Where do the choice of basis in QFT come from?

Question

Hi physics stack exchange.

I am new and a mathematicians - so go easy on me.

I have been trying to read up on QFT in the book Diagrammatica: The Path to Feynman Diagrams, and I have a question.

The construction there seems very dependent on the choice of basis. Indeed, there is often an incomming basis |0>,|p>,|p'>,|pp'>,... and a similar outgoing basis. Also when considering spin (and anti) structures it fixes some basis elements related to the representation of $\mathbb{R}^4$ to "fatten up" the basis for the Hilbert space.

My question is: Where does this choice of basis come from?

In this possibly related question it seems that the choice of basis maybe related to Eigen-vectors for the Hamiltonian, but maybe I misunderstood that question - indeed I don't understand the question but the answer seems to suggest this to me.

Now.. None of the two bases (plural basis?) seems to be Eigen-spaces for the Hamiltonian, but is it then true that the incoming basis is given by a trival Hamiltonian (by which I probably means something like purely kinetic energy), which corresponds to no interactions (and similar for the outgoing basis)? Does this even make sense? I can't make this compatible in my head with the fact that e.g. the Hamiltonian in the simplest $\sigma-\pi$-model is $2\sigma \pi^2$ does not seem trivial at time $\pm$infinity - indeed it seems time independent.

I have probably misunderstood something simple that makes some of this gibberish, and I would appreciate any clarifications.

or in other words; are all basis in QFT born equal as they do in QM? - i like this question, because choice of basis has deep implications for perturbative expansions, but those are not nearly made explicit enough in many QFT texts — lurscher, Jul 18 '11 at 20:04

score 4 · Answer 1 · answered Jul 18 '11 at 22:31

The basis at $t=-\infty$ and the basis at $t=+\infty$ are both the free field basis. The free field Hamiltonian has a kinetic energy term proportional to the square of the temporal gradient of the field, but it also has a potential energy term proportional to the sum of the squares of the spatial gradients of the field, and, in the case of $\mathrm{mass} > \mathrm{zero}$, a potential energy term proportional to the square of the amplitude of the field.

Perturbation theory calculates how a given state of the free field is modified by an additional interaction term (which after regularization and renormalization can be loosely thought of as infinitesimal) that acts for an infinite length of time, but that does not act at $t=-\infty$ or at $t=+\infty$. The interaction is modeled, at least in elementary textbooks, as being turned on and off by a smooth function $g(t)$ that we take to satisfy $g(t)=1$ for $|t|<T$ for some large time $T$, which is taken not to break translation invariance when we take the limit $T\rightarrow \infty$, although of course for $T<\infty$ it does. There are numerous other limits that have to be taken, all of which has to be done "right", which is the point of regularization and renormalization. Calculating what the eigenstates of the perturbed Hamiltonian are is an intractable problem, so we work with the eigenstates of the free field Hamiltonian, which are tractable — indeed they are elementary by comparison to the eigenstates of the perturbed Hamiltonian.

score 4 · Answer 2 · answered Jul 18 '11 at 23:01

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It works roughly as follows. Inspired by quantum mechanics, we'd like to find a space similar to $L^2({\mathbb R})$ that contains enough structure so that we can represent operators $\hat X$ and $\hat P$ (understood as elements of the Lie algebra of the Heisenberg group) on it as multiplication and derivation operators. One can generalize this to $n$ degrees of freedom by working on $L^2({\mathbb R}^n)$ and this also gives a clue about what to do for fields (by approximating those fields as piecewise constant configurations). Note that there is no discussion of basis whatsoever here due to the Stone-von Neumann theorem which claims that every two (strongly continuous) representations of the Heisenberg group are equivalent, i.e. it doesn't matter what basis one chooses.

