# What is the difference between the time dependent and time independent Schrödinger equation?

I've already gone through a couple of questions regarding the Schrödinger equation and none of them seem to solve my doubt. Some say that the Time Independent Schrödinger Equation (TISE) is just a simplification of the Time Dependent Schrödinger Equation (TDSE) while others don't agree on this.

From what I understood, if the system has time independent Hamiltonian then we use TISE to describe it. However in this case the wave functions which are the linear combination of the eigenfunctions don't satisfy the TISE? The system is described by the TISE so the wavefunctions should satisfy the TISE for this system.

I don't understand how and where to use the TDSE and TISE. I think TDSE can be used for any system because it's the general form of the SE, but in a system which is time independent, shouldn't the wavefunction be a solution to TISE also?

I also have trouble in grasping the importance of the Hamiltonian being time independent and time dependent.

• The name is a little confusing, the time independent Schrodinger equation is just the eigenvalue equation for the Hamiltonian. The idea being that some systems are "stationary" and their time evolution is such that if they are in an eigenstate of the Hamiltonian with eigenvalue $E$ at $t=0$ they will evolve in such a way that they remain in the same eigenspace with respect to the Hamiltonian. May 16, 2021 at 20:12
• "From what I understood if the system has time independent Hamiltonian then we use TISE to describe it." This is wrong, as I stated in my answer to your previous question. The time dependent Schrodinger equation is what you use to evolve state vectors forward in time, whether your Hamiltonian is explicitly time-dependent or not. May 16, 2021 at 21:09

• the so called time-depedent Schrödinger equation describes the time-evolution of a quantum system. It is, in a way, the true Schrödinger equation : $$i\hbar \frac{\text{d}}{\text dt} |\psi\rangle = \hat{H}(t)|\psi\rangle$$

• for a system whose Hamiltonian $$\hat{H}$$ has no explicit time dependence, it makes sense to try and find stationary states, whose time evolution is given by $$|\psi(t)\rangle = e^{-iEt/\hbar}|\psi_0\rangle$$. Plugging this ansatz in the (time-dependent) Schrödinger equation, you get the so-called time-independent Schrödinger equation : $$\hat{H}|\psi_0\rangle = E|\psi_0\rangle$$ This is an eigenvalue problem so solving for $$E$$ and $$|\psi_0\rangle$$ will give you the energies for which a stationary state exists, as well as the wavefunctions of the said states.

• Lastly, for a system with time-independent Hamiltonian, finding the stationary states allows to formally solve the full Schrödinger equation. Indeed, since $$\hat{H}$$ is hermitian, the spectral theorem ensures that we can find an orthonormal basis $$|n\rangle$$ of stationary states with energies $$E_n$$. Then, any state $$|\psi(0)\rangle$$ can be expanded over this basis : $$|\psi(0)\rangle = \sum_n a_n |n\rangle \qquad \text{with} \quad a_n = \langle n |\psi(0)\rangle$$ You can check that the solution of the Schrödinger equation is then : $$|\psi(t)\rangle = \sum_n a_ne^{-iE_nt/\hbar}|n\rangle$$

If $$\hat H(t)$$ is a time dependent Hamiltonian, then the spectral theorem still applies : at each instant $$t$$, we can find an orthonormal basis eigenvectors $$|n(t)\rangle$$ of $$\hat{H}(t)$$ with eigenvalues $$E_n(t)$$. Since the state $$|n(t)\rangle$$ may have an arbitrary time dependence, they are not called stationary states.
Since $$\{ |n(t)\rangle\}$$ is a basis, we can expand $$|\psi(t)\rangle$$ on it at each time $$t$$ : $$|\psi(t)\rangle = \sum_n a_n(t) |n(t)\rangle \qquad \text{with}\quad a_n = \langle n(t) |\psi(t)\rangle$$ The $$a_n$$ verify a set of differential equations. However, contrary to the time-independent case, those equations are coupled. Explicitely, by plugging the expansion of $$|\psi(t)\rangle$$ in the Schrödinger equation and take the product with $$\langle n(t) \rangle$$, we get : $$\dot{a}_n= -\frac{iE_n}{\hbar}a_n - \sum_k a_k \bigg\langle n(t)\bigg|\frac{\text d}{\text d t}\bigg|k(t) \bigg\rangle$$