Z複製Pauli — Z的eigenstates，∆X複製Pauli — X的eigenstates

此Protocol的步驟如下:

步驟一:

Alice有一個Qubit

透過∆Z傳送得到

方便後面使用，我們先進一步將其作整理，得到

步驟二:

將A通過∆X傳送得到

步驟三:

Bob對其使用CNOT，整個狀態變成

畫出Circuit會長得像這樣:

Alice和Bob最後都會擁有一個ebit。

其資源不等式為:

到這邊我們綜合一下3, 4最後得到的資源不等式，會發現將其結合得到

而這就是我們心心念念(X)的對偶!

以上，QIT系列第6篇結束~ 也是2月最後的一篇文章，希望大家2024的前兩個月都過得很快樂!