"QFT is simple harmonic motion taken to increasing levels of abstraction."

This is my memory of a quote from Sidney Coleman, which is probably in many textbooks.

What does it refer to, if he meant something specific?

If he did not, which is far most likely, then rather than asking a list question, if somebody can point to an example of how we move from SHM to, presumably some example of fields interacting, producing or destroying a particle.

I don't think I can go any further with asking a question, because of the restrictions when asking a specific recommendation, but if someone is able to say, "wait until you get to chapter X of Zee, Tong, P & S (please don't say Weinberg) and enlightenment will follow", that would be very much appreciated.

My apologies if I have mangled the quote, nobody has heard of it or I am trying to run way ahead of myself. No need to hold back on telling me that last part.

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    $\begingroup$ SHM plays an important role in QFT, just as in almost all physical theories. However, it would be going too far to state that QFT is SHM, in my opinion. The quote by Coleman is actually closer to: "The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction." $\endgroup$ – gj255 Sep 4 '17 at 22:09
  • $\begingroup$ @gj255 yes, isn't that funny that I left out the young part :) cheers and thank you. $\endgroup$ – user167453 Sep 4 '17 at 22:41
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    $\begingroup$ "Allez en avant, et la foi vous viendra" push on and faith will catch up with you. [advice of D'Alembert to those who questioned the calculus] Quoted in A L Mackay, Dictionary of Scientific Quotations (London 1994) $\endgroup$ – ZeroTheHero Sep 4 '17 at 23:19
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    $\begingroup$ Related: physics.stackexchange.com/q/127141/2451 , physics.stackexchange.com/q/159021/2451 and links therein. $\endgroup$ – Qmechanic Sep 5 '17 at 1:19
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    $\begingroup$ For reference, wait until you get to Chapter I.8 of Zee and enlightenment will follow ;) $\endgroup$ – J. Murray Sep 5 '17 at 1:24

If you quantize the oscillator you get a natural particle interpretation. If instead you extend to a field theory the oscillator becomes a classical Klein-Gordon field, with frequency become mass. If you now quantize that you again get a particle interpretation, but the field is a linear combination of ladder operators so particle number becomes another observable. (This is related to how relativity prevents you sticking to a theory of one particle.) Bogoliubov transformations provide different "perspectives" on particle number, in analogy with changes in a classical oscillator's choice of phase space coordinates.

If you now extend the theory to multiple particle species you can couple your oscillators to denote interactions that create or destroy particles, in analogy with energy transferred between classical oscillators. If you let your Lagrangian gain terms that make the EOMs nonlinear, you have nonzero VEVs as in the Higgs effect. This is analogous to anharmonic oscillation.

  • $\begingroup$ Thanks very much, I am semi familiar with some of the references in your post, thanks very much for taking the time to answer. Actually, on reading your answer again it, it's a mini roadmap, which most texts inexplicably omit. $\endgroup$ – user167453 Sep 4 '17 at 22:43
  • $\begingroup$ What is VEV please ? $\endgroup$ – magma Sep 5 '17 at 6:23
  • $\begingroup$ @magma Vacuum expectation value. $\endgroup$ – user167453 Sep 5 '17 at 6:46

The main idea is that you can take a complicated interacting or coupled system and write its solution as a sum of non interacting or free modes. Even in classical mechanics, if you have a linear chain of $N$ oscillators, one can show that the general solution is a sum of $N$ normal modes each of them being a simple harmonic oscillator. In Quantum Field Theory we can also write the fields as sums of modes, each mode behaving like a quantum harmonic oscillator so can accept energy in an integer number of lumps of size $\hbar\omega$. In the second quantization formalism, these lumps are created or annihilated by operators acting on vacuum and this is interpreted as particles being created or annihilated.

As a concrete example, consider a linear chain of (interacting) atoms whose Hamiltonian operator is $$H=\sum_i\left[\frac{p_i^2}{2m}+\frac 12k(x_{i+1}-x_i)^2\right]$$ After Fourier transforming both the positions and momenta, the Hamiltonian can be written in the reciprocal space as $$H=\sum_i\left[\frac{\tilde p_i\tilde p_{-i}}{2m}+\frac 12\omega_i^2\tilde x_i\tilde x_{-i}\right].$$ By defining the operators $$a_i=\sqrt{\frac{m\omega_i}{2\hbar}}(\tilde x_i+i\tilde p_i/m\omega_i),\quad a_i^\dagger=\sqrt{\frac{m\omega_i}{2\hbar}}(\tilde x_{-i}-i\tilde p_{-i}/m\omega_i),$$ the Hamiltonian can also be written as $$H=\sum_{i=1}^N\hbar\omega_i(\hat a^\dagger_i \hat a_i+1/2),$$ which is the sum of $N$ uncoupled quantum harmonic oscillators. Each $i$ in the above equation represent a mode or an oscillator and according to second quantization the $n$th excitation of this mode shall be interpreted as having $n$ particles, which in these example in particular we call phonons. The same approach can be carried out to a generic field. You Fourier transform the field and show that it can be written as a sum over modes. These modes decouples in the Hamiltonian and each of them gives a harmonic oscillator. The general solution is therefore a linear superposition of harmonic oscillators. If we quantize those oscillators we then quantize the field itself.

  • $\begingroup$ @ZeroTheHero Sorry, my mistake... it should read $\tilde p_i\tilde p_{-i}$ and $\tilde x_i\tilde x_{-i}$. $\endgroup$ – Diracology Sep 5 '17 at 2:30
  • $\begingroup$ then is your index $i$ on both $a_i$ and $a_i\dagger$ correct? $\endgroup$ – ZeroTheHero Sep 5 '17 at 2:40
  • $\begingroup$ Thanks very much for your answer, and the time you took to write it. I have accepted J.G.s because it fills in more gaps in my knowledge, with various areas I have read about here, but have not yet studied. $\endgroup$ – user167453 Sep 5 '17 at 10:16
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    $\begingroup$ @Countto10 No worries! $\endgroup$ – Diracology Sep 5 '17 at 11:12
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    $\begingroup$ @ZeroTheHero Sorry again for my sloppiness. You're totally correct. I did a mistake in the definition of $a^\dagger_i$, which I have corrected now. Thanks for pointing this out! $\endgroup$ – Diracology Sep 5 '17 at 17:08

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