# What is the difference between QM and non-relativistic QFT

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I have often heard that quantum mechanics can be considered to be 0-dimensional quantum field theory (QFT in 0 space and one dimension of time). I also know that there are classical field theories, such as fluid dynamics for example, that may depending on the regime considered relativistic or not.

But what are non-relativistic quantum field theories (examples?), and what is their difference to (non-relativistic) quantum mechanics?

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QM has finite degrees of freedom. Non-relativistic QFT has infinite degree of freedom. QM can be treated as an one dimensional Non-relativistic QFT.

So far, the zero dimensional QFT I've seen people discussing is the perturbation Gaussian integrals. An example of non-relativistic QFT is the Schroedinger field theory, which is introduced in its wikipedia page: https://en.wikipedia.org/wiki/Schr%C3%B6dinger_field
The Hamiltonian of a QFT can be obtained by the so-called 2nd quantization method starting from QM by sandwiching the first quantized Hamiltonian operator with the field operator and its Hermitian conjugate.

The term 2nd quantization is misleading because everything has been quantized already. For historical reason, people thought of it as if the "wavefunction" is quantized again, we now call it 2nd quantization.

answered Oct 2, 2016 by (50 points)
edited Oct 2, 2016

Ok, do you have some concrete examples of non-relativistic QFTs? Thinking about it, I have a rather hard time coming up with a specific example ...

@Dilaton You can find even more examples of non-relativistic QFT from many advanced quantum mechanics books and condensed matter field theory books.

1. Advanced Quantum Mechanics by Franz Schwabl

2. Quantum Field Theory of Point Particles and String by Hatfield B

3. Condensed Matte Field Theory by Altland Alexander Simons Ben.

As far as I can recall, the 2nd quantization is roughly using ladder operators to generalize a 1-particle QM to many particle QM. The Hilbert space for 1-particle QM is replaced by the Fock space for an arbitrary number of particles.

When the 2nd quantization method applied for non-relativistic QM is used for relativistic theory, it leads to unaccepted quantum theory. The reason is that in relativistic theory, it is not easy to define a well-behaved localized operator $\hat{x}^{\mu}$, and so you cannot find the relativistic generalization of the completeness $1=\int dx |x><x|$. In relativistic quantum field theories, the momentum representation is not equivalent (in the sense of Fourier transforms) to position representation. In fact, you cannot interpret a relativistic field $\phi(x)$ as a quantum mechanical wave. For a much clearer explanation, you may read some papers about "induced representation" of Poincare group.

In QM, we start from promoting $x(\tau)$ and $p(\tau)$ to quantum operators $\hat{x}$ and $\hat{p}$, where $\tau$ is a parameter, which is usually called time.

When the classical theory is relativistic, you may think the possibility that you promote time $t(\tau)$ as an operator. But there is a theorem which I cannot completely recall at the moment saying that the quantum commutation relation $[x^{\mu},p^{\nu}]=i\hbar\eta^{\mu\nu}$ for Lorentzian signature gives you a quantum theory whose energy is not bounded from below. This is why "relativistic quantum mechanics" does not work.

Then there are two approaches. One is called string theory, in which the quantum operators $x^{\mu}$ are parametrized by two variables $\tau$ and $\sigma$.

Another approach is the quantum field theory, in which the dynamical variables are fields $\phi$, which depends on 1+3 variables $t$ and $\vec{x}$. In this theory, space coordinates play a role as continuously labeling the dynamical variables. You may think of fields $\phi$ in QFT as $x$ in QM. In QFT, $\phi(t,\vec{x})$ is labeled by continuous indices $\vec{x}$; in QM $x_{i}$ is labeled by $i$. You may alternatively think of $\phi$ in QFT parametrized by four parameters $t,x^{i}$. but in QM $x(\tau)$ is parametrized by one variable $\tau$.

In its present formulation, your third sentence contradicts your first two sentences!

In QFT, there is still the Schroeding equation, but in the sense of state functional. For example, in QM, a wave function is $<x|\phi>=\phi(x)$. In QFT, you can define the wave functional $<\phi|\Omega>=\Omega[\phi]$, and the Schroedinger equation in QM is replaced by a functional differential equation. This functional approach of QFT is not very useful though.

@Arnold Neumaier Could you tell me which sentence you are referring?

''QM has finite degrees of freedom. Non-relativistic QFT has infinite degree of freedom. QM can be treated as an one dimensional Non-relativistic QFT.'' The third sentence contradicts your first two sentences, which assert two disjoint cases.

