Wave Function

We have already seen that the general function of waves propagating in space and time can be written as $$\psi(x,t) = \psi_0 e^{i(\omega t-kx)}$$ This does also apply to objects described within the quantum physics domain. we want to derive now two very important operators for the energy and momentum that will be used frequently in the following sections. Operators are mathematical objects that always have to be combined with the function to which they are going to be applied. And we know that the frequency $\nu = \omega/(2\pi)$ is proportional to the energy of the particle. In order to receive this value, we can then simply calculate the derivation of $\psi$ with respect to $t$: $$\frac{\partial}{\partial t} \psi(x,t) = \omega \psi(x,t)$$ For the conversion of $\omega$ to $E$, we only need to insert the factor $i\hbar$ derived from $$E = h\nu = \hbar \omega$$ This then leads to the fact that the energy of a quantum object, which can be described by a wave function, is calculated by applying the differential operator $$E = i\hbar\frac{\partial}{\partial t}$$ to the wave function itself. The same calculation can now be done for the momentum. In the first step, we calculate the derivation of $\psi$ with respect to $x$: $$\frac{\partial}{\partial x} \psi(x,t) = k\psi(x,t)$$ Now we take into account that the momentum is anti-proportional to the wavelength of a particle according to the previously given formula $p=h/\lambda$. Using the definition $k = 2\pi/\lambda$ results in following formula for the differential operator for $p$: $$p = -i\hbar\frac{\partial}{\partial x}$$ Both differential operators are very important in quantum mechanics.