There are various more or less formal ways of expressing the foundational principles of nonrelativistic quantum mechanics, including both nonmathmatical statements and more rigorous axiomatizations. Recently there have been various people who have treated this from a quantum-computing point of view. Others have explored whether QM can be bent without breaking it. I've given some references at the bottom of this question to some of this kind of work.

But more concretely, I think most physicists would consider the following to be some sort of consensus on an informal list of principles. (Actually, I'd be happy to hear criticisms of this list as well.)

  1. Wavefunction fundamentalism. All knowable information about a system is encoded in its wavefunction (ignoring phase and normalization).
  2. Unitary evolution of the wavefunction. The wavefunction evolves over time in a deterministic and unitary manner.
  3. Observables. Any observable is represented by a Hermitian operator.
  4. Inner product. There is a bilinear, positive-definite inner product on wavefunctions.
  5. Completeness. For any system of interest, there exists a set of compatible observables such that any state of the system can be expressed as a sum of eigenstates.

Question: Does this summary of principles have to be modified for QFT? If so, how? If not, then what is the core difference between these two theories?


Kapustin, https://arxiv.org/abs/1303.6917

Mackey, The Mathematical Foundations of Quantum Mechanics, 1963, p. 56ff

Aaronson, "Is Quantum Mechanics An Island In Theoryspace?," http://arxiv.org/abs/quant-ph/0401062

Masanes and Mueller, "A derivation of quantum theory from physical requirements," https://arxiv.org/abs/1004.1483

Hardy, "Quantum Theory From Five Reasonable Axioms," https://arxiv.org/abs/quant-ph/0101012

Dakic and Brukner, "Quantum Theory and Beyond: Is Entanglement Special?," https://arxiv.org/abs/0911.0695

Banks, Susskind, and Peskin, "Difficulties for the evolution of pure states into mixed states," Nuclear Physics B, Volume 244, Issue 1, 24 September 1984, Pages 125-134

Nikolic, "Violation of unitarity by Hawking radiation does not violate energy-momentum conservation," https://arxiv.org/abs/1502.04324

Unruh and Wald, https://arxiv.org/abs/hep-th/9503024

Ellis et al., "Search for violation of quantum mechanics," Nucl Phys B241(1984)381

Gisin, "Weinberg's non-linear quantum mechanics and supraluminal communications," http://dx.doi.org/10.1016/0375-9601(90)90786-N , Physics Letters A 143(1-2):1-2

Sebens and Carroll, "Self-Locating Uncertainty and the Origin of Probability in Everettian Quantum Mechanics," https://arxiv.org/abs/1405.7577

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    $\begingroup$ What if I told you that all these principles are exactly the same for QFT, with the exception that you have to substitute "wavefunction" by the more abstract "state in Hilbert space" (or "wavefunctional", if you're really a fan of waves)? That is, where exactly does the belief that QFT somehow differs in its fundamental principles from QM come from? To me, the sole difference seems to be that "QM" is ordinarily about finitely many degrees of freedom and QFT is not. $\endgroup$ – ACuriousMind Jul 13 '17 at 20:59
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    $\begingroup$ Contemporary relativistic quantum field theory is supposed to follow all basic principles of quantum theory and can be thought of as a subset of quantum theory. On the other hand, no mathematically rigorous definitions of a Hilbert space of states and observables as operators are known for interacting, relativistic quantum field theory in 3+1 dimensions. Hence the formalism ends up looking very different, perturbation expansions being a basis of the theory, rather than a method for getting approximate results from a fixed mathematical model, as is usually the case in quantum mechanics. $\endgroup$ – Adomas Baliuka Jul 13 '17 at 21:06
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    $\begingroup$ I prefer "state function" or even "state vector" to "wavefunction" when talking about quantum mechanics in the large, because "wavefunction" always puts me in mind of the Schrödinger picture in particular. $\endgroup$ – dmckee Jul 13 '17 at 21:06
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    $\begingroup$ @BenCrowell The non-relativistic particle in a box has as three degrees of freedom. To count degrees of freedom, we should count how many pairs of $x_i, p_i$ operators we have that satisfy the canonical commutation relations, $[x_i, p_j] = i\hbar\delta_{ij}$. For a particle in a box $i = 1,2,3$ so there are three d.of. For a field, we have instead $[\phi, p_\phi] = i\hbar\delta(\mathbf x-\mathbf x')$ -- the index now takes infinitely many values, so there are infinite d.o.f. There are infinitely many energy eigenstates, but that's not the same as the number of d.o.f. Any classical phase space $\endgroup$ – Robin Ekman Jul 14 '17 at 0:32
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    $\begingroup$ contains infinitely many points, too. Note that the quantum and classical cases are entirely analogous if you swap commutators with Poisson brackets: to count degrees of freedom classically you should count how many functions on phase space can verify $\{q_i, p_i \} = \delta_{ij}$. Consider also the equipartition theorem: if a particle in a box had infinitely many degrees of freedom, what would the heat capacity of an ideal gas be? $\endgroup$ – Robin Ekman Jul 14 '17 at 0:34
  1. All knowable information about a system is encoded in a ray in a Hilbert space. In QFT, and unlike non-relativistic QM, there is no $|x\rangle$ basis, so you cannot construct a wave-function $\varphi(t,x)=\langle x|\varphi(t)\rangle$ to encode this information. What you can do is encode this information in the so-called correlation functions (cf. Wightman Reconstruction Theorem). You need an infinite number of functions to encode all the information of the system. Equivalently, one may encode this same information in a single functional, either through a functional integral or as a wave functional (cf. 214552).

