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I am assuming that the wave functions contain all the possible information about the system, how is it then for some observables like spin, the wave function can be expressed as a linear combination of two eigenvectors, where as for observables like position, the wavefunction is an infinite dimensional vector?

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2 Answers 2

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Doing an experiment to measure spin in a certain direction (Stern-Gerlach) one only observes two results (spin-up, spin-down). Since for any state of the physical system (wave function) it must be possible to express it as a linear combination of the eigenstates corresponding to the observable (spin), the state (wave function) can be expressed as a sum of the two eigenvectors (spin-up,spin-down).

On the other hand, if the observable you're looking at is position, any vector $\vec{v} \in \mathbb{R}^3$ is a possible measurement, which are infinitely many. So in this case the wave-function is an infinite dimensional vector.

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  • $\begingroup$ The question was more regarding the mathematical consistency, if we take the wavefuction as a mathematical object,then what will it's dimension be?Since It can't be two different dimensions at the same time, how do we reconcile the infinite dimensional representation with the two dimensional one mathematically. $\endgroup$
    – veke
    Commented Aug 21, 2023 at 22:12
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It's not a matter of the choice of observable, but of the choice of system.

If you only plan to measure a particle's spin, and that spin is not coupled to the same particle's position, then you can treat the position as part of the environment. If you plan to measure the position, then you have to include the position in the system. Even if the particle is physically the same in both problems, you are splitting the world into system and environment in two different ways, so the Hilbert space is different.

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