A What is the relationship between QFT and QM?

jordi

I know this question has been asked, in several ways, many times before. I have read many of the posts. And still I do not fully understand the situation: is QM in any way a subset of QFT?

Apparently no: QM uses position variables, while QFT does not. QM has the Born rule and a wave function, while QFT does not.

But then, many of the postulates in both (unitarity, cluster decomposition ...) are common. It seems here that the difference is Galilean invariance on QM vs Poincaré invariance in QFT, but that is ok.

Classical mechanics is easily derived, in the path integral formalism, from QFT. But I have never seen the same for QM from QFT.

But then, there is a "second quantization" in QM, mostly in condensed matter, where QM becomes a field theory.

So, my question is then: are QM and QFT different theories? If so, how does this observation with Weinberg's "theorem" that "everything" is an effective field theory?

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hilbert2

Gold Member
In QFT there's an infinite number of mechanical degrees of freedom, similar to the position coordinates in ordinary QM. This is easy to understand by noting that to know the current state of a field in continuum space, you have to know the field "strength" at a very large number of densely spaced points in space. The same applies to the difference between classical point particle mechanics and classical field theory.

Another difference is that in QFT the canonical commutation relation is different for fields describing bosons and fermions. And also, the QFT has the problem of infinite quantities resulting from calculations, requiring renormalization before they can be made sense of. Well, actually perturbation series can diverge even in QM of finite number of degrees of freedom, but it's not as problematic there.

vanhees71

Gold Member
This is a strange question. To begin with there is only one quantum theory. The most comprehensive one is relativistic quantum field theory and the most comprehensive application of its methods is the Standard Model of Particle Physics.

Of course we do not always need the full machinery of relativistic QFT, and most physics problems can only be tackled in approximations adequate to the problem at hand. E.g., much of atomic physics (atoms with not too large $Z$), molecular physics, and condensed-matter physics can be treated in the non-relativistic approximation.

In non-relativistic physics (e.g., atomic physics) you often have a situation where you deal with a fixed number of particles (like the electrons around the nucleus), i.e., no particle creation and annihilation process (the normal thing in the relativistic realm). Then you can use quantum mechanics.

The name "second quantization" is a bit misleading. It comes from the fact that it is a simple heuristic to consider the Schrödinger equation for one particle ("first quantization") as a classical field theory, formulate it in terms of the Hamilton formalism and than "canonically quantize" it. Then you end up with creation and annihilation operators for fermions or bosons (depending on whether you use anticommutation or commutation relations for the field operators). All this is of course simply a heuristic way to motivate the formalism, and it's didactically not the best way, because it leads to this misunderstanding as if non-relativistic QFT were someting else than non-relativistic many-body QM.

Take as the most simple example a spin-0 particle. On basis you can use are the position eigenstates $|\vec{x} \rangle$. Now spin-0 particles are bosons. Thus a basis for for $N$ indistinguishable bosons is given by the totally symmetrized product states
$$|\vec{x}_1,\ldots,\vec{x}_N \rangle^{+} = \frac{1}{\sqrt{N!}} \sum_{P \in S_N} |\vec{x}_{P(1)},\ldots,\vec{x}_{P(2)} \rangle.$$
Here $P$ runs over all permutations $S_N$ of the indices of the positions of the particles. You can imagine that this becomes quite cumbersome if $N$ gets larger since the sum runs over $N!$ terms.

Now you can instead introduce bosonic annihilation and creation operators which create these basis states out of the "vacuum" state, which is defined as the situation that no particle is present. Then you define the creation operators recursively by
$$\hat{\psi}^{\dagger}(\vec{x}) |\vec{x}_1,\ldots,\vec{x}_N \rangle^{+}=|\vec{x},\vec{x}_1,\ldots,\vec{x}_N \rangle^{+}.$$
This makes a completely symmetrized product of position basis vectors for $N$ indisinguishable bosons a completely symmetrized product of positions basis vectors for $N+1$ indistinguishable particles. If you denote "the vacuum" with $|\Omega \rangle$ (which describes the situation that no particle is there), then this leads to
$$|\vec{x}_1,\ldots,\vec{x}_N \rangle^{+} = \hat{\psi}^{\dagger}(\vec{x}_1)\cdots \hat{\psi}^{\dagger}(\vec{x}_N) |\Omega \rangle.$$
The definition of the creation operators from the symmetrization of the bosonic basis states implies the commutation relation
$$[\hat{\psi}^{\dagger}(\vec{x}_1),\hat{\psi}^{\dagger}(\vec{x}_2)]=0,$$
so that with this definition the symmetrization is automatically guaranteed. That makes it much easier to deal with indistinguishable particles, even if you have a fixed number of states.

One can derive from this definition that the adjoint operator is an annihilation operator, particularly annihilating the vacuum state (i.e., it maps the vacuum state to the null vector).

