# Solutions to Schrodinger’s equation

### 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 $$\ket{\psi(t)} = e^{-\mathrm{i}\frac{H}{\hbar}t} \ket{\psi}$$