(E-V)stands for kinetic energy in schroedinger equattion because V is electrostatic potential energy.Kinetic energy is associated with motion but quantum interpretation negates the motion of electron which leads to the conclusion that (E-V) is energy without any specification of type of energy because there is no kinetic energy without motion Can you please explain what is (E-V) according to quantum mechanics?(E-V) is important part of schroedinger equation

In quantum mechanics observables like energy and momentum are replaced by operators. In particular, momentum is replaced by the operator:

 

[math]p=-i\hbar \frac{\partial }{\partial x}[/math]

 

This means that the kinetic energy operator is:

 

[math]T=\frac{p^2}{2m}=-\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}[/math]

 

The wavefunction [math]\Psi[/math] associated with a particle must satisfy the Schrodinger Equation (a partial differential equation), which simply tells us how the time derivative of the wavefunction is related to the Hamiltonian (total energy) of the particle:

 

[math]i\hbar \frac{\partial \Psi}{\partial t}=H\Psi =(T+V)\Psi=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}+V\Psi[/math]

 

What quantum mechanics also tells us is that, if we solve Schrodinger's Equation for the wavefunction, then the probability of finding the particle between the points x=a and x=b is given by:

 

[math]P(a\leq x\leq b)=\int_a^b \Psi^* \Psi~dx[/math]

 

where [math]\Psi^*[/math] is the complex conjugate of the wavefunction. Using this knowledge, we can determine the expectation value of kinetic energy. If you were to do a bunch of repeated identical experiments to determine the kinetic energy of the particle, you would find that the measurements "cluster" around the expectation value. It's a sort of average value. In line with that intuition that it's an average value, we determine the expectation value of the kinetic energy to be:

 

[math]\langle T \rangle =-\frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \Psi^* \frac{\partial^2 \Psi}{\partial x^2}~dx[/math]

 

 

If you're familiar with any linear algebra, QM also tells us that the value of a measured observable must always be one of the operator's eigenvectors (which are usually called "eigenstates" by most authors). So, for your example of kinetic energy, the values that it's allowed to take on are given by:

 

[math]-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi }{\partial x^2}=\tau \Psi [/math]

 

So the only values that the kinetic energy is allowed to have are the values [math]\tau[/math] which satisfy the above equation.