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I understand that a Qubit can be represented in the form of $$\vert\psi\rangle=\alpha \vert0\rangle+\beta\vert1\rangle$$ where $\alpha$ and $\beta$ are complex numbers and the $\alpha^2$ and $\beta^2$ are the probabilities of the state of the Qubit and $$\vert\alpha\vert^2+\vert\beta\vert^2=1$$ If $\alpha$ and $\beta$ were equal to $\frac{1}{\sqrt2}$, the measurement would give us a superposition of states where we have a 1/2 probability for 0 and 1/2 probability for 1: $$\vert\psi\rangle=\frac{1}{\sqrt{2}}\vert0\rangle+\frac{1}{\sqrt{2}}\vert1\rangle$$

I can't find an explanation for the difference between $\vert-\rangle$ and $\vert+\rangle$ and what's the use of putting a negative sign for the $\vert1\rangle$ state.

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  • $\begingroup$ What does that mean by $|0\rangle$ and $|1\rangle$ ? $\endgroup$ – Wang Yun Oct 17 '18 at 12:55
  • $\begingroup$ This is a perfectly good question about a perfectly reasonable source of confusion from a student. Future downvoters, please reconsider $\endgroup$ – TheEnvironmentalist Oct 18 '18 at 13:29
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In quantum computing, when dealing with a single spin-1/2 particle, the conventional definitions for $|+\rangle$ and $|-\rangle$ are

$$|+\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$$

$$|-\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle$$

where $|0\rangle$ and $|1\rangle$ are the spin-up and spin-down configurations for a measurement along the $z$-axis. If you measure spin in the $z$-direction, the $|0\rangle$ state gives you a positive number, and the $|1\rangle$ state gives you a negative number.

In this implementation, $|+\rangle$ and $|-\rangle$ are the spin-up and spin-down states along the $x$-axis. If you measure spin along the $x$-axis, the $|+\rangle$ state gives you a positive number, and the $|-\rangle$ state gives you a negative number. The relation between the two axes and the two sets of states is given by the way that the Pauli matrices interact, and the two observables (spin along the $z$-axis and spin along the $x$-axis) are related by the uncertainty principle, since the spin operators for those observables (the Pauli matrices $\sigma_z$ and $\sigma_x$) do not commute with each other.

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The Qubit has 2 basis states. In your case they are named $|1\rangle$ & $|0\rangle$. But there is no reason that they can't be named $|-\rangle$ & $|+\rangle$. It is just a notation.

What the two stand for you have to look in the actual experiment. For example in a spin qubit the $|-\rangle$ can stand for a spin pointing downwards and the $|+\rangle$ for a spin pointing upwards.

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In the literature on quantum computation, the vectors $|+\rangle,|-\rangle$ are normally defined as $$|+\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ and $$|-\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle).$$ This definition provides a different description of a qubit that is useful for understanding some issues in quantum computation and quantum information:

https://en.wikipedia.org/wiki/Mutually_unbiased_bases

and

https://en.wikipedia.org/wiki/Bell_state

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  • $\begingroup$ @ZeroTheHero The +/- states are not always used to refer to spin up/down. See the link about the Bell state. $\endgroup$ – alanf Oct 17 '18 at 11:21
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Some of the "magic" in quantum mechanics that no one really explains to budding students is this notation.

In a quantum system, your states are defined by probabilities, but what they actually represent is largely up to your choice of system. They can represent spin, but they can also represent any other quantum variable, that is to say a variable that follows the math you so aptly describe.

Regardless of what it is you're actually using as your quantum variable, you need some kind of notation for your two base states. Some prefer $|+\rangle$ and $|-\rangle$, some prefer $|1\rangle$ and $|0\rangle$, if you want to be really creative (rather unusual for us types) you can even go with $|😃\rangle$ and $|😭\rangle$, it's really just notation.

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    $\begingroup$ ... or $\vert $ Larry $\rangle$ and $\vert$ Moe $\rangle$, or $\vert $ creamy $\rangle$ and $\vert$ smooth$\rangle$... $\endgroup$ – ZeroTheHero Oct 17 '18 at 11:55

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