Steane kodundaki stabilizatör jeneratörleri ve eşlik kontrol matrisleri arasındaki bağlantı

Kendi kendine çalışma için Mike ve Ike (Nielsen ve Chuang) üzerinde çalışıyorum ve Bölüm 10'daki dengeleyici kodlarını okuyorum. Klasik bilgi teorisinde biraz arka plana sahip bir elektrik mühendisiyim, ama ben hiçbir şekilde cebirsel kodlama teorisinde bir uzman. Soyut cebirim aslında ekte olandan biraz daha fazla.

İki doğrusal klasik kodun kuantum kodu oluşturmak için kullanıldığı Calderbank-Shor-Steane yapısını tamamen anladığımı düşünüyorum. Steane kodu kullanılarak inşa edilir $C_1$ [7,4,3] Hamming kodu olarak (qbit döndürür kodu) ve $C_2^{\perp}$ aynı kod olarak (faz için kod çevirme). Eşlik kontrol matrisi [7,4,3] kodu:

$$\begin{bmatrix} 0&0&0&1&1&1&1 \\ 0&1&1&0&0&1&1 \\ 1&0&1&0&1&0&1 \end{bmatrix}$$ .

Steane kodu için stabilizatör jeneratörleri şu şekilde yazılabilir:

$$\begin{array} {|r|r|} \hline Name & Operator \\ \hline g_1 & IIIXXXX \\ \hline g_2 & IXXIIXX \\ \hline g_3 & XIXIXIX \\ \hline g_4 & IIIZZZZ \\ \hline g_5 & IZZIIZZ \\ \hline g_6 & ZIZIZIZ \\ \hline \end{array}$$ akıl sağlığım adına$IIIXXXX = I \otimes I\otimes I\otimes X \otimes X \otimes X \otimes X$vb.

O kitapta işaret ediyor $X$ ler ve $Z$ ler ile aynı pozisyonlarda bulunan $1$ orijinal parite kontrol kodunda s. Alıştırma 10.32, Steane kodunun kod kelimelerinin bu set tarafından stabilize edildiğini doğrulamanızı ister. Açıkçası bunu takıp elle kontrol edebilirim. Bununla birlikte, parite kontrol matrisi ile jeneratör arasındaki benzerliklerin gözlemlenmesi ile egzersizin "apaçık" olduğu belirtilmektedir.

Bu gerçeğin başka yerlerde de ( http://www-bcf.usc.edu/~tbrun/Course/lecture21.pdf ) kaydedildiğini gördüm , ancak bir çeşit (muhtemelen açık) sezgiyi kaçırıyorum. Ben klasik kod kelimeler kuantum kodlarına kodun yapımında temel unsurları endeksleme nasıl kullanılır dışında başka bir bağlantı eksik düşünüyorum (yani Bölüm 10.4.2).

Burada birkaç kongre ve sezgi var, belki de bunun dile getirilmesine yardımcı olabilir - $\def\ket#1{\lvert#1\rangle}\def\bra#1{\!\langle#1\rvert}$

• İşaret bitlerine karşı {0,1} bitler
  
  İlk adım, bazen 'büyük notasyonel kayma' olarak adlandırılan şeyi yapmak ve bitleri (klasik bitler bile) işaretlerde kodlanmış olarak düşünmektir. Bu, en çok ilgilendiğiniz şey, bit dizelerinin pariteleri ise, çünkü bit-döndürmeler ve işaret-döndürmeler temel olarak aynı şekilde hareket eder. $0 \mapsto +1$ ve $1 \mapsto -1$ eşleştiriyoruz , böylece bit dizisi $(0,0,1,0,1)$ işaret dizisi ( + 1 , + 1 ) ile temsil edilecek$(+1,+1,-1,+1,-1)$ .

  Bit dizilerinin pariteleri daha sonra işaret dizilerinin ürünlerine karşılık gelir. Örneğin, $0 \oplus 0 \oplus 1 \oplus 0 \oplus 1 = 0$ parite hesaplaması olarak tanıdığımız gibi, $(+1)\cdot(+1)\cdot(-1)\cdot(+1)\cdot(-1) = +1$ , işaret kuralını kullanarak aynı parite hesaplamasını temsil eden şekilde .
  
  Egzersiz yapmak.$(-1,-1,+1,-1)$ ve $(+1,-1,+1,+1)$ 'nin paritesini hesaplayın . Bunlar aynı mı?
  
