Can the initial wavefunction be any function?

Physics Asked by Anders Gustafson on November 29, 2020

Knowing the wavefunction $Psi(x,y,z,0)$ is enough to know the functions for $frac{partial^2Psi(x,y,z,0)}{partial{x^2}}$, $frac{partial^2Psi(x,y,z,0)}{partial{y^2}}$, $frac{partial^2Psi(x,y,z,0)}{partial{z^2}}$, seeing as how they are second derivatives of the wavefunction with respect to spatial coordinates, but in order to find the function for $frac{partialPsi(x,y,z,0)}{partial{0}}$ we need to also know the function for the potential operator function $V(x,y,z,0)$, implying that $frac{partialPsi(x,y,z,0)}{partial{0}}$ is free to take on the wavefunction needed to keep the Schrödinger Equation self consistent. So if the only requirement for the wavefunction is that it obeys the Schrödinger Equation then it looks like the wavefunction could be any function.

Does this mean that the initial wavefunction $Psi(x,y,z,0)$ can be any function of $x$, $y$, and $z$, or are there more requirements for the wavefunction in addition to being consistent with the Schrödinger Equation?

2 Answers

To answer the original question, it depends. In the generally accepted literature for non-relativistic quantum mechanics, quantum states are elements of the symmetric (or anti-symmetric) tensor algebra of a separable Hilbert space. There of course exists research considering more exotic spaces of states. Anyways, let's talk informally about the space of states $H$. If you want full rigor, this question has been asked and formally answered many times on math stackexchange and elsewhere; this answer is meant to give only a rough idea.

The space of states $H$ is larger than $L^2(Omega)$; I won't bother specifying where you want the functions to map to/from, $Omega$. As was noted in the comments, there are elementary examples of wave functions which are not square-integrable. Depending how far down the rabbit hole you want to go, separability is enough or you may want to look at what is formally called a Gelfand triple (aka a rigged Hilbert space) which, consists of a particular subspace of $L^2(Omega)$, $L^2(Omega)$ itself, and the dual of the smaller space which contains distributions.

The idea of distributions was introduced by Dirac. They are not formally functions as the Wikipedia page incorrectly notes for the Dirac Delta function. After mathematicians saw how useful distributions are, Dirac's idea essentially created an entire field of mathematics; distributions are now formalized, well-studied objects.

Relatedly, $L^2(Omega)$ itself is not a space of functions per se, but rather one of equivalence classes of functions, where square-integrable functions are defined as equivalent if they differ only by sets of measure zero. It is a nice space of functions for physicists, in particular, because it is self-dual. That is, the bra (a linear functional $L^2(Omega)rightarrowOmega$) of a particular ket in $L^2(Omega)$ is also in $L^2(Omega)$.

Answered by Antonino Travia on November 29, 2020

If a function is going to be a valid, Hilbert space solution to a Schrodinger equation, then it must be square integrable (or at least Dirac normalizable) and must satisfy the boundary conditions of the particular situation.

For example, for the 1D infinite square well of width $a$, with $V(x)=0, 0 < x < a$, the function $Acos( Bx)$, by itself, would not be a solution because it doesn't fit the boundary conditions at $x=0$. A combination of sin and cos would work.

Answered by Bill N on November 29, 2020

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