Violation of Heisenberg uncertainty principle

Question

When we measure the position of an object its wavefunction collapses to infinity at a particular point. So if we continuously measure the position of the object it will give same value continuously. This indicated the velocity of the object is $0$ and the value of position is the value we get after repeated measurements which is constant. So we have violated the Heisenberg principle by simultaneously measuring both with $100$% accuracy. I know I am wrong but where am I wrong?

Answer 1

So you measure it, and it collapses to $\delta(x_0)$, or, if you don't like it, have a normalized gaussian $\psi(x)$ with $\sigma \rightarrow 0$, either way:

Now in momentum space, that is a mix of all momenta and it immediately starts expanding in $x$ space.

Actually, it's best to start with a narrow gaussian and find $\phi(x)$ in momentum space.

The part where you say "now the velocity is zero" just doesn't apply to quantum mechanics.

Answer 2

When we measure the position of an object its wavefunction collapses to infinity at a particular point

Infinity is not a number, so the above statement is not really well defined. To think properly about this scenario you should consider a narrow, localized wavefunction with some width $\Delta x$, and then ask what happens at the limit $\Delta x \to 0$. Now due to Heisenberg's uncertainty principle the wavefunction will broaden at a rate inversely proportional to $\Delta x$, so if you measure the particle's location after some small time interval $\Delta t$ you will find it, on average, at a displacement of roughly

$$ \Delta L \sim \frac{\hbar}{m} \frac{\Delta t}{\Delta x}$$

so the measured "velocity" would be

$$ \frac{\Delta L}{\Delta t} \sim \frac{\hbar}{m \Delta x} \to \infty$$

in the limit $\Delta x \to 0$.