This is unfortunately not the case for fields. I'll not provide you with complete derivation here (which would be pedagogical but unfortunately too long) but instead I'll just state the result: the right space to work in is the space of $M \equiv L^2(X, \mu_{C, \eta})$ where $X$ is a suitable topological space (e.g. think of it as continuous functionals on ${\mathbb R}^d$ with some standard topology) and $\mu_{C, \eta}$ is a Gaussian measure (e.g. on a Borel $\sigma$-algebra associated to the space $X$) with covariance $C$ and expectation $\eta$. The physical significance of $C$ is that it is (naively) equal to $\Omega^{-1} / 2$ where $\Omega$ is a one-particle Hamiltonian and the physical significance of $\eta$ is that it gives a vacuum expectation value of the field $\hat \Phi(x)$.

Now the fun part starts. One can pick a representation of the field operators $\hat \Phi(x)$ and $\hat \Pi(x)$ on $M$ in the following way $\hat \Phi(x) \leftrightarrow \Phi(x) + \eta(x)$, $\hat \Pi(x) \leftrightarrow -i{ \delta \over \delta \Phi(x)} + \Omega^{-1}(\Phi(x) - \eta(x))$, find a vacuum state $V[\Phi]$ (of the $\Omega$ operator) and construct creation and annihilation operators in the usual fashion and eventually build a Fock space $F(V)$. Now consider a shift operation on the space of functionals $T_{\Psi} : \Phi(x) \mapsto \Phi(x) - \Psi_x$. It turns out that if $\Psi(x)$ is non-regular enough then $T_{\Psi} F(V)$ will be orthogonal to the original Fock space $F(V)$ (contrast this with the quantum mechanical case where the shift by $a$ is realized by the unitary operator $\exp(-ia \hat P)$).

Similarly, it is possible to find a Fock space associated to some other covariance operator $C'$ (or equivalently one-particles Hamiltonian $\Omega'$) that is orthogonal to $F(V)$. So we see that in contrast with quantum mechanics where it was enough to work on one space $L^2({\mathbb R}^n)$ and disregard things like choices of basis or even Hamiltonian until necessary, this is no longer the case when working with fields. The space of continuous functionals is simply too big and carries a huge amount of unitarily inequivalent representations (described as Fock spaces).

answered Jul 18 '11 at 23:01

Marek

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Rereading your question, I guess you were just asking where does the Fock space come from. In that case, the answer is that it comes from the Hamiltonian (the states $\left | p, \sigma \right>$ are its eigenstates representing particles with momentum $p$ and spin $\sigma$), that is generated from the vacuum by creation and annihilation + an arbitrary choice of the vacuum expectation value. But you won't see any discussion of the expectation value in any introductory course. Its importance lies in the possibility of naturally accomodating things like spontaneous symmetry breaking. – Marek Jul 18 '11 at 23:13
This sounds very interesting, but I don't understand it. When you write (e..g think of this...) do you mean think of X or M? Is some of this related to sections in a complex line bundle on $\mathbb{R}^d$ with some connection form and curvature? If $\Phi$-hat is a field, what is $\Phi$ then? I feel that I am missing some definitions.. – Thomas Kragh Jul 19 '11 at 00:59
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@Thomas: seems you are not familiar even with the quantum mechanics then; that is actually a big problem when trying to study QFT. But here goes: in QM you have a position $x \in {\mathbb R}^n$, an operator $\hat X$ that acts on $L^2({\mathbb R^n})$ and finally a wave function $\delta_x$ that is a (generalized) eigenstate of $\hat X$ corresponding to the eigenvalue $x$. Similarly with fields, you have operators $\hat \Phi(x)$ and their eigenstates $\widetilde \Phi_x[\Phi]$ (represented as a wave functional) corresponding to eigenvalues $\Phi(x)$. (cont.) – Marek Jul 19 '11 at 06:00
(cont.) Intuitively one can understand this by thinking about the field as a generalization of quantum mechanical system with $n$ degrees of freedom represented by a collection of $n$ $\hat X_i$ and $\hat P_j$ operators to an uncountable collection of $\hat \Phi(x)$ and $\hat \Pi(y)$ operators indexed by position. Of course, this is just a naive outlook because infinite dimensional spaces (which replace here ${\mathbb R}^n$ configuration space from the QM) are far more complicated, as we know from functional analyses. The formal treatment goes along the lines I described. – Marek Jul 19 '11 at 06:06
I am actually familiar with QM just not using the notation you apparently do in physics, but thanks for clarifying the hat-notation. – Thomas Kragh Jul 19 '11 at 13:36
@Thomas: ah, that's good then; sorry if I sounded disrespectful. And by the way, what notation do you use instead of hats? Personally I don't use them much either but sometimes they help distinguish similar objects as it is often the case that all of operator, its eigenvalues and eigenstates are denoted by the same symbol which can get confusing at times. Moreover, the operator can be involved both abstractly and in some representation and so can the states. – Marek Jul 19 '11 at 13:48
I have seen a lot of different very abstract definitions with operators satisfying the correct commutation relations, but one that comes to mind is $Q_i$ and $P_j$ for the operators associated to $q_i$ and $p_j$ in cotangent bundles, maybe this is because I am in symplectic topology - I think what confused me the most (because I did not know it) was this using of the same name for operators and eigen values. – Thomas Kragh Jul 19 '11 at 14:23
I am still wondering about the $L^2$ - just to be clear is the $X$ in your answer infinite dim - or were you referring to $L^2(X,\mu)$? Also if the operator has several eigenvalues and states how is this put into the notation? Indeed, in your representation above - which eigenvalue is used if $\Phi(x)$ has several? – Thomas Kragh Jul 19 '11 at 14:23
@ThomasKragh let us continue this discussion in chat – Marek Jul 19 '11 at 14:26

