Second Quantization vs Many-Particle QM

[stevendaryl]

Apparently, there are two different routes to get to quantum field theory from single-particle quantum mechanics: (I'm going to use nonrelativistic quantum mechanics for this discussion. I think the same issues apply in relativistic quantum mechanics.)

Route 1: Many-particle quantum mechanics
Start with single-particle QM, with the Schrodinger equation: $- \frac{1}{2m} \nabla^2 \psi = i \frac{\partial}{\partial t} \psi$

1. Now, extend it to many (initially, noninteracting) particles: $\Psi(\vec{r_1}, \vec{r_2}, ..., \vec{r_n}, t)$
2. Introduce creation and annihilation operators to get you from (a properly symmetrized) $n$-particle state to an $n+1$-particle state, and vice-verse.

Route 2: Second quantization
Once again, start with the single-particle wave function.

1. Instead of viewing $\psi$ as a wave function, you view it as a classical field.
2. Describe that field using a Lagrangian density $\mathcal{L} = i \psi^* \dot{psi} - \frac{1}{2m}|\nabla \psi|^2$
3. Derive the canonical momentum using $\pi = \dfrac{\partial}{\partial \dot{\psi}}$
4. Impose the commutation rule: $[\pi(\vec{r}), \psi(\vec{r'})] = -i \delta^3(\vec{r'} - \vec{r})$

Conceptually, these routes are very different. The first one is just many-particle quantum mechanics re-expressed in terms of creation and annihilation operators. The second is field theory in which the field is quantized. Is it just a coincidence that the result is the same, or is there some deeper reason?

Maybe it's inevitable because there is only one QFT possible for free nonrelativistic fields satisfying Galilean symmetry. But it still seems strange, because the first route doesn't start with a general principle about commutation relations; those rules are just consequences of the way creation and annihilation operators work on Fock space, together with the symmetry/anti-symmetry rules for identical particles. It doesn't presuppose any general commutation rule relating a field to its canonical momentum.

 

[atyy]

Wave-particle duality :)

In the case of QM, I don't know why this coincidence occurs.

But if one puts the particles on a lattice, then considers vibration modes of the lattice, the modes are wave like although the lattice is particle-like. This is Fourier series, which is conceptually close to the Fourier transform, which is closely related to wave-particle duality.

Edit: It could just be a periodic potential for Fourier series. Then one could use Bloch states and Wannier states, which are like Fourier counterparts. Then the question is how one would generalize to a periodic potentials.

 

[atyy]

Loosely speaking, the first quantized QM must be about many identical particles if it is to have a second-quantized counterpart.

In a way, the identicality of the particles is often said to be due to their being excitations of the same underlying field.

From that point of view, what is amazing is that relativistic QM is not a complete theory, whereas non-relativistic QM is a complete theory.

 

[HomogenousCow]

In a way, the identicality of the particles is often said to be due to their being excitations of the same underlying field.

From that point of view, what is amazing is that relativistic QM is not a complete theory, whereas non-relativistic QM is a complete theory.

Could you elaborate on this?

 

[Demystifier]

Is it just a coincidence that the result is the same, or is there some deeper reason?

It seems to me that it's a kind of coincidence. Indeed, the result is not the same for Fermi-Dirac statistics, which, before introducing any creation-destruction operators, means - antisymmetric wave function. For this case the "natural" canonical commutation relations for fields do not lead to the right result, so one must "artificially" replace canonical commutation relations with canonical anti-commutation relations.

An even greater mystery for me is why functional integrals for fermionic fields work? By starting from canonical field quantization for bosonic fields, one can derive the path-integral formulation. However, I have never seen a similar derivation for fermionic path integrals. Instead, one makes axioms for Grassmann integrals, which eventually turn out to give the right result, but it is not at all obvious (at least to me) that it should lead to the right result before checking it explicitly by deriving the Feynman rules.

 

[Jazzdude]

I believe at the core of this "coincidence" lies the interesting connection between the ladder operator algebra and the Heisenberg algebra of canonical conjugate observables. This connection seems to be very fundamental and is usually studied in the context of supersymmetry and the Wigner-Heisenberg algebra. I'm by no means an expert in this field, but it seems to me that there's a lot of insight to be gained from this question.

Cheers,

Jazz

 

[stevendaryl]

An even greater mystery for me is why functional integrals for fermionic fields work? By starting from canonical field quantization for bosonic fields, one can derive the path-integral formulation. However, I have never seen a similar derivation for fermionic path integrals. Instead, one makes axioms for Grassmann integrals, which eventually turn out to give the right result, but it is not at all obvious (at least to me) that it should lead to the right result before checking it explicitly by deriving the Feynman rules.

