What does $|<\psi|\psi>|^2=1$ physically mean ? Can we use born's rule for superpositioned states ? Does it carry any physical meaning ? Why don't wquwe ever observe a superosotioned state then?
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2$\begingroup$ @JohnRennie, why would OP mean that? It's a bra, not a wavefunction; there's no need to complex conjugate the bra-ket label. $|<\psi|\psi>|$ is the notation used in every single quantum mechanical text I've ever read. $\endgroup$– user27578Commented Mar 3, 2014 at 9:30
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The relation $|\langle \psi|\psi \rangle |^2=1$ is the normalization condition for quantum states - so by itself, it doesn't mean anything. It only means something if you put it together with the fact that time evolution is unitary and hence preserves this norm.
Then, if you define the process of measurements, it turns out that the probabilities can be inferred from the quantum state by Born's rule and the normalization of the probability density corresponds to the normalization of the state.
Born's rule (in the form most often seen in your standard QT course) states that for an arbitrary quantum state $|\psi\rangle$ and an observable $A$, the probability to measure the eigenvalue $a_i$ corresponding to the spectral projection $P_i$ of $A$ is given by $|\langle \psi| P_i|\psi\rangle|^2$. This is a postulate well proven by experiment and it holds for every state - also for superposition states.
However, this has nothing to do with the question, why you don't see superposition states in nature - in fact you do, quantum information wouldn't be possible without superposition. You only don't see macroscopic superposition states (or at least, seeing them is very difficult; superpositions have been achieved with buckyballs, quite big molecules, in double-slit experiments). Why do you normally not see superposition states in macroscopic objects? This is due to decoherence and has not much to do with Born's rule.
