# Why is a combination of stationary states not stationary, given the actual probability does seem to be so?

EDIT: So I thought about it and I found a trivial mistake, the eigenstates are only orthogonal over either all space or some specific interval (for example in an infinite square well), meaning that this would work only for said intervals where the eigenstates must be orthogonal, not in general.

Consider the general solution for a basis of stationary states:

$$\psi(x,t) = \sum_n \alpha_n \psi_n(x,t)e^{-\frac{i}{\hbar}E_nt}$$

The probability density function follows:

$$|\psi|^2=\psi^*\psi=\sum_{n,m} \alpha_m^* \alpha_n \psi_m^*\psi_n e^{-\frac{i}{\hbar}(E_n-E_m)t}$$

Where the phase term does not disappear as it would for a stationary state $$\psi_n$$. However, if we try to look for the probability of finding the particle between to arbitrary points $$a$$ and $$b$$:

$$\mathrm{Prob}(a

Where $$A_n$$ appears because the integral is not 1 for $$\int_{a}^{b}|\psi_n|^2dx$$.

The point is that the cross terms disappear because the eigenstates are orthogonal, so the phase term with $$E_n-E_m$$ just collapses to a 1 for every $$n=m$$ case and therefore the actual probability does not evolve with time.

¿Doesn't this mean that you could call a superposition of stationary states "stationary", even if the probability density is not?

• Indeed, the integrals only simplify to Kronecker deltas if you integrate over the entire support of the wave functions, which amounts to calculating the probability that the particle is anywhere at all. May 13, 2023 at 21:33

No it doesn't. $$\int_{a}^{b} \psi_m^* \psi_n = \delta_{nm}$$ only if $$a = -\infty$$ and $$b = \infty$$ which corresponds to the probability of the particle being anywhere which is always 1 and therefore always time-independent.
In fact, in case of a Hamiltonian that does not explicitely depend on t, the only stationary states are the eigenstates of said Hamiltonian ($$\cdot e^{-i E_n (t-t_0)/{\hbar}}$$):
In order for a state to be stationary, it has to have all time-dependence in a physically irrelevant phase. So we need $$\psi(\vec x,t) = \Phi(\vec x) \cdot e^{i \alpha(t)}$$ with real $$\alpha$$. If we plug that into Schrödinger's equation, we get $$- \hbar (\partial_t \alpha(t)) \Phi(\vec x) \cdot e^{i \alpha(t)} = \hat H \Phi(\vec x) \cdot e^{i \alpha(t)} \Rightarrow - \hbar \partial_t \alpha(t) = \frac{\hat H \Phi(\vec x)}{\Phi(\vec x)}$$ . The left hand side is independent of $$\vec x$$ and the right hand side is independent of $$t$$, so both sides must be constant, independent of $$\vec x$$ and $$t$$ (since they are equal - that argument is very common in theoretical physics and works exactly like that every time you have a separation ansatz).
Therefore (RHS=const.) $$\hat H \Phi(\vec x) = E_n \cdot \Phi(\vec x)$$ with some constant $$E_n$$. That is obviously an eigenvalue equation.
LHS: $$- \hbar \partial_t \alpha(t) = E_n$$ with the same constant $$E_n$$ $$\Rightarrow \alpha(t) = - \frac{E_n (t - t_0)}{\hbar}$$ with some $$t_0$$, which proves the statement.