Yeah, definitely the path-integral way of doing Fermi fields seems like hocus pocus. I think that the fact that it gives the right answer is explained by the fact that they started with the answer and worked backwards to find the axioms.

 

[Demystifier]

I think that the fact that it gives the right answer is explained by the fact that they started with the answer and worked backwards to find the axioms.

Maybe, but I never seen such a presentation in a textbook.

 

[atyy]

Maybe, but I never seen such a presentation in a textbook.

I think it's presented this way in Wen, Quantum Field Theory of Many-body Systems.

"The path integral for fermions is a bookkeeping formalism ... the physical interpretation of the fermion path integral is unclear."

 

[Demystifier]

I think it's presented this way in Wen, Quantum Field Theory of Many-body Systems.

Not really, or at least not at the technical level. Technically, he first defines the Grassmann integral, and then derives the correlation functions. I have never seen a reversed approach, in which one starts from the correlation functions (or something else which we already know to be correct from the canonical approach) and then derives a rule for the Grassmann integral.

 

[atyy]

Not really, or at least not at the technical level. Technically, he first defines the Grassmann integral, and then derives the correlation functions. I have never seen a reversed approach, in which one starts from the correlation functions (or something else which we already know to be correct from the canonical approach) and then derives a rule for the Grassmann integral.

Actually, would it be fair to say that even the bosonic path integral is a bookkeeping device, although it has a "physical" interpretation. After all, in field theory, the Wick rotation and imaginary time is not physically justified, is it?

 

[vanhees71]

Apparently, there are two different routes to get to quantum field theory from single-particle quantum mechanics: (I'm going to use nonrelativistic quantum mechanics for this discussion. I think the same issues apply in relativistic quantum mechanics.)

Route 1: Many-particle quantum mechanics
Start with single-particle QM, with the Schrodinger equation: $- \frac{1}{2m} \nabla^2 \psi = i \frac{\partial}{\partial t} \psi$

1. Now, extend it to many (initially, noninteracting) particles: $\Psi(\vec{r_1}, \vec{r_2}, ..., \vec{r_n}, t)$
2. Introduce creation and annihilation operators to get you from (a properly symmetrized) $n$-particle state to an $n+1$-particle state, and vice-verse.

Route 2: Second quantization
Once again, start with the single-particle wave function.

1. Instead of viewing $\psi$ as a wave function, you view it as a classical field.
2. Describe that field using a Lagrangian density $\mathcal{L} = i \psi^* \dot{psi} - \frac{1}{2m}|\nabla \psi|^2$
3. Derive the canonical momentum using $\pi = \dfrac{\partial}{\partial \dot{\psi}}$
4. Impose the commutation rule: $[\pi(\vec{r}), \psi(\vec{r'})] = -i \delta^3(\vec{r'} - \vec{r})$

Conceptually, these routes are very different. The first one is just many-particle quantum mechanics re-expressed in terms of creation and annihilation operators. The second is field theory in which the field is quantized. Is it just a coincidence that the result is the same, or is there some deeper reason?

Maybe it's inevitable because there is only one QFT possible for free nonrelativistic fields satisfying Galilean symmetry. But it still seems strange, because the first route doesn't start with a general principle about commutation relations; those rules are just consequences of the way creation and annihilation operators work on Fock space, together with the symmetry/anti-symmetry rules for identical particles. It doesn't presuppose any general commutation rule relating a field to its canonical momentum.

First we have to get a bit more precise between your two routes to non-relativistic QFT (NRQFT). Route 1 is extending many-body quantum theory, expressed in terms of wave mechanics, i.e., the position representation for a single particle to the many-body case by introducing wave functions for a fixed number of particles. In position representation you have wave functions with $N$ position arguments for an $N$-particle system.

Another important ingredient is the notion of indistinguishability of of particles of the same type, i.e., particles which have the same intrinsic quantum numbers (which usually are mass and electric charge in atomic physics) cannot be individually followed in their dynamics, which implies that it should not be possible to distinguish indidual particles. This leads, after some careful analysis, to the conclusion that in 3 spatial dimensions there can only be bosons or fermions, i.e., the wave functions must be symmetric or antisymmetric under exchange of any two spatial arguments of indistinguishable particles in the system.

