Representing Wavefunction as Superposition of Eigenstates

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Kvm90
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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?
 
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Answers and Replies

  • #2
cepheid
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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.
 

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