  

• İşaret bitlerini kullanarak parite kontrolleri
  
  {0,1} bit kuralında parite kontrolleri, iki boolean vektörün nokta ürünü olarak güzel bir temsile sahiptir, böylece doğrusal parite hesaplamalarını karmaşık parite hesaplamalarını gerçekleştirebiliriz. İşaret bitlerine geçerek kaçınılmaz olarak doğrusal cebirle bağlantıyı notasyonel bir seviyede kaybettik , çünkü toplamlar yerine ürünler alıyoruz. Hesaplamalı düzeyde, bu sadece gösterimde bir değişiklik olduğu için, gerçekten çok fazla endişelenmemiz gerekmiyor. Ancak saf matematiksel düzeyde, şimdi parite kontrolü matrisleri ile ne yaptığımızı tekrar düşünmek zorundayız.

  İşaret bitlerini kullandığımızda, ± 1 işaretleri yerine yine 0 ve 1'lerin matrisi olarak bir 'parite kontrol matrisi'ni temsil edebiliriz. Neden? Bir cevap, bitlerin eşlik kontrolünü tanımlayan bir satır vektörün bitlerin kendisinden farklı bir tipte olmasıdır: verilerin kendisinde değil, veriler üzerinde bir işlevi tanımlar. 0'lar ve 1'ler dizisi artık sadece farklı bir yorum gerektiriyor - toplamdaki doğrusal katsayılar yerine, bir üründeki üslere karşılık geliyorlar. İşaret bitlerimiz varsa $(s_1, s_2, \ldots, s_n) \in \{-1,+1\}^n$ , ve bir satır vektörü $(b_1, b_2, \ldots, b_n) \in \{0,1\}$ tarafından verilen bir parite kontrolünü hesaplamak istiyoruz , parite kontrolü daha sonra

  $$(s_1)^{b_1} \cdot (s_2)^{b_2} \cdot [\cdots] \cdot (s_n)^{b_n} \in \{-1,+1\},$$ burada geri çağırma $s^0 = 1$ for all $s$. As with {0,1}-bits, you can think of the row $(b_1,b_2,\ldots,b_n)$ as just representing a 'mask' which determines which bits $s_j$ make a non-trivial contribution to the parity computation.

  Egzersiz yapmak. Eşlik denetimi sonucunu $(0,1,0,1,0,1,0)$$(+1,-1,-1,-1,-1,+1,-1)$ üzerinde hesaplayın .
  
  

• Eşlik olarak özdeğerler.
  
  Sebebi neden veya daha fazla noktaya, yolu o bilgilere erişmesini açıklayabileceğinizi - kuantum bilgi teorisi işaretleri kodlamak bit istiyorum çünkü bilgi kuantum durumlarında depolanan biçimi taşımaktadır. Özellikle, standart temel hakkında çok konuşabiliriz, ancak bunun anlamlı olmasının nedeni, bu bilgiyi gözlemlenebilir bir ölçümü ölçerek çıkarabilmemizdir .

  Bu gözlemlenebilir sadece projektör olabilir $\ket{1}\bra{1}$, where $\ket{0}$ has eigenvalue 0 and $\ket{1}$ has eigenvalue 1, but it is often helpful to prefer to describe things in terms of the Pauli matrices. In this case, we would talk about the standard basis as the eigenbasis of the $Z$ operator, in which case we have $\ket{0}$ as the +1-eigenvector of Z and $\ket{1}$ as the −1-eigenvector of Z.

  So: we have the emergence of sign-bits (in this case, eigenvalues) as representing the information stored in a qubit. And better still, we can do this in a way which is not specific to the standard basis: we can talk about information stored in the 'conjugate' basis, just by considering whether the state is an eigenstate of $X$, and what eigenvalue it has. But more than this, we can talk about the eigenvalues of a multi-qubit Pauli operator as encoding parities of multiple bits — the tensor product $Z \otimes Z$ represents a way of accessing the product of the sign-bits, that is to say the parity, of two qubits in the standard basis. In this sense, the eigenvalue of a state with respect to a multi-qubit Pauli operator — if that eigenvalue is defined (i.e. in the case that the state is an eigenvalue of the Pauli operator) — is in effect the outcome of a parity calculation of information stored in some choice of basis for each of the qubits.

  Exercise. What is the parity of the state $\ket{11}$ with respect to $Z \otimes Z$? Does this state have a well-defined parity with respect to $X \otimes X$?
  
