1.3 Conditions for Valid Wavefunctions

From what we have derived so far, there are some constraints we have on what type of wavefunctions can represent physical states. In classical mechanics, the trajectory $x(t)$ which is the solution of Newton’s second law is the object which we impose constraints on smoothness on. For quantum mechanics, the role of the trajectory is replaced by the wavefunction, the square magnitude of which represents probability density. This means that while it also has the smoothness requirements determined by the Schrodinger equation, the wavefunction needs some additional constraints which we will see below.

1.3.1 Smoothness

As stated above, the form of Newton’s second law requires the trajectory to be not only continuous, but twice-differentiable.

The wavefunction must be continuous, because a discontinuity in $\psi$ at $x_{0}$ would like to ambiguous probabilities near $x_{0}$ . It must also have a continuous slope, except at points where the potential energy is infinite. To see why, we can rearrange the TISE to

\frac{\operatorname{\mathrm{d}}\!^{2}u(x)}{\operatorname{\mathrm{d}}\!x^{2}}=% \frac{2m}{\hbar^{2}}(V(x)-E)u(x),

(1.24)

and then integrate over a small interval $[x_{0},x_{0}+\varepsilon]$ to get

\left.\frac{\operatorname{\mathrm{d}}\!u(x)}{\operatorname{\mathrm{d}}\!x}% \right|_{x_{0}}-\left.\frac{\operatorname{\mathrm{d}}\!u(x)}{\operatorname{% \mathrm{d}}\!x}\right|_{x_{0}+\varepsilon}=\frac{2m}{\hbar^{2}}\int_{x_{0}}^{x% _{0}^{\varepsilon}}(V(x)-E)u(x)\operatorname{\mathrm{d}}\!x.

(1.25)

If the left-hand side is not equal to zero in the limit $\varepsilon\to 0$ , then the slope $\frac{\operatorname{\mathrm{d}}\!\psi}{\operatorname{\mathrm{d}}\!x}$ is nonzero at $x_{0}$ . When can the right-hand side be nonzero? The term $Eu(x)$ is finite so the integral goes to zero in the limit $\varepsilon\to 0$ , and the same will be true for the term $V(x)u(x)$ even if $V(x)$ is discontinuous at $x_{0}$ . Only if $V(x)$ is infinite at $x_{0}$ will the integral be nonzero in the limit $\varepsilon\to 0$ .

1.3.2 Normalisation

As alluded to above, the fact that the square magnitude of $\psi(x,t)$ represents a probability density necessitates an extra constraint on the wavefunction. Namely, the integral of the probability density over all space must be equal to one. This represents the fact that the particle must be located somewhere.

\int_{-\infty}^{\infty}\psi^{\ast}(x,t)\psi(x,t)\operatorname{\mathrm{d}}\!x=% \int_{-\infty}^{\infty}\lvert\psi(x,t)\rvert^{2}\operatorname{\mathrm{d}}\!x=1.

(1.26)

This constraint is called the normalisation condition. If wavefunctions do not have this property, then the probabilities calculated from them will not make sense.

This gives us a requirement that the physically realisable wavefunction must be square integrable, which implies that the integral of the square magnitude over all space is finite:

\int_{-\infty}^{\infty}\lvert\psi(x,t)\rvert^{2}\operatorname{\mathrm{d}}\!x=N% <\infty.

(1.27)

If a given wavefunction $\psi(x,t)$ is square integrable, then $\frac{1}{\sqrt{N}}\psi(x,t)$ where $N$ is given by the equation above will be a correctly normalised wavefunction.

Square integrability imposes that $\psi(x,t)$ decays to zero sufficiently rapidly as $x\to\pm\infty$ . This makes physical sense, because it would not make sense for a particle to have a nonzero probability to be located infinitely far away. Wavefunctions which are not square integrable are called unnormalisable.

In Dirac notation, the normalisation condition is written as

\langle\psi\mathclose{}|\mathopen{}\psi\rangle=1.

(1.28)