I will present a question which already is buzzing in my head for quite a time. Actually quantum physics developed as a interplay of empirical results and theoretical developments where it is difficult to find the basic principles. Obviously Planck's relation giving quantized energies $E=n\hbar\omega$ ($E$, $\omega$ are the energy and angular frequency of a photon, $n$ the number of photon quanta and $\hbar$ Planck's constant) plays a very important role, it appears as a kind of postulate since Planck did not derive it, but only used it to explain empirical data on black-body radiation. The same did Einstein to explain the photo-electrical effect. Later it was (at least through conclusion by analogy) used to draw up Schroedinger's equation. However, it seems to be a kind of corollary of second quantization. This is actually the point I am interested in. Is it really a corollary, or can it only be postulated ? Because in order to get this corollary, quantum mechanics has to be used, which is already based on $E=n\hbar\omega$.

In order to give a bit more substance to this post I will recall the steps to quantize the electromagnetic (EM) field (in a non-Lorentz-covariant way because this is not the point now) according to Landau/Lifschitz Vol.2 (chap.52) and Vol.4 (chap. 2). In the following up all the steps I try to recap the assumptions which are used for getting $E=n\hbar\omega$.

The vector potential of the free EM-field can developed into transverse plane waves( $\mathbf{a_\mathbf{k}}\mathbf{k}=0$) $$\mathbf{A}=\sum_\mathbf{k} (\mathbf{a}_\mathbf{k} e^{i\mathbf{k r}} +\mathbf{a}^{\ast}_\mathbf{k} e^{-i\mathbf{k r}})$$

Here a discrete summation for the modes is used for simplicity, it could also be formulated continously with an integral. The discrete summation of the modes has nothing to do with the later quantisation. As we are interested in the energy respectively the Hamiltonian of the free EM-field we recall its formula (everything here is in cgs-units):

$$H \equiv E =\frac{1}{8\pi}\int (\mathbf{E}^2 +\mathbf{H}^2) dV $$

The free EM-fields can be computed with the vector potential:

$$\mathbf{E}=-\frac{1}{c}\dot{\mathbf{A}}$$ and $$\mathbf{H}=rot\mathbf{A}$$

After a longer calculation the following the result can be obtained:

$$E= \sum_\mathbf{k} \frac{k^2V}{2\pi} \mathbf{a}_\mathbf{k} \mathbf{a}^{\ast}_\mathbf{k}$$

I recall that in this approach the EM-field is considered to be inside a large, but finite volume $V$, the volume $V$ yields simply by the integration over $dV$. The quantity $k$ is given by $k^2 =\mathbf{k}^2$. In order to make use of the classical Hamiltionian formalism the canonical variables $\mathbf{Q_\mathbf{k}}$ and $\mathbf{P_\mathbf{k}}$ are introduced: $$\mathbf{Q_\mathbf{k}} = \sqrt{\frac{V}{4\pi c^2}}(\mathbf{a}_\mathbf{k} + \mathbf{a}^\ast_\mathbf{k})$$ and

$$\mathbf{P_\mathbf{k}} = -i\omega_\mathbf{k}\sqrt{\frac{V}{4\pi c^2}}(\mathbf{a}_\mathbf{k} -\mathbf{a}^\ast_\mathbf{k}) = \mathbf{\dot{Q}_\mathbf{k}}$$ So upon replacement of the $\mathbf{a}_\mathbf{k}$ by $\mathbf{Q_\mathbf{k}}$ and $\mathbf{P_\mathbf{k}}$ we get:

$$H \equiv \sum_\mathbf{k} H_\mathbf{k}= \sum_\mathbf{k}\frac{1}{2}( \mathbf{P^2_\mathbf{k}} + \omega^2_\mathbf{k}\mathbf{Q^2_\mathbf{k}})$$

The Hamiltonian equations $\frac{\partial H}{\partial\mathbf{Q_\mathbf{k}}}=- \mathbf{\dot{P}_\mathbf{k}}$ can be used to get to the following set of equations:

$$\mathbf{\ddot{Q}_\mathbf{k}} + \omega_\mathbf{k}\mathbf{Q_\mathbf{k}}=0$$

which are actually identical to the Maxwell-equations for the free EM-field. Each of the vectors $\mathbf{Q_\mathbf{k}}$ and $\mathbf{P}_\mathbf{k}$ is orthogonal to the wave vector $\mathbf{k}$ (because $\mathbf{a_\mathbf{k}}\mathbf{k}=0$ was assumed in the expression of the vector potential $\mathbf{A}$), so has 2 independent components. If both components of $\mathbf{Q_\mathbf{k}}$ are denoted as $Q_{\mathbf{k}j}$ with $j=1,2$, i.e. $\mathbf{Q}^2_\mathbf{k}= \sum_j Q^2_{\mathbf{k}j}$. The same can be done for $\mathbf{P}_\mathbf{k}$. Using this denotation we get:

$$H = \sum_{\mathbf{k}j} H_{\mathbf{k}j} = \sum_{\mathbf{k}j}\frac{1}{2}(P^2_{\mathbf{k}j} +\omega^2_kQ^2_{\mathbf{k}j} )$$

