Recent content by shankk

An energy eigenstate is a wavefunction ##\psi(x)## for which ##\hat{H}\psi = E\psi##. On solving for free particle in one dimension, we get ##\psi(x) = Ae^{iKx}## with ##K = \pm \frac{\sqrt{2mE}}{\hbar}##. Now, we apply the momentum operator ##-i\hbar \frac{\partial}{\partial x} (Ae^{iKx}) =...

Okay, but for the free particle case, we first obtained the energy eigenstate and found out that it is also an eigenstate of the momentum operator. So, for this particular case of a free particle, every eigenstate of energy is also an eigenstate of momentum, right?

I see, I just had to apply the momentum operator to ##\psi## and I would have received all my answers. I had thought about this problem for quite some time but this simple step just didn't cross my mind. I am a complete beginner at QM. Thanks for clearing things up.

To me, the ##K## obtained by solving the Schrodinger equation and the de broglie wavelength seem two completely unrelated quantities. Can someone explain why have we equated ##K## and ##\frac{2\pi}{\lambda}##. Also, isn't writing ##p = \hbar K## implying that eigenstate of energy is also an...

Okay. So that cleared up some of my doubts. Now I think that the relation of ##T = E - V = \frac{p^2}{2m}## still holds but here E is now the eigenvalue of the hamiltonian, p is the positive square root of the eigenvalue of ##\hat{p}^2## and V is the potential energy. But now here, p would...

Here we are talking about non-relativistic quantum physics. So we all know kinetic energy T = E - V = \frac{1}{2}mv^2 in classical physics. Here V is the potential energy of the particle and E is the total energy. Now what I am seeing is that this exact same relation is being used in quantum...