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The basic qubit is written as $|\psi\rangle= \alpha |0\rangle + \beta |1\rangle$ where $p= |\alpha|^2 +|\beta|^2=1$ and $\alpha,\beta \in \mathbb{C}$. But I was reading about interference and that $p=|\alpha|^2+|\beta|^2+|\alpha||\beta|cos\theta$ where there is an interference term.

Now my question is:

Are we assuming that the qubit is not interfering with itself when we say $p= |\alpha|^2 +|\beta|^2=1$?

Or are the phases orthogonal to each other and $\alpha,\beta \in \mathbb{R}$?

Refer to the links here that I saw: link 1 link 2

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    $\begingroup$ This question might be more relevant to quantum physics forum, because once you are talking about quantum computing, the assumption is that all the wave function of the paricle looks like the regularknown one $\endgroup$
    – Ron Cohen
    Mar 3, 2022 at 5:34
  • $\begingroup$ Can you put a reference link to where you saw it? $\endgroup$
    – Ron Cohen
    Mar 3, 2022 at 8:00
  • $\begingroup$ Sure. I have been reading Prof Ekert's ebook. qubit.guide/2.2-quantum-bits-called-qubits.html qubit.guide/… $\endgroup$ Mar 3, 2022 at 11:50

2 Answers 2

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We don't tend to say that a qubit is interfering with itself,, but there is a strong parallel with what's going on with that initial example, so you could say that it's interfering with itself.

Let me replicate the example expressed using qubit terminology. Your qubit starts in a state $$ |\psi\rangle=\alpha|0\rangle+\beta|1\rangle. $$ The state is normalised (this is nothing to do with interference) such that $|\alpha|^2+|\beta|^2=1$. This simply says "the probability of getting either 0 or 1 when we measure the qubit is 1".

Now, let us apply a phase gate to it $$ |\psi\rangle\rightarrow \alpha|0\rangle+\beta e^{i\phi}|1\rangle. $$ Next, we apply a gate called Hadamard (equivalently, you could apply a beamsplitter). This is the gate that makes the different components of the qubit state (when viewed from a particular perspective) interfere with each other. $$ |\psi\rangle\rightarrow \frac{1}{\sqrt{2}}(\alpha+e^{i\phi}\beta)|0\rangle+\frac{1}{\sqrt{2}}(\alpha-e^{i\phi}\beta)|1\rangle. $$

So, you should now notice two things. The first is that we we measure the qubit, the probability that we get answer 0 is $$ p_0=|\alpha+e^{i\phi}\beta|^2=\frac{1}{2}\left(|\alpha|^2+|\beta|^2+\alpha\beta^*e^{-i\phi}+\alpha^*\beta e^{i\phi}\right). $$ If we write $\alpha\beta^*=|\alpha| |\beta| e^{i\gamma}$, then this simplifies to $$ \frac{1}{2}\left(|\alpha|^2+|\beta|^2+2|\alpha||\beta|\cos(\phi-\gamma)\right)=\frac{1}{2}+|\alpha||\beta|\cos(\phi-\gamma). $$ So, it's exactly that interference term that you were looking for.

The second thing to notice is that this doesn't contradict the normalisation condition. If you look at the probability of getting the 1 answer, that also shows an interference term, $$ p_1=\frac{1}{2}-|\alpha||\beta|\cos(\phi-\gamma) $$ and so $p_0+p_1=1$.

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  • $\begingroup$ Two questions: 1. Do $\alpha$ and $\beta$ have a phase term when you start out? ie are they real part of complex number or are they complex numbers? 2. After the Hadamard gate, the equation you wrote is a superposition of superpositions? This is a mixed state right? $\endgroup$ Mar 4, 2022 at 9:01
  • $\begingroup$ 1: $\alpha$ and $\beta$ are complex. 2. Yes, you could call it a superposition of superpositions. But that's still just a superposition (the information about the original superposition is largely lost). It's not mixed, it's still just a pure quantum state in superposition, as it's a qubit $\alpha'|0\rangle+\beta'|1\rangle$, just with $\alpha'=(\alpha+\beta e^{i\gamma})/\sqrt{2}$. $\endgroup$
    – DaftWullie
    Mar 4, 2022 at 9:08
  • $\begingroup$ Sorry but why does $|\beta|^2$ become zero when you simplify $p_0$? Also is this a typo? $p_0+p+1=1$ Shouldn't it be $p_0+p_1=1$? $\endgroup$ Mar 5, 2022 at 1:05
  • $\begingroup$ Nevermind the first part of my question figured it out. Its because $|\alpha|^2+|\beta|^2=1$. $\endgroup$ Mar 5, 2022 at 1:08
  • $\begingroup$ Thank you. I think your answer cleared up a lot of confusion I've been troubled with. So to summarize the points, a single qubit by definition does not interfere with itself until it is put into a Hadamard gate. I think whats been really confusing to me in general is keeping track of whether the state is before or after measurement. $\endgroup$ Mar 5, 2022 at 3:46
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You are confusing what you saw with the double-slit experiment.

The whole idea of qubits, is finding a physical system with 2 orthogonal bases:

$$|\psi\rangle=a|0\rangle+b|1\rangle$$

They imply that

$$\langle0|1\rangle=\langle0|1\rangle=0$$

In the case of double slit experiment, they are not orthgonal, and than:

$$\langle \psi|\psi\rangle=|a|^2\langle0|0\rangle+|b|^2\langle1|1\rangle+ab^*\langle1|0\rangle+a^*b\langle0|1\rangle$$

This is not correct in the case of 2 paths of light in this experiment, they are not orthogonal to each other.

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  • $\begingroup$ So to answer my question: The qubit by definition is not interfering with itself? $\endgroup$ Mar 3, 2022 at 13:02
  • $\begingroup$ No, or in another words - the result of interference between |0> and |1> is zero $\endgroup$
    – Ron Cohen
    Mar 3, 2022 at 14:10
  • $\begingroup$ Great. That last equation was helpful. Thx $\endgroup$ Mar 3, 2022 at 14:47
  • $\begingroup$ ⟨𝜓∗|𝜓⟩ : do you mean conjugate of conjugate here? $\endgroup$ Mar 4, 2022 at 4:25
  • $\begingroup$ No, my mistake. Thank you, Fixed $\endgroup$
    – Ron Cohen
    Mar 4, 2022 at 7:03

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