Does the eigenbasis associated with an observable changes after measuring a different observable?

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 $$\{|\phi_i\rangle\}$$?

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

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

• 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 ${e^iHt_1/\bar h}$? – Leo L. Jan 11 at 0:05
• 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. – Francesco Bernardini Jan 11 at 0:42
• (obviously there's an imaginary unit in all those exponentials.......) – Francesco Bernardini Jan 11 at 0:54