Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My understanding from my QFT class (and books such as Brown), is that many-particle QM is equivalent to field quantization. If this is true, why is it not an extremely surprising coincidence? The interpretation of particles being quanta of a field is -- at least superficially -- completely different from the quantum mechanical description of N point particles.

share|improve this question
    
Consider each excited state of Hydrogen as a new particle, so in transients an old one disappears and a new one appears. –  Vladimir Kalitvianski Feb 23 '13 at 16:31
3  
I detail the correspondence here: fizix.quora.com/What-is-a-Quantum-Field –  hwlin Feb 23 '13 at 17:16
1  
hwlin, yes, in your post you describe precisely the correspondence I am asking about. The correspondence I think has been well understood, but I find it to be an amazing coincidence that I wish was more often remarked upon. –  user1247 Feb 23 '13 at 18:47

3 Answers 3

It took a lot of work to show that the computational power of topological quantum field theories and the computational power of quantum computers were equivalent. See Friedman, Kitaev, Wang and Friedman, Larsen, Wang. If your book's claim had been mathematically rigorous, this equivalence would have taken one line.

In some loose sense, this may be true, as many-body quantum systems and quantum field theories behave similarly in many respects and appear to have the same computational power, but it is certainly not an equivalence at the level of mathematical rigor.

share|improve this answer

As remarked in many references, QFT and QM (many-particle or not) are disjoint theories. The textbooks by Mandl & Shaw, Landau & Lifshitz, Sakurai... report some of the fundamental differences between QFT and QM.

The "interpretation of particles being quanta of a field" is misleading. There is a partial formal analogy [*] for some simple cases --e.g. a collection of independent particles in an external harmonic potential $V(x)$-- but this analogy is lost in the general case, generating well-known difficulties for QFT [Quantum Mechanics: Myths and Facts]:

Whereas the formal mathematical analogy between first quantization and QFT (which implies the irrelevance of the number operator $\hat{N}$) is clear, there is one crucial physical difference: Whereas in first quantization there is really no reason to attribute a special meaning to the operator $\hat{N}$, there is an experimental evidence that this is not so for QFT. The existence of particles is an experimental fact! Thus, if one wants to describe the experimentally observed objects, one must either reject QFT (which, indeed, is what many elementary-particle physicists were doing in the early days of elementary-particle physics and some of them are doing it even today [51]), or try to artificially adapt QFT such that it remains a theory of particles even with general interactions (such as those in (98)).

[*] Partial because only some aspects of the particle approach are covered by the analogy whereas others are ignored.

share|improve this answer

Many body quantum mechanics is the non-relativistic limit of an underlying relativistic quantum field theory (QED in the case of electrons in atoms or metals, QCD in the case of nucleons in a nucleus). This can be made manifest by constructing a non-relativistic effective field theory which describes the low energy limit of the underlying theory. The Dyson-Schwinger equations of the non-rel QFT are easily seen to be equivalent to the schroedinger equations.

There is nothing fundamentally different about non-rel QFTS. The main difference is that in a Gallilean invariant field theory the number of particles is always conserved. As a consequence, sectors with different numbers of particles decouple, and the state with zero particles is trivial (there is no vacuum polarization in a non-relativistic field theory).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.