What would be the meaning of an $i$ in a qubit state $i\alpha|0\rangle+\beta|1\rangle$?

Question

I do not know if the question is not too easy, but I'll put it here, because I'm interested in it.

So the state of a qubit is often stated in this form: $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$ An example would be: $$|\psi\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$$ So now to my actual question. Suppose that in front of the $\alpha$ and/or $\beta$ would be an $i$.

About as:

$$|\psi\rangle=i\alpha|0\rangle+\beta|1\rangle$$

What would that say, what would be the meaning of it. In short what does that mean? What does this "$i$" say?

I hope my question is understandable.

Answer 1

Let's consider the state $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle.$$ As you said, the magnitudes $|\alpha|^2$ and $|\beta|^2$ give you the relative probabilities of finding the state in $|0\rangle$ and $|1\rangle$ if you make a measurement in the $\{|0\rangle,|1\rangle\}$ basis.

There is more to it than the magnitudes however, because quantum mechanics is based on the idea of waves, and waves have both a magnitude and a phase. Imagine two sine waves of equal amplitude, what happens when you interfere them with each other? The answer depends on if they are in-phase (they sum to something with double the amplitude), out-of-phase (the result has zero amplitude), or something in between (somewhere between zero and double amplitude).

Similarly, when you see $\alpha|0\rangle$, imagine $\alpha$ as a ray in the 2D complex plane. This ray has magnitude $|\alpha|$, but it also has a phase. If we interfere our qubit $|\psi\rangle$ with another $|\phi\rangle=\alpha'|0\rangle+\beta'|1\rangle$, the result will be $$\frac{1}{\sqrt{2}}\left(|\psi\rangle+|\phi\rangle\right)=\frac{\alpha+\alpha'}{\sqrt{2}}|0\rangle+\frac{\beta+\beta'}{\sqrt{2}}|1\rangle.$$ The coefficient of $|0\rangle$ is found by adding together the two rays $\alpha$,$\alpha'$, and similarly for $|1\rangle$ and $\beta$,$\beta'$. Even if $|\alpha|=|\alpha'|$, $|\beta|=|\beta'|$, the result will be very different depending on the phases.

Answer 2

You have applied a

$$ U = \begin{pmatrix} i &0\\ 0&1 \end{pmatrix} $$

gate.

You have not affected the probabilities of measuring $0$ or $1$ in the computational basis but you have affected other observables.

For example, consider the case you described already as

$$ \mid \psi \rangle = \frac{1}{\sqrt{2}} \mid 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 \rangle\\ U \mid \psi \rangle = \frac{i}{\sqrt{2}} \mid 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 \rangle\\ $$

Before if you applied $X$ you would just get the same state back.

$$ X \mid \psi \rangle = X \frac{1}{\sqrt{2}} \mid 0 \rangle + X \frac{1}{\sqrt{2}} \mid 1 \rangle\\ = \frac{1}{\sqrt{2}} \mid 1 \rangle + \frac{1}{\sqrt{2}} \mid 0 \rangle\\ = \frac{1}{\sqrt{2}} \mid 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 \rangle\\ = \mid \psi \rangle\\ X U \mid \psi \rangle = X \frac{i}{\sqrt{2}} \mid 0 \rangle + X \frac{1}{\sqrt{2}} \mid 1 \rangle\\ = \frac{i}{\sqrt{2}} \mid 1 \rangle + \frac{1}{\sqrt{2}} \mid 0 \rangle\\ \neq U \mid \psi \rangle $$

So before measuring X gave 1 with probability 1, but after applying $U$ it did not.

So even though it didn't affect the probabilities in the eigenbasis of the Z operator which gives the computational basis, if you do other transformations, you can see the effect.