• Support PF! Buy your school textbooks, materials and every day products Here!

Representing Wavefunction as Superposition of Eigenstates

  • Thread starter Kvm90
  • Start date
  • #1
28
0

Homework Statement



A particle in the infinite square well with V(x)=0 for 0<x<a and V(x)=infinity otherwise has the initial (t=0) wave function:

psi(x,0)=Ax for 0<x<a/2
psi(x,0)= A(a-x) for a/2<x<a

1) Sketch psi and psi^2 (DONE)
2) Determine A [DONE - 2*sqrt(3)*a^(-3/2)]
3) Find psi(x,t) (HELP!)
4) What is the probability that a measurement of the energy yields the first eigenenergy level E1 of this infinite square well?
5) Find the expectation value of the energy.


Homework Equations



Psi(x,t)=SUMMATION[Cn*Psi n] = SUM[Cn sin((n*pi*x)/a)

Cn=int(psi(x,0)sin((n*pi*x)/a)
Cn=int(A*x*sin((n*pi*x)/a)) + int(A(a-x)*sin((n*pi*x)/a))

Questions!

Do I replace the 'n's in sin((n*pi*x)/a) with 1 and 2 when I'm solving for Cn? Am I trying to represent the wavefunction as a sum of sin((pi*x)/a) and sin((2*pi*x)/a) .... or do I just leave the quantum number n in my equations?
 
Last edited:

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
Questions!

Do I replace the 'n's in sin((n*pi*x)/a) with 1 and 2 when I'm solving for Cn? Am I trying to represent the wavefunction as a sum of sin((pi*x)/a) and sin((2*pi*x)/a) .... or do I just leave
the quantum number n in my equations?
Nope. In general, an infinite number of eigenfunctions is required to represent an arbitrary wavefunction. Once you carry out your integral (which splits up into two integrals since the initial wavefunction is defined piecewise), you will obtain an expression for cn that depends on n (of course).

Don't forget that the summation you have expressed is actually for [itex] \Psi(x,0) [/itex] and that the time dependence comes in as an extra exponential factor multiplying each eigenfunction.
 

Related Threads on Representing Wavefunction as Superposition of Eigenstates

Replies
1
Views
2K
Replies
2
Views
2K
Replies
3
Views
701
Replies
2
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
2
Replies
27
Views
3K
  • Last Post
Replies
0
Views
2K
Top