Consider a free particle in one-dimension. By the Planck and de Broglie hypotheses, we know that we want plane waves of the form $\psi(x,t) = Ae^{i(kx-\omega t)}$ to be solutions of the equation of motion. These solutions have definite momentum $p = \hbar k$ and energy $E = \hbar \omega$. One equation that we already know about that has plane wave solutions is the classical wave equation \[\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2},\] where $v$ is the wave speed.

Substituting $\psi(x,t) = Ae^{i(kx-\omega t)}$ into the wave equation gives \[-Ak^2 e^{i(kx-\omega t)} = -\frac{A}{v^2}\omega^2 e^{i(kx-\omega t)}.\] Cancelling the common terms on both sides and rearranging gives \[\omega^2 = k^2v^2,\] or \[\omega = kv.\] We can convert this dispersion relation into an equation relating energy and momentum using $p = \hbar k$ and $E = \hbar \omega$, which gives \[E = pv.\] This is the correct equation for a massless relativistic particle, e.g. a photon, but for a nonrelativistic massive particle we should have \[E = \frac{p^2}{2m},\] which implies that the classical wave equation cannot be the correct equation of motion for such a particle.

How should the wave equation be modified to get the correct energy-momentum equation? Notice that we get one factor of $E$ from each $\frac{\partial \psi}{\partial t}$ and one factor of $p$ from each $\frac{\partial \psi}{\partial x}$. Therefore, the simplest way to get $E = p^2/2m$ would be to have an equation that is first order in $\frac{\partial \psi}{\partial t}$ and second order in $p$ from each $\frac{\partial \psi}{\partial x}$, so let's try \[\frac{\partial \psi}{\partial t} = B \frac{\partial^2 \psi}{\partial x^2},\] where $B$ is a constant to be determined.

In an in class activity, you will show that to get the relation $E = p^2/2m$$, we need B = \frac{i\hbar}{2m}$. Therefore, we get \[\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m} \frac{\partial^2 \psi}{\partial x^2},\] and multiplying both sides by $i\hbar$ gives \[i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}.\] Now, in the position representation, the Hamiltonian of a free particle is \[\hat{H} = \frac{\hat{p}^2}{2m} \rightarrow -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}.\] Therefore, we have \[i\hbar \frac{\partial \ket{\psi}}{\partial t} = \hat{H} \ket{\psi},\] for a free particle.

The same argument works in three dimensions, which gives the same equation with \[\hat{H} = \frac{\har{p}^2}{2m} \rightarrow -\frac{\hbar^2}{2m} \nabla^2.\]

Of course, this argument only works for a free particle. For situations with a nonzero potential energy term, $V(x)$ or $V(\vec{r})$ in the Hamiltonian $\hat{H}$, we cannot derive the Schrödinger equation, but only argue that just adding the potential term to $\hat{H}$ in the equation is the simplest possible generalization.

The first step of this argument is to assume the superposition principle. This means that if $\ket{\psi_1(t_0)}$ and $\ket{\psi_2(t_0)}$ are solutions to the equation of motion at time $t_0$, and $\ket{\psi_1(t)}$ and $\ket{\psi_2(t)}$ are the corresponding time evolved states at $t > t_0$ then $a\ket{\psi_1(t_0)} + a\ket{\psi_1(t_0)}$ is another possible solution at time $t_0$ and it evolves in time to $a\ket{\psi_1(t)} + a\ket{\psi_1(t)}$ at time $t$.

This means that the time evolution of the system between $t_0$ and $t$ is described by a linear operator $\hat{U}(t,t_0)$, i.e. for any initial state $\ket{\psi(t_0)}$, \[\ket{\psi(t} = \hat{U}(t,t_0)\ket{\psi(t_0)}.\]

The second thing to note is that the dynamics ought to preserve the norm of the state. To see why, consider a nondegenerate observable \[\hat{A} = \sum_n a_n \proj{a_n}.\] Since it is nondegenerate, the eigenstates $\ket{a_n}$ form a complete orthonormal basis.

The Born rule says that the probability of getting the outcome $a_n$ when measuring $\hat{A}$ at time $t_0$ is \[p(a_n,t_0) = \QProb{a_n}{\psi(t_0)} = \braket{\psi(t_0)}{a_n}\braket{a_n}{\psi(t_0)}.\] This implies, the probability of obtaining any outcome at all, which should be $=1$, is \[\sum_{n}p(a_n,t_0) = \sum_n \braket{\psi(t_0)}{a_n}\braket{a_n}{\psi(t_0)} = \braket{\psi(t_0)}{\psi(t_0)} = 1,\] where we have used the Dirac Notaty to remove the identity $\hat{I} = \sum_n \proj{a_n}$.

