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The home page concentrates on registers with three qubits because three is the smallest number that will give clear results, for two reasons: (a) 3 is relatively prime to 23 = 8, so the student will not falsely partition the states among the component qubits, and (b) 3 is large enough to allow successive nontrivial partial measurements on a qureg.

Here are again the generative ratios for the three-qubit register:

• in binary:
• c001 ÷ c000 = c011 ÷ c010 = c101 ÷ c100 = c111 ÷ c110 (III)
• c010 ÷ c000 = c011 ÷ c001 = c110 ÷ c100 = c111 ÷ c101 (II)
• c100 ÷ c000 = c101 ÷ c001 = c110 ÷ c010 = c111 ÷ c011 (I)
• in octal:
• c1 ÷ c0 = c3 ÷ c2 = c5 ÷ c4 = c7 ÷ c6
• c2 ÷ c0 = c3 ÷ c1 = c6 ÷ c4 = c7 ÷ c5
• c4 ÷ c0 = c5 ÷ c1 = c6 ÷ c2 = c7 ÷ c3

With two qubits, the generative ratios shrink to:

• in binary:
• c01 ÷ c00 = c11 ÷ c10
• c10 ÷ c00 = c11 ÷ c01
• in quartal:
• c1 ÷ c0 = c3 ÷ c2
• c2 ÷ c0 = c3 ÷ c1

With only one qubit, the ratios degenerate, and the entanglement is zero.

In the four-qubit case, there will be four eight-member equations — incorporated whole are the three-qubit ratios, here in boldface:

• in binary:
• c0001 ÷ c0000 = c0011 ÷ c0010 = c0101 ÷ c0100 = c0111 ÷ c0110 = c1001 ÷ c1000 = c1011 ÷ c1010 = c1101 ÷ c1100 = c1111 ÷ c1110
• c0010 ÷ c0000 = c0011 ÷ c0001 = c0110 ÷ c0100 = c0111 ÷ c0101 = c1010 ÷ c1000 = c1011 ÷ c1001 = c1110 ÷ c1100 = c1111 ÷ c1101
• c0100 ÷ c0000 = c0101 ÷ c0001 = c0110 ÷ c0010 = c0111 ÷ c0011 = c1100 ÷ c1000 = c1101 ÷ c1001 = c1110 ÷ c1010 = c1111 ÷ c1011
• c1000 ÷ c0000 = c1001 ÷ c0001 = c1010 ÷ c0010 = c1011 ÷ c0011 = c1100 ÷ c0100 = c1101 ÷ c0101 = c1110 ÷ c0110 = c1111 ÷ c0111
• c1 ÷ c0 = c3 ÷ c2 = c5 ÷ c4 = c7 ÷ c6 = c9 ÷ c8 = cB ÷ cA = cD ÷ cC = cF ÷ cE
• c2 ÷ c0 = c3 ÷ c1 = c6 ÷ c4 = c7 ÷ c5 = cA ÷ c8 = cB ÷ c9 = cE ÷ cC = cF ÷ cD
• c4 ÷ c0 = c5 ÷ c1 = c6 ÷ c2 = c7 ÷ c3 = cC ÷ c8 = cD ÷ c9 = cE ÷ cA = cF ÷ cB
• c8 ÷ c0 = c9 ÷ c1 = cA ÷ c2 = cB ÷ c3 = cC ÷ c4 = cD ÷ c5 = cE ÷ c6 = cF ÷ c7

Quantum digits need not be binary; they can for instance be ternary — then we have qutrits. Here is for example a two-qutrit register:

StateAmplitudeProbability
│000 ⟩−0.05931 + 0.34072i0.11961
│001 ⟩+0.20682 + 0.24024i0.10049
│002 ⟩−0.09873 − 0.02121i0.01020
│010 ⟩−0.30612 − 0.30631i0.18753
│011 ⟩+0.42147 + 0.17293i0.20755
│012 ⟩−0.23422 + 0.37470i0.19526
│020 ⟩−0.01559 − 0.02411i0.00082
│021 ⟩+0.23885 + 0.30264i0.14864
│022 ⟩−0.00347 − 0.17289i0.02990

The generative ratios are:

• c21 ÷ c20 = c11 ÷ c10 = c01 ÷ c00
• c22 ÷ c21 = c12 ÷ c11 = c02 ÷ c01
• c20 ÷ c22 = c10 ÷ c12 = c00 ÷ c02
• c12 ÷ c02 = c11 ÷ c01 = c10 ÷ c00
• c22 ÷ c12 = c21 ÷ c11 = c20 ÷ c10
• c02 ÷ c22 = c01 ÷ c21 = c00 ÷ c20

One can even mix radices. The next example shows a register of one qubit and one qutrit:

StateAmplitudeProbability
│00 ⟩−0.18028 + 0.29657i0.12045
│01 ⟩+0.25434 − 0.40179i0.22612
│02 ⟩−0.36540 − 0.22754i0.18529
│10 ⟩+0.03971 + 0.28238i0.08132
│11 ⟩−0.17964 − 0.30279i0.12395
│12 ⟩+0.39310 + 0.32914i0.26286

The generative ratios are:

• c11 ÷ c10 = c01 ÷ c00
• c12 ÷ c11 = c02 ÷ c01
• c10 ÷ c12 = c00 ÷ c02
• c12 ÷ c02 = c11 ÷ c01 = c10 ÷ c00
• c02 ÷ c22 = c01 ÷ c21 = c00 ÷ c20