# Something special about energy eigenstates when it comes to time evolution?

A particle is subject to an infinite square well potential with

$$V(x)= \begin{cases} 0 & −a \lt x \lt a\\ \infty & \,\,\,\,\text{otherwise} \end{cases}$$

At a time $t=0$ its wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)$$

(a) What are the possible results of an energy measurement at $t = 0$, and with what probabilities?

Re-writing the terms in $\psi(x,t=0)$ as $$u_1(x)=a^{-1/2}\cos\left(\frac{\pi x}{2a}\right)\qquad \text{&} \qquad u_2(x)=a^{-1/2}\sin\left(\frac{\pi x}{a}\right)$$ such that $$\psi(x,t=0)= \frac{1}{\sqrt{5}}u_1(x)+\frac{2}{\sqrt{5}}u_2(x)$$ with energies $$E_1=\frac{\hbar^2 \pi^2}{8ma^2}\qquad \text{&} \qquad E_2=\frac{\hbar^2 \pi^2}{2ma^2}$$ and probabilities $$P(E_1)=\frac15 \qquad \text{&} \qquad P(E_2)=\frac45$$

(b) If no measurement is performed, what is the wavefunction $\psi(x, t)$ at all times $t$? What are the possible energies and their probabilities if a measurement is first performed at time $t$?

At a time $t$ the wavefunction is given by $$\psi(x, t=0)=\frac{1}{\sqrt{5a}}\cos\left(\frac{\pi x}{2a}\right)\exp\left(\frac{-i E_1 t}{\hbar}\right)+\frac{2}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right)\exp\left(\frac{-i E_2 t}{\hbar}\right)$$

The problem is that I cannot answer the second part of the question (b).

The answer states that:

Because the two terms are energy eigenstates, the relative probabilities do not change and are as in the previous part.

I have some questions about the above statement: why does an "energy eigenstate" mean that the probablities don't change with time? Or, put in another way, does a "momentum eigenstate" (for example) have constant relative probabilities when measured at any time?

The answer seems to be implying there is something special about energy eigenstates when it comes to time evolution. The eigenvalues of energy measurements don't seem to change regardless of how many times you measure them, and no matter how long you wait until you measure them. I'm not sure if this is the case, but could someone please explain why all quantum operators (except Hamiltonian) do not exhibit this behaviour.

## 2 Answers

The reason as to why the energy eigenstates are special is because of the Schrodinger's equation $$i \hbar \frac{\partial}{\partial t} \Psi(x,t) = \hat{H} \Psi(x,t)$$ If $\hat{H}$ is time independent, the solution of the equation is given by $$\Psi(x,t)= e^{-\frac{i}{\hbar}\hat{H} t} \, \Psi(x,0)$$

The fact that energy eigenstates are special with respect to the time evolution is precisely due to the fact that the Hamiltonian controls the time evolution of the system as given by the above equation.

If you expand your initial state $\Psi(x,0)$ in terms of the eigenstates of the Hamiltonian, then the action of the exponential operator on those states is easily given. Suppose $\psi_n(x)$ represents a state of energy $E_n$ and suppose you can expand your initial state as $$\Psi(x,0) = \sum_n c_n \psi_n(x)$$ where $c_n$ are constants (independent of time). Then the full solution to the Schrodinger's equation is $$\Psi(x,t) = \sum_n c_n e^{-\frac{i}{\hbar} E_n t} \psi_n(x)$$ Now because $c_n$ are independent of time, the probability of the system being in a particular energy eigenstate is same for all time. You can see this by calculating $|\langle \psi_m | \Psi(x,t) \rangle|^2=|c_m|^2$

The reason why for other operators this may not be true for a random operator, say $\hat{B}$ is because the eigenstates of this operator may not evolve independently. Upon time evolution, they may mix into each other. In other words, if you expand a random initial state in terms of the eigenstates of the operator $\hat{B}$ and then look at the projection onto an eigenstate after some time evolution, the coefficient will change. With a little work you can figure out that the condition that the coefficients remain time independent is that the operator $\hat{B}$ commutes with the hamiltonian, i.e., $[\hat{H},\hat{B}]=0$.

P.S - As pointed out in the comments by @ACuriousMind, although $[\hat{H},\hat{B}]=0$ is necessary, it may not be sufficient to ensure that the coefficients are time independent. Ultimately what you need is a common eigenbasis for the operators $\hat{B}$ and $\hat{H}$ to ensure that the time evolution does not mix the states.

An energy eigenstate is an eigenstate of the Hamiltonian $H$. If the Hamiltonian is not time-dependent, then the time evolution operator is $U(t) = \mathrm{e}^{\mathrm{i}Ht}$, so every energy eigenstate is also an eigenstate of the time evolution operator, meaning it remains the same state under time evolution.

No other operator exhibits this peculiar feature for its eigenstates because the Hamiltonian is special in being connected to time evolution through the Schrödinger equation.