The definition of a stationary state in my notes is:

A stationary state is a quantum state with all observables independent of time. It is an eigenvector of the Hamiltonian. This corresponds to a state with a single definite energy (instead of a quantum superposition of different energies).

Is the statement 'it is eigenvector of the Hamiltonian' part of the original definition or is it something we proved and thus included.

Does being the eigenvector of the Hamiltonian means that a state is a superposition?

My questions is related the answer given in this post: Is the superposition of stationary states a stationary state? If not, then why not?.

In the proof the person doesn't uses the $|\psi(x,t)^2|$ to show time independence or dependence but only say that the superposition state are not solution of TISE (eigenstate of the Hamiltonian) and (??) this means they are not stationary state.


3 Answers 3


To look at the answer in a different way, let's say that you had a system in some state $|\psi\rangle$. The Schrodinger equation tells us that this state evolves with time according to the equation:

$$|\psi(t)\rangle = \exp\left(- \frac{i}{\hbar}\hat{H} t\right) |\psi(0)\rangle,$$

where to obtain the above equation I just formally solved the differential equation $$i\hbar\frac{\text{d}}{\text{d} t}|\psi(t)\rangle =\hat{H} |\psi(t)\rangle.$$

As you can see, the "time evolution operator" $\hat{U} = \exp\left(- \frac{i}{\hbar}\hat{H} t\right)$ is intricately linked to the Hamiltonian $\hat{H}$ because of the Schrodinger Equation. It turns out that formally $\hat{U}$ is just defined using the power series of the exponential: $$\hat{U}(t) = \sum_{n=0}^\infty \left(-\frac{i}{\hbar}\right)^n\frac{\hat{H^n}}{n!}.$$ We can now ask ourselves what happens when $|\psi\rangle$ is some arbitrary state. The problem is, the action of $\hat{U}$ on any arbitrary state is not easy to define. However, there is a certain class of states for which the action of $\hat{U}$ is easy to define, and these are the eigenstates of the Hamiltonian. This is because if we consider a state of definite energy $|\phi_E\rangle$, meaning that $$\hat{H} |\phi_E\rangle = E |\phi_E\rangle \quad \quad \implies \quad \quad \hat{H}^n |\phi_E\rangle = E^n |\phi_E\rangle.$$

Using this fact, it's easy to show that $$\hat{U} |\phi_E\rangle = e^{-iEt/\hbar}|\phi_E\rangle.$$

In other words, its time evolution is very easy to describe:

$$|\psi(0) \rangle = |\phi_E\rangle \quad \quad \implies \quad \quad|\psi(t)\rangle = \hat{U}|\psi(0)\rangle = e^{-iEt/\hbar}|\phi_E\rangle.$$

Notice how the time dependence of the state $|\psi(t)\rangle$ is only in an overall phase factor! As a result, the probability amplitude of finding the particle at any position $x$ is just $$\Psi(x,t)\equiv \langle x | \psi(t)\rangle = e^{-iEt/\hbar}\phi_E(x),$$ where $\phi_E(x) \equiv \langle x | \phi_E\rangle$. Clearly, from this, you should be able to see that the probability distribution is independent of time, or rather, it's "stationary"!

However, suppose you had a superposition of two energy eigenstates $| E_1\rangle$ and $|E_2\rangle$. Then, $$|\psi(0)\rangle = \frac{1}{\sqrt{2}} \left(|E_1\rangle + |E_2\rangle \right) \quad \quad \implies \quad \quad |\psi(t)\rangle = \frac{1}{\sqrt{2}} \left(e^{-iE_1t/\hbar}|E_1\rangle + e^{-iE_2t/\hbar} |E_2\rangle \rangle \right).$$

In this case, you can see that the time dependence of each of energy eigenstates is different. Because of this, when you calculate the probability distribution, you can show that: \begin{aligned}|\Psi(x,t)|^2 = |\langle x | \psi(t)\rangle|^2 &= \Big|\phi_{E_1}(x) + e^{i(E_2-E_1)t/\hbar} \phi_{E_2}(x)\Big|^2 \\ &= |\phi_{E_1}(x)|^2 +|\phi_{E_2}(x)|^2 + 2 \,\,\phi^*_{E_1}\phi^{\,}_{E_2} \cos\left( \frac{E_2 - E_1}{\hbar} t\right)\end{aligned}

Notice how, because the states $|E_1\rangle$ and $|E_2\rangle$ are different, the probability density now has a time dependence and is therefore no longer stationary. To illustrate this, I've used the example of a particle in a box, with $m=1$, $\hbar =1$, and $L=1$, and plotted the above examples to show how the probability distributions vary with time:

                         enter image description here

You can extend the above argument to show how any arbitrary state evolves in time by realising that the states of definite energy form a complete basis, and so any arbitrary state can be written as \begin{aligned}|\psi(0)\rangle &= c_0 |E_0\rangle + c_1 |E_1\rangle + c_2 |E_2\rangle + ... = \sum_{n=0}^\infty c_n |E_n\rangle \implies |\psi(t)\rangle = \sum_{n=0}^\infty c_n e^{-iE_n t/\hbar} |E_n\rangle \end{aligned}

Repeating the same analysis as above, you can show that the only states that are "stationary" are states of definite energy (eigenstates of the Hamiltonian). Any arbitrary superposition of them will no longer be stationary (provided, of course, there is no degeneracy etc. which would need to be addressed separately).


