Question

Which of the following conditions is incorrect for a well behaved wave function ( $\psi$ )?
A. $\psi$ must be finite
B. $\psi$ must be single valued
C. $\psi$ must be infinite
D. $\psi$ must be continuous

Answer 1

The square of wave function $\psi$ (i.e., ${\psi^2}$ ) in chemistry gives us the probability of finding an electron/or electron cloud density (orbitals) in space. Also, the wave function of different atoms interfere with each other to form molecular orbitals and the atomic orbitals.

Complete answer:
There are some requirements a wave function must follow to be acceptable as a well behaved (or meaning full) wave function, which are;
1. The wave function must be single valued in any given coordinate(x, y, z), because there can be only one probability value at a given position.
2. The wave function must be continuous, in order for its second derivative ( $\dfrac{{{\delta ^2}y}}{{\delta {x^2}}}$ ) to exist and be well behaved.
3. And the last requirement is that the wave function must be finite, to be able to get a normalized wave function, $\psi$ should be integrable.
Option (C) says that the wave function should be infinite, which is an incorrect statement.

**Therefore, the correct answer to the question is option (C) i.e, $\psi$ must be infinite.

Additional information:**
If we integrate the probability of finding a particle ( ${\psi^2}$ towards ‘x’ coordinate in 2- dimension) in the entire space (taking limit from $- \infty$ to $\infty$ ), then the result of that integration will be unity i.e., ${\int {{{|\psi }}({\text{x}})|} ^2}{\text{dx = 1}}$. And those $\psi$ function for which | $\psi_{\text{x}}^2$ | = 1, they are called normalized wave functions.

Note:
While solving for Schrondinger’s wave function ( ${{\psi }}$ ) remember that this wave function is different from its square valued ${{{\psi }}^2}$, to avoid confusion between these two, remember that ${{\psi }}$ has no physical significance, while ${{{\psi }}^2}$ gives the probability of finding electron cloud density in space.