  2. This is unchanged, except perhaps for the fact that it is usually much more convenient to evolve operators instead of states, because covariance becomes manifest. The abstract Schrödinger equation, $\frac{\mathrm d}{\mathrm dt}|\psi\rangle=-iH|\psi\rangle$ is as valid in non-relativistic QM as it is in QFT (and so is the Heisenberg equation, $\dot A=i[H,A]$). In this sense, the evolution is still unitary, but it is expressed in terms of operators instead of states.

  3. This is unchanged.

  4. This is unchanged, except perhaps for the fact that it is sometimes convenient to artificially enlarge the Hilbert space so as to include "negative norm states", that is, the inner product is relaxed into a sesquilinear form (which agrees with the positive-definite "true" inner product in the "true", physical Hilbert space).

  5. This is unchanged.

  • $\begingroup$ So if I replace "wavefunction" with "state vector," can I keep everything unchanged (ignoring issues of convenience)? $\endgroup$ – Ben Crowell Jul 13 '17 at 22:36
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    $\begingroup$ Yes. To be more precise, you should speak of rays instead of state vectors (because, as you already know, the phase of the vector is irrelevant). $\endgroup$ – AccidentalFourierTransform Jul 13 '17 at 22:39
  • $\begingroup$ Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." $\endgroup$ – tparker Jul 16 '17 at 21:33
  • $\begingroup$ @tparker good question. I don't really know the answer. The reason might be that "it excludes the $0$, and therefore is a "ray" emanating from the origin". Or perhaps the concept was introduced by someone from a non-English-speaking country and "ray" was a literal translation. We might end up asking about the historical origin of the name on History of science and mathematics. I'll let you know. $\endgroup$ – AccidentalFourierTransform Jul 16 '17 at 21:42
  • $\begingroup$ That may indeed be the reason, but if so it's not a very good reason. By analogy, why would $\mathbb{R} \setminus \{0\}$ be more like a ray than like a line? $\endgroup$ – tparker Jul 16 '17 at 21:49

Quantum field theory is quantum mechanics applied to Lorentz covariant causal systems. That is, quantum field theory is simply quantum mechanics plus special relativity. Demanding Lorentz covariance and causality constrains the systems you can talk about. For example, a crystal lattice completely breaks Lorentz symmetry, so that's out.

The systems that you can talk about turn out to be those made from Lorentz covariant local quantum fields. This is basically the message of the first 250 pages of Weinberg's The Quantum Theory of Fields. Here is the beginning of Ch.2:

The point of view of this book is that quantum field theory is the way it is because (with certain qualifications) this is the only way to reconcile quantum mechanics with special relativity. [...] First, some good news: quantum field theory is based on the same quantum mechanics that was invented by Schrödinger, Heisenberg, Pauli, Born, and others in 1925-26, and has been used ever since in atomic, molecular, nuclear, and condensed matter physics. [...] [T]his section provides only the briefest of summaries of quantum mechanics [...]

(i) Physical states are represented by rays in Hilbert space. [...]

(ii) Observables are represented by Hermitian operators. [...]

These -- in the full form in the book -- more or less cover your points 1 through 5.

I also recommend Weinberg's talk, What is quantum field theory and what did we think it is?