score 2 · Answer 3 · answered Jul 19 '11 at 14:37

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The basis for QFT is normally the eigenvectors of some exactly solvable problems, free particles, as example. These eigenvectors (in scattering problems) are exact ones in asymptotic regions where the interaction is negligible. The latter is not always the case so the one encounters mathematical and conceptual problems.

These basis eigenvectors can describe bound states too, see atom-atomic scattering in atomic physics.

answered Jul 19 '11 at 14:37

Vladimir Kalitvianski

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@ Vladimir Kalitvianski-Asymptotic states are eigenstates of free theory or they are the dressed particles? – SRS Dec 03 '16 at 08:40
Both at the same time. I explain it on atomic collision phenomena. Consider two atoms in some initial state. Each atom is a compound system of interacting particles, but it ca be and is represented as a direct product of the center of mass free motion and the internal motion. In other words, separated variables of a compound system have their own "independent " equations similar to free (non interacting) equations. After collision the atoms can be found again in some states for which we can use the separated variables. – Vladimir Kalitvianski Dec 03 '16 at 09:33
An excited atom in the final state is similar to QED electron accompanied with soft photons in the final state. There are indications that photon oscillators are "internal degrees of freedom" of a compound system called (by me) electronium. – Vladimir Kalitvianski Dec 03 '16 at 09:33

score 1 · Answer 4 · answered Jul 19 '11 at 13:32

I don't quite understand what you are asking, but I interpret your question to be the following:

Is the construction of the Fock space $F(V)$ via multi-particle states $|0\rangle, |n_1\rangle, |n_1n_2\rangle, \dots$ independent of the basis? In other words, does the Fock space depend on the choice of basis for the single particle Hilbert space, $V \simeq \bigoplus_n \mathbb C|n\rangle$?

The answer is "no".

It is often very convenient to choose a basis of the single particle Hilbert space and use it for labeling states from the Fock space. For instance, the notation $|n_1,n_2\rangle$ is often denotes a state with $n_1$ particles in the (single-particle) quantum state $|1\rangle$ and $n_2$ particles in quantum state $|2\rangle$.

However, choosing a basis is not mandatory for constructing the Fock space. It's simply the direct sum of the tensor powers of $V$,

$$ F(V) = \bigoplus_{n=1}^\infty S(V^{\otimes n}) ,$$

where $S$ denotes symmetrization (for bosons) or antisymmetrization (for fermions). No choice of basis to be seen anywhere.

