Let $N\equiv \mathrm{dim}(V)$ and let $i,j,k,n$ range from $1$ to $N$. Let $\{|E_n\rangle\}$ as well as $\{|Q_n\rangle\}$ be two orthonormal sets of basis vectors where in particular $H|E_n\rangle=E_n|E_n\rangle$. Fix a number $m\in\mathbb N$ and let $\Delta t\equiv \frac{1}{m_\Delta}(t-s)$.

With the representation of the identity operator as sum of projections, e.g. $1=\sum_i\left|Q_i\right\rangle\left\langle Q_i\right|$, we find

$\exp\left(i\,\Delta t\frac{1}{\hbar}H\right)=$

$=\sum_{{i_0}=1}^N\sum_{j=1}^N\sum_{{i_1}=1}^N\left|Q_{i_0}\left\rangle\left\langle Q_{i_0}\right|\exp\left(i\,\Delta t\frac{1}{\hbar}H\right)\left|E_j\right\rangle\left\langle E_j\left|\right. Q_{i_1}\right\rangle\right\langle Q_{i_1}\right|$

$=\sum_{{i_0,i_1}=1}^N\left|Q_{i_0}\right\rangle\left(\sum_{j=1}^N \mathrm{e}^{i\,\Delta t\frac{1}{\hbar}E_j} \left\langle Q_{i_0}\right|\left.E_j\right\rangle\left\langle E_j\left|\right. Q_{i_1}\right\rangle\right)\left\langle Q_{i_1}\right|$

Remark: This is an example where $H$ actually has a continuous spectrum.

The heat equation

$\frac{\partial}{\partial t}\psi(t,{\bf{Q}})=H\,\psi(t,{\bf{Q}}),\ t>0$

with $H=-\kappa\nabla^2$ has the solution $\psi(t,{\bf{Q}})=\mathrm{e}^H\psi(0,{\bf{Q}})$, but how to evaluate this expression of arbitrarily high powers of a differential operator? Now the $H$-eigenfuctions coincides with the momentum eigenfunctions and we have $E_{\bf{p}}=-\kappa\,{\bf{p}}^2$ and $\langle {\bf{Q}}|{\bf{p}}\rangle=\frac{1}{(2\pi)^{3/2}}\mathrm{e}^{i\,{\bf{p}}\cdot{\bf{Q}}}$. Therefore sum (integral) in brackets above becomes

$\frac{1}{(2\pi)^3}\int_{\mathbb R^3}\mathrm{e}^{-\kappa\,{\bf{p}}^2+i({\bf{Q}}_0-{\bf{Q}}_1)}\,\mathrm{d}^3{\bf{p}} = (4\pi t^{-1/2})^3\,\mathrm{e}^{-\tfrac{1}{4\kappa t}({\bf{Q}}_0-{\bf{Q}}_1)^2}\equiv \mathcal K({\bf{Q}}_0,t,{\bf{Q}}_1,0)$

which is the Fourier transform of the Gaussian curve. The evaluation is explained in magic Gaussian integral. This is called the heat kernel and now

$\psi(t,{\bf{Q}}) = \int_{\mathbb R^3}\,\mathcal K({\bf{Q}},t,{\bf{q}},0)\,\psi(0,{\bf{q}})\,\mathrm{d}^3{\bf{q}}$,

which solves the equation.

If the operator $H$ is a sum of a differential operator $D_{\bf q}$ and a potential $U{\bf q}$, then finding the eigenvalue is tricky. However, since for small times $\Delta t$ we have $\mathrm{e}^{\Delta t\,(D_{\bf q}+U({\bf q}))}=\mathrm{e}^{\Delta t\,D_{\bf q}}\mathrm{e}^{\Delta t\,U({\bf q})}+\mathcal O(\Delta t)$, we can just use another set of eigenfunctions to compute the kernels $\mathcal K_{\Delta t}$ and try to pass to a complete one $\mathcal K$. In the next discussion section, we are going to work out such a time discretization in theory.

Back to the general calculation. To express the sum over energy states in terms of two real quantities, we write the whole term $(\sum\dots)$ in polar form as $\mathrm{e}^{i\frac{1}{\hbar}\Delta S_{i_0i_1}}\Delta\mu_{i_0i_1}$. Note that $m_\Delta$, $\Delta\mu$ and $\Delta S$ depend on the discretization $\Delta t$.

Via successive multiplications and by using $\left\langle Q_i\right|\left.Q_i\right\rangle=\delta_{i,j}$ we find

$\exp\left(i\,(t-s)\frac{1}{\hbar}H\right)$

$=\exp\left(i\,\Delta t\frac{1}{\hbar}H\right)^{m_\Delta}$

$=\sum_{{i_0,i_{m_\Delta}}=1}^N\left|Q_{i_0}\left\rangle\left(\sum_{{i_1,i_2,\dots,i_{{m_\Delta}-1}}=1}^N\mathrm{e}^{i\frac{1}{\hbar}\sum_{k=1}^{m_\Delta}\Delta S_{i_{k-1}i_{k}}}\prod_{k=1}^m\Delta\mu_{i_{k-1}i_{k}}\right)\right\langle Q_{i_{m_\Delta}}\right|$

Note that if $Q\left|Q_n\right\rangle=q_n\left|Q_n\right\rangle$, then the above implies

$\left\langle Q_{i_\text{out}}\right|\,P(s,t)\,\left|Q_{i_\text{out}} \right\rangle = \delta_{i_\text{in}}^{i_0}\,\delta_{i_\text{out}}^{i_m}\sum_{{i_1,i_2,\dots,i_{{m_\Delta}-1}}=1}^N\mathrm{e}^{i\frac{1}{\hbar}\sum_{k=1}^{m_\Delta}\Delta S_{i_{k-1}i_{k}}}\prod_{k=1}^m\Delta\mu_{i_{k-1}i_{k}}$

where we use the Einstein summation convention for the two deltas. Or also (with all indices dropped for readability)

$\left\langle Q_{\text{out}}\right|\,P(s,t)\,Q^3\,P(t,\tau)\,Q\,P(\tau,\sigma)\,\left|Q_{\text{out}} \right\rangle =\ " \sum q^3\,q\,\mathrm{e}^{i\frac{1}{\hbar}\Delta S}\prod\Delta\mu"$

To define the continuum limit of this expression is a difficult task. In quantum field theory, an explicit expression for the $\lim_{\Delta t\to 0}\sum_{k=1}^{m_\Delta}\Delta S_{i_{k-1}i_{k}}$ is taken as a starting point and identified with the action corresponding to the Hamiltonian $H$. The measure, in essence given by the product $\prod_{k=1}^m\Delta\mu_{i_{k-1}i_{k}}$, is generally ill-defined.

Hilbert space

Observable, Linear operator algebra

Hamiltonian, Linear first-order ODE system