Quantum Computing: Qubelet amplitudes and probabilities (p. 40 and 142 ff.)

Given a qubit with 3 pentagon |0> qubelets and 1 triangle |1> qubelet. As described on p. 40 and in the answer to question 1.b.ii, " …the relative frequencies of recording the binary state 0 to 1 is 3:1" As per my understanding, this means that with a probability of 75% the qubit collapses to the binary state 0.

On page 142 ff. it is shown that the squared (normalized) amplitudes represent the probabilities of the qubit collapsing to a certain state.

For the above mentioned qubit, 3 |0> + 1 |1>, the normalized amplitude w_0 is 3 / sqrt(3^2 + 1^2) = 3 / sqrt(10). This means, the probability of this qubit collapsing to |0> is w_0^2, meaning (3/sqrt(10))^2 = 9/10 = 90%.

As per my current understanding, this contradicts the statement mentioned earlier from page 40, stating the probability would be 75%. I was also under the impression, that the qubelet model tried to convey that the ratio of pentagon vs triangle qubelets are the propabilties of the qubit collapsing to one of these qubelets.

Any help and clarification is very much appreciated!

You are correct. The answer to the statement 1.b.ii on Page 40 should be False, not True as stated in the answers. The rest of your understanding that the probability of the qubit collapsing to |0> is 9/10 is correct.

Thank you for pointing this out.

Thank you very much for your swift reply! And I just wanted to add that the qubelet model has helped me tremendously in getting an intuitive understanding of what is going on in a quantum circuit.