How does one decide whether a wave function is a physically acceptable solution of the Schrödinger equation? For example: $\tan x$ , $\sin x$, $1/x$, and so on.

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    $\begingroup$ You mean besides sticking it into the Schrodinger equation and seeing if it is a valid solution? $\endgroup$
    – Kyle Kanos
    Commented Nov 26, 2014 at 15:00
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    $\begingroup$ physicspages.com/2011/01/25/wave-function-borns-conditions contains a nice summary of the conditions you're looking for. Most of the conditions are a consequence of interpreting the wavefunction as a probability amplitude. $\endgroup$
    – Endulum
    Commented Nov 26, 2014 at 15:04
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    $\begingroup$ Shove it into the Schrodinger equation, and try to solve for the potential $V$; that will give you some idea of what system it is describing. $\endgroup$
    – JamalS
    Commented Nov 26, 2014 at 15:09
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    $\begingroup$ Also, note that physically unacceptable solutions of the Schrödinger equation can still be useful. For example, the eigenfunctions of the position operator (delta function) and momentum operator (plane wave) cannot be created experimentally. But it is still very useful to describe the physically realizable states in terms of these eigenfunctions. $\endgroup$
    – jabirali
    Commented Nov 26, 2014 at 16:14

2 Answers 2


The very minimum that a wavefunction needs to satisfy to be physically acceptable is that it be square-integrable; that is, that its $L_2$ norm, $$ \int |\psi(x)|^2\mathrm d x, $$ be finite. This rules out functions like $\sin(x)$, which have nonzero amplitude all the way into infinity, and functions like $1/x$ and $\tan(x)$, which have non-integrable singularities.

In the most rigorous case, however, one needs to impose additional conditions. The physically preparable states of a particle denote functions which are continuously differentiable to any order, and which have finite expectation value of any power of position and momentum. Thus:

  • $\psi$ must be continuous everywhere.
  • All of $\psi$'s derivatives must exist and they must be continuous everywhere.
  • The expectation value $\int\psi^*(x) \:\hat x^n \hat p^m \psi(x)\:\mathrm dx$ must be finite for all $n$ and $m$.

This rules out discontinuous functions like $\theta(x)$, functions with discontinuous derivatives, and functions like $(1+x^2)^{-1/2}$, which decay too slowly at infinity. States which satisfy these conditions are called 'physical' because they are the states that can be prepared with finite energy in finite time. The way to implement these states rigorously is by using a construction known as a Rigged Hilbert Space (see also Galindo & Pascual's QM textbook).

In everyday practice, most people adopt a bit of a mixed approach. The requirements that a function be continuous is never dropped, and one requires it to be differentiable at least almost everywhere. If the Hamiltonian is not a nice function of position, though, such as with $\delta$-function or square-well potentials, the requirements are sometimes slackened to only those; this is in the understanding that a truly discontinuous potential is not physical, and that any problems brought into the higher derivatives of $\psi$ can be fixed by using a smoother Hamiltonian.

  • $\begingroup$ Realize that for the condition of finite energy, one just needs the restriction that the second momentum moment $\langle p^2\rangle_\psi$ is finite, not all position and momentum moments. $\endgroup$ Commented Nov 26, 2014 at 15:53
  • $\begingroup$ @MateusSampaio The statement is about the preparation of the state, using a Hamiltonian of the form $\hat H=H(\hat x,\hat p,t)$, within finite time and starting from a reference physical state. (You would also ask for all moments of $\hat H$ to be finite if the function $H$ was not well-behaved enough.) $\endgroup$ Commented Nov 26, 2014 at 16:00
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    $\begingroup$ I don't see why the preparation of the state would require all moments of $H$ to be finite. $\endgroup$ Commented Nov 26, 2014 at 16:03
  • $\begingroup$ Why should a physical state be on the nuclear space of the rigged hilbert space? I can't see any physical reason for that $\endgroup$
    – Hydro Guy
    Commented Nov 26, 2014 at 16:16
  • $\begingroup$ @MateusSampaio I'm trying to find references and will post them here when I find a proof. I probably first read the statement here and in related papers. $\endgroup$ Commented Nov 26, 2014 at 22:08

If you are talking about the time independent Schrödinger equation, it's not a trivial question as it may seem, as the comments suggest. I will restrict the answer to the one-dimensional case, since multiply connected domains in higher dimensions give some additional problems. Not all functions $\psi$ that are solutions of the equation $$-\frac{\hbar^2}{2m}\psi''+V\psi=E\psi$$ are valid ones. The first condition is that $\psi\in L^2(\Omega)$, where $\Omega\subset \Bbb{R}$ is the domain of the function, since it must be an element of the Hilbert space, otherwise it would not be a quantum state.

More subtle conditions are required when you look at the domain of the Hamiltonian $$H=-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}.$$ In general, this will depend on the conditions satisfied by the potential $V(x)$. Usually one ends up with subsets of the Sobolev space $\mathcal{H}^2(\Omega)$, which restricts the original space to functions such that their second order (weak-)derivative is in $L^2(\Omega)$. If $\Omega$ is an interval (which is the usual setup) this can also be put in an equivalent way as the functions that are, along with their derivative, absolutely continuous and whose second derivative is also in $L^2(\Omega)$. Also, when the domain $\Omega$ is a proper subset of $\Bbb{R}$, the boundary conditions, which are established via physical arguments, play a decisive role in choosing the right domain $\mathcal{D}(H)$ of self-adjointness, and so they must be considered too. For example, the condition that $\psi(0)=\psi(a)=0$ for the infinite square well, rules out some solutions of the Schrödinger equation that would satisfy the other conditions.

If we look otherwise to the time dependent Schrödinger equation $$H\psi(t)=i\hbar\partial_t\psi(t), \qquad \psi(0)=\psi_0$$ any function $\psi_0\in L^2(\Omega)$ can be an initial condition for the system. But for $\psi_0 \notin\mathcal{D}(H)$ the trajectory given by $\psi(t)=e^{-iHt/\hbar}\psi_0$ is only a weak solution, in the sense that it's not a differentiable path and that the average energy is not defined (can be considered infinite) for all $t$, so these solutions can be considered non-physical.


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