5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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? $\endgroup$ – Ruvi Lecamwasam Jun 9 at 23:12
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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 yesterday
  • 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 22 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.