# Difference between the energy eigenstates and the eigenstates of other physical variables

I am having difficulties understanding the difference between the energy eigenstates and the eigenstates of other physical variables. I was told that if a system is an energy eigenstate at $t = 0$ then it stays in that state until an external perturbation is applied. But if the system of any operator is in an eigenstate that corresponds to a physical observable, it can change without any perturbation. Why is that the case? Where does this difference come from? Also why does a set of eigenstates of an operator has to commute with $H$ in order to be a stationary state?

• Its simply that $H$ is the operator which appears in the Schrodinger equation, which describes the time evolution of the system – By Symmetry May 8 '18 at 10:37
• @BySymmetry sorry but I am struggling to find a connection between $H$ appearing in the SE and stationary states. – daljit97 May 8 '18 at 12:17

This "speciality" of Energy eigenstates comes from the definition of energy itself. We know from Noether's theorem that every symmetry of a system gives us a conserved charge. Now if your system has a time-translation symmetry, the conserved quantity associated to that is called energy. In other words, when you look at a system at two different times, the only thing guaranteed to be same in both the measurements is energy of the system.

Now every symmetry also has a generator associated to it (whose eigenvalue gives you the conserved charge). For time translation symmetry, the generator is the Hamiltonian itself and energy is its eigenvalue. That's why a system which is in an eigenstate of Hamiltonian(energy eigenstate) will stay just like that. If any other generator commutes with hamiltonian, that implies that the eigenstates of these two are the same and the corresponding states will also be the stationary states.

Mathematically speaking, $$H \, |E(0) \rangle = E \; |E\rangle$$

$$|E(t) \rangle = e^{-iHt}|E \rangle \implies |E(t) \rangle = e^{-iEt}|E \rangle$$ where in the last step, the state has changed only by a irrelevant phase.

The difference is in the time-evolution.

Eigenstates of the Hamiltonian, which are states of definite energy since $\Delta E=0$ for those states, evolve in time according to $$\vert \Psi_n(t)\rangle =e^{-i E_nt/\hbar}\vert\psi_n\rangle\, . \tag{1}$$ This follows immediately from separation of variables in the time-dependent Schrodinger equation: $$i\hbar\frac{\partial d}{\partial t}\Psi(x,t)=\hat H\Psi(x,t)$$ and the ansatz $\Psi(x,t)=e^{-i Et/\hbar}\psi(x)$ with $\hat H\psi(x)=E\psi(x)$. This makes the evolution of eigenstates of $\hat H$ particularly simple.

If the state $\vert\Phi(t)\rangle$ is not an eigenstate of the Hamiltonian but an eigenstate of some other operator $\hat{A},$ its evolution is NOT of the type given in Eq(1). The evolution of this state is more complicated and obtained by first expressing $\vert\Phi(t)\rangle$ in terms of the $\vert \Psi_n(t)\rangle$'s, for which the evolution is simple: \begin{align} \vert\Phi(0)\rangle&=\sum_n \vert \Psi_n(0)\rangle \langle \Psi_n(0)\vert \Phi(0)\rangle \, ,\\ &=\sum_n\vert \Psi_n(t)\rangle b_n\, , \tag{2} \end{align} with $b_n=\langle \Psi_n(0)\vert \Phi(0)\rangle$ determined at $t=0$. Then \begin{align} \vert\Phi(t)\rangle&=\sum_n \vert \Psi_n(t)\rangle b_n \, ,\\ &=\sum_n e^{-iE_nt/\hbar} \vert \psi_n\rangle b_n\, . \tag{3} \end{align} Note that, if $\hat A$ is taken to be the Hamiltonian, then by orthogonality of the eigenstates of the Hamiltonian, Eq.(2) will contain only a single term - say the term where $n=m$ and $E_m$ the energy of the state at $t=0$.

If $\hat A$ commutes with $\hat H$ and $\vert\psi_n\rangle$ is an eigenstate of $\hat H$ with eigenvalue $E_n$, then write $\vert\chi_A\rangle = \hat A\vert\psi_n\rangle$ \begin{align} \hat H\vert\chi_A\rangle&= \hat H\hat A\vert\psi_n\rangle = \hat A \hat H\vert\psi_n\rangle =E_n \hat A\vert\psi_n\rangle \, ,\\ &=E_n\vert\chi_A\rangle \tag{4} \end{align} meaning that $\vert\chi_A\rangle$ is also an eigenstate of $\hat H$. In fact, if the eigenvalue $E_n$ occurs only one, $\vert\chi_A\rangle$ must be proportional to $\vert \psi_n\rangle$ since this is the only eigenstate of $\hat H$ with eigenvalue $E_n$. If, on the other hand $\hat A$ does not commute with $\hat H$, then $\hat A\vert\psi_n\rangle$ is not an eigenstate of $\hat H$ and so will not evolve "simply" as per (1).