  Exercise. What is the parity of the state $\ket{+-}$ with respect to $X \otimes X$? Does this state have a well-defined parity with respect to $Z \otimes Z$?
  
  Exercise. What is the parity of $\ket{\Phi^+} = \tfrac{1}{\sqrt 2}\bigl(\ket{00} + \ket{11}\bigr)$ with respect to $Z \otimes Z$ and $X \otimes X$?
  
  

• Stabiliser generators as parity checks.
  
  We are now in a position to appreciate the role of stabiliser generators as being analogous to a parity check matrix. Consider the case of the 7-qubit CSS code, with generators

  $$\begin{array} {|r|ccccccc|} \hline \scriptstyle\text{Generator} & & & & \!\!\!\!\!\!\!\!\!\scriptstyle\text{Tensor factors}\!\!\!\!\!\!\!\!\! & & & \\[-0.5ex] & \scriptstyle1 & \scriptstyle2 & \scriptstyle3 & \scriptstyle4 & \scriptstyle5 & \scriptstyle6 & \scriptstyle7 \\ \hline \hline g_1 & & & & X & X & X & X \\ \hline g_2 & & X & X & & & X & X \\ \hline \hline g_3 & X & & X & & X & & X \\ \hline g_4 & & & & Z & Z & Z & Z \\ \hline g_5 & & Z & Z & & & Z & Z \\ \hline g_6 & Z & & Z & & Z & & Z \\ \hline \end{array}$$ I've omitted the identity tensor factors above, as one might sometimes omit the 0s from a {0,1} matrix, and for the same reason: in a given stabiliser operator, the identity matrix corresponds to a tensor factor which is not included in the 'mask' of qubits for which we are computing the parity. For each generator, we are only interested in those tensor factors which are being acted on somehow, because those contribute to the parity outcome.

  Now, the 'codewords' (the encoded standard basis states) of the 7-qubit CSS code are given by

  $$\begin{align} \ket{0_L} \propto{}&{} \ket{0000000} + \ket{0001111} + \ket{0110011} + \ket{0111100} \\&+ \ket{1010101} + \ket{1011010} + \ket{1100110} + \ket{1101001} = \sum_{y \in C} \ket{y}, \\[1ex] \ket{1_L} \propto{}&{} \ket{1111111} + \ket{1110000} + \ket{1001100} + \ket{1000011} \\&+ \ket{0101010} + \ket{0100101} + \ket{0011001} + \ket{0010110} = \sum_{y \in C} \ket{y \oplus 1111111}, \end{align}$$ where $C$ is the code generated by the bit-strings $0001111$, $0110011$, and $1010101$. Notably, these bit-strings correspond to the positions of the $X$ operators in the generators $g_1$, $g_2$, and $g_3$. While those are stabilisers of the code (and represent parity checks as I've suggested above), we can also consider their action as operators which permute the standard basis. In particular, they will permute the elements of the code $C$, so that the terms involved in $\ket{0_L}$ and $\ket{1_L}$ will just be shuffled around.

  The generators $g_4$, $g_5$, and $g_6$ above are all describing the parities of information encoded in standard basis states. The encoded basis states you are given are superpositions of codewords drawn from a linear code, and those codewords all have even parity with respect to the parity-check matrix from that code. As $g_4$ through $g_6$ just describe those same parity checks, it follows that the eigenvalue of the encoded basis states is $+1$ (corresponding to even parity).

  This is the way in which

  'with the observation about the similarities between the parity check matrix and the generator the exercise is "self evident"'

  — because the stabilisers either manifestly permute the standard basis terms in the two 'codewords', or manifestly are testing parity properties which by construction the codewords will have.
  
  

• Moving beyond codewords

  The list of generators in the table you provide represent the first steps in a powerful technique, known as the stabiliser formalism, in which states are described using no more or less than the parity properties which are known to hold of them.

  $\ket{\Phi^+} \propto \ket{00} + \ket{11}$$\ket{\Psi^-} \propto \ket{01} - \ket{10}$$\ket{\Phi^+}$$X \otimes X$$Z \otimes Z$$\ket{\Psi^-}$ is the only one which is a −1-eigenvector of both these operators.) These are known as stabiliser states, and one can consider how they are affected by unitary transformations and measurements by tracking how the parity properties themselves transform. For instance, a state which is stabilised by $X \otimes X$ before applying a Hadamard on qubit 1, will be stabilised by $Z \otimes X$ afterwards, because $(H \otimes I)(X \otimes X)(H \otimes I) = Z \otimes X$. Rather than transform the state, we transform the parity property which we know to hold of that state.

