The quantum state just after a position measurement The wave function of a free particle is given as,
$$\psi(x) ~=~ e^{-{ x }^{ 2 }/{ a }^{ 2 }}.$$
Then a position measurement is made and the position of the particle is found 
to be at $x=a$.
My question is: What is the state of the particle just after this measurement? Is it equal to $\psi (x=a)$  ?
 A: In a true measurement procedure of position, the outcome is an interval $(a-\delta,a+\delta)$, $\delta>0$ being the precision  of the instrument. In view of Luders-von Neumann's postulate on the reduction of the state, if the state before the measurement was  described by the normalized vector $\psi \in L^2(R)$, immediately after the measurement the state is represented by the vector: $$\psi'(x) = \frac{\chi_{(a-\delta,a+\delta)}(x) \psi(x)}{\sqrt{\int_{a-\delta}^{a+\delta}|\psi(s)|^2 ds} }$$
where
$$\chi_{(a-\delta,a+\delta)}(x) =1 \quad \mbox{if $x \in (a-\delta,a+\delta)$}\qquad \chi_{(a-\delta,a+\delta)}(x) =0 \quad \mbox{if $x \not\in (a-\delta,a+\delta)$}$$
This is coherent with the fact that: 
$$\int_{a-\delta}^{a+\delta}|\psi(s)|^2 ds$$
is the probability to find the particle in $(a-\delta,a+\delta)$.
The new wavefunction belongs to $L^2(R)$ again, as it is due.
A: When the position is measured on this state, the wavefunction collapses to a position eigenstate which is a Dirac-delta function $\delta(x-a)$ spiked about $x=a$. The measurement of position on any wavefunction collapses the state a position eigenstate. This is the case immediately after the measurement. This is not a stationary state and with time it will evolve in a non-trivial fashion which can easily be worked out if the governing Hamiltonian is known.
A: Every measurement of a continuous variable, although it may give one definite result $a$, is loaded with uncertainty (error) which means that the actual value $x^*$ could be different than the result of the measurement $a$. This means that after the measurement, the state of the particle is such that it is somewhere around $a$. The uncertainty in knowledge of how far from $a$ is the actual $x^*$ can be quantified by variance $\langle (x^* - a)^2 \rangle$. This is the lower the higher the accuracy of the original measurement.
A: So $\psi(x = a)$ is not a state, because it isn't a function --- it's just a number. In fact, evaluating this number, you've just asked "is the state of the particle 0.368?" Hopefully you see that this is a nonsensical statement!
One of the central rules of quantum mechanics is that, when a measurement of an observable $O$ is made, the outcome is one of the eigenvalues of $\hat{O}$, and the system collapses into the corresponding eigenstate of $\hat{O}$.
So if position is measured, and found to be $x=a$, then the system is going to collapse into the eigenstate of position with eigenvalue $a$. Informally, this state is the delta function.
$$\psi(x) = \delta(x-a) \qquad \mathrm{after\ measurement}$$
To see this, note that we're looking for a function $\psi$ that, when multiplied by $x$, returns the same function scaled by a factor $a$. At first glance it might appear that there is no such function: $\sin x$ is certainly not proportional to $x \sin x$, and $1/x^2$ looks rather different to $1/x$ etc. The delta function is the only thing that works, because it's only non-zero at one specific point. Thus, multiplying it by $x$ has the effect of changing the height of the spike by a certain amount $a$ corresponding to where the spike is, leaving everything else zero. This is equivalent to multiplying the whole function by $a$. Hence it is an eigenstate.
A: Thanks. So, I understand that after this measurement the wave function can be written as $\psi (x)=\delta (x-a)$. What if a second measurement is made after some time t, to find an eigenvalue of another observable? What's the form of the wave function that's going to evolve with time? Is it $\psi (x)={ e }^{ -\frac { { x }^{ 2 } }{ { a }^{ 2 } }  }$ or $\psi (x)=\delta (x-a)$ ?. In other words, what the wave function after some time t?
$\psi (x,t)={e}^{-iHt/\hslash}{ e }^{ -\frac { { x }^{ 2 } }{ { a }^{ 2 } }  }$ or
$\psi (x,t)= {e}^{-iHt/\hslash}\delta (x-a)$. Here, H is the Hamiltonian of the above free particle.  I think the latter is true.
