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

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?

  • $\begingroup$ A measurement apparatus which correctly measures the energy of the system must correspond to the observable called "the" Hamiltonian, which continues to exist and govern the evolution of the system even if no one ever observes it. $\endgroup$ Sep 22, 2019 at 19:43
  • 2
    $\begingroup$ Every measurement process can be described by a different set of eigenvectors and eigenvalues, as you said. When those match with those of the system's Hamiltonian, it is called a measurement of energy. If they don't match, then you're measuring something, but it isn't energy. $\endgroup$
    – knzhou
    Sep 22, 2019 at 19:43
  • 1
    $\begingroup$ Note that the system operator that a measurement apparatus effectively measures is not the same thing as the Hamiltonian operator of the apparatus. That wouldn't even make sense, since these operators are not even defined in the same spaces. $\endgroup$
    – knzhou
    Sep 22, 2019 at 19:45
  • $\begingroup$ You need to make a distinction between the Hamiltonian of the measured system and the Hamiltonian of the measuring apparatus (and the Hamiltonian of the combined system-apparatus system). Also, I do not understand what exactly is the contradiction that you point out. $\endgroup$
    – oleg
    Sep 22, 2019 at 21:03
  • $\begingroup$ Thanks for the comments; they answered my question. $\endgroup$ Sep 27, 2019 at 11:17

1 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}

  • $\begingroup$ do not now mean $c=\hbar$? $\endgroup$ Jul 24, 2022 at 0:29
  • $\begingroup$ $\hbar$ is the reduced Plank constant. $\endgroup$
    – Cal Abel
    Jul 24, 2022 at 0:51
  • $\begingroup$ exactly! $Ht/c$ is not dimensionless so your exponential makes no sense… $\endgroup$ Jul 24, 2022 at 0:52
  • $\begingroup$ maybe my initial comment should be $c\to\hbar$ rather than the equal sign… $\endgroup$ Jul 24, 2022 at 0:53
  • $\begingroup$ @ZeroTheHero How can a velocity converge to an action? $\endgroup$
    – Filippo
    Jul 24, 2022 at 6:30

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