So the Hamiltonian decomposes into a series of independent terms each depending on a particular pair $Q_{\mathbf{k}j}$ and $P_{\mathbf{k}j}$. Each $H_{\mathbf{k}j}$ has the form of a one-dimensional harmonic oscillators, therefore the free EM-field can be interpreted as an assembly of many independent harmonic oscillators (HQs). So we actually recover the interpretation of the free EM-field used by Planck 1900. So far everything only classical physics was used.

Now we can make a big jump into quantum mechanics and consider the assembly of HQs each of them quantum mechanically. Accordingly we obtain as Hamiltonian for the quantum assembly of HQs:

$$H = \sum_{\mathbf{k}j} (N_{\mathbf{k}j} + \frac{1}{2})\omega_k$$

So we indeed get Planck's relation with a discrete energy spectrum! Which assumptions were used to reach this result? Apparently, some kind of Schroedingers' equation for the QM harmonic oscillator. So it seems at the end that everything is based on Schroedingers' equation. But there is a little detail which draws my particular attention. Why do we get a discrete energy spectrum, the actual cause for quantization ?

It seems that the differential operator

$$-\frac{\hbar^2\Delta}{2m} + \frac{m\omega^2}{2}Q^2 $$

having discrete eigenvalues causes the quantisation of fields. I think this observation is not trivial as a similar Hamiltonian

$$-\frac{\hbar^2\Delta}{2m} - \frac{e}{r} $$

can also have a continous spectrum, at least for eigenvalues $E>0$.

Therefore my final and most important question is: Is the reason for a microscopic world being quantized the fact that self-adjungated (or similar with less good analytical properties) operators as above can have (apart from occasionally also having continous spectra) discrete spectra ?

Of course one can also quantize fields without using the result of the quantum-mechancial harmonic oscillator and only use (a generalization of )Heisenberg's commutation rule for conjugated variables and also get quantized quantum states. So what is eventually the reason for quantization ?


You don't even have to go to Quantum Field Theory to "justify" quantisation, which is why QFT is usually referred as second quantisation.

In first quantisation, i.e. Quantum mechanics and Schrödinger equation, the quantisation of energy levels and allowed momenta follows from the (very classical) fact that differential equations require their solutions to exist and make sense - i.e. converging to something (not just tending to infinity) and obeying boundary/initial conditions. These two facts usually lead to a discrete set of allowed solutions.

A string obeying the wave equation, held at two ends, will only undergo oscillations at $nf_0$, $f_0$ being the fundamental frequency and $n \in \mathbb{N}$, a condition which arises from $\Psi(0) = \Psi(a) = 0$ on the solution $\Psi(x) \propto \sin(k x)$.

An atom confined in a potential well (e.g. Coulomb attraction from the nucleus) obeys the Schrödinger equation, a type of wave equation. The solution of the problem (see online) requires series solutions, and lead to the recurrence relation $a_n \propto f(a_{n-1})$, which requires to be truncated in order for the solution not to blow up.
From this truncation you basically get the quantum numbers $\ell, m$, and the solutions are given names like "Laguerre" or "Legendre" polynomials.

Jumping to QFT, qualitatively I would say that:

  • in the low energy limit you need to recover quantum mechanics, which is quantised. So you'd expect QFT to be quantised as well.
  • we know from experiments that there are elementary particles, so one can only get "stuff" (be it matter or light) only in discrete quantities. So why not encoding it in the theory?

Finally, let me ask you this.
You claim that recovering $H = \sum_{\mathbf{k}j} (N_{\mathbf{k}j} + \frac{1}{2})\omega_k$ is the smoking gun for your theory to be quantum.
But is it really?
Did you prove that $[x,p] \neq 0$ ? Did you show that there is entanglement?
After all, I just showed you above that even a classical string displays discrete energies... quantum in your sense just means "obeying" quantum theories.

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  • $\begingroup$ I was not my intention to know here where non-commutative algebra or entanglement in QM comes from, I was actually intrigued most by discrete energy eigenvalues and their possible mathematical origin. $\endgroup$ – Frederic Thomas Jan 20 '19 at 22:33
  • $\begingroup$ But that's my point. Nowadays for an effect to be "quantum" it has to show effects that are not present in classical physics at all - e.g. entanglement or commutators. Quantised energy levels, like quantum tunnelling and the uncertainty principle, are all consequences of applying wave optics to particles. $\endgroup$ – SuperCiocia Jan 20 '19 at 22:35

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