# The Schrödinger Equation in Dirac Notation

The Schrödinger equation, as introduced in the previous chapter, is a special case of a more general equation that is satisfied by the abstract state vector $|\Psi(t)\rangle$ describing the system. More specifically, it is an equation describing the components of the state vector in position space. We will now introduce this more general equation, written in terms of the state vector itself, and show how one can recover the wave equation from the previous chapter.

In Dirac notation, the Schrödinger equation is written as

$i\hbar \frac{d}{dt}|\Psi(t)\rangle=\hat{H}(t)|\Psi(t)\rangle.$

We see that the Hamiltonian of the system determines how a given initial state will evolve in time.

To show how to recover the equation for the wave function, let us consider the Hamiltonian for a particle moving in one dimension,

$\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x},t).$

We now write our state vector in position space. Since the position space is continuous, rather than discrete, the state vector as a linear superposition of position eigenstates must now be written as an integral:

$|\Psi(t)\rangle=\int dx\,\Psi(x,t)|x\rangle,$

where $\langle x'|x\rangle=\delta(x'-x)$ and $\delta(x)\!$ is the Dirac delta function.

With the aid of the identity,

$\frac{\partial}{\partial x}\delta(x'-x)=\frac{\delta(x'-x)}{x'-x},$

one may verify that

$\langle x'|\hat{p}|x\rangle=i\hbar\frac{\delta(x'-x)}{x'-x}=i\hbar\frac{\partial}{\partial x}\delta(x'-x)$

and that

$\hat{p}|\Psi(t)\rangle=-i\hbar\int dx\,\frac{\partial\Psi}{\partial x}|x\rangle.$

If we now substitute the above form of the Hamiltonian into the Schrödinger equation and project the resulting equation into position space, we will arrive at the wave equation stated in the previous chapter,

$i\hbar\frac{\partial\Psi(x,t)}{\partial t}=\left [-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x,t)\right ]\Psi(x,t).$

The above procedure can be generalized to multiple dimensions, again recovering the multi-dimensional wave equation given in the previous chapter:

$i\hbar\frac{\partial \Psi(\textbf{r},t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\Psi(\textbf{r},t)$

We could also have chosen to work in momentum space; a similar procedure yields

$i\hbar\frac{\partial \Phi(\textbf{p},t)}{\partial t} = \left[ \frac{\textbf {p}^{2}}{2m} + V\left ( i\hbar \frac{\partial}{\partial \textbf{p}}\right)\right]\Phi(\textbf{p},t).$

Here, $\Phi(\textbf{p},t)$ and $\Psi(\textbf{r},t)$ are related through a Fourier transform as described in a previous section.

Let us now consider a time-independent Hamiltonian. As described previously, we may solve the Schrödinger equation in this case by first assuming that the state vector has the form,

$|\psi_n(t)\rangle=e^{-iE_n t/\hbar}|\psi_n\rangle,$

where $|\psi_n\rangle$ is independent of time. Substituting this form into the Schrödinger equation yields the equation for stationary states in Dirac notation:

$E_n|\psi_n\rangle=\hat{H}|\psi_n\rangle.$

The eigenfunctions are replaced with eigenvectors. Use of this notation makes solution of the Schrödinger equation much simpler for some problems; if we write the eigenvectors in a convenient basis, we may project the above eigenvalue equation onto all states in the basis, thus reducing the problem to diagonalizing a matrix.

We now ask how an arbitrary state $|\Psi(t)\rangle$ evolves in time for a time-independent Hamiltonian. Let us expand this state in terms of an orthonormal basis $|1\rangle,|2\rangle,\ldots,$ obtaining

$|\Psi(t)\rangle=\sum_{n}c_{n}(t)|n\rangle.$

If we now substitute this into the Schrödinger equation and project the result onto each of the basis states, we obtain

$i\hbar\frac{dc_{m}(t)}{dt}=\sum_{mn}\langle m|H|n\rangle c_{n}(t).$

If, in particular, we work in the basis $\left\{|\psi_n\rangle\right\}$ of eigenstates of the Hamiltonian, the above equation reduces to a set of decoupled equations for the coefficient of each eigenstate,

$i\hbar\frac{dc_{m}(t)}{dt}=E_{m}c_{m}(t),$

whose solutions are $c_{m}(t)=c_{m}(0)e^{-iE_{m}t/\hbar}.$ Therefore, the time evolution of a general state is given by

$|\Psi(t)\rangle=\sum_{n}c_{n}(0)e^{-iE_{n}t/\hbar}|\psi_n\rangle,$

which is just a linear superposition of the time-dependent eigenvectors obtained previously. One could also have obtained this from the fact that any linear superposition of solutions of the Schrödinger equation is itself a solution, and thus any state may be written in the above form.

We will now show that the Schrödinger equation in this form preserves the normalization of the state vector; i.e., if the vector is normalized initially, then it will remain normalized at all times. We start by writing the dual of the Schrödinger equation,

$-i\hbar\frac{d}{dt}\langle\Psi(t)|=\langle\Psi(t)|\hat{H}(t).$

We now act on the Schrödinger equation the left with $\langle\Psi(t)|$ and on its dual from the right with $|\Psi(t)\rangle$ and subtract the two results, obtaining

$\langle\Psi(t)|\frac{d}{dt}|\Psi(t)\rangle+\frac{d}{dt}\left [\langle\Psi(t)|\right ]|\Psi(t)\rangle=0,$

or

$\frac{d}{dt}\langle\Psi(t)|\Psi(t)\rangle=0.$

As asserted, $\langle\Psi(t)|\Psi(t)\rangle=\text{const.},$ so that we only need to normalize the state vector at $t=0.\!$