Thedepolarizing channelis a very natural channel to consider. In this channel, with probability 1−p, each qubit is left alone. In addition, there are equal probabilities p/3 that σx, σy, or σz affects the qubit. We can apply similar methods to the depolarizing channel as with the erasure channel to place upper and lower bounds on its capacity. However, currently these bounds do not meet, so the actual capacity of the depolarizing channel is unknown.
The depolarizing channel can also simulated by imagining Charlie is ran- domly stealing some qubits from the channel. If Charlie steals a qubit with probability q and replaces it with a random qubit (not telling Bob which one was stolen), there is still a 1/4 chance that Charlie happens to replace the stolen qubit with one in the same state. There is only a chanceq/4 of Charlie applying each ofσx, σy, and σz. Therefore, this situation corresponds to the depolariz- ing channel withp= 3q/4. We can make a cloning argument just as with the erasure channel to set an upper bound on the capacity. Again we find that the capacity is limited by 1−2q= 1−8p/3. Whenp >3/8, the rate of transmission is necessarily zero.
Actually, we can set a tighter upper bound than this. Randomly stealing qubits is not the best eavesdropping method available to Charlie that will look like the depolarizing channel. The best eavesdropping method actually allows him to produce the same state as Bob whenever p > 1/4 [60]. This means that the rate is limited to 1−4p. This is the asymptotic form of the Knill- Laflamme bound, which was derived for codes with a fixed minimum distance in section 7.1.
We can set a lower bound for the achievable rate by again considering the rate for a random stabilizer code. If we encode k qubits in n qubits using a random stabilizer S, the expected number of errors is pn. We need measure one of the errors to be distinguishable from each other. The errorsE and F are distinguishable ifE†F anticommutes with some elements ofS, and are not if they do not. The typical productE†F actually does not have weight 2pn. There is a chancep2 that E andF will both have nontrivial action on a given qubit. If they act as different Pauli matrices, the product will still act on that qubit. If they act as the same Pauli matrix, the product will not act on that qubit at all. The probability of having both act as the same Pauli matrix is p2/3. Therefore, the expected length of the productE†F is (2p−4p2/3)n. Let x= 2p−4p2/3.
Let the number of errors of weightwbeN(w). Then the number of different products of weightxnisN(xn), and therefore the number of typical products that commute with everything inSisN(xn)/2n−k. Now, there areN(pn) likely errors, so the number of ways we can pair them into products isN(pn)[N(pn)− 1]/2. This means that the number of ways of getting any given operatorO of
weightxnis µ
N(pn) 2
¶Á
N(xn). (7.60)
For each of the pairs that gives one of theN(xn)/2n−k products that commute withS, we must remove one of the errors in the pair from the group of likely errors. Therefore, we must remove
µ N(pn)
2 ¶Á
2n−k (7.61)
number to be small compared toN(pn) for largen. Thus,
N(pn)/2n−k+1 ¿ 1 (7.62)
N(pn) ¿ 2n−k+1 (7.63)
k/n < 1−1
nlog2N(pn) = 1−plog23−H(p). (7.64) This is just the quantum Hamming bound (7.3). In other words, a random code saturates the quantum Hamming bound.
However, the quantum Hamming bound only limits the efficiency of non- degenerate codes. The typical element of a random stabilizer will have weight 3n/4, which is much larger than pn for any p where the rate could possibly be nonzero. Therefore, a random code will have a negligable number of de- generate errors, and the quantum Hamming bound will still apply. However, if we choose the stabilizer to be of a restricted form rather than totally random, we can choose it to have very many degeneracies, and the quantum Hamming bound may be exceeded [61], although existing codes only allow us to exceed the rate of a random code by a very small amount. Shor and Smolin showed that by concatenating a random code with a simple repetition code (|0ibecomes the tensor product of|0i’s and|1ibecomes the tensor product of|1i’s), the rate of the code is improved slightly near the zero-rate limit. The optimum block size for repetition turns out to be five.
We can still set an upper bound on the efficiency of a degenerate stabilizer code using similar arguments to those that gave us the capacity of a random stabilizer code. Note that this upper bound does not necessarily apply to all codes, so it may not be a strict upper bound on the capacity. However, non- stabilizer codes are very difficult to work with, so it does provide a practical upper bound on the capacity.
