Schrödinger equation in energy basis

Question

Does there exist a Schrödinger equation for the energy space, like for momentum? I would say no, because the energy basis is countable, but are there any other reasons?

Answer 1

The basis-independent form of the Schrödinger equation is $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\lvert\alpha\rangle = H\lvert\alpha\rangle.$$

If we wish to express this in a particular basis, we simply multiply by the corresponding basis bras. Thus, in the energy basis, we get \begin{align} \langle n\rvert i\hbar\frac{\mathrm{d}}{\mathrm{d}t}\lvert\alpha\rangle &= \langle n\rvert H\lvert\alpha\rangle\\ i\hbar\dot c_\alpha(n) &= E_n c_\alpha(n), \end{align}

where $c_\alpha(n)$ is the energy-space wavefunction for the state $\lvert\alpha\rangle$, and $\lvert n\rangle$ are the energy eigenstates with eigenvalues $E_n$. Here, we can see that the energy basis is actually quite nice to work with, due to the fact that the time derivative of $c_\alpha(n)$ only depends $c_\alpha(n)$ and not on the value of $c_\alpha$ at other "points" (energy levels).

Answer 2

Let's first recall the process of emergence of the momentum schrödinger equation, strating from the time-independent schrödinger equation. We make a Fourier transformation in time, then obtain a spatial equation with eigen energy $E = \hbar \omega$: \begin{align} i\hbar\frac{\partial \Psi(\vec r,t)}{\partial t} &= -\frac{\hbar^2}{2m} \nabla^2 \Psi(\vec r,t) + V(\vec r) \Psi(\vec r, t). \\ \text{Define }\,\psi_{\omega}(\vec r) &= \int dt \,\Psi(\vec r,t)\, e^{i\omega t}.\\ \hbar\omega \,\psi_{\omega}(\vec r) &= -\frac{\hbar^2}{2m} \nabla^2 \psi_{\omega}(\vec r) + V(\vec r) \psi_{\omega}(\vec r). \end{align} The final expression is the one you called the momentum space equation, an equation in space.

Can we do it in the other way by first taking Fourier transformation in space, and leave an equation in time. Yes, but it comes with some complexity. \begin{align} i\hbar\frac{\partial \Psi(\vec r,t)}{\partial t} &= -\frac{\hbar^2}{2m} \nabla^2 \Psi(\vec r,t) + V(\vec r) \Psi(\vec r, t). \\ \text{Define }\,\xi_{\vec k}(t) &= \int d^3r \,\Psi(\vec r,t)\, e^{-i\vec k \cdot \vec r}.\\ \int d^3r\, e^{-i\vec k \cdot \vec r} & \left\{ i\hbar\frac{\partial \Psi(\vec r,t)}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \Psi(\vec r,t) + V(\vec r) \Psi(\vec r, t). \right\}\\ i\hbar\frac{\partial\xi_{\vec k}(t)}{\partial t} &= +\frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \Psi(\vec r, t).\\ \end{align}

A trick is needed to convert the last term. Remind that $$ \delta^3(\vec r - \vec r') = \frac{1}{(2\pi)^3}\int\, d^3q\, e^{i\vec q \cdot(\vec r - \vec r')} $$ Apply this to the previous equation: \begin{align} i\hbar\frac{\partial \xi_{\vec k}(t)}{\partial t} &= +\frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \Psi(\vec r, t).\\ &= \frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \int d^3\vec r' \delta^3(\vec r - \vec r') \Psi(\vec r', t).\\ &= \frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \frac{1}{(2\pi)^3}\int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \int d^3\vec r' \int\, d^3q\, e^{i\vec q \cdot (\vec r - \vec r')} \Psi(\vec r', t).\\ &= \frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \frac{1}{(2\pi)^3}\int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \int\, d^3q\, e^{i\vec q \cdot\vec r}\left\{\int d^3\vec r' \, e^{-i\vec q \cdot \vec r'}\,\Psi(\vec r', t).\right\}\\ &= \frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \frac{1}{(2\pi)^3}\int d^3r\, e^{-i\vec k \cdot \vec r} V(\vec r) \int\, d^3q\, e^{i\vec q \cdot \vec r} \xi_{\vec q}(t).\\ &=\frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \frac{1}{(2\pi)^3} \int d^3q \left\{\int d^3r e^{-i (\vec k -\vec q) \cdot \vec r} V(\vec r) \right\} \xi_{\vec q}(t).\\ &=\frac{\hbar^2 k^2}{2m} \xi_{\vec k}(t) + \frac{1}{(2\pi)^3} \int d^3q V_{\vec k - \vec q} \xi_{\vec q}(t).\\ \end{align} The result is a first order differential equation in time, but with a pay at the non-local connection in the momentum space.