# Considering an arbitrary wavefunction for a free particle, are all normalizable functions valid?

One can show that a possible solution to a wavefunction with constantly zero potential is equal to, only considering the spacial piece:

$$\psi(x) = \int_{-\infty}^{\infty} A(k) \ e^{ikx} dk$$

This is partially due to the fact that, since there are no boundary conditions applying a rule to $$k$$, such as $$k = \frac{n \pi}{L}$$ like in the case of the infinite square well. That merely explains why a sum of eigenstates cannot represent an arbitrary ket vector. This is, however, mainly due to the fact that a sum is not only unnecessary but inoperable, as a Fourier sum without boundary conditions like the infinite square well will be periodic throughout space and hence will not be normalizable and have weird implications on its position and momentum uncertainty I would think. This is why a Fourier integral is necessary.

However, if I am to postulate that $$A(k)$$ is the Fourier transform of $$\psi(x)$$, then could I not contend that this is just the Fourier transform of some arbitrary function $$f(x)$$ in disguise, and the solution to Schrodinger's equation given constant zero potential? In this case, since $$f(x)$$ is arbitrary, any function would be a solution given this case, as long as it is normalizable. Why or why not?

You are asking two very different questions here.

Is $$\psi$$ a valid wavefunction?

$$\psi$$ is a valid wavefunction if it is an element of the Hilbert space underlying your quantum mechanical system. In the case of a free particle on a line, that Hilbert space is $$L^2(\mathbb{R})$$, which is (roughly) the space of square-integrable functions.

Therefore, if $$\int_{-\infty}^\infty |\psi|^2 dx$$ is finite, then $$\psi$$ is a perfectly reasonable wavefunction. It need not even be continuous, though this requirement is often imposed anyway.

If this is your question, then the answer is yes - any square-integrable function is an acceptable wavefunction, but this is true for every system which has $$L^2(\mathbb{R})$$ as the underlying Hilbert space - the free particle, the harmonic oscillator, etc.

Is $$\psi$$ a solution to the time-independent Schrödinger equation?

This question is asking whether $$\psi$$ is a solution to the eigenvalue equation $$\hat H \psi = E\psi$$ for some eigenvalue $$E$$. This is far more restrictive; in order for this to be true, then

1. $$\psi$$ must be square-integrable
2. $$\hat H$$ must be able to act on $$\psi$$ to produce another square-integrable function, which means in this case that (a) $$\psi$$ must be at least twice-differentiable and (b) $$\psi''$$ must be square-integrable
3. $$\psi$$ satisfies the equation $$\psi'' = -k^2 \psi$$ for some constant $$k$$

As it turns out, there are no functions which obey all three requirements, which means that in a strict sense, there are no states of constant energy (also called stationary states) for this system.

Introductory courses typically sidestep this issue by noting that functions of the form $$e^{ikx}$$ satisfy the necessary differential equation and correspond to states of definite energy, and that even though they themselves are not valid physical states, you can create linear combinations of them which are. This is where the Fourier transform comes into play. One can write a state $$\psi$$ in the following way:

$$\psi(x) = \frac{1}{2\pi}\int_{-\infty}^\infty A(k) e^{ikx} dk$$

Physically, one says that $$\psi$$ is made up of an infinite linear combination of (individually unphysical) states of definite energy. The function $$A(k)$$ essentially tells you how much of state $$k$$ goes into the mix.

It can be shown (see Plancherel's Theorem) that if $$A(k)$$ is square-integrable, then the corresponding $$\psi(x)$$ is as well, and vice-versa. Additionally, one finds that they can invert the transform to yield $$A(k) = \int_{-\infty}^\infty \psi(x) e^{-ikx} dx$$ which means that any arbitrary square-integrable function $$\psi$$ can be written in this way.

Remember, though - this $$\psi$$ is still not itself a solution to the time-independent Schrödinger equation, which in the case of the free particle either (a) has no solutions at all, or (b) has solutions, but they are all individually unphysical. Which answer you choose depends on your point of view.

• Ah right. This seems legit to me, as so far all the solutions of various potentials have included an orthogonal set (sines/cosines, Hermite polynomials) which means that any old function that satisfies the boundary conditions can be legit. So, any function can be a solution to Schrodinger's equation if the solution forms an orthonormal basis, as long as it fulfills the stuff you mentioned in your "Is $\psi$ a solution to the TISE?" bit? Or am I misconstruing things. – sangstar Nov 4 '18 at 0:17
• No - any square-integrable function can be a wavefunction, but not all wavefunctions (and in certain cases, like this one, no wavefunctions at all) are solutions to the TISE. It's possible to extend our notion of solutions of the TISE to include objects which don't lie in the Hilbert space, but such objects do not constitute physical states. For example, the position and momentum operators generically do not have any eigenfunctions that live in $L^2(\mathbb{R})$. – J. Murray Nov 4 '18 at 4:53