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Hallo I'm trying to understand the concept of representation in the position space. I read that $|x\rangle$ are the eigenstates of the position operator, but I think this states should evolve in time since there aren't stationary states with a precise position?

What does $|x\rangle$ really mean?

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The states $|x\rangle$ are eigenstates of the position operator, and they do not change with time.

That means that they are not solutions of the Schrödinger equation. This is fine: not every state in the Hilbert space* needs to evolve with time or obey the Schrödinger equation.

If you do take a position eigenstate as the initial state for a Schrödinger-equation evolution, then the state will obviously evolve, since it is not in an eigenstate of the hamiltonian. For a free particle, it will immediately spread over all of space; for the details, see The Dirac-delta function as an initial state for the quantum free particle.

*That's a minor cheat - the position eigenstates are not actually in the Hilbert space. That doesn't affect the conclusions, though.

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  • $\begingroup$ Does this mean that in representation theory I can use states that phisically doesn't exist? The important is that their combination satisfy the Schrodinger equation (and so it exist) $\endgroup$
    – SimoBartz
    Commented Mar 14, 2019 at 16:47
  • $\begingroup$ Be careful about the term 'representation theory' - you're almost certainly not using it in the usual sense of the term. $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2019 at 16:52
  • $\begingroup$ There are two different aspects here - whether the states "physically exist", and whether they are Schrödinger-equation solutions - the two are not synonymous. The answer to the latter is yes - it is perfectly fine to construct your state as a linear combination of states in a basis that does not satisfy the Schrödinger equation, so long as the combination as a whole does. $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2019 at 16:54
  • $\begingroup$ The "physical existence" of the position eigenstates is a more complicated matter, though, and it is in question not because they don't satisfy the Schrödinger equation, but because they are not normalizable, which means that they do not live in the usual Hilbert space. This can be fixed in a rigorous way (the gory details are here) but you don't really need to worry about that at this stage. $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2019 at 16:56
  • $\begingroup$ @EmilioPisanty The position eigenstate is not physically meaningful because it would contradict the Heisenberg uncertainty principle (a quantum object cannot have a fixed position). That it is not normalizable is more of a consequence. $\endgroup$
    – yuggib
    Commented Mar 15, 2019 at 6:59

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