Haorong Wu said:
Summary:: Can we physical realize unbound states?
In principle, unbound states exist when ##E \gt V## at infinity. Are these unbound states just for calculation, or can we physical realize unbound states? After all, an unbound state means its possibility at infinity is not zero, and it seems impossible to realize such a state.
Eigenstates of a self-adjoint operator to an eigenvalue in the continuum of its spectrum are usually not normalizable, i.e., they do not belong to the Hilbert-space of square integrable functions, and only such wave function represent (pure) quantum states. The generalized eigenvectors for eigenvalues in the continuous spectrum are distributions. The math is somewhat involved, if you want to get it rigorous. For physicists, if they bother at all about mathematical rigorous formulations, the socalled "rigged Hilbert space formalism" is the most convenient one.
For the practitioner the non-rigorous treatment of standard textbooks are enought, though one should be aware that trouble can occur when one is not carefully remembering that there's some non-rigorous handwaving used.
The most simple example is a particle moving in one dimension and considering the momentum "eigenstates". The momentum operator is ##\hat{p}=-\mathrm{i} \hbar \partial_x##. The eigenvalue problem thus reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solution for any ##p \in \mathbb{C}## is obviously
$$u_p(x)=A_p \exp(\mathrm{i} p x/\hbar).$$
Now for any ##p \notin \mathbb{R}## no "scalar-product integral" of two such solutions makes sense, because if diverges for sure when integrated over the real axis. Only for ##p \in \mathbb{R}## it makes sense within the theory of distributions since then you get
$$\langle p_1|p_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p_1}^*(x) u_{p_2}(x)=A_{p_1}^* A_{p_2} \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i}(p_1-p_2)/\hbar]=A_{p_1}^* A_{p_2} 2 \pi \hbar \delta(p_1-p_2)=|A_{p_1}|^2 2 \pi \hbar \delta(p_1-p_2),$$
where ##\delta## is the Dirac-##\delta## distribution. For ##p_1=p_2## the integral of course diverges, and thus a momentum eigenfunction cannot represent a proper state of the particle. It's convenient to "normalize" the eigenfunctions "to a ##\delta## distribution", i.e., you make
$$A_{p}=\frac{1}{\sqrt{2 \pi \hbar}}.$$
Nevertheless these generalized eigenstates are very important, because you can represent any proper wave function as a "generalized superposition" of them, because of the completeness relation,
$$\int_{\mathbb{R}} \mathrm{d} p u_{p}^*(x_1) u_p(x_2)=\frac{1}{2 \pi \hbar} \int_{\mathbb{R}} \mathrm{d} p \exp[\mathrm{i} p (x_2-x_1)/\hbar]=\delta(x_2-x_1).$$
In this case the entire spectrum of the momentum operator is continuous, and thus the "superposition" is not a sum but an integral:
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}(p) u_p(x).$$
The "momentum-space wave function" is uniquely defined because of the "normalization to a ##\delta## distribution"
$$\tilde{\psi}(p)=\int_{\mathbb{R}} \mathrm{d} x u_{p}^*(x) \psi(x).$$
If you write out the eigenfunctions you see that this is of course nothing else than the well-known Fourier transformation (with some conventions concerning the constants appearing in the Fourier integral and its inverse Fourier integral adapted to the needs of quantum mechanics).
