In quantum mechanics, we usually write the Hamiltonian as:


But in classical mechanics, there are several reasons why it would not have this form:

  • We've chosen some particular coordinate system where there is no conservation of energy (we're neglecting the source of external forces).
  • We can modify the Hamiltonian without changing the equations of motion:

    The Lagrangian can be written as: $L=\frac{1}{3}T^2+2TV-V^2$. The Hamiltonian would be:

    $\hat{H}=p\left[\frac{\left(\sqrt{9p^2m^8 +32m^9V^3}+3pm^4\right)^{1/3}}{\sqrt[3]{2}m^2}-\frac{2\sqrt[3]{2}mV}{\left(\sqrt{9p^2m^8 +32m^9V^3}+3pm^4\right)}\right]-L=(T+V)^2$

    The eigenvalues of this operator are $E^2$, not the energy.

Do we consider this possibilities in quantum mechanics? In the negative case, why do we discard them?

And (besides the necessity to be self-adjoint and bounded from below) does the Hamiltonian have some restriction on the operators it includes (position, angular momentum,...)?

  • 1
    $\begingroup$ Self-adjointness guarantees a unitary time evolution and boundedness from below stability. I think that this is sufficient for an acceptable Hamitonian. The decomposition H=T+V, although quite often the case, does not seem to be necessary. In addition, to have a proper Lagrangian (from which the Hamiltonian is derived, a convexity property is required. $\endgroup$
    – Urgje
    Mar 18, 2014 at 10:29
  • $\begingroup$ @Urgje Is convexity relavant here, since this Hamiltonian is not derived from the Lagrangian? I know that in QM there are Lagrangian, but are there Legendre trasforms? $\endgroup$
    – jinawee
    Mar 18, 2014 at 10:40
  • $\begingroup$ I am not an expert on this matter so let me refer to physics.stackexchange.com/q/103997 math.stackexchange.com/q/482553 $\endgroup$
    – Urgje
    Mar 18, 2014 at 18:25
  • $\begingroup$ Just a side remark, there is less restriction of Hamiltonian in non-relativistic quantum mechanics (Hermitian, bounded from below). It is much restricted in QFT by gauge invariance and Lorentz symmetry in the Lagrangian. $\endgroup$
    – user26143
    Mar 18, 2014 at 19:09
  • $\begingroup$ You end up with ambiguities from the expression $(T+V)^2$ for, e.g., the simple harmonic oscillator. This is because we cannot consistently quantize terms involving higher than quadratic powers, and the cross terms would be a nightmare to deal with adequately... $\endgroup$ Mar 29, 2014 at 0:03

1 Answer 1


Myth 1: We Need Unitary Time Evolution.

Problem a: Self-Adjointness. This would require the Hamiltonian to be self-adjoint. But there are interesting situations when this won't be true, consider the following papers:

  1. F. Bagarello, A. Inoue, C. Trapani, "Non-self-adjoint hamiltonians defined by Riesz bases". Eprint arXiv:1402.6199
  2. Fabio Bagarello, Miloslav Znojil, "The dynamical problem for a non self-adjoint Hamiltonian". Operator Theory: Advances and Applications 221 (2012) 109--119, arXiv:1105.4716

Problem b: Constrained Systems. For topological field theory, parametrized fields, or even the relativistic particle (taking the action to be its proper time), the Hamiltonian operator no longer controls time-evolution.

The Hamiltonian instead becomes a constraint for these systems. It vanishes when acting on the physical states. Asking the time evolution operator to be "unitary" is meaningless for physical states.

For more on constrained systems, see, e.g.,

  1. Henneaux and Teitelboim's Quantization of Gauge Systems
  2. Andreas W. Wipf, "Hamilton's Formalism for Systems with Constraints". Eprint arXiv:hep-th/9312078

Myth 2: We Can Quantize Anything.

Not quite true. For example, as I noted in my comment, if we consider the simple harmonic oscillator (dropping constants for simplicity) $$ T = v^{2},\quad\mbox{and}\quad V=x^{2}$$ then $$ H = (p^{2}+x^{2})^{2}. $$ But look, when we try to quantize this, we can't put hats on everything and hope it all works for the best. Why? Well, how do we quantize $x^{2}p^{2}$? Observe $$ (\hat{x}\hat{p})^{2}\neq \hat{x}^{2}\hat{p}^{2}\neq\left(\frac{\hat{p}\hat{x}+\hat{x}\hat{p}}{2}\right)^{2}$$ They give inequivalent commutators.

This ambiguity is well-known in the literature. See, for example:

  1. S. Twareque Ali, Miroslav Engliš, "Quantization Methods: A Guide for Physicists and Analysts". Rev.Math.Phys. 17 (2005) 391-490, arXiv:math-ph/0405065

So...what DO we do?

"We do what works" is the motto! It'd be nice to have some algorithm that always works no matter what...but instead we restrict our focus to the physically relevant situations.

And even then, we only use the quantized systems that make sense.

In a sense, this is the proper route: instead of asking for a procedure to turn something classical into something quantum, we should think that nature already is quantized and we need to seek out the quantum model that emulates the natural phenomena we're interested in.


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