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it would be great if someone could help me understand the following quote regarding wavefunctions :)

"$$\psi(x)=\sum_n C_nu_n(x)+\int dE C(E)u_E(x)$$ The expression is schematic because we have not specified exactly which form of the continuum eigenfunctions we are supposed to use."

Why is it "schematic"? What would the expression look like when we have specified the "forms" of the eigenfunctions?

Thanks! :)

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Could you tell us from what source the quote was taken? –  AdamRedwine Aug 27 '11 at 13:54
Sure. From here :) –  bra-ket Aug 27 '11 at 15:24
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2 Answers

up vote 3 down vote accepted

I think this has to do with the integration measure $dE$ which is the only thing that makes the continuum part of the expansion possibly more "schematic" than the discrete one. In general you expand your wave function in a basis of common eigenstates of a (complete) set of commuting operators, which means that every eigenfunction is uniquely specified by a set of quantum numbers. In the discrete part of the spectrum, you just sum over all different eigenstates, labelled here "schematically" by $n$, while in the continuous part of the spectrum you have to integrate over certain domain of eigenvalues (for energy typically over positive values). The integration measure depends on the choice of the eigenstates as well as on their normalization, that is why it is schematic if you just write it formally as $dE$. For example, for a free spinless particle, you can choose as a basis either simultaneous eigenstates of all three components of momentum (plane waves), or simultaneous eigenstates of the Hamiltonian, square of angular momentum and one of its components (spherical waves). In the former case the quantum numbers are $p_1,p_2,p_3$ and the integration measure becomes $\int{d^3p}/{(2\pi)^3}$, while in the latter case the quantum numbers are $E,l,m$ and the "integration measure" would be $\sum_{l,m}\int dE$, modulo factors depending on the normalization of the eigenfunctions.

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Thanks a lot, Tomas! This is very well-explained :) –  bra-ket Aug 27 '11 at 13:15
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Some eigenstates may be degenerate so it is not always the energy that is a "summation subscript". In such cases there must be degeneracy factor introduced in the integrand, like $d^3p=4 \pi p^2dp \propto \sqrt{E}dE$ .

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Thanks, Vladimir :) –  bra-ket Aug 27 '11 at 13:15
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