### The Hamiltonian and energy

Given two postulates - linearity and unitarity - we have discovered that for *every* system, there exists a Hermitian operator, called the **Hamiltonian**, which describes the time evolution of a given state in the system's state space.

It is generally different for each system, but given a Hamiltonian $H$, all states ,$\ket{\psi}$, in the system's state-space satisfy Schrodinger's equation, $$\frac{\partial}{\partial t}\ket{\psi} = -\mathrm{i}\frac{H}{\hbar}\ket{\psi}$$

The Hamiltonian must represent some kind of observable, since all Hermitian operators do. In fact it represents the **energy** associated with the system and its eigenvalues are the discrete energy bands that the system can attain.

In general, the Hamiltonian is always a conserved quantity, and in particular - for certain very well-defined situations at least - it does correspond to the energy, both theoretically and experimentally.

The Hamiltonian could, in principle, also be a function of time but that would generally mean that there is some influence on the system from outside. We could then just define a larger system - including the outside influence - and the Hamiltonian of this larger system would not depend on time.

Hence, we normally consider the Hamiltonian to be fixed, and not vary over time.

### Solution for a particular eigenvector

Suppose $E$ is an eigenvalue of $H$ with associated eigenvector $\ket{\psi_E}$, meaning that $$H\ket{\psi_E} = E\ket{\psi_E}$$

$\ket{\psi_E}$ can depend on time, but the postulates *require* that $\ket{\psi_E}$ evolves to states $\ket{\psi_E(t)}$ that are proportional to $\ket{\psi_E}$, which means that we can assume that only the *coefficient* depends on time.

That is, there exists a function $f_E(t)$ such that $$\ket{\psi_E(t)} = f_E(t)\ket{\psi_E}$$

But we already have shown that there exists an evolution operator, $U$ such that $$\ket{\psi_E(t)} = U(t)\ket{\psi_E}$$ so we must have $$U(t)\ket{\psi_E(t)} = f_E(t)\ket{\psi_E(t)}$$

This is equivalent to saying that $\ket{\psi_E}$ - an eigenvector of $H$ - is also an eigenvector of $U(t)$ with eigenvalue $ f_E(t)$ (at time $t$). We note that, since $U(0) = \mathbf{I}$, we must have $$f_E(0) = 1$$

Taking the derivative, we get $$\frac{\partial}{\partial t} \ket{\psi_E(t)} = \frac{\mathrm{d} f_E}{\mathrm{d} t}\ket{\psi_E}$$

We can compare this to the result from Schrodinger's equation, where $$\begin{align*}\frac{\partial}{\partial t} \ket{\psi_E(t)}&= -\mathrm{i}\frac{H}{\hbar}\ket{\psi_E(t)}\\&= -\mathrm{i}\frac{H}{\hbar}\left(f_E(t)\ket{\psi_E}\right) & \text{ by definition}\\&= -\mathrm{i}\frac{1}{\hbar}f_E(t)\left(H\ket{\psi_E}\right)\\&= -\mathrm{i}\frac{1}{\hbar}f_E(t)\left(E\ket{\psi_E}\right)\\&= -\mathrm{i}\frac{E}{\hbar}f_E(t)\ket{\psi_E}\end{align*}$$

Thus, we find that $$\frac{\mathrm{d} f_E}{\mathrm{d} t} = -\mathrm{i}\frac{E}{\hbar}f_E(t)$$ which can be wriiten as a first-order linear differential equation $$\frac{\mathrm{d} f_E}{\mathrm{d} t} + \mathrm{i}\frac{E}{\hbar}f_E = 0$$ which has the general solution $$f_E(t) = f_E(0)e^{-\mathrm{i}\frac{E}{\hbar}t}$$

Since $f_E(0) = 1$, the particular solution is $$f_E(t) = e^{-\mathrm{i}\frac{E}{\hbar}t}$$

In conclusion, we have found a particular solution to **Schrodinger's equation** - an eigenvector of the Hamiltonian which evolves with time by being multiplied by a *pure-phase*, time-dependent function $$\ket{\psi_E(t)} = e^{-\mathrm{i}\frac{E}{\hbar}t}\ket{\psi_E}$$

Nb. Notice that the *angular frequency* associated with the phase is $\omega = \frac{E}{\hbar}$. This is a very famous equation in physics, relating energy to angular frequency. It is normally written as $$E = \hbar \omega$$ or even $$E = h \nu$$ where $2\pi\hbar = h$ and $\nu = \tfrac{\omega}{2\pi}$ is the frequency (cycles per second).

### General solution

Since the Hamiltonian $H$ is Hermitian, it will have a full set of eigenvalues, $E_n$, for $n=1,2,\cdots,N$, and associated eigenvectors $\ket{\psi_n}$.

We can then express any state of the system, $\ket{\psi}$, as a superposition of these vectors (at time $t=0$), $$\ket{\psi} = \sum_{n=1}^{N} \alpha_n \ket{\psi_n}$$ and then this state will evolve over time according to $$\ket{\psi(t)} = \sum_{n=1}^{N} \alpha_n f_n(t) \ket{\psi_n} = \sum_{n=1}^{N} \alpha_n e^{-\mathrm{i}\frac{E_n}{\hbar}t} \ket{\psi_n}$$

We can put this into matrix form, which allows to write the solution in a coordinate-free form.

Note that, for any matrix $M$, $$e^M = \sum_{k=0}^{\infty}\frac{1}{k!} M^k$$ and for any *diagonal* matrix, $$M = \diag{M_1}{M_2}{M_N}$$ we have $$\begin{align*}e^M &= \sum_{k=0}^{\infty}\frac{1}{k!} M^k\\&= \sum_{k=0}^{\infty}\frac{1}{k!} \diag{M_1}{M_2}{M_N}^k\\&= \sum_{k=0}^{\infty}\frac{1}{k!} \diag{M_1^k}{M_2^k}{M_N^k}\\&= \diag{\sum_{k=0}^{\infty}\frac{1}{k!} M_1^k}{\sum_{k=0}^{\infty}\frac{1}{k!} M_2^k}{\sum_{k=0}^{\infty}\frac{1}{k!} M_N^k}\\&= \diag{e^{M_1}}{e^{M_2}}{e^{M_N}}\end{align*}$$

Thus, we can write $$\begin{align*}\ket{\psi(t)} &= \sum_{n=1}^{N} \alpha_n e^{-\mathrm{i}\frac{E_n}{\hbar}t} \ket{\psi_n}\\&= \diag{e^{-\mathrm{i}\frac{E_1}{\hbar}t}}{e^{-\mathrm{i}\frac{E_2}{\hbar}t}}{e^{-\mathrm{i}\frac{E_N}{\hbar}t}}\begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_N \end{bmatrix}\\&= e^{-\mathrm{i}\frac{H}{\hbar}t} \ket{\psi}\end{align*}$$ where we note that since $H$ is diagonal, so is $e^{-\mathrm{i}\frac{H}{\hbar}t}$.

Thus, finally the general solution to Schrodinger's equation is given by $$\ket{\psi(t)} = e^{-\mathrm{i}\frac{H}{\hbar}t} \ket{\psi}$$