# Complex number in the representation of a qubit

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 $$i$$.

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

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.

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.

• Thank you for your answer. Could you make something more understandable for me, as how I can imagine the imaginary part. Put simply, what does it mean when I write an "i" in teh representation of a qubit? Jun 8 '19 at 12:00
• Maybe think instead about waves. We can write a one-dimensional wave as $A(x,t)=A_0e^{i(kx-\omega t)}$. Here $A_0\in\mathbb{R}$ is the max amplitude, $k$ the wavenumber, $\omega$ the frequency, and $\mathrm{Re}\left(A(x,t)\right)$ gives you the visible wave. The phase of the wave at a given point in space is $kx-\omega t$, so $i=e^{i\pi/2}$ gives you a phase of $\pi/2$. Imagine two sine waves, a phase of $i$ means that one is $\pi/2$ ($1/4$ wave) out of phase with the other. As we see in the example, this phase difference becomes important when we interfere two qubits. Does that make sense? Jun 9 '19 at 23:12

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.

• Thank you for your explanation. To be honest, I have not even thought of a special gate in my question. I was actually just interested in what the imaginary part of a qubit says (regardless of any gate). Jun 8 '19 at 11:54
• In addition to your last equation: $$\frac{i}{\sqrt{2}}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle \neq \frac{i}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$$ holds. But if I measure in base $$\{|0\rangle,|1\rangle\}$$ the probability for a measurement in $|0\rangle$ is still at 50% for both sides of the equation. So how can I imagine the $i$, since it does not effect the measurement in my case? Oct 16 '19 at 17:43
• The last equation is about measuring in the eigenbasis of the X operator. The point is that there are other observables besides the Z being used for computational basis. Oct 16 '19 at 22:39