Can the precise form of a quantum field Hamiltonian be determined by harmonic oscillator analogy?

Question

In one of my other questions What is the precise formal correspondance between an oscillator and a quantum field? , I was helpfully given the exact form of the well discussed analogy between the quantum harmonic oscillator and the quantum field mode.

The answers there give the analogy in terms of classical quantities. However, there is happily an obvious identification between classical quantities and quantum quantities, which let me understand how quantum field modes will look.

However, I think between the classical equations of motion and the quantum solutions, there is the quantum Hamiltonian. It seems this carries an extra degree of freedom. The general classical equation of motion contains a term $\omega$ . However in the harmonic Schrödinger equation this splits out into $\omega=\sqrt{\frac{k}{m}}$ to give $$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{k\hat{x}^2}{2}$$ or equivalently $$i\frac{\partial}{\partial t}=\frac{-1}{2m}\frac{\partial^2}{\partial\hat{x}^2}+\frac{m\omega^2\hat{x}^2}{2}$$ in "natural units". I realise my notation might not be formal. I hope it's clear enough to see what I mean and glad for any corrections.

So, this could give the Schrödinger equation for a single field mode if we replace $\hat{x}$ with field "value" and know how to split $\omega$ into $k$ and $m$. Now as physicists we know both systems have a mass, so maybe $m$ is mass in both cases. But I think that's not correct because the mass of each system occupies a different formal position/role in the answer I linked to. Or maybe $m$ has no effect in the second equation I wrote, because it does not affect frequency or energy (??) of the system (apart from through $\omega$) so it can be discarded? How do I determine $m$ in the second equation to get a complete understanding of the mathematics of a field mode?

Answer 1

How do I determine m in the second equation to get a complete understanding of the mathematics of a field mode?

I think it might be more fruitful to start with a uniform string of length $L$ with mass per length $\rho$ and under tension $T$.

It's then straightforward to write the total energy in terms of the amplitude $A_r(t)$ of the $r_{th}$ normal mode

$$E = \frac{1}{2}\sum_{r=1}^\infty\left( \frac{\rho L \dot{A}^2_r}{2} + \frac{\rho L\omega^2_rA^2_r}{2} \right)$$

where

$$\omega_r = r\,\omega_0\,\quad\omega_0\equiv\pi\sqrt{\frac{T/L}{\rho L}}$$

See that the quantity $T/L$ has dimensions of force per distance (like a spring constant $k$) while the quantity $\rho L$ has dimension of mass. So, to make the analogy explicit, we could make the following identifications

$$k \equiv T/L$$

$$m \equiv \rho L$$

and then write the energy as

$$E = \frac{1}{2}\sum_{r=1}^\infty\left( \frac{(m\dot{A}_r)^2}{2m} + \frac{m \omega^2_rA^2_r}{2} \right),\quad \omega_r = r\pi\sqrt{\frac{k}{m}}$$

See now that this has the form of an (infinite) sum of harmonic oscillators. When quantized, the energy eigenvalues will be of the form

$$E = \sum_{r=1}^\infty\left(n_r+\frac{1}{2}\right)\hbar\omega_r$$

where $n_r$ is the number of $r$ mode quanta present.