What is the fundamental probabilistic interpretation of Quantum Fields?

Question

In quantum mechanics, particles are described by wave functions, which describe probability amplitudes. In quantum field theory, particles are described by excitations of quantum fields. What is the analog of the quantum mechanical wave function? Is it a spectrum of field configurations (in analogy with QM wave functions' spectrum of particle observables), where each field configuration can be associated with a probability amplitude? Or is the field just essentially a superposition of infinitely many wave functions for each point along the field (as if you quantized a continuous mattress of infinitesimal coupled particles)?

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user user1247

Answer 1

It seems you are asking "of what variables is the wave function in quantum field theory a function of?". However, this question doesn't have a unique answer; in contrast with your implicit assertion, it doesn't have a unique answer in ordinary quantum mechanics of particles, either.

The wave function is a set of complex numbers that determine the complex coefficients in front of basis vectors of a particular basis (or a "continuous basis") of the Hilbert space. So quantum mechanics' wave function may be encoded in $\psi(x)$; but it may also be described in the momentum representation, $\tilde \psi(p)$ which is the Fourier transform of $\psi(x)$.

We may also choose different eigenvectors, a different basis. For example, the harmonic oscillator (and many other systems) has a discrete set of energy eigenstates labeled by $n=0,1,2,3,\dots$. In that case, the coefficients $a_n$ completely encode the state vectors – the wave function – as well.

There is a hugely infinite number of possible bases, and even a very large number of bases that are commonly used.

In quantum field theory, the most natural basis (or the most widely used one) is the basis analogous to the harmonic oscillator energy eigenstate example. For each type of a quantum field, each polarization, and each $\vec k$ (the wavenumber, i.e. the frequency and the direction of motion), we define the contribution of this mode of the quantum field to the non-interacting part of the energy. This turns out to be just a rescaled harmonic oscillator with the spectrum $E=n\hbar\omega$ where $n$ is a non-negative integer that may be interpreted as the number of particles.

In this basis, to specify the wave function, we need to determine the complex amplitude in front of the $n_i=0$ "vacuum state", in front of the numerous $n_i=(0,0,0,1,0,0,...)$ states which is equivalent to the non-relativistic wave function (e.g. of positions or, more often, momenta) $\tilde \psi(p_1)$, the complex numbers determinining the complex amplitudes for all two-excitation eigenstates $\tilde \psi(p_1,p_2)$, and so on, indefinitely (increasing number of particles).

However, it's also true that one may choose a "functionally super continuous" basis of all configurations of the quantum fields. In this language, the state vector is a functional $$ \Psi [\phi(x,y,z), A_i(x,y,z),\dots] $$ which depends on all the fields at $t=0$ but not their time derivatives. A functional is morally equivalent to a function of infinitely many variables (continuously infinitely many, at least in this case). It's not terribly practical to work with such a description of the wave function and it's hard to "measure" the values of the functional at different points (for different configurations of the fields) but it is possible.

The particle-occupation basis mentioned above is more physical because Nature evolves according to the Hamiltonian so all states (objects and their configurations) that tend to be at least a little bit "lasting" are close to energy eigenstates. That's why bases that are close to bases of energy eigenstates (or eigenstates of a "free" part of the Hamiltonian etc.) are generally more useful and natural to work with, especially when one thinks about applications.

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user Luboš Motl

Answer 2

(Lubos just posted an answer, but I think this is sufficiently orthogonal to post too). The usual wavefunction for a bosonic field is a complex number for each field configuration at all points of space:

$$\Psi(\phi(x))$$

This wavefuntion(al) obeys the Schrodinger equation with the field Hamiltonian, where the field momentum is a variational derivative operator acting on $\Psi$. This formulation is fine in principle, but it is not useful to work with this object directly under usual circumstances for the following reasons:

• You need to regulate the field theory for this wavefunctional to make mathematical sense. If you try to set up the theory in the continuum right from the beginning, to specify a wavefunction over each field configuration you need to work just as hard as to do a rigorous definition for the field theory. For example, just to normalize the wavefunction over all constant time slice field values, you need to do a path integral over all the constant time field configurations. This is a path integral in one dimension less, but the thing you are integrating is no longer a local action, so there is no gain in simplicity. Even after you normalize, the expectation value of operators in the wavefunctional is a field theory problem in itself, in one dimension less, but with a nonlocal action.
• Once you regulate on a lattice, the field wavefunctional is just an ordinary wavefunction of all the field values at all positions. But even when you put it on an infinite volume lattice, a typical wavefunctional in infinite volume will have a divergent energy, because you will have a certain energy density at each point when the wavefuntional is not the vacuum, a finite energy density. Infinite energy configurations of the field theory, those with a finite energy density, are very complicated, because they do not decompose into free particles at asymptotic times, but keep knocking around forever.
• The actual equations of motion for the wavefuntional are not particularly illuminating, and do not have the manifest Lorentz symmetry, because you chose a time-slice to define the wavefunction relative to.

