Let’s ask the question: do the wavefunctions which take the particular form \(\psi(x,t)=F(t)G(x)\) satisfy Schrödinger’s time-dependent equation? We can answer this question by substituting \(\psi(x,t)\) into Schrödinger’s equation and check to see if \(\psi(x,t)\) satisfies Schrödinger’s equation. Schrödinger’s time-dependent equation can be viewed as a machine where if you give me the initial wavefunction \(\psi(x,0)\) as input, this machine will crank out \(\psi(x,t)\) and tell you how that initial wavefunction will evolve with time. We are asking the question: if some wavefunction starts out as \(\psi(x,0)=F(0)G(x)\), is \(\psi(x,t)=F(t)G(x)\) a valid solution to Schrödinger’s equation? We know that \(\psi(x,0)=F(0)G(x)\) is a valid starting wavefunction. It is easy to imagine wavefunctions \(\psi(x,0)=\psi(x)=F(0)G(x)=(constant)G(x)\); this is just a wavefunction \(\psi(x)\) that is a constant times some function of \(x\). But does \(\psi(x,t)=F(t)G(x)\) satisfy Schrödinger’s equation? Let’s substitute it into Schrödinger’s equation and find out

$$iℏ\frac{∂}{∂t}(F(t)G(x))=\frac{-ℏ^2}{2m}\frac{∂^2}{∂x^2}(F(t)G(x))+V(x)F(t)G(x).\tag{1}$$

This can be simplified to

$$iℏG(x)\frac{d}{dt}(F(t))=F(t)(\frac{-ℏ^2}{2m}\frac{d^2}{dx^2}G(x)+V(x)G(x)).\tag{2}$$

The term \(\frac{-ℏ^2}{2m}\frac{d^2}{dx^2}G(x)+V(x)G(x)\) is just the energy operator \(\hat{E}\) acting on \(G(x)\); thus we can rewrite Equation (2) as

$$iℏG(x)\frac{d}{dt}F(t)=F(t)\hat{E}(G(x)).\tag{3}$$

Let’s divide both sides of Equation (3) by \(G(x)F(t)\) to get

$$\frac{iℏ}{F(t)}\frac{d}{dt}F(t)=\frac{\hat{E}(G(x))}{G(x)}.\tag{4}$$

Since the left hand side of Equation (4) is a function of time and the right hand side of Equation (4) is a function of position, it follows that

$$\frac{iℏ}{F(t)}{d}{dt}F(t)=\frac{\hat{E}(G(x))}{(G(x)}=C\tag{5}$$

where \(C\) is a constant. From Equation (5), we see that

$$\hat{E}(G(x))=CG(x).\tag{6}$$

In other words, \(G(x)\) must be an eigenfunction of the energy operator \(\hat{E}\) with eigenvalue \(C=E\). Thus, we can rewrite Equation (6) as

$$\hat{E}(\psi_E(x))=E\psi_E(x).\tag{7}$$

What this means is that the initial wavefunction \(\psi(x,0)=F(0)G(x)=(constant)\psi_E(x)\) must be an energy eigenfunction to satisfy Schrödinger’s equation. (Note that a constant times a wavefunction corresponds to that same wavefunction.) We also see from Equation (5) that

$$\frac{d}{dt}(F(t))=\frac{E}{iℏ}F(t).\tag{8}$$

The solution to Equation (8) is

$$F(t)=Ae^{Et/iℏ}.\tag{9}$$

It is easy to verify this by plugging Equation (9) into Equation (8). \(\psi(x,t)=F(t)G(x)\) is indeed a valid solution to Schrödinger’s time-dependent equation if \(G(x)=\psi_E(x)\) and \(F(t)=Ae^{Et/iℏ}\). If a system starts out in an energy eigenstate, then the wavefunction will change with time according to the equation

$$\psi(x,t)=Ae^{Et/iℏ}\psi_E(x).\tag{10}$$

Schrödinger’s equation describes how a wavefunction evolves with time if the system is “undisturbed” without measuring the system. If we measure the energy \(E\) of a system and collapse its wavefunction to the definite energy eigenstate \(\psi_E(x)\), and then simply just “leave the system alone” for a time \(t\) without disturbing it, the wavefunction \(\psi_E(x)\) will change to \(Ae^{Et/iℏ}\psi_E(x)\). But multiplying \(\psi_E\) by \(Ae^{Et/iℏ}\) does not change the expectation value \(〈\hat{L}〉\). Also, if \(|\phi⟩\) is any arbitrary state, then \(|〈\phi|\psi_E(x)〉|^2\) is the same as \(|〈\phi|Ae^{Et/iℏ}\psi_E(x)〉|^2\). Thus the probability of measuring any physical quantity associated with the system does not change with time. For an atom in its ground state with a definite energy \(E_0\) which is not disturbed, \(〈\hat{L}〉\) will not change with time. You are equally likely to measure any particular position, momentum, angular momentum, and so on at any time \(t\).