To give this bound, assume that every element of S actually has weight xn. This bound is unlikely to be achievable, since the product of two oper- ators of weight xn will only rarely have weight xn again. There are at least N(xn)/2n−k operators of weightnthat commute withS, but 2n−k of them are inS. Therefore, in the best case, there are onlyN(xn)/2n−k−2n−k operators that can potentially cause a problem. In the limit where n and k = rn are both large, eitherN(xn)/2n−k will dominate the number of troublesome oper- ators, orN(xn)/2n−k ¿ 2n−k. In the first case, the calculation goes through as for a completely random stabilizer, giving us a capacity only at the quantum Hamming bound. In the second case,
N(xn) ¿ 22(n−k) (7.65) r=k/n < 1− 1 2nlog2N(xn) = 1− x 2log23− 1 2H(x). (7.66) Sincex= 2p−4p2/3, this is higher than the quantum Hamming bound. Equa- tion (7.66) gives an upper bound on the capacity of the depolarizing channel achievable using stabilizer codes. It is shown in figure 7.1 along with the Knill- Laflamme bound and the quantum Hamming bound. Cleve has also proved a
0.05 0.1 0.15 0.2 0.25 Error Rate 0.2 0.4 0.6 0.8 1 Transmission Rate
Figure 7.1: The quantum Hamming bound (dashed), the Knill-Laflamme bound (dotted), and the bound from equation (7.66) (solid).
bound on the capacity achievable using degenerate stabilizer codes [58], but it is slightly worse than (7.66) everywhere in the region of interest, so it is not shown in the figure.
Chapter 8
Examples of Stabilizer
Codes
There are many known stabilizer codes [10, 11, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 42]. I will not attempt to list them all here, but will instead concentrate on a few interesting individual codes and classes of codes. In a number of cases, I will not just describe the stabilizers of the codes, but will also discuss the normalizers and automorphism groups of the stabilizers, since these are important to realizing fault-tolerant computation in the most efficient possible way.
8.1
Distance Two Codes
For even n, there is always an [n, n−2,2] code. The stabilizer S has two generators, one the product of all n σx’s and one the product of all the σz’s. For evenn, these commute. N(S) consists of tensor products inGthat contain an even number ofσx’s, an even number ofσy’s, and an even number of σz’s. We can write
Xi = σx1σx(i+1) (8.1)
Zi = σz(i+1)σzn, (8.2)
fori= 1, . . . , n−2.
The automorphism group A(S) contains all possible permutations of the qubits and the Hadamard rotation R applied to all n qubits at once. If n is a multiple of four, any single-qubit operation in N(G) applied to all the qubits gives an element ofA(S). The order of A(S) is thus either 2n! or 6n!. Swapping qubit i with qubit j switches the (i−1)th encoded qubit with the (j−1)th encoded qubit (for 1< i, j < n). Swapping qubit 1 with qubit i+ 1 (i= 1, . . . , n−2) transforms
Xj → XiXj (i6=j)
Zi → Z1Z2· · ·Zn−2 (8.3) Zj → Zj (i6=j).
Similarly, swapping qubitnwith qubiti+ 1 (i= 1, . . . , n−2) transforms Xi → X1X2· · ·Xn−2
Xj → Xj (i6=j)
Zi → Zi (8.4)
Zj → ZiZj (i6=j).
Swapping the first qubit with thenth qubit performs the transformation Xi → X1· · ·Xi−1Xi+1· · ·Xn−2
Zi → Z1· · ·Zi−1Zi+1· · ·Zn−2. (8.5) Performing R on every qubit performs the same transformation as swapping the first andnth qubits, but also performsR on every encoded qubit. Forna multiple of four, performingP on every qubit performs the following operation:
Xi → −XiZ1· · ·Zi−1Zi+1· · ·Zn−2
Zi → Zi. (8.6)
Because these codes are of the CSS form, a CNOT applied to every qubit transversally between two blocks is also a valid fault-tolerant operation, and performs CNOTs between the corresponding encoded qubits.
The case of n= 4, the smallest distance two code, is of particular interest. The code from figure 3.5 can be converted into the form of the codes currently under consideration using single-qubit rotations, although theX andZ oper- ators will need to redefined. It can be used to detect a single error [18] or to correct a single erasure [19]. In this case,
X1 = σx1σx2 X2 = σx1σx3
Z1 = σz2σz4 (8.7)
Z2 = σz3σz4.
Switching the second and third qubits or switching the first and fourth qubits both swap the two encoded qubits. Swapping the first and second qubits or the third and fourth qubits produces the transformation
X1 → X1 X2 → X1X2
Z1 → Z1Z2 (8.8)
This is just a CNOT from the second encoded qubit to the first encoded qubit. Similarly, swapping the first and third qubits or the second and fourth qubits performs a CNOT from the first encoded qubit to the second encoded qubit. The transversal Hadamard rotation in this case performs the Hadamard rotations on both qubits and switches them. ApplyingP to all four qubits performs the gate
X1 → −X1Z2 X2 → −Z1X2
Z1 → Z1 (8.9)
Z2 → Z2.
We can recognize this as the encoded conditional sign gate followed by an en- codedσz1σz2.
A more extensive discussion of the properties of distance two codes (and a few codes of greater distances) appears in [62].