''This functional approach of QFT is not very useful though.'' Though it is played down in textbooks, it is essential for the semiclassical analysis of instantons and solitons.

@Arnold Neumaier Thanks a lot. I don't think that the two sentences "QM has finite degrees of freedom. Non-relativistic QFT has infinite degree of freedom." are correct anymore. But I think QM is a 1-dim QFT. Take path integral quantization as an example, QM is a 1-dim QFT since we are summing over paths $x(t)$ parametrized by a single variable.

According to the customary terminology in nonrelativistic QFT, and consistent with the definition given in the question,

''0-dimensional quantum field theory (QFT in 0 space and one dimension of time)''

QM is a 0-dimensional QFT. You count the dimensions in the relativistic way, which is inappropriate for QM.

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A traditional quantum mechanical system of $N$ particles has a finite number of $3N$ position coordinates as degrees of freedom. The Hamiltonian can be expressed for any $N$; hence one can form the corresponding system where $N$ is treated as uncertain. This is done by working in the corresponding Fock space, defined as the direct sum of all $N$-particle Hilbert spaces. The resulting system has now an infinite number of degrees of freedom. That this constitutes a nonrelativistic QFT becomes apparent when changing the notation to that of second quantization, with creation and annihilation operators that change the number of particles.

Thus all statistical mechanics done in the grand canonical ensemble is nonrelativistic quantum field theory.

Other examples are given by the statistical mechanics of crystals, which are infinitely extended periodic systems. The periodicity, however, allows one to map the problem on the 3-dimensional torus, which is a compact manifold; so that the total number of particles on the torus is finite (but huge).

Both classes of examples together constitute the main examples of nonrelativistic quantum field theories.

In general, one considers the infinite number of degrees of freedom as the essence of a quantum field theory, but this is the case only when the underlying position space is non-compact. In the compact case, and in particular for a torus (e.g. crystal) or a single point (= 0-dimensional field theory; e..g, ordinary quantum mechanics in the rest frame in a basis of harmonics), there are only finitely many degrees of freedom.

0-dimensional QFT is QM in the same sense as 0-dimensional classical field theory is described by ODEs rather than PDEs for higher dimensions.

The difference between relativistic and nonrelativistic is the same in QFT as in classical field theory - namely whether the theory is invariant under the Poincare group or the Galilei group.

answered Oct 2, 2016 by (13,637 points)
edited Oct 2, 2016
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QM is certainly not a $0$-dim QFT.

Starting from the partition function of a quantum field living on a connected manifold $M$,
$$\int D[\phi]e^{iS[\phi]},$$
where the field $\phi$ is regarded as a map $\phi$: $M\rightarrow\mathbb{R}$, if the manifold is of $0$-dimensional, the isometry group of $M$ is then trivial since $M$ is a single point.

Furthermore, this field cannot have spin degree of freedom, i.e. it indeed carries no spinoral indices. Lastly, there can be no spacetime derivative acting on $\phi$ in the action $S[\phi]$.

We then can conclude that the classical action $S[\phi]$ can only be a real-valued function $S(\phi)$, which can be Taylor expanded as
$$S(\phi)=\frac{m^2}{2}\phi^{2}+\frac{g}{3!}\phi^3+\cdots.$$
This partition function then is simply a perturbative Gaussian integral in calculus
$$\int_{\mathbb{R}}e^{\frac{m^2}{2}\phi^{2}+\frac{g}{3!}\phi^3+\cdots}d\phi.$$
Then the $0$-dim QFT is simply a theory of perturbative Gaussian integrals in calculus.

If the manifold $M$ is $1$-dim, then by following a similar procedure, we can find the partition function given by the Feynman path-integral in QM.

answered Oct 4, 2016 by (50 points)

What you describe is a field theory with a single field only and $-1$ space dimensions, so it doesn't even have time. With a zero-dimensional space, the fields are functions of time, the action may involve the first derivative with respect to time, and the path integral is over all histories (time-dependent fields). What one gets is precisely the same as an anharmonic quantum oscillator in the Heisenberg picture.

Moreover, you forgot that in QFT $\phi$ is a single label for all the fields, which are components of $\phi$. If you look at the the 0-dimensional case of a QFT with $3N$ scalar fields invariant under a SO(3) acting on the components in sets of three, and rename the triples as $q_k~(k=1,\ldots,n)$ you get precisely the quantum mechanics of $N$ particles.

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