I think the nicest way to look at the definition of stationary state is that there are two different notions that turn out to be equivalent.

Definition (Stationary State) A stationary state $|\psi\rangle$ is one in which the expectation value of any operator $\hat{Q}$ (without explicit time-dependence) is not time dependent. That is:

$$\frac{d}{dt}\langle \psi | \hat{Q} | \psi \rangle = 0 $$

Definition (Energy Eigenstate) An energy eigenstate $|\phi \rangle $ is an eigenvector of the Hamiltonian operator: $\hat{H} |\phi_m\rangle = E_m |\phi_m\rangle$ (where I've put in some subscripts to indicate that there are a collection for any given $\hat{H}$).


Now as it turns out, the energy eigenstates are stationary states. This is easy to prove since it turns out that:

$$\frac{d}{dt} \langle \psi | \hat{Q} | \psi \rangle \propto \langle \psi | \hat{Q}\hat{H}-\hat{H}\hat{Q} |\psi \rangle $$

and if you take $|\psi\rangle = |\phi \rangle$ one immediately gets the right hand side to be zero by acting to the right with the first $\hat{H}$ and left with the second.

It also turns out that stationary states are energy eigenstates. The eigenstates form a basis so we may write $|\psi \rangle = \sum_m \psi_m |\phi_m \rangle$. We then have:

$$\frac{d}{dt}\langle \hat{Q} \rangle \propto \psi_m^*\psi_n \langle \phi_m |\hat{Q} E_n - E_m \hat{Q} |\phi_n \rangle $$ $$\frac{d}{dt} \langle \hat{Q} \rangle \propto \psi_m^* \psi_n (E_n-E_m) Q_{mn} $$

Now we want this to hold for any $Q_{mn}$ so we need the prefactor $\psi_m^*\psi_n (E_m-E_n)$ to vanish. However the only way to achieve this is to select the co-efficients such that they are non-zero for only one value of $m$. That is, we pick $|\psi\rangle = |\phi_{n_0}\rangle$ corresponding to $\psi_m=\delta_{m,n_0}$. (If two energy levels coincide then we could also make it zero by just picking a linear combination of those energy eigenstates, however this will still be an energy eigenstate with the same energy so degeneracy doesn't cause any issues here.)

Thus energy eigenstates are stationary states (which is pretty simple) but crucially (and this seems to be more what OP was asking) stationary states are energy eigenstates. This explains why the terminology is often blurred. So yes, the first sentence in OP's quoted notes can be taken as a definition and the second as a theorem (or vice versa).


A stationary state may be defined as one that allows for separation of variables:

$$ \Psi(x, t) = \psi(x)\phi(t), $$

where $\phi(t) = e^{-iEt/\hbar}$ and $\psi(x)$ satisfies the time-independent Schrödinger equation $\hat{H}\psi(x) = E\psi(x)$, with the Hamiltonian operator

$$ \hat{H} = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2} + V. $$

We can find the expectation value of $\hat{H}$:

$$ \langle \hat{H} \rangle = \int \psi^*\hat{H}\psi dx = E\int|\psi|^2dx = E\int|\Psi|^2 dx = E $$

as well as $\hat{H}^2$:

$$ \langle \hat{H}^2 \rangle = \int \psi^*\hat{H}^2\psi dx = E^2\int|\psi|^2dx = E^2\int|\Psi|^2 dx = E^2 $$

and then find that the standard deviation of $\hat{H}$ is

$$ \sigma = \sqrt{\langle \hat{H}^2 \rangle - \langle \hat{H} \rangle ^2} = \sqrt{E^2 - E^2} = 0. $$

A standard deviation of zero means that every measurement produces an identical value. Thus, every measurement of $\hat{H}$ on a stationary, and therefore separable, state must always give the same result. This makes it a determinate state, and determinate states are eigenvectors of their operator.

Returning to your original question, notice that we did not need to require that this state be an eigenvector of the Hamiltonian in order to arrive at this result. It is rather something that we proved and concluded.

  • $\begingroup$ Would it be correct to assume that if a quantum state is an eigenvector of the Hamiltonian then the quantum state is a stationary state $\endgroup$
    – Jack Jack
    Jan 31, 2021 at 10:32
  • $\begingroup$ @JackJack Yep! That's the definition of a stationary state. $\endgroup$
    – Philip
    Jan 31, 2021 at 10:34
  • $\begingroup$ Correct me if I am wrong, but you assumed that $\psi(x)$ is an eigenstate of $\hat{H}$, no? $\endgroup$ Jan 31, 2021 at 10:35

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