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Am I using Bra-ket notation correctly? I want to define a superposition of the states, $|0\rangle$ and $|1\rangle$. Is it simply?

$|0\rangle + |1\rangle$

I can't find any non-technical information on it. I'm just a hobbyist.

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    $\begingroup$ Well, any $a\lvert 0 \rangle + b \lvert 1 \rangle$ will be a "superposition" of the two states for arbitrary $a,b \in \mathbb{C}/\{0\}$. You've written down the superposition that is equally likely to be a one of the two if you are measuring the observable of which $\lvert 0 \rangle$ and $\lvert 1 \rangle$ are eigenstates. What is the question, exactly? $\endgroup$
    – ACuriousMind ♦
    Commented Aug 21, 2014 at 3:00
  • $\begingroup$ hi @ACuriousMind, I know this question is a bit old, but am I right in saying that |0> + |1> is "bad" in the sense that a^2 + b^2 should equal 1? Is that normalization just something from quantum computing, or is it always necessary? $\endgroup$
    – Mohammad Athar
    Commented Feb 21, 2020 at 19:04

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Yes, that is correct. A more general form of the superposition of the stationary state would be $$a|0\rangle + b|1\rangle$$ where $a,b$ describes the probability of each state. The state : $$|0\rangle + |1\rangle$$ assumes that the state $|0\rangle$ and $|1\rangle$ are equally probable.

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Close, but not quite. Since quantum mechanics deals in probabilities, it is necessary to "normalize" the state in order to use it in later calculations. The most general state for the two-state quantum system you're considering would be \begin{equation} |\psi\rangle = \alpha\,|0\rangle+\beta\,|1\rangle \end{equation} where the quantities $\alpha$ and $\beta$ are NOT probabilities, but are complex numbers which satisfy the "normalization condition" \begin{equation} |\alpha|^2+|\beta|^2=1 \end{equation} In case you are not familiar with it, the notation $|\alpha|^2$ means $\alpha$ times its complex conjugate.

The value of $|\alpha|^2$ is a real number, and is equal to the probability that a measurement of the system will find it the state corresponding to $|0\rangle$. Similarly, the value of $|\beta|^2$ is equal to the probability that the system will be found in state $|1\rangle$.

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    $\begingroup$ Particularly, if one wants to construct a state with equal probabilities, s/he can set $|\psi\rangle=\tfrac{1}{\sqrt{2}}|0\rangle+\tfrac{1}{\sqrt{2}}|1\rangle$ or $|\psi\rangle=\tfrac{1}{\sqrt{2}}|0\rangle-\tfrac{1}{\sqrt{2}}|1\rangle$ or $|\psi\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle+\exp(i\phi)|1\rangle)$ for any angle $\phi$. $\endgroup$
    – firtree
    Commented Aug 21, 2014 at 14:29

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