Example: for $n_1=1$, $n_2=1$, we have

$$ |n_1,n_2\rangle = \frac{1}{\sqrt{2}}(|1\rangle\otimes|2\rangle - |2\rangle\otimes|1\rangle)$$

As you can see, the notation on the left-hand side depends on a choice of basis, but the expression on the right-hand side gives it a basis-independent meaning.

This is correct but it should be noted that physicists usually work in a bigger space than the Fock space (in a space of fields, for whatever it means precisely) of which Fock space is some subspace that describes the particle approximation and then there is a choice involved. In particular, the Fock space always has a countable basis (as it is spanned by states representing finite number of particles) while sometimes we also need spaces with an uncountable basis that contain states with an infinite number of particles (which are better thought of as genuine fields). — Marek, Jul 19 '11 at 13:54
@Marek: Ah, right. For instance, the coherent states $|\phi\rangle = e^{\sum a^+_i \phi_i} |0\rangle$ used for the field integral have an infinite number of particle. Isn't this related to some theorem by Haag, which I don't know much about? — Greg Graviton, Jul 19 '11 at 17:40
I don't know anything about axiomatic QFT (besides its existence and that physicists ignore it more or less completely). I wish I would though :) — Marek, Jul 19 '11 at 17:44

Ron Maimon · Answer 5 · 2011-08-14T23:34:23.067

I think that the simple thing that the OP missed is that infinite plane wave states are always non-interacting. No matter how big the cross section for collisions is, if particles are smeared out over all space, they do not ever collide. This is the reason you can choose "in" and "out" free states to parametrize the collisions.

The actual collisions describe what happens when a localized superposition of asymptotic states collides with another localized superposition of other asymptotic states. This is reflected in the fact that the subleading corrections to the S-matrix are small, not in the perturbation theory sense, but in the sense of writing S=I + i A , where A is the collision amplitude.

When you are using a plane wave with standard relativistic normalization $<f|I|i>$ is divergent for in=out, because the normalization is infinite. On the other hand, $<f|A|i>$ is finite, which reflects the small probability of particles finding each other. The results become finite when you make wave-packets, because you smear in and out, and the delta function when in=out just gives you the intensity of collisions based on the width of the beam.

score 0 · Answer 6 · edited Apr 13 '17 at 12:40

In a lot of physics experiments, one arranges for initial and final states and these conditions are what one uses for the basis states. For example, if an electron is detected with momentum $p'$, then one of the basis states in the theoretical explanation for the experiment will be $|p'\rangle$.

An example of this is the CKM matrix, which is a unitary transformations between two basis sets, each with three elements. One of the bases consists of the quarks with charge +2/3, i.e. the up, charm, and top quarks. The other basis is the charge -1/3 quarks, the down, strange and bottom quarks. In this case, again mother nature chose the bases for us.

More generally, there are situations where experiment doesn't choose the basis set. For two electrons with anti-correlated spin-1/2, our choice of basis (i.e. spin direction) is arbitrary and choosing different bases does not change the mathematics or the prediction.

score 0 · Answer 7 · answered Aug 17 '11 at 05:57

There's no preferred field coordinatization in quantum field theory. In fact, it has been shown there are infinitely many Borchers classes possible in algebraic quantum field theory. Even using path integrals, what your professor didn't tell you was fixing the path integral measure to be translationally invariant, and the leading kinetic term to be diagonal doesn't fix the field coordinatization at all, once you start to consider all the possible Wilson terms you can add. The only invariant is the S-matrix. If you don't trust me, meditate upon quantum field theory in curved spacetime. Now, there's really no canonical field coordinatization. The idea of a canonical field coordinatization is a prejudice of theoretical physicists. They like to think there's a canonical coordinatization for everything, something God-given and objectively better than all the other choices with no room for arbitrariness. Guess what, their prejudice is very mistaken. There always has to be an arbitrary choice imposed from the outside.

Where do the choice of basis in QFT come from?

7 Answers7