  You can use this also to characterise how subspaces characterised by these parity properties will transform. For instance, given an unknown state in the 7-qubit CSS code, I don't know enough about the state to tell you what state you will get if you apply Hadamards on all of the qubits, but I can tell you that it is stabilised by the generators $g_j' = (H^{\otimes 7}) g_j (H^{\otimes 7})$, which consist of

  $$\begin{array} {|r|ccccccc|} \hline \scriptstyle\text{Generator} & & & & \!\!\!\!\!\!\!\!\!\scriptstyle\text{Tensor factors}\!\!\!\!\!\!\!\!\! & & & \\[-0.5ex] & \scriptstyle1 & \scriptstyle2 & \scriptstyle3 & \scriptstyle4 & \scriptstyle5 & \scriptstyle6 & \scriptstyle7 \\ \hline \hline g'_1 & & & & Z & Z & Z & Z \\ \hline g'_2 & & Z & Z & & & Z & Z \\ \hline \hline g'_3 & Z & & Z & & Z & & Z \\ \hline g'_4 & & & & X & X & X & X \\ \hline g'_5 & & X & X & & & X & X \\ \hline g'_6 & X & & X & & X & & X \\ \hline \end{array}$$ This is just a permutation of the generators of the 7-qubit CSS code, so I can conclude that the result is also a state in that same code.

  There is one thing about the stabiliser formalism which might seem mysterious at first: you aren't really dealing with information about the states that tells you anything about how they expand as superpositions of the standard basis. You're just dealing abstractly with the generators. And in fact, this is the point: you don't really want to spend your life writing out exponentially long superpositions all day, do you? What you really want are tools to allow you to reason about quantum states which require you to write things out as linear combinations as rarely as possible, because any time you write something as a linear combination, you are (a) making a lot of work for yourself, and (b) preferring some basis in a way which might prevent you from noticing some useful property which you can access using a different basis.

  Still: it is sometimes useful to reason about 'encoded states' in error correcting codes — for instance, in order to see what effect an operation such as $H^{\otimes 7}$ might have on the codespace of the 7-qubit code. What should one do instead of writing out superpositions?

  The answer is to describe these states in terms of observables — in terms of parity properties — to fix those states. For instance, just as $\ket{0}$ is the +1-eigenstate of $Z$, we can characterise the logical state $\ket{0_L}$ of the 7-qubit CSS code as the +1-eigenstate of

  $$Z_L = Z \otimes Z \otimes Z \otimes Z \otimes Z \otimes Z \otimes Z$$ and similarly, $\ket{1_L}$ as the −1-eigenstate of $Z_L$. (It is important that $Z_L = Z^{\otimes 7}$ commutes with the generators $\{g_1,\ldots,g_6\}$, so that it is possible to be a +1-eigenstate of $Z_L$ at the same time as having the parity properties described by those generators.) This also allows us to move swiftly beyond the standard basis: using the fact that $X^{\otimes 7}$ anti commutes with $Z^{\otimes 7}$ the same way that $X$ anti commutes with $Z$, and also as $X^{\otimes 7}$ commutes with the generators $g_i$, we can describe $\ket{+_L}$ as being the +1-eigenstate of $$X_L = X \otimes X \otimes X \otimes X \otimes X \otimes X \otimes X,$$ and similarly, $\ket{-_L}$ as the −1-eigenstate of $X_L$. We may say that the encoded standard basis is, in particular, encoded in the parities of all of the qubits with respect to $Z$ operators; and the encoded 'conjugate' basis is encoded in the parities of all of the qubits with respect to $X$ operators.

  By fixing a notion of encoded operators, and using this to indirectly represent encoded states, we may observe that

  $$(H^{\otimes 7}) \,X_L\, (H^{\otimes 7}) = Z_L, \quad (H^{\otimes 7}) \,Z_L\, (H^{\otimes 7}) = X_L,$$ which is the same relation as obtains between $X$ and $Z$ with respect to conjugation by Hadamards; which allows us to conclude that for this encoding of information in the 7-qubit CSS code, $H^{\otimes 7}$ not only preserves the codespace but is an encoded Hadamard operation.
  
