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 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
- in hexadecimal:
-
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:
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:
-
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:
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:
-
c11 ÷ c10 =
c01 ÷ c00
-
c12 ÷ c11 =
c02 ÷ c01
-
c10 ÷ c12 =
c00 ÷ c02
-
c12 ÷ c02 =
c11 ÷ c01 =
c10 ÷ c00
-
c02 ÷ c22 =
c01 ÷ c21 =
c00 ÷ c20