Route 2 is not different at all. All you do is to represent the states a bit differently. You introduce another Hilbert space, a socalled Fock space, where the particle number is not fixed a priori, and then the natural basis are occupation-number states with respect to a given single-particle basis. Thus this approach is a somewhat more general setup, because you can describe processes, where in principle interactions can change the particle number, i.e., the creation and destruction of particles. In many cases of NR QT the interactions are, however such that the particle number is conserved, and thus given by the initial condition of the dynamics. Then everything can be described in a subspace of the Fock space with a fixed total number of particles, and in this case the occupation-number representation and the position representation for a fixed number of particles are precisely the same theory.

However, NRQFT can be very convenient when describing many-body systems in terms of socalled quasiparticles, which are not conserved. This happens all the time in condensed-matter physics and is a concept that goes back to Landau when working out the theory of liquid helium (in his case first He4, i.e., bosons and superfluidity). There you find that the system can be described most elegantly not in terms of the real particles (electrons and nuclei) but in terms of the relevant collective excitations above the ground state of the many-body system. In thermal equilibrium these quantum collective excitations behave like an ideal gas dilute gas of quasiparticles, which can be destroyed and created in interactions. You end up with Feynman rules as in relativistic QFT. Only the meaning of these symbols are different.

The final question is not specific to the one or the other route (often misleadingly called "first and second quantization") to many-body NRQT. This is applied group theory, and in my opinion, the very basis of NRQT to begin with. Usually one doesn't do this in QM1, because it you need a quite long time to introduce the necessary mathematical concepts, namely the ray representations of groups in Hilbert space, and for NRQT it's even a bit more cumbersome than in relativistic QT, because the Galileo group has a somewhat different structure than the Poincare group. You can derive this in a very similar way as for the Poincare group, and it was done in both cases by Wigner. The conclusion is that in quantum theory the classical Galileo group is substituted by a quantum version, where in addition to the 10 generators of the original group (temporal and spatial translations (4 generators for energy and momentum), rotations (3 operators for angular momentum), and Galilei boosts (3 generators)) the mass occurs as an additional central charge of the group. Instead of the classical rotation group you have to use the covering group SU(2) to get the complete realizations. An elementary non-relativistic particle is then defined such that its quantum mechanics deals with a Hilbert space where a specific irreducible representation of this quantum-Galileo group is realized by the corresponding operator algebra. These representations are classified by the Casimir operators of the quantum Galileo group, i.e., mass and spin of the particle as in the relativistic case. The only difference is that for the Galileo group the mass is a central charge, while for the Poincare group the mass squared is a Casimir operator of the Poincare group, i.e., $p_{\mu} p^{\mu}$. Thus in non-relativistic QT you have a mass-superselection rule, which you do not have in the relativistic case. Also in NRQT particles with mass 0 do not lead to useful quantum dynamics.

Whether you describe the physics of a non-relativistic many-body system with fixed total particle number (implying that this total particle number is conserved and the corresponding operator thus commutes with the Hamiltonian) in the first or second quantization formalism doesn't matter, you always deal with the same representations of the quantum Galileo group.

For a good introduction to the group-theoretical aspects, see the textbook by Ballentine, Quantum Mechanics.

 

[king vitamin]

I'll add that the Fock space without particle number conservation is still useful in many-body physics when particle number is conserved, provided the Gibbs ensemble coincides with the canonical ensemble. You simply need to do the usual process in stat mech: work with a variable number of particles and a chemical potential, tune the chemical potential to set <N> to the actual number of particles in your system, and check that number fluctuations <N^2>/<N> vanish in the thermodynamic limit.

 

[vanhees71]

The grand canonical ensemble describes a subsystem of a closed larger system, where both energy and particles can be exchanged, e.g., by looking at a subvolume of the larger volume containing the closed system. The total particle number is still conserved since otherwise the equilibrium statistical operator couldn't be a function of it, for the statistical operator must obey the von Neumann equation of motion
$$\partial_t \hat{R}+\frac{1}{\mathrm{i} \hbar} [\hat{R},\hat{H}].$$
If $\partial_t \hat{R}$ (equilibrium condition), the statistical operator must commute with $\hat{H}$, i.e., it must be a function of the conserved quantities of the theory.

In relativistic many-body theory, you cannot use the particle number but net conserved charges (baryon number, electric charge,...).

