# How initial states of definite energy change with time

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 $$t$$.