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:


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.


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – P_Gate
    Jun 8, 2019 at 12:00
  • 1
    $\begingroup$ 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? $\endgroup$ Jun 9, 2019 at 23:12

You have applied a

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


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.

  • $\begingroup$ 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). $\endgroup$
    – P_Gate
    Jun 8, 2019 at 11:54
  • $\begingroup$ 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? $\endgroup$
    – P_Gate
    Oct 16, 2019 at 17:43
  • 1
    $\begingroup$ 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. $\endgroup$
    – AHusain
    Oct 16, 2019 at 22:39

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