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Probability of measuring specific energy of particle (in a box)
Wow...okay, I guess I'll end up with an expression in terms of Ci's. Also, when you said "these remaining lines are all wrong" did you also mean that I was wrong to make the following assumption: since normalized, => <ϕi|ϕi> = 1- doorknob
- Post #5
- Forum: Advanced Physics Homework Help
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Probability of measuring specific energy of particle (in a box)
Vela, Thank you for your response. However, can I use the fact that I know: Σ|ci|^2 = 1 (i.e., the state is normalized) to obtain an expression without Ci? For example, when calculating mean energy <H>: Expectation energy <H> is defined as: <H> = <Ψ|H|Ψ> *Note, (my mistake) my...- doorknob
- Post #3
- Forum: Advanced Physics Homework Help
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Probability of measuring specific energy of particle (in a box)
Homework Statement Consider a particle-in-a-box problem, involving a particle of mass m subject to a potential: V(x) = +∞ for X≤0 V(x) = 0 for 0<X<L V(x) = +∞ for X≥L |ϕn> are the eigenstates of the Hamiltonian H with the corresponding eigenvalues: En = (n^2)*(hbar^2)*(Pi^2) / 2mL^2...- doorknob
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- Box Energy Measuring Particle Probability Specific
- Replies: 5
- Forum: Advanced Physics Homework Help