I have one qubit and I apply two gates to it: H and T, which yields the following superposition:

$$ \frac{1}{\sqrt{2}} |0\rangle + \frac{1+i}{2}|1\rangle $$

Now I want to calculate probability of 0 and 1 state:

$$ \left(A_1\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} $$ $$ \left(A_2\right)^2 = \left(\frac{i+1}{2}\right)^2 = \frac{i}{2} $$ $$ S = \left(A_1\right)^2 + \left(A_2\right)^2 = \frac{1+i}{2} $$ $$ P(A_i) = \frac{A_i^2}{S} $$ according to prior equations i receive following values $$ P(0) = \frac{1}{1+i} \land P(1) = \frac{i}{i+1} $$

I was expecting to get $A_i=0.5$ however $P(0) + P(1) = 1$

How do I interpret this results?


2 Answers 2


You forgot to take the absolute value.

The Born rule for computing measurement outcome probability from the state vector amplitudes says that the probability is the square of the magnitude of the amplitude.

In your case, we get

$$ \left(A_1\right)^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} $$ $$ \left(A_2\right)^2 = \left|\frac{i+1}{2}\right|^2 = \left(\frac{\sqrt{2}}{2}\right)^2 =\frac{1}{2} $$

as expected.


Just to note that the complex number also specify a quantum phase of a qubit. In this particular case $$ \frac{1+i}{2} = \frac{1}{\sqrt{2}}\mathrm{e}^{i\frac{\pi}{4}}, $$ so the relative phase is $\frac{\pi}{4}$.

You can rewrite your qubit state as $$ |q\rangle = \frac{1}{\sqrt{2}}(|0\rangle + \mathrm{e}^{i\frac{\pi}{4}} |1\rangle). $$

Now, you easily see that probability of $|0\rangle$ and $|1\rangle$ is $\frac{1}{2} = \Big(\frac{1}{\sqrt{2}}\Big)^2$ in both cases and the phase is $\pi/4$.


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