# What the wave function looks of a particle in the infinite square well looks like after collapse for measurements of position and energy

Consider a particle in a the infinite square well from x=0 to x=L. At t=to, I make a measurement of position and get x=L/2. What is the resulting wave function at t=to? My understanding, from reading, is that it is the dirac delta function with the spike at L/2. This makes sense to me, but I just want to confirm my logic. And then I can treat this function as the initial state and use schrodinger's equation to see how it evolves with time, right?

Additionally, if I make a a measurement of energy some time after - what is the function going to look like? This idea is more blurry to me. Will the collapsed function be expressed in a different vector space?

Thanks in advance.

## 2 Answers

In general, when you make any measurement on a system, the wave function collapses to the eigenstate of the measured observable with the eigenvalue corresponding to the measured value.

So for a position measurement, it is the delta function in space; for energy measurements, it is one of the eigenstates of the Hamiltonian (in this case, sinusoidal functions of position that go to zero at the walls of the well, $x=0$ and $x=L$).

The fact that you measured the position (or any other observable) previously does not influence the kind of functions the wave function collapses to, but, together with time evolution, only the probability of collapse to each of the eigenfunctions.

it is the dirac delta function with the spike at L/2.

if you make a a measurement of energy some time after - the function is still a delta function in spacetime with a definite energy.