I think on a pedagogical level thinking of quantum field theory as different from and not just a subset of quantum mechanics may have something to do with that students are first exposed to the Schrödinger equation in the wrong form. The Shrödinger equation is, fundamentally, not a PDE in real-space. It's an ODE in Hilbert space. Correspondingly, one should not start with wavefunctions, but with statevectors, as other answers and comments have pointed out.

  • $\begingroup$ For example, a crystal lattice completely breaks Lorentz symmetry, so that's out. This seems wrong to me. Lorentz invariance is an invariance of the laws of physics. It isn't violated just because a physical state of a system lacks that symmetry. $\endgroup$ – Ben Crowell Jul 16 '17 at 2:57
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    $\begingroup$ @BenCrowell yes, a crystal is a symmetry-breaking state of a Lorentz invariant theory. What I mean is that the Hamiltonian describing perturbations around this state -- in the jargon, the low-energy effective field theory -- isn't Lorentz covariant, so that's not a system we are interested in as an explicitly relativistic theory. I use it as an example of Lorentz invariance restricting the Hamiltonians we can use. The question is then is then how to construct ones that we can use. We can't use the regular x,p-operators because they are 3-vectors. Classically, we use Lorentz tensor fields, each $\endgroup$ – Robin Ekman Jul 16 '17 at 3:39
  • $\begingroup$ component of which is a number. In QM, we should figure out how the Lorentz group acts on the Hilbert space, and go from there. Skipping 200 pages of Weinberg, it turns out that the way to guarantee a Lorentz invariant theory is to construct it from local products of causal quantum fields, each component of which is an operator. $\endgroup$ – Robin Ekman Jul 16 '17 at 3:46

All of these postulates continue to hold in relativistic QFT, except that the time-evolution operator is no longer defined by the Schrodinger equation with a nonrelativistic Hamiltonian.

The only one that requires significant new elaboration in the relativistic context is the existence of an inner product. In nonabelian gauge theory, it often a useful calculational trick to formally expand your Hilbert space to a larger state space that includes negative-norm "ghosts." Such a state space is no longer a Hilbert space because its sesquisymmetric bilinear form is no longer positive definite, and is therefore no longer an inner product. But the key point is that you never have to introduce ghosts; they are merely a useful calculation trick, but do not physically exist. You can always do any calculation without invoking ghosts; See here.


QFT is just QM...

To expand on Adomas Baliuka's comment: all QFTs that we know how to construct in a mathematically rigorous way do fulfill your 5 axioms. As nicely summarized in AccidentalFourierTransform's answer, the differences to the standard quantum mechanical formulation of a single particle are more in the physical interpretation of the state vector and in the kind of observables you can define on your Hilbert space (eg. you will be measuring the number of particles in a given region of space, rather than measuring the position of a given particle).

In addition, you may want to restrict the term QFT to quantum theories which fulfills additional axioms beyond the basic QM ones, such as locality, causality, (local) Lorentz invariance...

...or some suitable generalization thereof?

But in any case it is important to keep in mind that there aren't that many QFTs that we know how to construct: free fields in any dimension, polynomially-interacting fields in 1+1 and 2+1 dimensions, a couple of other interacting theories in low dimensions, some topological theories (ie. theories that looks like field theories at first glance but turn out to have only finitely many true, physical degrees of freedom),... There are many QFTs that we would like to construct but don't know how, so it remains pretty much an open question in mathematical physics what the "right" axiomatic framework to do QFT should be.

Specifically, the axioms you ask for are likely to break down at least for QFT on a curved, non static spacetime: there, it may no longer be possible to come up with a Hilbert space on which the time evolution could be represented as a unitary transformation. What happens is that when you try write down the evolution you find that it kicks you out of your Hilbert space )-; So you may need to somewhat loosen your definition of what a "state" of your quantum theory is, to guarantee that all states remain valid states as they evolve in time.

A proposal for a more general notion of quantum states are so-called algebraic states (see eg this answer of mine for an elementary introduction). They can be thought as the natural mathematical generalization of mixed states (aka density matrices). Note that even in the context of finitely many degrees of freedom, there are axiomatizations of quantum mechanics in which mixed states are the fundamental object, rather than an afterthought as in the standard formalism. For example, this is the case in the so-called "generalized probability theories" approach, which aims to rederive quantum mechanics from very basic assumptions on measurement processes.


protected by Qmechanic Jul 14 '17 at 3:35

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