 

[king vitamin]

The Gibbs ensemble doesn't always have that interpretation (consider the photon gas). I agree that particle number still commutes with the Hamiltonian (I did state things badly above), I was just trying to stress that it "fluctuates" in the usual time-independent sense, and is described by the multi-particle Fock space in stevendaryl's "Route 2."

My confusion now gets into how this is applied in cases where the hamiltonian does not conserve particle number. There was an interesting thread about this last year which you participated in: https://www.physicsforums.com/threads/grand-canonical-ensemble-n-operator-problem.776948/ .That topic got derailed with the subject of gauge invariance (on that topic, I'm most familiar with the Hansson, Oganesyan, Sondhi paper linked by atyy), but there didn't seem to be an answer for the question: If we are given a Hamiltonian which does not conserve N (say a certain variational Hamiltonian that BCS wrote down in 1957), and we do not even know anything about its derivation (whether it's gauge-fixed or we have integrated out other dynamical fields), we usually just plug this into the GCE. I can find this exact statement in some textbooks, but your argument above implies that this is wrong at some level. The closest to a resolution to this is in post #28 by DrDru; I'm not sure if I understand their first explanation, but I think their second involves the number non-conservation $[H,N]$ to vanish macroscopically. If this isn't correct, maybe I should start a new thread addressing this question.

 

[Demystifier]

Actually, would it be fair to say that even the bosonic path integral is a bookkeeping device, although it has a "physical" interpretation. After all, in field theory, the Wick rotation and imaginary time is not physically justified, is it?

The path integral can be calculated even without Wick rotation.

 

[vanhees71]

The Gibbs ensemble doesn't always have that interpretation (consider the photon gas). I agree that particle number still commutes with the Hamiltonian (I did state things badly above), I was just trying to stress that it "fluctuates" in the usual time-independent sense, and is described by the multi-particle Fock space in stevendaryl's "Route 2."

My confusion now gets into how this is applied in cases where the hamiltonian does not conserve particle number. There was an interesting thread about this last year which you participated in: https://www.physicsforums.com/threads/grand-canonical-ensemble-n-operator-problem.776948/ .That topic got derailed with the subject of gauge invariance (on that topic, I'm most familiar with the Hansson, Oganesyan, Sondhi paper linked by atyy), but there didn't seem to be an answer for the question: If we are given a Hamiltonian which does not conserve N (say a certain variational Hamiltonian that BCS wrote down in 1957), and we do not even know anything about its derivation (whether it's gauge-fixed or we have integrated out other dynamical fields), we usually just plug this into the GCE. I can find this exact statement in some textbooks, but your argument above implies that this is wrong at some level. The closest to a resolution to this is in post #28 by DrDru; I'm not sure if I understand their first explanation, but I think their second involves the number non-conservation $[H,N]$ to vanish macroscopically. If this isn't correct, maybe I should start a new thread addressing this question.

In the case of a photon gas, the Gibbs ensemble (grand canonical ensemble) has precisely this interpretation. To have radiation in thermal equilibrium you have to couple it to a heat bath, e.g., the walls of a cavity. There is no conserved number or charge-like quantity for photons. Thus there's no chemical potential for photons. The mean photon number is determined by the temperature of the walls and fluctuates.

If you don't have some conserved number- or charge-like quantity, you cannot introduce any (together with the corresponding chemical potentials) into the equilibrium statistical operator. Then all number-like quantities you may be able to still define (like photon number in the case of em. radiation) are determined by temperature and fluctuate.

There is no problem with gauge invariance in thermodynamics. In case of BCS theory (as in any Higgsed gauge theory) there is some sloppy talk about "spontaneous symmetry breaking of local gauge symmetries" although in fact there is none. This is important, because it explains that there are no Nambu-Goldstone bosons but massive gauge fields in such a theory. The most elegant way to see this is the path-integral formalism. For the Abelian case, see

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

 

[atyy]

The path integral can be calculated even without Wick rotation.

Even in relativistic field theory?

 

[king vitamin]

There is no problem with gauge invariance in thermodynamics. In case of BCS theory (as in any Higgsed gauge theory) there is some sloppy talk about "spontaneous symmetry breaking of local gauge symmetries" although in fact there is none. This is important, because it explains that there are no Nambu-Goldstone bosons but massive gauge fields in such a theory. The most elegant way to see this is the path-integral formalism. For the Abelian case, see

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

Sorry, but I don't see how this addresses my question, which didn't involve gauge invariance at all:

If we are given a Hamiltonian which does not conserve N... we usually just plug this into the GCE. I can find this exact statement in some textbooks, but your argument above implies that this is wrong at some level. The closest to a resolution to this is in post #28 by DrDru; I'm not sure if I understand their first explanation, but I think their second involves the number non-conservation [H,N] to vanish macroscopically. If this isn't correct, maybe I should start a new thread addressing this question.