  

Thus we see that the idea of observables as a way of describing information about a quantum states in the form of sign bits — and in particular tensor products as a way of representing information about parities of bits — plays a central role in describing how the CSS code generators represent parity checks, and also in how we can describe properties of error correcting codes without reference to basis states.

One way that you could construct what the codeword is is to project on the $+1$ eigenspace of the generators,

$$|C_1\rangle=\frac{1}{2^6}\left(\prod_{i=1}^6(\mathbb{I}+g_i)\right)|0000000\rangle.$$ Concentrate, to start with, one the first 3 generators $$(\mathbb{I}+g_1)(\mathbb{I}+g_2)(\mathbb{I}+g_3).$$ If you expand this out, you'll see that it creates all the terms in the group ($\mathbb{I},g_1,g_2,g_3,g_1g_2,g_1g_3,g_2g_3,g_1g_2g_3)$. Corresponding it to binary strings, the action of multiplying two terms (since $X^2=\mathbb{I}$) is just like addition modulo 2. So, contained within the code are all of the words generated by the parity check matrix (and this is a group, with group operation of addition modulo 2).

Now, if you multiply by one of the $X$ stabilizers, that's like doing the addition modulo 2 on the corresponding bit strings. But, because we've already generated the group, by definition every group element is mapped to another (unique) group element. In other words, if I do

$$g_1\times\{\mathbb{I},g_1,g_2,g_3,g_1g_2,g_1g_3,g_2g_3,g_1g_2g_3\}=\{g_1\mathbb{I},g_1g_2,g_1g_3,g_2,g_3,g_1g_2g_3,g_2g_3\},$$ I get back the set I started with (using $g_1^2=\mathbb{I}$, and the commutation of the stabilizers), and therefore I'm projecting onto the same state. Hence, the state is stabilized by $g_1$ to $g_3$.

You can effectively make the same argument for $g_4$ to $g_6$. I prefer to think about first applying a Hadamard to every qubit, so the Xs are changed to Zs and vice versa. The set of stabilizers are unchanged, so the code is unchanged, but the Z stabilizer is mapped to an X stabilizer, about which we have already argued.

What follows perhaps doesn't exactly answer your question, but instead aims to provide some background to help it become as 'self-evident' as your sources claim.

The $Z$ operator has eigenstates $|0\rangle$ (with eigenvalue $+1$) and $|1\rangle$ (with eigenvalue $-1$). The $ZZ$ operator on two qubits therefore has eigenstates

$$|00\rangle, |01\rangle, |10\rangle, |11\rangle$$ . The eigenvalues for these depend on the parity of the bit strings. For example, with $|00\rangle$ we multiply the $+1$ eigenvalues of the individual $Z$ operators to get $+1$. For $|11\rangle$ we multiply the $-1$ eigenvalues together and also get $+1$ for $ZZ$. So both these even parity bit strings have eigenvalue $+1$, as does any superposition of them. For both odd parity states ($|01\rangle$ and $|10\rangle$) we must multiply a $+1$ with a $-1$, and get a $-1$ eigenvalue for $ZZ$.

Note also that superpositions of bit strings with fixed parity (such as some $\alpha |00\rangle + \beta |00\rangle$) are also eigenstates, and have the eigenvalue associated with their parity. So measuring $ZZ$ would not collapse such a superposition.

This analysis remains valid as we go to higher number of qubits. So if we want to know about the parity of qubits 1, 3, 5, and 7 (to pick a pertinent example), we could use the operator $ZIZIZIZ$. If we measure this and get the outcome $+1$, we know that this subset of qubits has a state represented by an even parity bit string, or a superposition thereof. If we get $-1$, we know that it is an odd parity state.

This allows us to write the [7,4,3] Hamming code using the notation of quantum stabilizer codes. Each parity check is turned into a stabilizer generator which has $I$ on every qubit not involved in the check, and $Z$ on every qubit that is. The resulting code will protect against errors that anticommute with $Z$ (and therefore have the effect of flipping bits).

Of course, qubits do not restrict us to only working in the $Z$ basis. We could encode our qubits for a classical Hamming code in the $|+\rangle$ and $|-\rangle$ states instead. These are the eigenstates of $X$, so you just need to replace $Z$ with $X$ in everything I've said to see how this kind of code works. It would protect against errors that anticommute with $X$ (and so have the effect of flipping phase).

The magic of the Steane code, of course, is that it does both at the same time and protects against everything.