These problems are overcome by working with the path integral. In the path integral, if you are adamant that you want the wavefuntion, you can get it by doing the path integral imposing a boundary condition on the fields at a certain time. But a path integral Monte-Carlo simulation, or even with just a little bit of Wick rotation, will make the wavefunction settle to be the vacuum, and insertions will generally only perturb to finite energy configurations, so you get the things you care about for scattering problems.

Still the wavefunction of fields is used in a few places for special purposes, although, with one very notable exception, the papers tend to be on the obscure side. There are 1980s papers which attempted to find the string formulation of gauge theories which tried to work with the field Hamiltonian in the Schordinger representation, and these were by famous authors, but the name escapes me (somebody will know, maybe Lubos knows immediately).

The best example of where this approach bears fruit is when the reduction in dimension gives a field theory which has a relationship with known solvable models. This is the example of the 2+1 gauge vacuum, which was analyzed in the Schrodinger representation by Nair and collaborators in the past decade.

A recent paper which reviews and extends the results is here: http://arxiv.org/PS_cache/arxiv/pdf/1109/1109.6376v1.pdf . This is, by far, the most significant use of Schrodinger wavefunctions in field theory to date.

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user Ron Maimon

Answer 3

In nonrelativistic quantum mechanics you can think (modulo technicalities with rigged Hilbert spaces) of the coordinate representation of the wavefunction as the projection of the state vector $\Psi$ in the "direction" of the position eigenvector, i.e. $\Psi(x)=\langle x|\Psi \rangle$.

Quantum field theory is traditionally (in textbooks) formulated in the interaction picture. It can, however, also be formulated in the Schroedinger picture, in which a natural notion of wavefunction emerges. Instead of being a complex valued function on the space of particle positions, it now becomes a complex valued functional on the space of field configurations. So the wavefunction(al) is the projection of a Schroedinger picture state $\Psi$ in the "direction" of a field configuration $\phi(\mathbf{x})$

$|\Psi(t) \rangle = \int \cal{D}\phi \Psi[\phi,t]|\phi \rangle$

It satisfies the functional Schroedinger equation

$i\frac{\partial}{\partial t} \Psi [\phi ,t] = H(\phi (\mathbf{x}), -i\frac{\delta}{\delta \phi (\mathbf{x})})\Psi [\phi ,t]$

The only textbook reference treating this that I know of is Brian Hatfield's book.

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user twistor59

Comments

D'oh I see some more detailed answers appeared while I was typing mine....

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user twistor59

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Maybe a less detailed answer, but I think you most directly address my intended question. Thanks! I will check out the Hatfield book.

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user user1247

Answer 4

Ron and Luboš's Answers, +1. FWIW, however, I have found it worthwhile to take QFT to be a stochastic signal processing formalism in the presence of Lorentz invariant (quantum) noise.

The devil is in the details, and I cannot claim to be able to say much, or even anything, about interacting quantum fields, but it is possible to construct random fields that are empirically equivalent, in a specific sense, to the quantized complex Klein-Gordon field (EPL 87 (2009) 31002, http://arxiv.org/abs/0905.1263v2) and to the quantized electromagnetic field (http://arxiv.org/abs/0908.2439v2, completely rewritten a few weeks ago). Needless to say, the fact that a random field satisfies the trivial commutation relation $[\hat\chi(x),\hat\chi(y)]=0$ instead of the nontrivial commutation relation $[\hat\phi(x),\hat\phi(y)]=\mathrm{i}\!\Delta(x-y)$ plays out in numerous ways, and this is not for anyone who wants to stay in the mainstream.

