Why do we tend not to use the time equation for bound states in non-relativistic QM?

Question

The Time Dependent Schrodinger Equation is generally written as

$$i\frac{\partial\Psi}{\partial{t}}=-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi$$

however it can also be written as

$$i\frac{\partial\Psi}{\partial{t}}+\frac{\hbar^2}{2m}\nabla^2\Psi-V\Psi=0$$

and if $\Psi$ is separable in space and time then I can set $\Psi=XT$, to get

$$i\frac{\partial{XT}}{\partial{t}}+\frac{\hbar^2}{2m}\nabla^2XT-VXT=0$$

$$iX\frac{\partial{T}}{\partial{t}}+T\frac{\hbar^2}{2m}\nabla^2X-VXT=0.$$

Then I can divide by $XT$ to get

$$\frac{1}{T}i\frac{\partial{T}}{\partial{t}}+\frac{1}{X}\frac{\hbar^2}{2m}\nabla^2X-V=0.$$

Next I can separate the equation into a time equation

$$\frac{1}{T}i\frac{\partial{T}}{\partial{t}}-E=0$$

and a space equation

$$\frac{1}{X}\frac{\hbar^2}{2m}\nabla^2X-V+E=0.$$

The space equation can be changed to

$$E=-\frac{1}{X}\frac{\hbar^2}{2m}\nabla^2X+V$$

then I can multiply the space equation by $X$ to get

$$E=-\frac{1}{X}\frac{\hbar^2}{2m}\nabla^2X+V$$

$$EX=-\frac{\hbar^2}{2m}\nabla^2X+VX.$$

The space equation is the same as the Time Independent Schrodinger Equation, but with $X$ in place of $\Psi$, which means that the time Independent Schrodinger Equation can be derived from the Time Dependent Schrodinger Equation using separation of variables.

So why is it that we tend not to use the time equation $i\frac{\partial{T}}{\partial{t}}=ET$ for bound states?

Answer 1

This is used all the time. What you are doing is deriving the time-evolution for an eigenstate of the Schrödinger operator (ie a solution of the time-independent Schrödinger equation).

Such solutions are usually written $\psi$ (rather than $X$ which is reserved for things like the position operator). The equation on the time-dependent factor is not often explicitely written, because it's solution is (using your notations) : $$T(t) =e^{-iEt}$$

What often appears in quantum mechanics texts is that, if $\psi(x,0) = \psi_E(x)$ is an eigenstate with energy $E$ at time $t = 0$, then it's time-evolution is given by : $$\psi(x,t) = e^{-iEt} \psi(x,0)$$