helpppmeee
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Edit: Problem solved please disregard this post
A particle in the harmonic oscillator potential has the initial wave function [itex]\Psi[/itex](x, 0) = ∑(from n = 0 to infinity) Cnψn(x) where the ψ(x) are the (normalized) harmonic oscillator eigenfunctions and the coefficients are given by the expression Cn = 1/(√(2^(n=1))). What is the probability that a measurement of the oscillator's energy at an arbitrary time t>0 will yield a result greater than 2(hbar)ω.
En = (n + 1/2)(hbar)ω
I believe I can attempt the answer. The P(E>2(hbar)ω) is when En = 2(hbar)ω so ∴, n > 2 is when En = 2(hbar)ω. So, from 2 to infinite integers, we have the potential energies according to the equation [itex]\Psi[/itex](x, 0) = ∑(from n = 0 to infinity) Cnψn(x). So therefore, I believe that the integral of ([itex]\Psi[/itex](x, 0))^2 will give me my probability of finding the energy. So therefore, P(E>2(hbar)ω) = 1 - P(E≤2(hbar)ω) where P(E≤2(hbar)ω) = C1ψ1 + C0ψ0. I just can't seem to figure out what ψ is.
Homework Statement
A particle in the harmonic oscillator potential has the initial wave function [itex]\Psi[/itex](x, 0) = ∑(from n = 0 to infinity) Cnψn(x) where the ψ(x) are the (normalized) harmonic oscillator eigenfunctions and the coefficients are given by the expression Cn = 1/(√(2^(n=1))). What is the probability that a measurement of the oscillator's energy at an arbitrary time t>0 will yield a result greater than 2(hbar)ω.
Homework Equations
En = (n + 1/2)(hbar)ω
The Attempt at a Solution
I believe I can attempt the answer. The P(E>2(hbar)ω) is when En = 2(hbar)ω so ∴, n > 2 is when En = 2(hbar)ω. So, from 2 to infinite integers, we have the potential energies according to the equation [itex]\Psi[/itex](x, 0) = ∑(from n = 0 to infinity) Cnψn(x). So therefore, I believe that the integral of ([itex]\Psi[/itex](x, 0))^2 will give me my probability of finding the energy. So therefore, P(E>2(hbar)ω) = 1 - P(E≤2(hbar)ω) where P(E≤2(hbar)ω) = C1ψ1 + C0ψ0. I just can't seem to figure out what ψ is.
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