What is the Hamiltonian operator, and is it unique?

Question

$$\hat V=\sum_i v_i |v_i\rangle \langle v_i| $$

An observable in quantum mechanics is defined as above, with {$| v_i \rangle$} being an orthonormal basis, so the observable $\hat V$ is a Hermitian operator.

From what I understand, an observable describes a piece of measurement apparatus, with its eigenstates being the basis the apparatus performs the measurement in and the eigenvalues being the readings corresponding to each eigenstate measurement. For example, if we are using a polarising beam splitter (PBS) that reflects vertically polarised photons and transmits horizontally polarised photons to measure the polarisation of a photon, and we choose to assign a value of $1$ to a horizontal photon detection and a value of $-1$ to a vertical photon detection, our observable will be as follows:

$$\hat V= |H\rangle \langle H| -|V\rangle \langle V|$$

Where {$| H \rangle,| V \rangle$} is the orthonormal basis of the photon polarisation Hilbert space that the PBS measures in (horizontal and vertical polarisation, respectively).

The Hamiltonian $\hat H$ is defined as an energy observable, with energy eigenstates and corresponding eigenvalues:

$$\hat H=\sum_i E_i |E_i\rangle \langle E_i| $$

From the understanding elucidated above, this implies that any measurement apparatus designed to measure the energy of a quantum system will have its own Hamiltonian. However, the Hamiltonian is commonly defined as the sum of potential and kinetic energies in the system and the operator from which the future evolution of the system can be derived, implying that the Hamiltonian is unique.

Where have I gone wrong?

Answer 1

In the axiomatic formulation of quantum theory, there is an axiom that is used to define the time evolution of quantum states in an isolated system, \begin{equation} |\psi(t)\rangle = \hat{U}(t) |\psi(0)\rangle.\tag{1}\label{eq:1} \end{equation} We expand the time-dependent operator around an arbitrary time $\epsilon$, \begin{equation} \hat{U}(\epsilon) = \hat{I} - i\frac{\hat{H}}{c} \epsilon + O(\epsilon^2).\tag{2}\label{eq:2} \end{equation} Since $\hat{U}(t)$ is a unitary operator, $\hat{U}^\dagger(t)\hat{U}(t) = \hat{I}$. This results in the new operator, $\hat{H}$, being Hermitian, $\hat{H}^\dagger = \hat{H}$. Neglecting the higher order terms of \eqref{eq:2} and making the substitution back into equation \eqref{eq:1} we have, \begin{eqnarray} |\psi(\epsilon + t)\rangle & = & (\hat{I} - i\frac{\hat{H}}{c} \epsilon) |\psi(t)\rangle, \\ \psi(\epsilon + t)\rangle - |\psi(t)\rangle & = & (\hat{I} - i\frac{\hat{H}}{c} \epsilon) |\psi(t)\rangle - |\psi(t)\rangle, \\ ic\frac{|\psi(\epsilon + t)\rangle - |\psi(t)\rangle}{\epsilon} & = & \hat{H} |\psi(t)\rangle, \\ \lim_{\epsilon \rightarrow 0} \frac{|\psi(\epsilon + t)\rangle - |\psi(t)\rangle}{\epsilon} & = & \frac{d}{dt}|\psi(t)\rangle, \text{ and} \\ ic \frac{d}{dt}|\psi(t)\rangle & = & \hat{H} |\psi(t)\rangle. \tag{3}\label{eq:3} \end{eqnarray}

Equation \eqref{eq:3} is clearly the Schrödinger wave equation and $\hat{H}$ is the Hamiltonian operator. Thus, the definition of the Hamiltonian is the expansion coefficient, $\frac{d}{dt}\hat{U}(t)$, of a first order Taylor series expansion of time dependent unitary operator. It is unique up to a linear transformation, $c$, which determines the units. If $c=1$, then the Hamiltonian has units of inverse time. If $c=\hbar$ then the Hamiltonian has units of energy, e.g. $eV$, $J$, etc. Note: the linear transformation also includes an additive constant. This constant defines the potential to which the Hamiltonian is be referenced.

To complete the derivation the time dependent unitary operator is, \begin{equation} \hat{U}(t) = e^{-i\frac{\hat{H}}{c}t}. \end{equation}