The wikipedia says on measurement in quantum mechanics that:
Repeating the same measurement without any evolution of the quantum
state will lead to the same result.
On the other hand, doesn't uncertainty (in momentum) entail that I can't expect to measure the same position of a particle twice?
EDIT – I found the following excerpt by Feynman in the second chapter of the first volume of his lectures; it seems related to the question and possibly at odds with some of the answers:
What keeps the electrons from simply falling in? This principle: If
they were in the nucleus, we would know their position precisely, and
the uncertainty principle would then require that they have a very
large (but uncertain) momentum, i.e., a very large kinetic energy.
With this energy they would break away from the nucleus. They make a
compromise: they leave themselves a little room for this uncertainty
and then jiggle with a certain amount of minimum motion in accordance
with this rule. (Remember that when a crystal is cooled to absolute
zero, we said that the atoms do not stop moving, they still jiggle.
Why? If they stopped moving, we would know where they were and that
they had zero motion, and that is against the uncertainty principle.
We cannot know where they are and how fast they are moving, so they
must be continually wiggling in there!)
Best Answer
The Projection Postulate states that if we have an observable $O$ with discrete spectrum $\{\lambda_i\}$, after a measure in a system resulting in the eigenvalue $\lambda_a$, the initial system $|\psi_i\rangle$ is reducted to the state $$|\psi_f\rangle=\frac{P_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}$$ where $P_a$ is the projection on the eigenspace of the eigenvalue $\lambda_a$. Any repeated measurement of $O$ in this state without time evolution will yield the same result $\lambda_a$, since $$O|\psi_f\rangle=\frac{OP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\frac{\lambda_aP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\lambda_a|\psi_f\rangle$$ i.e., $|\psi_f\rangle$ is an eigenvector of $O$ with eigenvalue $\lambda_a$.
In the case of continuous spectrum, there is some complications, since the eigenvectors are not normalizable and hence are not acceptable states. But as the measurement of an exact value of position, for example, cannot be made with arbitrary precision, what is really being measured is the projection over some region of the spectrum, like the dimension of the detector. In that case, say the particle is measured in the region $[a,b]$. By the projection postulate, the final state will be (in position representation, ignoring the normalization): $$\psi_f(x)=P_{[a,b]}\psi_i(x)=\chi_{[a,b]}\psi_i(x)$$ where $\chi_{[a,b]}$ is the characteristic function of $[a,b]$. That is, the final wavefunction will be the restriction of the initial one to the interval $[a,b]$. Realize that we still have an uncertainty in the momentum as expected, but if the system does not evolve with time, any other measurement of position, will still be restricted to $[a,b]$, since the final wavefunction support is in that region.