Probabilistic

 

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Quantum computing; I am given Statevector v = ([(1 + 2.0j) / 3, -2 / 3]) What does the negative amplitude for |1> do

In quantum computing, a negative amplitude in a statevector represents a relative phase difference between the corresponding basis state and the other states in the superposition.

Given the statevector v = ([(1 + 2j) / 3, -2 / 3]), we can break it down as follows:

The first element (1 + 2j) / 3 represents the amplitude of the basis state |0>. The complex number 1 + 2j has a magnitude of sqrt(1^2 + 2^2) = sqrt(5) and an argument (phase) of arctan(2/1) = arctan(2).

The second element -2 / 3 represents the amplitude of the basis state |1>. The negative sign indicates a relative phase difference of π (180 degrees) between |1> and |0>.

When a qubit is in this state, the negative amplitude of |1> means that if the qubit is measured, the probability of collapsing to |1> is still positive (since probabilities are calculated as the square of the magnitude of the amplitude), but the phase of the resulting state will be opposite to that of |0>.

In other words, the negative amplitude introduces a phase flip between the two basis states, which can be important in quantum algorithms and interference patterns.

Which of the two states has the higher probability

To determine the probability of each state, we need to calculate the square of the magnitude of their respective amplitudes.

State |0>:

Amplitude = (1 + 2j) / 3

Magnitude = sqrt((1/3)^2 + (2/3)^2) = sqrt(1/9 + 4/9) = sqrt(5/9)

Probability = (sqrt(5/9))^2 = 5/9

State |1>:

Amplitude = -2/3

Magnitude = |-2/3| = 2/3

Probability = (2/3)^2 = 4/9

Since 5/9 > 4/9, state |0> has a higher probability of being measured.

                                                                      

Moving on with operators, albeit all on the same qubit: