Home.
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 2^{3} = 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 threequbit register:
 in binary:

c_{001} ÷ c_{000} = c_{011} ÷ c_{010} =
c_{101} ÷ c_{100} = c_{111} ÷ c_{110} (III)

c_{010} ÷ c_{000} = c_{011} ÷ c_{001} =
c_{110} ÷ c_{100} = c_{111} ÷ c_{101} (II)

c_{100} ÷ c_{000} = c_{101} ÷ c_{001} =
c_{110} ÷ c_{010} = c_{111} ÷ c_{011} (I)
 in octal:

c_{1} ÷ c_{0} = c_{3} ÷ c_{2} =
c_{5} ÷ c_{4} = c_{7} ÷ c_{6}

c_{2} ÷ c_{0} = c_{3} ÷ c_{1} =
c_{6} ÷ c_{4} = c_{7} ÷ c_{5}

c_{4} ÷ c_{0} = c_{5} ÷ c_{1} =
c_{6} ÷ c_{2} = c_{7} ÷ c_{3}
With two qubits, the generative ratios shrink to:
 in binary:
 c_{01} ÷ c_{00} = c_{11} ÷ c_{10}
 c_{10} ÷ c_{00} = c_{11} ÷ c_{01}
 in quartal:
 c_{1} ÷ c_{0} = c_{3} ÷ c_{2}
 c_{2} ÷ c_{0} = c_{3} ÷ c_{1}
With only one qubit, the ratios degenerate, and the entanglement is zero.
In the fourqubit case, there will be four eightmember equations — incorporated whole are the threequbit ratios, here in boldface:
 in binary:

c_{0001} ÷ c_{0000} = c_{0011} ÷ c_{0010} =
c_{0101} ÷ c_{0100} = c_{0111} ÷ c_{0110} =
c_{1001} ÷ c_{1000} = c_{1011} ÷ c_{1010} =
c_{1101} ÷ c_{1100} = c_{1111} ÷ c_{1110}

c_{0010} ÷ c_{0000} = c_{0011} ÷ c_{0001} =
c_{0110} ÷ c_{0100} = c_{0111} ÷ c_{0101} =
c_{1010} ÷ c_{1000} = c_{1011} ÷ c_{1001} =
c_{1110} ÷ c_{1100} = c_{1111} ÷ c_{1101}

c_{0100} ÷ c_{0000} = c_{0101} ÷ c_{0001} =
c_{0110} ÷ c_{0010} = c_{0111} ÷ c_{0011} =
c_{1100} ÷ c_{1000} = c_{1101} ÷ c_{1001} =
c_{1110} ÷ c_{1010} = c_{1111} ÷ c_{1011}

c_{1000} ÷ c_{0000} = c_{1001} ÷ c_{0001} =
c_{1010} ÷ c_{0010} = c_{1011} ÷ c_{0011} =
c_{1100} ÷ c_{0100} = c_{1101} ÷ c_{0101} =
c_{1110} ÷ c_{0110} = c_{1111} ÷ c_{0111}
 in hexadecimal:

c_{1} ÷ c_{0} = c_{3} ÷ c_{2} =
c_{5} ÷ c_{4} = c_{7} ÷ c_{6} =
c_{9} ÷ c_{8} = c_{B} ÷ c_{A} =
c_{D} ÷ c_{C} = c_{F} ÷ c_{E}

c_{2} ÷ c_{0} = c_{3} ÷ c_{1} =
c_{6} ÷ c_{4} = c_{7} ÷ c_{5} =
c_{A} ÷ c_{8} = c_{B} ÷ c_{9} =
c_{E} ÷ c_{C} = c_{F} ÷ c_{D}

c_{4} ÷ c_{0} = c_{5} ÷ c_{1} =
c_{6} ÷ c_{2} = c_{7} ÷ c_{3} =
c_{C} ÷ c_{8} = c_{D} ÷ c_{9} =
c_{E} ÷ c_{A} = c_{F} ÷ c_{B}

c_{8} ÷ c_{0} = c_{9} ÷ c_{1} =
c_{A} ÷ c_{2} = c_{B} ÷ c_{3} =
c_{C} ÷ c_{4} = c_{D} ÷ c_{5} =
c_{E} ÷ c_{6} = c_{F} ÷ c_{7}
Quantum digits need not be binary; they can for instance be ternary — then we have qutrits. Here is for example a twoqutrit register:
State  Amplitude  Probability


│000 ⟩  −0.05931 + 0.34072i  0.11961

│001 ⟩  +0.20682 + 0.24024i  0.10049

│002 ⟩  −0.09873 − 0.02121i  0.01020

│010 ⟩  −0.30612 − 0.30631i  0.18753

│011 ⟩  +0.42147 + 0.17293i  0.20755

│012 ⟩  −0.23422 + 0.37470i  0.19526

│020 ⟩  −0.01559 − 0.02411i  0.00082

│021 ⟩  +0.23885 + 0.30264i  0.14864

│022 ⟩  −0.00347 − 0.17289i  0.02990

The generative ratios are:

c_{21} ÷ c_{20} =
c_{11} ÷ c_{10} =
c_{01} ÷ c_{00}

c_{22} ÷ c_{21} =
c_{12} ÷ c_{11} =
c_{02} ÷ c_{01}

c_{20} ÷ c_{22} =
c_{10} ÷ c_{12} =
c_{00} ÷ c_{02}

c_{12} ÷ c_{02} =
c_{11} ÷ c_{01} =
c_{10} ÷ c_{00}

c_{22} ÷ c_{12} =
c_{21} ÷ c_{11} =
c_{20} ÷ c_{10}

c_{02} ÷ c_{22} =
c_{01} ÷ c_{21} =
c_{00} ÷ c_{20}
One can even mix radices. The next example shows a register of one qubit and one qutrit:
State  Amplitude  Probability


│00 ⟩  −0.18028 + 0.29657i  0.12045

│01 ⟩  +0.25434 − 0.40179i  0.22612

│02 ⟩  −0.36540 − 0.22754i  0.18529

│10 ⟩  +0.03971 + 0.28238i  0.08132

│11 ⟩  −0.17964 − 0.30279i  0.12395

│12 ⟩  +0.39310 + 0.32914i  0.26286

The generative ratios are:

c_{11} ÷ c_{10} =
c_{01} ÷ c_{00}

c_{12} ÷ c_{11} =
c_{02} ÷ c_{01}

c_{10} ÷ c_{12} =
c_{00} ÷ c_{02}

c_{12} ÷ c_{02} =
c_{11} ÷ c_{01} =
c_{10} ÷ c_{00}

c_{02} ÷ c_{22} =
c_{01} ÷ c_{21} =
c_{00} ÷ c_{20}