I take it as a significant ingredient that we treat quantum fields as (linear) functionals from a Schwartz space $\mathcal{S}$ of window functions into a $\star$-algebra $\mathcal{A}$ of operators, $\hat\phi:\mathcal{S}\rightarrow\mathcal{A};f\mapsto\hat\phi_f$, instead of dealing with the operator-valued distribution $\hat\phi(x)$ directly, even though we may construct $\hat\phi_f$ directly from $\hat\phi(x)$, by "smearing", $\hat\phi_f=\int\hat\phi(x)f(x)\mathrm{d}^4x$. In terms of these operators, the algebraic structure of the quantized free Klein-Gordon field algebra is completely given by the commutator $[\hat\phi_f,\hat\phi_g]=(f^*,g)-(g^*,f)$, where $(f,g)=\int f^*(x)\mathrm{i}\!\Delta_+(x-y)g(y)\mathrm{d}^4x\mathrm{d}^4y$ is a Hermitian (positive semi-definite) inner product. A window function formalism is part of the Wightman axiom approach to QFT. Note that what are called "window functions" in signal processing are generally called "test functions" in QFT. The signal processing community works with Fourier and other transforms in a way that has close parallels with quantum theory and implicitly or explicitly uses Hilbert spaces.

In the vacuum state of the quantized free Klein-Gordon field, we can compute a Gaussian probability density using the operator $\hat\phi_f$, $$\rho_f(\lambda)=\left<0\right|\delta(\hat\phi_f-\lambda)\left|0\right>=\frac{e^{-\frac{\lambda^2}{2(f,f)}}}{\sqrt{2\pi(f,f)}},$$ which depends on the inner product $(f,f)$ [take the Fourier transform of the Dirac delta, use Baker-Campbell-Hausdorff, then take the inverse Fourier transform]. The reason it's good to work with smeared operators $\hat\phi_f$ instead of with operator-valued distributions $\hat\phi(x)$ is that the inner product $(f,f)$ is only defined when $f$ is square-integrable, which a delta function at a point is not. We can compute a probability density using an operator $\hat\phi_f$ in all states, but, of course, we can only compute a joint probability density such as $$\rho_{f,g}(\lambda,\mu)=\left<0\right|\delta(\hat\phi_f-\lambda)\delta(\hat\phi_g-\mu)\left|0\right>$$ if $\hat\phi_f$ commutes with $\hat\phi_g$; in other words, whenever, but in general only when, the window functions $f$ and $g$ have space-like separated supports. At space-like separation, the formalism is perfectly set up to generate probability densities, which is why QFT is like stochastic signal processing, but at time-like separation, the nontrivial commutation relations prevent the construction of probability densities.

I take it to be significant that the scale of the quantum field commutator, the imaginary component of the inner product $(f,g)$, is the same as the scale of the fluctuations in the probability density $\rho_f(\lambda)$, determined by the diagonal component, $(f,f)$. It's possible, indeed, to construct a quantum field state in which the two scales are different (Phys. Lett. A 338, 8-12(2005), http://arxiv.org/abs/quant-ph/0411156v2), so the equivalence could be thought as surprising as the equivalence of gravitational and inertial mass.

Needless to say, there many people who are working away at QFT. The approach I've outlined is only one, with one person working on it, in contrast to string theory, supersymmetry, noncommutative space-time geometry, etc., all of which have had multiple Physicist-decades or centuries of effort poured into them. It's probably better to follow the money. Also, please note that I have hacked this out in an hour, which I have done because rehearsal is always good. Did anyone read all of this?

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user Peter Morgan

Comments

This is the description of the vaccuum state of a free field theory, which is always given by a Gaussian wavefunctional, and in the case of the electromagnetic field dates back to Schwinger (or Feynman, I think only Schwinger gave it explicitly). The commutation relation for stochastic fields is not trivial, even though they are classical-looking. I find the most interesting versions are the stochastic 2d fields which reproduce the N=2 SUSY theory by Nicolai map. This is an active field now.

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user Ron Maimon

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's true, Ron. To construct a nontrivial Lorentz invariant vacuum state for a random field that has trivial commutators for the observables $\hat\chi(x)$ in a quantum field formalism requires nontrivial commutators for creation and annihilation operators, which is what makes the connection into QFT. Such an approach effectively puts the dynamical evolution and the description of measurements into the state instead of into the algebra ($[\hat\chi(x),\hat\chi(y)]=0$ gives no information, so information has to come from some other part of the structure).

This post imported from StackExchange Physics at 2015-04-11 10:30 (UTC), posted by SE-user Peter Morgan