# Is non-relativistic quantum field theory equivalent with quantum mechanics?

Some books of many-body physics, e.g. A.L.Fetter and J.D.Walecka in Quantum theory of many-particle systems, claimed that at non-relativistic level, quantum mechanics (QM) and quantum field theory (QFT) are equivalent. They proved the second quantized operators $$T= \sum_{rs} \langle r | T | s \rangle a_r^{\dagger} a_s$$ $$V= \frac{1}{2} \sum_{rstu} \langle r s | T | t u \rangle a_r^{\dagger} a_s^{\dagger} a_u a_t$$

could obtain the same matrix elements as the "first-quantized" ones. Here $T$ and $V$ stand for kinetic and interaction operators, respectively.

However, some book, H. Umezawa et al Thermo field dynamics and condensed states in chapter 2, claimed that even at non-relativistic level, QFT is not equivalent with QM. They used a series of derivations (too long to present here, I may add a few steps if necessary), showed that Bogoliubov transformation with infinity space volume yields unitary inequivalent representations. In QM, all representations are unitary equivalent. Therefore, QM and non-relativistic QFT are not equivalent. However, as they said in p32

This might suggests that, in reality, the unitary inequivalence mentioned above may not happen because every system has a finite size. However, this point of view seems to be too optimistic. To consider a stationary system of finite size, we should seriously consider the effects of the boundary. As will be shown in later chapters, this boundary is maintained by some collective modes in the system and behaves as a macroscopic object with a surface singularity, which itself has an infinite number of degrees of freedom.

Nevertheless, Surface is an idealized concept. In reality, the boundary between two phases is a microscopic gradually changing of distribution of nuclei and electrons. My question is about, is the argument of surface singularity from Umezawa et al a pure academic issue? The academic issue here means if I have sufficient computational power, I compute all electrons and nuclei by quantum mechanics, I could very well reproduce the experimental results up to relativistic corrections.

P.S. The terminology "second-quantization" may not be appropriate, since we quantize the system only once. Nevertheless I could live with it.

## 1 Answer

''They [...] showed that Bogoliubov transformation with infinity space volume yields unitary inequivalent representations.'' Yes, this is an important effect and the source for phenomena like phase transitions and superconductivity. The isotropic limit of infinite space volume is called the thermodynamic limit. It figures everywhere in the derivation of classical thermodynamics from classical or quantum statistical mechanics and erases all surface effects.

Without thermodynamic limit, there would be no phase transitions! Statistical mechanics of a system with finitely many particles always leads to an equation of state without discontinuities in the response functions. The latter (i.e., the phase transitions in the sense of thermodynamics) appear only in the thermodynamic limit. (Indeed, a system can be defined as being macroscopic if the thermodynamic limit is an appropriate idealization. Note that the Avogadro number $N$ is well approximated by infinity.)

The equivalence with ordinary quantum mechanics suggested by the derivation of second quantization only holds at a fixed number of particles. But quantum field theory is the formulation at an indefinite number of particles. This already requires (even without the thermodynamic limit and independent of surface effects) a Hilbert space with an infinite number of degrees of freedom, where the canonical commutation relations have infinitely many inequivalent unitary representations.

• "Without thermodynamic limit, there would be no phase transitions!" Phase transitions are observed in my freezer, which is not infinitely large... – DanielSank May 30 '18 at 13:36
• @DanielSank: If you model the water in your freezer by statistical mechnaics as a system with finitely many particles, you'll find that there are no discontinuities. The latter (i.e., the phase transitions in the sense of thermodynamics) appear only in the thermodynamic limit. – Arnold Neumaier May 30 '18 at 15:54
• Yeah yeah, I know that the are no discontinuities in a finite system. It just sounds weird to say that there are "no phase transitions" in the thermodynamic limit, given that we see things go from one phase to another in daily life. – DanielSank May 30 '18 at 16:29
• @DanielSank: Well, macroscopic matter is FAPP matter in the thermodynamic limit. Instead of letting the volume go to infinity you can also let the atom sizes go to zero in an appropriate way and get the same result at finite volume. – Arnold Neumaier May 30 '18 at 17:09