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When a particle like an electron is at x position, it is at a quantum state for that position. If it has more of these states, can it be at more positions?

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Until we measure it, and therefore fix it's position (for the time of measurement only), the idea of it having "a position or positions" is not really meaningful in quantum mechanics.

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A particle can be in quantum superposition, which can place it in multiple quantum states simultaneously. However, if the particle is "observed", that is if information is extracted from it, the quantum states will decohere, and the particle then will occupy only the state in which it is "observed".

Researchers at the University of Bonn designed a clever experiment which may have succeeded in pulling a single Caesium atom to two places at once. The had to use indirect measurements to avoid quantum decoherence.

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Short answer is that when you do a measurement to find a particles position you will always measure it to be at one position $x$ with a measurement uncertainty of $\Delta x$.

Every wave function $\Psi(x)$ will already give you a multitude of possible valid $x$ positions with different probabilities to occur. (note: $\delta(x)$ is not a valid wave function as it is not square-integrable)So you do dot need the superposition for the basis of your question. When you do your measurement to determine the current position of the particle, the previous state $\lvert \Psi \rangle$ breaks down to (i.e. will be projected to) a state in a basis which is given by your measurement $\lvert\Phi\rangle= \sum_{i} \phi_i \lvert\Phi\rangle $.

The wiki page about the Measurement in quantum mechanics is a good follow up to read.

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    $\begingroup$ Nitpick: The "almost" is unnecessary, every wavefunction has uncertainty in position. The only imaginable "function" that does not - $\delta(x)$ - is neither square-integrable nor a function to begin with, so it is not an allowed state. $\endgroup$
    – ACuriousMind
    Feb 19 '17 at 20:35
  • $\begingroup$ In fact I was was using almost because of $\delta(x)$. Thank you for refreshing my memory that it is not square-integrable @ACuriousMind. $\endgroup$
    – user_na
    Feb 21 '17 at 18:26
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It is only at one place at any given time which we find only by detection. It certainly is not at more than one positions at a time because that has never been detected experimentally.

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  • $\begingroup$ No, the issue is not "smallness" at all. It's that it is inherently meaningless to talk of "the position" of a quantum object prior to position measurement unless in terms of expectation values and standard deviations. $\endgroup$
    – ACuriousMind
    Feb 19 '17 at 20:33
  • $\begingroup$ @ACuriousMind: changed. $\endgroup$
    – kpv
    Feb 19 '17 at 21:23
  • $\begingroup$ "certainly not" and "because" does not compute. $\endgroup$ Feb 19 '17 at 23:01
  • $\begingroup$ @RussellMcMahon: If "experimentally" does not compute, then nothing computes. $\endgroup$
    – kpv
    Feb 19 '17 at 23:19