Wave function/Infinite square well confusion

[Salviati]

Homework Statement

"A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a.
\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a}),
E₁=\frac{h^2\pi ^2}{2ma^2}*

What is the initial wave function \Psi(x,0)?

*h is supposed to be h bar, I just couldn't find it)

Homework Equations

The Attempt at a Solution

My attempt: If the general solution is a superposition of all stationary states, \Psi(x,t)=\sum c_n\psi_ne^\frac{-iE_nt}{h}, at t=0, \Psi(x,0)=\sum c_n\psi_n. Also, at this time, the particle is in the ground state (n=1), so: \Psi(x,0)=c_1\psi_1. Do I assume c₁=1 at this point, because the wave function "collapses" once the energy becomes known? I'm just not sure if I understand exactly what happens when the known data is given.

The solution itself is supposed to be \Psi(x,0)=\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a}).

 

[Avodyne]

c₁=1 by normalization, assuming that \Psi(x,t) is normalized. For a more general initial state,
\sum_{n=1}^\infty |c_n|^2=1.

 

[turin]

"A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a.
\psi_1(x) =\sqrt{\frac{2}{a}}sin(\frac{\pi x}{a}),
E₁=\frac{h^2\pi ^2}{2ma^2}*

What is the initial wave function \Psi(x,0)?

You are right to be confused. You cannot know the answer to this question; that is the whole idea behind boundary conditions/initial values: they are INPUTS. For a first-order differential equation (like (d/dt)psi=iHpsi), you need one input BC for each degree of freedom.

*h is supposed to be h bar, I just couldn't find it)

Try \hbar.

 

[Salviati]

Thanks guys.

Try \hbar.

\hbar