
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
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Hint: 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.
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.
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