It's also possible to define all kinds of operators of the first-quantization formalism to an equivalent operator in this second-quantization formalism. Now, any Hamiltonian of the first-quantization formalism keeps the particle number the same. You cannot even easily formulate the creation and annihilation of particles in the first-quantization formalism. This implies that if you express any model of the first-quantization formalism into its equivalent formulation in the second-quantization formalism your Hamiltonian consists of a sum of field-operator expressions where you always have as many creation and annihilation field operators in each term, i.e., applying the Hamiltonian to any completely symmetrized $N$-particle basis vector doesn't lead into a space with a different particle number. So for this cases the 2nd-quantization formalism is indeed completely equivalent with the 1st-quantization formalism. It then only makes the task to always stay in the properly symmetrized (or for fermions antisymmetrized) many-particle spaces much easier.

jordi

Thank you for your detailed explanation.

Questions:

1. Is QM in the second quantization identical to "traditional" QM? If true, is this explicitly shown in some textbook?
2. Is QM in the second quantization the explicit non-relativistic limit of a QFT? If true, is this explicitly shown in some textbook?

If 1 and 2 are answered positively, I have a final question: why on Earth QM is usually explained in the "traditional" way? Why not explaining all quantum theories (QM or QFT) as just QFT, i.e. in the Fock states formalism, unlike now, when there are two clearly different ways to quantize? (of course, there are other quantization methods, like geometric quantization).

hilbert2

Gold Member
I think that in David Tong's QFT lecture notes there was an explanation how to extract a Schrödinger wave function from a QFT state vector.

jordi

Yes. And from:

http://hitoshi.berkeley.edu/221B/QFT.pdf

one can see if by "second quantizing" a classical field equation that is formally identical to the Schroedinger equation (but of course, without the probabilistic interpretation), one gets a field theory that reproduces the Schroedinger equation for single particle, two particles and three particles (and gets the symmetry well automatically).

So, it seems as if the second quantization is the same as the "normal" QM.

Then, my question is why teaching QM in the way it is done, since it seems QM and QFT are two different things. Instead, by "second quantizing" the "classical Schroedinger equation", one can see QM and QFT are "the same" (except for the Galileo/Poincaré symmetry).

Maybe the reason is that it is more intuitive to canonically quantize classical mechanics, by making X and P are operators, rather than to quantize a "classical Schroedinger equation", which nobody knows why is presented? This motivation is the only advantage I see.

But one could do everything in second quantization, and then show, as Hitoshi, that this is equivalent to the "traditional" Schroedinger equation, and then state that this equation can be interpreted intuitively as the "promotion" of X and P from c-numbers into operators. But this is of historical interest only.

Is it like this? Or are there more subtleties?

hilbert2

Gold Member
The way how hermitian operators (position, momentum, total energy) and unitary opetators (time evolution, spatial rotations) are handled in QFT, as well as how you represent a composite system as a tensor product of its components, is exactly the same as in QM. It's just easier to practice that formalism with nonrelativic QM where the infinite-dimensionality is only in the state vector space and not also in the classical phase space of the physical system.

jordi

I understand that reason. But the cost is to have two different formalisms, such that it looks like that QM and QFT are different.

I am browsing Greiner's Field Quantization, third chapter, and yes, it seems clear the canonical quantization of the classical field theory that looks like the Schroedinger equation is equivalent to QM.

Then, only one question remains:

Is this non-relativistic field theory (the second quantization of QM), the non-relativistic limit of a given QFT?

A. Neumaier

Is this non-relativistic field theory (the second quantization of QM), the non-relativistic limit of a given QFT?
yes.

Which QFT?

hilbert2

Gold Member
I understand that reason. But the cost is to have two different formalisms, such that it looks like that QM and QFT are different.
If you take a look at how to form the Dyson series for the time evolution of a QM system with explicitly time dependent hamiltonian, you will encounter the same concepts of time-ordered operators that are everywhere in QFT calculations. An introductory QM course just doesn't go that far.

jordi

Is there a path integral quantization version, using the Lagrangian that gives, through the Euler-Lagrange equations, the "classical Schroedinger equation"?

A. Neumaier

Which QFT?
For symmetrized wave functions $\psi(x_1,\ldots,x_n)$, a scalar field theory.

charters

Two good papers on this:

jordi

7.14 of Kleinert book on path integrals gives a path integral description of the second quantization of QM.

As a consequence, it seems clear that yes, QM is just a type of non-relativistic QFT. And the traditional presentation of QM is done for simplicity purposes (it is easier to promote X and P to operators, rather than the second quantization formalism), with the risk of hiding the fact that QM is a QFT.

Or in other words: there are either classical field theory or quantum field theory, but nothing else (and classical field theory is a limit case of QFT).

meopemuk

Here is my understanding: both QM and QFT are examples of the general "quantum theory". Any quantum theory consists of two major components - a Hilbert space of states and a unitary representation of the symmetry group (Galilei or Poincare). The generator of time translations - the Hamiltonian - has a special status, because its interacting part defines the specific dynamics of the theory.

1. Traditional QM is a quantum theory in which the number of particles is not allowed to change. The corresponding Hilbert space is a tensor product of N 1-particle spaces.

