The one-dimensional quantum HO can be solved in Schrodinger representation by getting Hermite Differential Equation

$$ \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + \lambda y = 0 $$ with solutions $$ y(x) = \mathcal{H_n(x)} $$ which is true for integer values of $\lambda$. In the case of non-integer values of $\lambda$ the solutions are given by hyper-geometric functions

$$ y(x) = c_1 H_{\frac{\lambda }{2}}(x)+c_2 \, _1F_1\left(-\frac{\lambda }{4};\frac{1}{2};x^2\right) $$ and it seems like these functions are not square integrable and hence not valid/physically acceptable wavefunctions.

In this perspective, I tend to understand that quantisation comes up only as a feature of restricting the solutions to integer values, which alone are square integrable.

I am looking forward to explanations concerning this aspect, since there are so many cases where the polynomial solutions occur (which are solutions for some integer value of the eigen-value in differential equation). (For instance, the solutions of Hydrogen atom).

PS : In the case of infinite square well potential, the quantisation of the energy seems to come from the boundary conditions. So what is an analogy to that in the case of Harmonic Oscillator and Hydrogen atoms cases.


2 Answers 2


If a function is a solution of the Schroedinger equation, it is a solution. The adjective "valid" does not seem to be very helpful in such situation.

The word "valid" is better used in "valid solution of the eigenvalue problem" which has some additional requirements to being a solution of the Schr. equation - most often one requires that the function decays to zero at infinity or that it is integrable, for example to make it susceptible to the Born interpretation of $|\psi|^2$.

This way, only some of the solutions of the Schr. equation are also solutions of the eigenvalue problem; they need to satisfy also prescribed boundary conditions and normalization, which are regarded as part of the eigenvalue problem. The situation is somewhat analogous to the situation in mechanics: not all solutions of the equations of motion are interesting; only those satisfying the prescribed initial conditions are.

  • $\begingroup$ I am sorry about that, I meant not a quantum mechanically acceptable wavefunction !! I have made some edits. Thanks !! $\endgroup$
    – user35952
    Feb 2, 2014 at 14:31
  • $\begingroup$ So in this case, the boundary condition is restricting the solutions to integer values ? $\endgroup$
    – user35952
    Feb 2, 2014 at 14:35
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    $\begingroup$ @user35952 yes. Without boundary conditions you have general solution for any $\lambda$, it has two free parameters. When you impose boundary conditions, one of these parameters is used to satisfy one boundary condition, and another one is normalization coefficient. To satisfy second boundary condition you have to restrict $\lambda$ to some spectrum of values. $\endgroup$
    – Ruslan
    Feb 2, 2014 at 14:56
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    $\begingroup$ @user35952 In general quantization appears as a property of the operator you select. When you define an operator, you must define its domain and thus boundary conditions. Eigenvectors of that operator are the solutions of time-independent Schroedinger equation. If the spectrum of the operator has discrete part, then you have some quantized states. $\endgroup$
    – Ruslan
    Feb 2, 2014 at 15:00
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    $\begingroup$ Discrete indexing may occur only for some part of the spectrum. For hydrogen atom, spectrum is discrete only for negative energies. For positive energies, there are continuously-indexed positive eigenvalues associated with improper eigenfunctions - functions that satisfy Schroedinger equation but not the normalization condition. These are important too to obtain good generalized basis for expressing integrable functions. $\endgroup$ Feb 2, 2014 at 15:02

In the present case it is square integrability rather than boundary conditions that restricts the set of eigenvalues.

  • $\begingroup$ Is not the square integrability itself a way of boundary condition. i.e. $ |\psi(x)|^2 \rightarrow 0 $ as $ x \rightarrow \infty $. $\endgroup$
    – user35952
    Feb 3, 2014 at 15:23
  • $\begingroup$ I do not think so. Square integrability does not necessarily imply that |ψ(x)|^2→0 as x→∞. Think of a function that vanishes everywhere except for the positive integers n and a small interval I_n around them where it has the value 1. If the length of I_n tends to 0 sufficiently fast as n→∞ then the function is square integrable but does not tend to 0. $\endgroup$
    – Urgje
    Feb 4, 2014 at 13:25

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