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quasar987

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"We consider a particle in an infinite 1-dimensional well. The particle is described by an arbitrary wave function [itex]\psi(x)[/itex]. (a) For this particle, show that [itex]<E> \geq E_1[/itex]. (b) What is the condition on [itex]\psi(x)[/itex] such that [itex]<E> = E_1[/itex]?"

__My solution:__First of all, I'm assuming that by <E> he means <H>. But <H> = E. And E is the energy associated with the superposition of the sine waves that make up [itex]\psi(x)[/itex]. So

[tex]E = \sum_{i=1}^{\infty}c_i E_i[/tex]

where the c_i are either 1 or 0 depending on wheter or not [itex]\psi_i(x) = sin(k_ix)[/itex] is present in the Fourier expansion of [itex]\psi(x)[/itex]. This said, if [itex]\psi_1(x)[/itex] is in the expansion of [itex]\psi(x)[/itex], then [itex]E = E_1 +... \geq E_1[/itex]. If not, then [itex]E = 0\cdot E_1 + ... + 1\cdot E_j +... \geq E_1[/itex].

And for (b), the condition on psi is that psi be exactly psi_1.

That looks pretty neat to me. The only point I'm unsure of is wheter or not, it is true that the energy of a psi made up of a linear combination of other psi_i is the sum of the energy of the psi_i. Actually, this seems

*FALSE*to me. For suppose psi_1 and psi_2 are solution of the time ind. SE with respective energy eigenvalues E_1 and E_2. Then let's see if psi = psi_1 + psi_2 is a solution with eigenvalue E_1 + E_2.

[tex](E_1+E_2)\psi = (E_1+E_2)(\psi_1 + \psi_2) = E_1\psi_1 + E_2\psi_2 + E_1\psi_2 + E_2\psi_1 = H\psi_1 + H\psi_2 + E_1\psi_2 + E_2\psi_1 = H\psi + E_1\psi_2 + E_2\psi_1 \neq H\psi [/tex]

So the energy of psi_1 + psi_2 is not E_1 +E_2. Correct? Hopefully I'm missing something.