The unit of its probability density is L^(-n) n is the no. of axes.
If a wave function's[psi(x, y, z)] probability density has an unit of L^(-3), then naturally psi(x, y, z) also should possess an unit of L^(-3/2). A good contradition. How?
The dimension of the wave function is set such that the scalar product in state space is dimensionless (since probability is dimension-less by definition).
A wave function in quantum physics is a mathematical description of the quantum state of an isolated quantum system. The wave function is a complex-valued probability amplitude, and the probabilities for the possible results of measurements made on the system can be derived from it. The most common symbols for a wave function are the Greek letters ψ and Ψ (lower-case and capital psi, respectively).
Hello, sir. I had used a term called "probability density". I didn't use "probability". The fundamental expression for probability density of a wavefunction in a system is defined as modulus square of Ψ. It is defined as the probability of finding the particles in the system within its volume {probability/∂V}. The S.I. unit of it in this 3d case of system is [L]^(-3). The S.I. unit of the wave function of those particles is [L]^(-3/2).
Similarly we have surface probability density of Ψ(xy or yz or zx) and linear probablity density of Ψ(x or y or z) of the praticles. Their S.I. units are [L]^(-1) and [L]^(-1/2).
Dear Muthumanickam Venkatachalam, in addition to the previous interesting answer, I would like to add that the wave function is defined on the 3N-dimensional configuration space, rather than on a 3D physical space.
This gives the impression that QM cannot have a 3D space interpretation in terms of a pure wave function.
Article Representation of the wave function on the three-dimensional space
On the other hand, the probability current j * which is an operator and therefore is observable in QM has more physical meaning than the same wave function: