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I have an issue, perhaps with normalization with the following state. For $\alpha^2 + \beta^2 =1 $, the probabilities in this state does not sum up to 1.

$$|\psi\rangle := \frac{1}{2}\left[\alpha\left(|0\rangle(|x\rangle+|x'\rangle\right) + \beta \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$

as $p(0)=\frac{2\alpha^2+2\alpha^2\langle x|x'\rangle}{4}=\alpha^2(\frac{1+\langle x|x'\rangle}{2})$ and $p(1)=\frac{2\beta^2 - 2\beta^2\langle x | x'\rangle}{4} = \beta^2\frac{1-\langle x | x'\rangle}{2}$.

And $p(0)+p(1) = \frac{\alpha^2 + \beta^2 + \alpha^2\langle x| x'\rangle- \beta^2\langle x | x'\rangle }{2} = \frac{1 + (\alpha^2 - \beta^2) \langle x | x'\rangle}{2}$

Long version:

I created the state as follows:

  • $|0\rangle |0\rangle$
  • Hadamard on first qubit $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|0\rangle$
  • Controlled operations: if 0 on first register I create state $|x\rangle$ otherwise I create state $|x'\rangle$ on second qubit. $\frac{1}{\sqrt{2}}(|0\rangle|x\rangle+|1\rangle|x'\rangle)$

  • Now a perform a second Hadamard gate and re-group the terms:

$$ \frac{1}{2}\left[\left(|0\rangle(|x\rangle+|x'\rangle\right) + \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$

  • Now I am ready to do a rotation on the $Y$-axis of $\alpha$ on the first qubit. This leads to: $$ \frac{1}{2}\left[\alpha\left(|0\rangle(|x\rangle+|x'\rangle\right) + \beta \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$

Perhaps I am doing something wrong with the normalization? But I don't need any normalization factor as $\alpha^2+\beta^2=1$

Thank you.

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    $\begingroup$ The problem is at the step where you claim to do a $y$ rotation. All you've done is added extra coefficients, which is clearly not going to maintain normalisation. You need to actually calculate the effect of the $Y$ rotation (which will not keep the state of the second qubit as nicely as you're obviously hoping). $\endgroup$
    – DaftWullie
    Apr 24 '20 at 10:32
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$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$

Let's start where the quantum world is still in order:

Now I perform a second Hadamard gate and re-group the terms: $$ \frac{1}{2}\left[\left(|0\rangle(|x\rangle+|x'\rangle\right) + \left(|1\rangle(|x\rangle-|x'\rangle \right)\right] $$ Now I am ready to do a rotation on the $Y$-axis of $\alpha$ (i.e. $\tiny\pmatrix{\cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&\cos(\alpha)}$) on the first qubit. This leads to:

$$ \frac{1}{2}\left[\left((\cos(\alpha)\ket0 +\sin(\alpha)\ket1)(|x\rangle+|x'\rangle\right) + \left((-\sin(\alpha)\ket0+\cos(\alpha)\ket1)(|x\rangle-|x'\rangle \right)\right] $$

Since I'm not sure where you want to go from here, I end...

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