We can extend the stabilizers of the codes from section 8.3 to get distance four codes. The parameters of these distance four codes will be [2j,2j−2j−2,4]. The first two generators ofS will again beMX andMZ. The nextj generators of S are the generators M1 through Mj from section 8.3, so S includes the stabilizer for a distance three code. The lastj generators ofS areNi=RMiR for i = 1, . . . , j, where R is applied to all 2j qubits. As with the codes of section 8.3, the error occuring for these codes can be efficiently determined from the error syndrome.
We can summarize this by writing the error syndromes forσxi andσzi: f(σxi) = 01⊕i⊕σ(i) (8.22) f(σzi) = 10⊕σ(i)⊕i. (8.23) Since S includes the stabilizer of a distance three code, it automatically has distance at least three. We need to check thatf(E)6= 0 for any weight three operatorE. The only form of an operatorEfor which the first two bits off(E) could be 00 isE=σxaσybσzc. Then
f(E) = 00⊕(a+σ(b) +b+σ(c))⊕(σ(a) +b+σ(b) +c) (8.24) = 00⊕(a+b+σ(b+c))⊕(b+c+σ(a+b)). (8.25) Ifr=a+bands=b+c, thenf(E) is nonzero as long asr6=σ(s) ors6=σ(r). This means that we need
s6=σ(σ(s)) =σ2(s) (8.26) for all nonzeros (whenr =s= 0, E =I). To see that this is true, note that for evenj, σ2(0. . .0001) = 10. . .00 σ2(0. . .0010) = 11. . .11 σ2(0. . .0100) = 0. . .001 (8.27) .. . σ2(1000. . .0) = 001. . .0.
If shas a 0 in the next-to-last bit, it cannot have σ2(s) = s unless s = 0. If shas a 1 in the next-to-last bit, it must have a 0 for the fourth-from-the-last bit, and so on. Ifj is a multiple of four, we find that the first bit must be a 0, which means that the last bit ofs must be a 1. This in turn implies that the third-from-the-last bit is 0, and so on until we reach the second bit ofs, which must be 0, sos= 001100. . .11. However, the second bit of σ2(s) is 1 because
MX σx σx σx σx σx σx σx σx σx σx σx σx σx σx σx σx MZ σz σz σz σz σz σz σz σz σz σz σz σz σz σz σz σz M1 I σx I σx I σx I σx σz σy σz σy σz σy σz σy M2 I σx I σx σz σy σz σy σx I σx I σy σz σy σz M3 I σx σz σy σx I σy σz I σx σz σy σx I σy σz M4 I σy σx σz I σy σx σz I σy σx σz I σy σx σz N1 I σz I σz I σz I σz σx σy σx σy σx σy σx σy N2 I σz I σz σx σy σx σy σz I σz I σy σx σy σx N3 I σz σx σy σz I σy σx I σz σx σy σz I σy σx N4 I σy σz σx I σy σz σx I σy σz σx I σy σz σx
Table 8.2: The stabilizer for a [16,6,4] code.
the next-to-last bit is. Therefore,σ(s)6=sin this case. If j is even, but not a multiple of four, the first bit ofs must be 1, which means that the last bit is 0. Again we follow the chain of logic back to the second bit ofsand again find that it must be 0, again giving a contradiction. Therefore σ2(s) 6= s for any nonzerosfor any even j. An example for even j is the [16,6,4] code given in table 8.2. Ifj is odd, σ2(0. . .0001) = 0111. . .11 σ2(0. . .0010) = 1111. . .11 σ2(0. . .0100) = 000. . .001 σ2(0. . .1000) = 000. . .010 (8.28) .. . σ2(010. . .00) = 0001. . .00 σ2(1000. . .0) = 0101. . .11.
In order to haveσ2(s) =s, we cannot have the first bit and last two bits ofsall 0. If the first bit ofsis 1, then the next-to-last bit ofsmust also be 1. Then if the last bit is 0, the third-from-the-last bit must be 0 and the fourth-from-the-last bit must be 1. Also, the second bit is 0 and the third bit is 1. After the third bit, they must continue to alternate 0 and 1 until the next-to-last bit. This means odd numbered bits are 1 and even numbered bits are 0. However, the fourth- from-the-last bit is an even numbered bit, giving a contradiction. Therefore, if the first bit ofs is 1, the last two bits must both be 1 also. That means the third-from-the-last and fourth-from-the-last bits must both be 0. However, it also means that the second bit ofsis 1 and the third bit ofs is 0. The fourth bit is 0 again, but the fifth bit is 1, and after that they alternate until the last two bits. This contradicts the fact that the third- and fourth-from-the-last bits must both be 0.
That leaves the possibility that the first bit ofsis 0. Then the next-to-last bit is 0 too, so the last bit must be 1. That means the third-from-the-last bit
MX σx σx σx σx σx σx σx σx MZ σz σz σz σz σz σz σz σz M1 I σx I σx σy σz σy σz M2 I σx σz σy I σx σz σy M3 I σy σx σz σx σz I σy N1 I σz I σz σy σx σy σx N2 I σz σx σy I σz σx σy N3 I σy σz σx σz σx I σy Table 8.3: The stabilizer for the [8,0,4] code.
is 0 and the fourth-from-the-last bit is 1. Also, the second and third bits ofs are both 1. The next two bits are both 0, and the two after that are both 1. The bits pair up to be the same, with the pairs alternating between 0 and 1. However, the fourth- and third-from-the-last bits form one of these pairs, and they are different, giving another contradiction. Therefore, σ2(s)6=s for any nonzerosfor odd j as well as for evenj. An example for odd j is the [8,0,4] code shown in table 8.3.
To show that this set of generators forms the stabilizer for a code, we still have to show that they all commute. From the fact thatMr andMs commute with each other and MX and MZ, we can immediately conclude that Nr and Ns commute with each other and MX and MZ. Also, Mr and Nr commute, since they get one sign of−1 for eachσx or σz in Mr, and there are an even number ofσx’s andσz’s. We must show thatMrcommutes with Nsforr6=s. Now, fMr(Ns) = nX−1 i=0 h i(r)i(s)+σ(i)(r)σ(i)(s)i. (8.29)
Here, x(r) is the rth bit of x. Now, σis a permutation of 0 through n−1, so the second term in the sum is equal to the first term in the sum. Therefore, the sum is automatically zero, and these generators do form a stabilizer.