I am currently going through Griffiths over the summer but I am a bit confused by one point and I don't have any instructor to ask, so I was wondering if you could help clarify. In Section 2.3, the harmonic oscillator, he writes: "it suffices to solve the time-independent Schrödinger equation."

Clearly, this is not sufficient in every case. I was wondering how we know a priori that

  1. it is sufficient and

  2. we are not missing some information by only solving the time independent case.

  • 1
    $\begingroup$ In any circumstance where something more than steady-state solutions are needed, v.g. scattering of wavepackets, time evolution of specific initial states... $\endgroup$ Commented May 26, 2018 at 22:59
  • $\begingroup$ These solutions form a basis for the state space (function space, i.e. Hilbert space). All time evolution can do is mix them up. You can expand any function in this basis so it is very useful. The same thing is true in acoustics and optics w/r to solving the Helmholtz equation. You can write the Green's function as a series over modes. $\endgroup$
    – user196418
    Commented May 26, 2018 at 23:28

3 Answers 3


The time-independent Schrödinger equation is just separation of variables acting on the “true” Schrödinger equation. The eigenvalues (the separation constants) of such equation just so happen to represent the energy of our quantum system. As such, if our interest is solely on the available and accessible states of or system, the time-independent version does just fine. If, however, we seek to model the system’s time evolution, then we need to evoke the time part of the Schrödinger equation.

If our time-independent equation has normalized solutions $\psi_1(x), \psi_2(x),\dots$, with $\int_{-\infty}^\infty\psi^*_m(x)\psi_n(x)=\delta_{mn},$ then we write


Where $H$ is the Hamiltonian and $E_n$ are the corresponding energies.

The time-dependent equation is

$$i\hbar\frac{\partial}{\partial t}\Psi(t,x)=H\Psi(t,x).$$

As such, we can write our time dependent quantum state in terms of a superposition of the independent states:


where the $A_n$ are a normalized set of constants, $\sum_n|A_n|^2=1$.


If you can solve the time-independent case, it always suffices to do so, since the time-evolution of a stationary state is simply $\psi_n(t)=e^{i\omega t}\psi_n(0)$, and any state can be written as a superposition of stationary states.

You can be sure you are not missing anything because every self-adjoint operator$^1$ (like the Hamiltonian) has a complete orthonormal basis of eigenvectors. This is called the spectral theorem.

On the other hand, the time-independent Schrodinger equation is intractable in all but the simplest of cases, so you are forced to rely on approximations in most cases. This often involves the time-dependent Schrodinger equation.

$^1$There's some fiddly details here for a completely general self-adjoint operator, but they don't really make a difference from a physicist's standpoint.

  • $\begingroup$ Specifically the operator needs to be compact. $\endgroup$
    – shalop
    Commented May 27, 2018 at 2:16
  • 1
    $\begingroup$ @Shalop For there not to be any fiddly bits, yes. A bounded operator is sufficient for there to be a complete state of eigenvectors if you expand what you mean by "eigenvector." Physicists tend to be perfectly happy to use a Dirac delta function as an "eigenvector" even though, formally, a Dirac delta is not in $\mathcal L^2$. $\endgroup$
    – Chris
    Commented May 27, 2018 at 3:29
  • $\begingroup$ Sure, and there's also the general version of the spectral theorem which reduces to this one in the compact case. $\endgroup$
    – shalop
    Commented May 27, 2018 at 3:53

The time-independent Schrodinger equation is the eigenvalue equation for the Hamiltonian operator, so it gives you the states with definite Hamiltonian, which by the time-dependent Schrodinger equation are the same as those of definite energy, since the thing on the right hand side of the Schrodinger equation is the energy operator. In particular, the time-dependent equation is:

$$\hat{H} \psi = \hat{E} \psi$$

and the time-independent equation is

$$\hat{H} \psi = E \psi$$

where now $E$ is a constant, but before it was an operator. The operator $\hat{E} = i\hbar \frac{\partial}{\partial t}$ is the generator of temporal translation in the same regard that the momentum operator $\hat{p}$ is the generator of spatial translation (associated to the famous Noether's theorem that energy is the conserved quantity associated to the symmetry of temporal translation, as momentum is likewise associated to spatial translation. Note also the two have effectively the same form - e.g. in one dimension $\hat{p} = -i\hbar \frac{\partial}{\partial x}$.). The first equation tells you that for a physically valid solution, acting on the wave function with the Hamiltonian operator will have the same result as the energy operator, and thus its eigenvalues would be the same as if we could solve the "legendary", metaphorical eigenvalue equation $\hat{E}\psi = E\psi$ (which we can't, because it yields garbage and there is no term for the system structure), but $\hat{H}$ sets up the structure of the system, so this one is actually useful, and to carry the same physical meaning as being a generator of time translation for real states and so an energy operator too - and thus solving the time-independent equation lets you get the energy levels and states of definite energy: energy eigenstates.

However, once you do that, you actually have effectively solved the first equation, because these will form a basis for the state space and so you can represent any solution as a (possibly infinite, possibly even integral) linear combination of these, which the first equation would evolve as though each component evolved separately, and the evolution of a single energy eigenstate forward in time is trivial (it just has a complex exponential out in front in the time).


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