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Suppose a system is initially in a superposition: $$\psi(x) = \sum\limits_{i}|c_i\phi_i(x)\rangle$$ After a position measurement, the wave function collapses to one of the position eigenfunctions,$\phi_i(x).$ Geometrically, I understand this as projecting the wave function to one of its components along its position eigenbasis in Hilbert space.

If I then measure momentum, the wavefunction is projected to one of its component along its momentum eigenbasis. If I measure position again, would the set of position eigenbasis change? Or is it still the same set of position eigenbasis $\{|\phi_i\rangle\}$?

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The state, your initial state collapses on, is always one of the eigenstates of the observable you are measuring.

these eigenstates are defined a priori, and don't change as long as the observable doesn't change.

So, the formal answer to your question depends on the picture you are working in:

  • if you are in Schroedinger picture, where operators don't change and states evolve, the eigenvalues of the operator $\hat X$ wouldn't evolve in time and would be the same at every instant $t$

  • if you are in Heisenberg picture, and you make your measurements in two different instants $t=0$ and $t=t_1$, your position operator would be $\hat X(t_1)=e^{iHt_1/\hbar}X(0)e^{-iHt_1/\hbar}$ so the set of eigenstates of this operator are related to the initial one by

$|\phi(x_i,t_1)\rangle=e^{iHt_1/\hbar}|\phi(x_i,0)\rangle$

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  • $\begingroup$ Thanks for the answer! Geometrically, how does the set of eigenstates (in the Heisenberg picture you mentioned) changes in time? Are the eigenbasis rotating due to the factor ${e^iHt_1/\bar h}$? $\endgroup$ – Leo L. Jan 11 at 0:05
  • $\begingroup$ yes: basically you want to determine which are the eigenvectors of $X(t)$ at some time $t$, so you start from the definition: $X(0)|x\rangle= x|x\rangle$ at time $t=0$. Then, multiply both sides from the left by $e^{-Ht_1/\hbar}$ and insert $\mathbb I=e^{-Ht_1/\hbar}e^{Ht_1/\hbar}$ between $X(0)$ and $|x\rangle$: you reach the equation $e^{Ht_1/\hbar}X(0)e^{-Ht_1/\hbar}e^{Ht_1/\hbar}|x\rangle=e^{Ht_1/\hbar}|x\rangle$. But $e^{Ht_1/\hbar}X(0)e^{-Ht_1/\hbar}=X(t_1)$, so $e^{Ht_1/\hbar}|x\rangle$ is eigenvector of $X(t_1)$ with eigenvalue $x$, as we wanted. $\endgroup$ – Francesco Bernardini Jan 11 at 0:42
  • $\begingroup$ (obviously there's an imaginary unit in all those exponentials.......) $\endgroup$ – Francesco Bernardini Jan 11 at 0:54

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