I have seen Zetilli's QM book deals with $E>V$ and $E< V$ (tunnelling) in case of the potential wells deliberately avoiding the E=V case,So I thought maybe something is intriguing about this and made this up.
Suppose the total energy of the particle is equal to its potential energy.Then its kinetic energy should be zero,(Speaking non relativistically). But Kinetic energy operator is $\hat{T}=\hat{p}^2/2m$ (where $\hat{p}=-i\hbar\frac{\partial}{\partial x}$), So clearly since Kinetic energy is 0 here,momentum eigenvalue will also vanish.
Now, Putting $E=V$ in time-independent Schrodinger equation (1D) we get, $$\frac{\partial^2\psi}{\partial x^2}=\frac{2m(E-V)}{\hbar^2}\psi\implies\frac{d^2\psi}{d x^2}=0\implies\psi=Ax+B$$ where $A$ and $B$ are arbitrary constants. Since,the Wave function must vanish at $\pm\infty$, $A=0$,hence the wave function equals a constant=$B$ and is not normalizable. So,a particle with no momentum(or kinetic energy),gives a physically unrealizable wave function!
Does this imply $E=V$ is a restricted critical case or momentum cant be zero in quantum mechanics or did i just go wrong somewhere?