• Start by writing the Schrodinger equation on the board.

$$i\, \hbar \, \frac{d}{dt}\, \vert \psi (t)\rangle =\hat{H} \, \vert\psi (t)\rangle \; \; . $$

• We don't know what $\vert\psi (t)\rangle$ is, but we do know what a basis is for it. Ask the class what basis we would want $\vert\psi (t)\rangle$ in. So, we want $\vert\psi (t)\rangle $ in the operator's basis, meaning the basis for the eigenstates of the Hamiltonian.
• Ask the class how we should name the eigenstates of the Hamiltonian. After fielding answers, write on the board

$$\hat{H}\vert E_{i} \rangle=E_{i}\vert E_{i} \rangle \; \; .$$

• We can then write our wavestate $\vert\psi (t)\rangle$ as a superposition of the Hamiltonian's eigenstates each scaled with some constant. Write on the board:

$$\vert\psi (t)\rangle \: = \: \sum_{n}c_{n}\vert E_{n} \rangle \; \; .$$

However, we need to add some time dependence to the term on the right since $\psi$ is dependent on time. Let us assume the Hamiltonian has no time dependence; this means that the eigenvectors of the Hamiltonian must also be time dependent. Ask the class, “Where then should we put the time dependence in our expression?” A student should say that $c_{n}$ must have the time dependence. Write in on the board that $c_{n}$ has time dependence.

• Now, we can insert our expression for $\vert\psi (t) \rangle$ into the Schrodinger equation to find more information. This gives

$$i\, \hbar \, \frac{d}{dt}\, \left(\sum_{n}c_{n}(t)\vert E_{n} \rangle\right)=\hat{H} \, \left(\sum_{n}c_{n}(t)\vert E_{n} \rangle\right) \; \; . $$

• It's important to keep track of what we know and what we don't know. At this point, we assume we know:

1. The Hamiltonian
2. The eigenvectors
3. The Hamiltonian's eigenvalues

So, all we really need at this point are to find the unknown constants $c_{n}(t)$.

• Now, we have a time derivative on the left side and the Hamiltonian on the right side. To do some bookkeeping, notice that

1. $\frac{d}{dt}$ will only operate on $c_{n}(t)$, since we know that $\vert E_{n} \rangle$ is time independent.
2. $\hat{H}$ will distribute through and only operate on each $\vert E_{n} \rangle$ since operators only act on kets.

Write on the board that these operations change the expression to

$$i\, \hbar \, \sum_{n}\frac{d\, c_{n}(t)}{dt}\,\vert E_{n} \rangle= \, \sum_{n}c_{n}(t)\, \hat{H}\vert E_{n} \rangle \; \; . $$

• Where do we go now? Typically a student will notice that $\hat{H}\vert E_{n} \rangle$ can be exchanged with $E_{n}\vert E_{n} \rangle$ using the energy eigenvalue equation. The equation then changes to

$$i\, \hbar \, \sum_{n}\frac{d\, c_{n}(t)}{dt}\,\vert E_{n} \rangle= \, \sum_{n}c_{n}(t)\, E_{n}\vert E_{n} \rangle \; \; . $$

• We need to find information about the coefficients of each $\vert E_{n} \rangle$ component on both sides of the equation. Ask the class, “how can we do this?” This is done by projecting onto the basis. So, to find out about the particular constant $c_{k}(t)$, we can project each side onto the corresponding energy state $\vert E_{k}(t) \rangle$. So, first we project onto our states as so:

$$\langle E_{k}\vert \, i\, \hbar \, \sum_{n}\frac{d\, c_{n}(t)}{dt}\,\vert E_{n} \rangle= \langle E_{k}\vert \, \sum_{n}c_{n}(t)\, E_{n}\vert E_{n} \rangle \; \; . $$

Then, we are projecting each component of the sum of $\vert E_{n} \rangle$ onto $\vert E_{k} \rangle$; do not be disturbed about moving $\langle E_{k}\vert$ through the sum! Remember, the sum is just a shorthand for a bunch of terms that are just added together. In short, the bra distributes through the sum to all of the kets. So, our expression becomes

$$\, i\, \hbar \, \sum_{n}\frac{d\, c_{n}(t)}{dt}\,\langle E_{k}\vert E_{n} \rangle= \, \sum_{n}c_{n}(t)\, E_{n}\langle E_{k}\vert E_{n} \rangle \; \; . $$

• We now have a single bra $\langle E_{k}\vert$ distributing over a bunch of eigenvectors $\vert E_{n} \rangle$. We know that if these states are orthogonal (if $k\neq n$), then $\langle E_{k}\vert E_{n} \rangle=0$, and that if they are the same (if $k=n$), then $\langle E_{k}\vert E_{n} \rangle=1$.

(As a sidenote, write down that this kind of relation can be written as the Kronecker delta, or $\langle E_{k}\vert E_{n} \rangle= \delta_{k,n}$.)

If the only non-zero terms are those where $n=k$, we can substitute every n for a k in our equation, remove the sum (since only the one term with k is left), and receive

$$\, i\, \hbar \, \frac{d\, c_{k}(t)}{dt}= \, c_{k}(t)\, E_{k} \; \; . $$

• We're left with an ordinary differential equation we can solve. Prompt the students with, “How can we solve this?” Separation of variables is the easiest strategy for this case, so isolate the $c_{k}$ terms from the $t$ terms and both sides of integrate both sides. This should look like

$$\int_{0}^{t}\frac{dc_{k}(t)}{c_{k}(t)}= \int_{0}^{t}\frac{E_{k}}{i\, \hbar}dt \; \;. $$

Carrying out this integral and factoring an i out on the right hand side will leave us with

$$ln\,c_{k}(t)-ln\,c_{k}(0)=\frac{-i\, E_{k}}{\hbar}t \; \; . $$

Check to see if students know how to use logarithmic rules to simplify the left side. After simplifying by combining the terms on the left side of the equal sign, taking the exponential of both sides will give

$$\frac{c_{k}(t)}{c_{k}(0)}=e^{\frac{-i\;E_{k}}{\hbar}t} \; \; .$$

Multiplying over the value $c_{k}(0)$ will give that the time-evolved constant is

$$c_{k}(t)=c_{k}(0)e^{\frac{-i\;E_{k}}{\hbar}t} \; \; .$$

• We did a large amount of computations and quantum mechanical reasoning to find this expression. Now, what do we do with it? Recall that the goal was to find what the coefficient was in $|\psi(t)\rangle$. Now, we can insert the answer we received back into what that coefficient would be. So, $c_{k}(t)$ is the constant for a particular energy eigenstate, but we can return the generality to this expression by replacing $k$ with $n$ once again, and then we find that our quantum state as a function of time is

$$\vert\psi (t)\rangle \: = \: \sum_{n}c_{n}(0)e^{\frac{-i\;E_{n}}{\hbar}t}\vert E_{n} \rangle \; \; . $$

• As an example, write out how $|\psi (t) \rangle$ would look in the two-dimensional case. If given some state whose coefficients have time dependence, represented as

$$\vert \psi (t) \rangle=c_{1}(t)\vert E_{1} \rangle \, + \, c_{2}(t)\vert E_{2} \rangle \; \; , $$

we know that the coefficients will explicitly look like

$$\vert \psi (t) \rangle=c_{1}(0)e^{\frac{-i\;E_{1}}{\hbar}t}\vert E_{1} \rangle \, + \, c_{2}(0)e^{\frac{-i\;E_{2}}{\hbar}t}\vert E_{2} \rangle \; \; . $$