Now consider the situation at time $t$. The probability of getting the outcome $a_n$ when $\hat{A}$ is measured at time $t$ is \begin{align*} p(a_n,t) & = \Qprob{a_n}{\psi(t)}) \\ & \Abs{\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)}}^2 \\ & = \sand{\psi(t_0)}{\hat{U}(t,t_0)}{a_n}\sand{a_n}{\hat{U}(t,t_0)}{\psi(t_0)}, \end{align*} so, applying the Dirac Notaty again, the probability of obtaining any outcome at all is \[sum_n p(a_n,t) = \sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)}.\] We want this to equal $1$ as well, so, in particular, comparing with $\sum_{n}p(a_n,t_0)$ gives \[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = braket{\psi(t_0)}{\psi(t_0)},\] which we can also write as \[\sand{\psi(t_0)}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = sand{\psi(t_0)}{\hat{I}}{\psi(t_0)}.\]

Now, if $\ket{\psi(t_0)}$ can be any vector in the Hilbert space then we have that \[\sand{\psi}{\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0)}{\psi(t_0)} = sand{\psi}{\hat{I}}{\psi},\] for all vectors $\ket{\psi}$. In an in class activity you will show that this implies that $\hat{U}^{\dagger}(t,t_0)\hat{U}(t,t_0) = \hat{I}$, i.e. the operator $\hat{U}(t,t_0)$ must be unitary.

Now, we also want that $\hat{U}(t_0,t_0) = \hat{I}$, i.e. evolving for no time does nothing, and we also want the time evolution to be continuous, so that the state stays close to the initial state if the period of time evolution is small $\Delta t = t - t_0 \ll 1$. Therefore, we will assume that, to first order in $\Delta t$, \[\hat{U}(t,t_0) = \hat{I} + \delta t \hat{A}.\]

In an in class activity, you will show that the unitarity of $\hat{U}(t,t_0)$ implies that $\hat{A}^{\dagger} = -\hat{A}$, i.e. $\hat{A}$ is anti-Hermitian. Therefore, we have \begin{align*} \frac{\partial \ket{\psi(t)}}{\partial t} & = \lim_{\Delta t \rightarrow 0} \left ( \frac{\ket{\psi(t +\Delta t)} - \ket{\psi(t)}}{\Delta t}\right ) \\ & = \lim_{\Delta t \rightarrow 0} \left ( \frac{\hat{U}(t+\Delta t,t)\ket{\psi(t)} - \ket{\psi(t)}}{\Delta t}\right ) \\ & = \lim_{\Delta t \rightarrow 0} \left ( \frac{\Delta t \hat{A}\ket{\psi(t)}}{\Delta t}\right ) \\ & = \hat{A}\ket{\psi(t)}. \end{align*}

Now, if we define $\hat{H} = i\hbar \hat{A}$ then $\hat{H}$ is a Hermitian operator and \[\frac{\partial \ket{\psi}}{\partial t} = \frac{\hat{H}}{i\hbar}\ket{\psi},\] or \[i\hbar\frac{\partial \ket{\psi}}{\partial t} = \hat{H}\ket{\psi}.\]

This shows that the Schrödinger equation must be correct for some Hermitian operator $\hat{H}$, but not that $\hat{H}$ must be the Hamiltonian (the operator representing the energy). More sophisticated arguments can show that the generator of time translations, i.e. $\hat{H}$ must be the Hamiltonain, and you will learn about this if you go to grad school for physics.

The previous argument shows that the total probability is conserved in time by Schrödinger evolution, i.e. \[\braket{\psi(t)}{\psi(t)} = \braket{\psi(0)}{\psi(0)},\qquad\text{so }\frac{\D}{\D t}\left ( \braket{\psi}{\psi}\right ) = 0.\]

In fact, we can derive a more detailed result describing the local conservation of probability. This is normally done for the probability density of a position measurement, which is what we will do here, but note that you could derive a similar equation for the conservation of probability for any observable.

The basic idea should be familiar if you have studied continuity equations in fluid dynamics or electromagnetism. In a fluid, consider the mass of fluid contained in a small volume. The increase in mass in that region over short period of time is the net current density flux into the region over that time. Similarly, in electromagnetism the increase in electric charge in a small region is the net electric current flux into that region over that time. The general form of a continuity equation is \[\frac{\partial \rho(\vec{r},t)}{\partial t} + \vec{\nabla}\cdot \vec{J}(\vec{r},t),\] where $\rho(\vec{r},t)$ is the density and $\vec{J}(\vec{r},t)$ is the current.

In quantum mechanics, we are going to derive an equation like this for the probability density $\rho(\vec{r},t) = \Abs{\psi(\vec{r},t)}^2$. This means that $\vec{J}(\vec{r},t)$ will be a probability current.

It is worth stopping to consider the meaning of this before we proceed. We normally think about continuity equations being about the flow of density of actual particles. For example, a fluid is made out of a large number of small particles. The density describes the amount of particle mass within a small volume and the current describes the rate of flow of those particles. In quantum mechanics, we are dealing with the wavefunction of a single particle. It is the probability density of this particle being found in a small volume that we are describing in the continuity equation and $\vec{J}(\vec{r},t)$ represents the rate of flow of this probability density. No physical particles are necessarily flowing into or out of the region, just the likelihood of finding the particle there if we make a measurement.

Having understood this slightly weird concept, we know that $\rho(\vec{r},t) = \Abs{\psi(\vec{r},t)}^2$, so our aim is to use the Schrödinger equation to find a probability current $\vec{J}(\vec{r},t)$ that satisfies the continuity equation.