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$\newcommand\ket[1]{\left\vert #1\right\rangle}$ Let $\ket\phi = 12 \ket{0} + 1 + 2\sqrt{i2}\ket{1}$. Write $\ket\phi$ in the form $\alpha_0\ket{+} + \alpha_1\ket{-}$. What is $\alpha_0$?

I came across this problem in a course i am doing, i have been struggling writing things in sign basis, much appreciated.

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I suspect the $12$ and the $2 \sqrt i 2$ need some punctuation. Please type them in $LaTeX$ (you might see… for how) so we know what you mean. – Ross Millikan Jul 23 '12 at 20:44
Did you possibly mean $2\sqrt{2}\mathrm{i}$? While $2\sqrt{\mathrm{i}}2$ is a valid expression, it would be a very unusual one. – celtschk Jul 23 '12 at 20:44
I hope you do not want to break the rules of the online course" quantum mechanics and quantum info." delivered by Umesh Vazirani! – Ehsan M. Kermani Jul 23 '12 at 20:55
im not asking for them to do the quesiton, im asking for help on the topic – fosho Jul 23 '12 at 21:02
I've converted the math display to MathJax (to see whether it helps with . I tried to transcribe the mathematics faithfully as written, that is preserving the slightly odd expressions (adding a scalar to a state and taking the square root of $i$). It'd be great if the OP can clarify. – Willie Wong Jul 24 '12 at 8:29
up vote 1 down vote accepted

The "sign basis" is defined as $$\begin{aligned} \lvert+\rangle &= \frac{1}{\sqrt{2}}\left(\lvert0\rangle+\lvert1\rangle\right)\\ \lvert-\rangle &= \frac{1}{\sqrt{2}}\left(\lvert0\rangle-\lvert1\rangle\right) \end{aligned}$$ Those equations can be solved for $\lvert0\rangle$ and $\lvert1\rangle$. Then it's just a matter of inserting in the given state and comparing the coefficients.

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could you show a worked example if possible, would be a great help! – fosho Jul 23 '12 at 20:53
@Daniel: What is your problem: Solving a system of linear equations, or inserting values and comparing? – celtschk Jul 23 '12 at 20:56
inserting values... sorry im new to this – fosho Jul 23 '12 at 20:58
@Daniel: Well, that one is simple: After solving the linear equation system you've got equations $\lvert0\rangle=a\lvert+\rangle+b\lvert-\rangle$ and $\lvert1\rangle=c\lvert+\rangle+d\lvert-\rangle$. Your state has the form $e\lvert0\rangle+f\lvert1\rangle$. So you just replace the $\lvert0\rangle$ with $(a\lvert+\rangle+b\lvert-\rangle)$ and $\lvert1\rangle$ with $(c\lvert+\rangle+d\lvert-\rangle)$. Then you just have to use standard arithmetic to get it in simple form, and can read off the coefficients. Basically, forget that $\lvert*\rangle$ are quantum states and use standard arithmetics. – celtschk Jul 23 '12 at 21:04

Hint: All you need to know is that $|0>=\frac{1}{\sqrt{2}}(|+> +|->)$ and $|1>=\frac{1}{\sqrt{2}}(|+>-|->).$

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you can write this in general $A\lvert0\rangle+B\lvert1\rangle=\frac{(A+B)}{2}\lvert+\rangle + \frac{(A-B)}{2}\lvert-\rangle$ where A and B are complex numbers.

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