How to apply Schrödinger equation to superposition with time dependent factors?

Question

The question comes from reading through either of these two papers:

https://doi.org/10.1103/PhysRevB.35.3629

https://arxiv.org/abs/1811.05886

The question is on the time dependence of a state like: $$|\psi\rangle = \sum_n c_n(t)|n\rangle,$$ where the states $|n\rangle$ are not Eigen states of the time independent Hamiltonian, i.e. they could be a state in a tight binding model where the n'th site is occupied. In the paper they state that it is an ansatz that the state satisfies the Schrödinger equation like $i\hbar \frac{d}{dt}|\psi\rangle = H|\psi\rangle$, but isn't the Schrödinger equation universally true in quantum mechanics no matter how you create your state? How am I supposed to understand this?

Also it made me question some basic things about time evolution, like how am I supposed to use the Schrödinger equation for this state that is not written in terms of basis states, as I can write: \begin{align} i\hbar \partial_t|\psi\rangle &= H|\psi\rangle \\ \Rightarrow i\hbar \sum_n \dot c_n|n\rangle+i\hbar\sum_n c_n \partial_t|n\rangle &= i\hbar\sum_n c_n H|n\rangle, \end{align} but from the Schrödinger equation I would just get $i\hbar \partial_t |n\rangle = H|n\rangle$ so the last term on the LHS and the RHS would cancel and leave me with $i\hbar \sum_n \dot c_n|n\rangle = 0$?

Answer 1

This depends entirely on what you mean by your collection of states $|n⟩$.

The natural reading is that these states (i) are time-independent and (ii) form a basis for the space.

If they are time-independent, then your assertion that they obey a Schrödinger equation $i\hbar \partial_t |n\rangle = H|n\rangle$ is not (cannot) be right. Instead, they are likely to just be some convenient set of states with useful physical interpretation.

If they form a basis for the space, the form $|\psi\rangle = \sum_n c_n(t)|n\rangle$ is not really an Ansatz — it is just a basis decomposition of your state. When properly used, the term Ansatz refers to an artificial narrowind down of possibilities (which will later on be shown to be acceptable). Here you are just choosing a basis and representing your state in that basis.

So, what happens in this standard case? If you have a time-independent basis set $\{|n⟩\}$, you indeed go from \begin{align} i\hbar \partial_t|\psi\rangle &= H|\psi\rangle \end{align} to \begin{align} i\hbar \sum_n \dot c_n|n\rangle+i\hbar\sum_n c_n \partial_t|n\rangle &= i\hbar\sum_n c_n H|n\rangle, \end{align} which simplifies to \begin{align} i\hbar \sum_n \dot c_n|n\rangle = i\hbar\sum_n c_n H|n\rangle. \end{align} You can then project from the left with a basis state $⟨m|$ to get \begin{align} i\hbar \dot c_n= i\hbar\sum_m c_m H_{nm} \end{align} where $H_{nm}=\langle n|H|m\rangle$. This is just the time-dependent Schrödinger equation expressed in the $\{|n⟩\}$ basis.

Answer 2

You're almost there. The point is that $\vert n\rangle$ is a solution to the time independent Schrödinger equation for some Hamiltonian $H_0$ so $\partial_t\vert n\rangle=0$, just like the wavefunction $\psi_n(x)$ has no explicit $t$-dependence but are eigenstates of some $H_0$ and thus satisfy $H_0\psi_n(x)=E_n\psi_n(x)$.

To remove the confusion, it would be better to write $$ i\hbar \partial_t\vert\psi(t)\rangle = H\vert\psi(t)\rangle $$ with explicit time-dependence or better yet use a symbol different from $\psi$, v.g. $$ i\hbar \partial_t\vert\Psi(t)\rangle = H\vert\Psi(t)\rangle $$ which makes it clear that $\vert \psi\rangle$ does not depend explicitly on $t$. In this notation, time-dependent eigenstates of $\hat H_0$ are just $$ \vert\Psi_n(t)\rangle= e^{-i E_nt/\hbar}\vert n\rangle $$ with $\vert n\rangle$ time-independent and $\hat H_0\vert n\rangle=E_n\vert n\rangle$.

If you do this you actually start with: \begin{align} i\hbar \sum_n \dot c_n|n\rangle &= \sum_n c_n H|n\rangle \tag{1} \end{align} (and no $i\hbar$ in front of $H$ on the right), close with $\langle k\vert$ to get \begin{align} i\hbar \dot c_k &= \sum_n c_n H_{kn} \, , \qquad \langle k\vert H\vert n\rangle=H_{kn} \end{align} In the special case where $H=H_0$ is diagonal, i.e. where $\vert n\rangle$ are eigenstates of $H_0$, then $(H_0)_{kn}=E_{k}\delta_{kn}$ and $c_k(t)=c_k(0)e^{-i E_k t/\hbar}$.