2. One can build a slightly more complicated theory by constructing the Fock space as a direct product of all N-particle spaces and keeping the above QM description within each Fock sector. This will look like QFT, but without particle creation and destruction.

3. A more general approach is possible which has the Fock space as the Hilbert space of the theory, but the symmetry group representation is allowed to cross the boundaries between sectors. This can be achieved by building interactions as polynomials in particle creation and annihilation operators. There is a huge number of quantum theories that can be constructed in this way. Many of them can be both relativistically invariant and cluster separable. It is even possible to make such theories finite, i.e., avoid the need for renormalization.

4. In my understanding, QFT is just a subset of general Fock-space quantum theories from point 3. The distinctive property of QFT is that their interactions (in the Hamiltonian) are build as polynomials of quantum fields -- which are quite specific linear combinations of particle creation and annihilation operators. These linear combinations are carefully chosen (see Weinberg) to ensure the relativistic invariance and the cluster separability of the Hamiltonian. The huge disadvantage is that all QFT Hamiltonians have built-in self-interactions of particles. This is the reason for the divergences. One can suppress divergences in the S-matrix by renormalization, but this makes the Hamiltonian divergent and ill-defined.

All examples 1.-4. are versions of the generic quantum theory. So, all basic quantum ingredients are present there: wave functions, superpositions, unitary time evolution, particle observables (position, momentum) uncertainty relationships, etc.

Eugene.

jordi

Yes, I agree with this point of view. For this reason, it is strange that usually QM and QFT are taught as different animals. For sure there are analogies, but it was never clear to me that QM could be understood as a QFT.

In fact, there is an even bigger analogy: statistical mechanics and QFT share the same "generating functional" / correlations framework. For sure, statistical mechanics has real fields, so correlations are natural there, and QFT has Hilbert spaces, so the "Green's functions" need to be identified as correlations.

I am surprised that this conceptual unity is not emphasized more.

vanhees71

Gold Member
The way how hermitian operators (position, momentum, total energy) and unitary opetators (time evolution, spatial rotations) are handled in QFT, as well as how you represent a composite system as a tensor product of its components, is exactly the same as in QM. It's just easier to practice that formalism with nonrelativic QM where the infinite-dimensionality is only in the state vector space and not also in the classical phase space of the physical system.
The point is that in QFT you have to construct the self-adjoint (hermitean is not enough!) position, momentum, and angular-momentum operators from the field algebra. To that end you analyze the space-time symmetries and use the corresponding relations to conservation laws via Noether's theorem. For non-relativistic and massive relativistic particles that's straight forward, while for massless particles you find that one cannot define position observables for spin greater or equal 1.

atyy

I think its nice to have the traditional QM formalism before seeing it reformulated as QFT, because strictly speaking QFT is about an infinite number of particles (or whatever), but maybe our theories should work for large but finite number of "things". The analogy is that it is nice to have the classical theory of a few particles, even though it is only the thermodynamic limit of classical statistical physics which has phase transitions, since it everyday life we think we have a finite number of particles and no true phase transitions.

jordi

I agree that "its nice to have the traditional QM formalism before seeing it reformulated as QFT". In fact, learning that two things that you thought were separated, and you have learned very much each of them, they are in fact "the same" (or one is a particular case of the other), is intellectually appealing. This "flash" is a burst of endorphins. And people that ask "why?" like this burst.

But one thing is this, and the other is to keep it is a mystery forever, unless one reads Greiner or a similar book. I think this fact should be stated initially, but only explicitly stated and proven once the student has learned QM "alone".

But say at the beginning, for God's sake!

vanhees71

Gold Member
I think its nice to have the traditional QM formalism before seeing it reformulated as QFT, because strictly speaking QFT is about an infinite number of particles (or whatever), but maybe our theories should work for large but finite number of "things". The analogy is that it is nice to have the classical theory of a few particles, even though it is only the thermodynamic limit of classical statistical physics which has phase transitions, since it everyday life we think we have a finite number of particles and no true phase transitions.
QFT is not about an infinite number of particles. In contradistinction to the first-quantization formalism, which by construction deals only with situations where particle number is conserved, is applicable in cases where the particle number is not conserved.

A. Neumaier

In my understanding, QFT is just a subset of general Fock-space quantum theories
But there are many examples of QFTs (e.g., conformal field theories in 1+1 dimension) whose Hilbert space does not have such a Fock space structure. The latter is specific to the nonrelativistic case and to free relativistic theories.

A. Neumaier

The point is that in QFT you have to construct the self-adjoint (hermitean is not enough!) position, momentum, and angular-momentum operators from the field algebra. To that end you analyze the space-time symmetries and use the corresponding relations to conservation laws via Noether's theorem. For non-relativistic and massive relativistic particles that's straight forward, while for massless particles you find that one cannot define position observables for spin greater or equal 1.
But in a quantum field theory, symmetry principles give no position operator. They only give total momentum, and angular-momentum operators, not even (the nonexisting) momentum operators for single particles!

A. Neumaier

strictly speaking QFT is about an infinite number of particles (or whatever)
Strictly speaking, QFT is about an infinite number of degrees of freedom.