The procedure is often done for mean-field hamiltonians of Bogoliubov form for superfluids and superconductors. I suppose one could also consider the approximations made when deriving these mean-field hamiltonians and check that the results make sense from the point of view of a particle-number conserving microscopic hamiltonian.

 

[Demystifier]

Even in relativistic field theory?

Yes, at least for free relativistic fields, where the path integrals can be computed exactly without Wick rotation.

For interacting fields, the two main methods are (i) analytic perturbative approach and (ii) numerical lattice approach. For (i) one calculates loop integrals where it is convenient to perform Wick rotation, but it is well justified because the positions of propagator singularities are known. For (ii) one uses Wick rotation to speed up the convergence. Quantities of physical interest such as energy of the ground state are not affected by the Wick rotation.

 

[vanhees71]

Sorry, but I don't see how this addresses my question, which didn't involve gauge invariance at all:
The procedure is often done for mean-field hamiltonians of Bogoliubov form for superfluids and superconductors. I suppose one could also consider the approximations made when deriving these mean-field hamiltonians and check that the results make sense from the point of view of a particle-number conserving microscopic hamiltonian.

Then I don't understand the question, because if there are no conserved charges or particle numbers, then you have to use the canonical ensemble, where only the mean energy is fixed. According to the maximum-entropy principle you have
$$\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}),$$
or to write it covariant, introducing the four-velocity of the heat bath
$$\hat{R}=\frac{1}{Z} \exp(-\beta u_{\mu} \hat{P}^{\mu},$$
where $\hat{P}^{\mu}$ is the operator of the total four-momentum of the system, which is conserved.

You find all this in the mentioned manuscript.

 

[DrDu]

The procedure is often done for mean-field hamiltonians of Bogoliubov form for superfluids and superconductors. I suppose one could also consider the approximations made when deriving these mean-field hamiltonians and check that the results make sense from the point of view of a particle-number conserving microscopic hamiltonian.

For these mean field hamiltonians, you are imposing a given mean particle number through the chemical potential in the grand canonical ensemble.

 

[DrDu]

In case of BCS theory (as in any Higgsed gauge theory) there is some sloppy talk about "spontaneous symmetry breaking of local gauge symmetries" although in fact there is none.

The relevant point here seems to be rather the breaking of global gauge symmetry (or phase symmetry or how you like to call it) which goes in hand with the non-conservation of particle number in irreducible representations of the operator algebra.

 

[vanhees71]

For these mean field hamiltonians, you are imposing a given mean particle number through the chemical potential in the grand canonical ensemble.

This depends on the system. Let's take the example of a relativistic "QED plasma" (i.e., a system of photons, electrons, and positrons). There you don't have conserved particle numbers at all but of course electric-charge conservation, i.e., "net-electron number", i.e., the quantity $N_{e^+}-N_{e^-}=Q$ is conserved. Then the only possible chemical potential refers to electric charge rather than particle numbers and discribes an asymmetric abundance of electrons over positrons, when you start with a system that has more electrons than positrons present. There is no conserved photon number, i.e., there's no chemical potential for photons.

 

[DrDu]

Of course. I was referring to the nonrelativistic BCS example.

 

[Demystifier]

My confusion now gets into how this is applied in cases where the hamiltonian does not conserve particle number.

Let me try to answer your question with a specific example - photons in thermodynamic equilibrium. They are a good example because the number of photons is not conserved.

To provide an equilibrium, let us enclose the photons inside the box of volume $V$. The walls of the box are at temperature $T$, so they constantly create new photons and absorb the old ones. Thus the number of photons fluctuates $\Delta N\neq 0$, but the average number of photons $\bar{N}$ is fixed. If $\bar{N}$ is very large, then $\Delta N \ll \bar{N}$.

Now let us adiabatically move the walls, so that the new volume of the box is $V'=2V$. The temperature $T$ of the walls is the same. What will happen with the number of photons and their density? The answer is that their density will remain the same, while their number will double $\bar{N}'=2\bar{N}$.

I am not sure that this answers your question, but I suspect that it might help.

 

[stevendaryl]

Yes, at least for free relativistic fields, where the path integrals can be computed exactly without Wick rotation.

It seemed to me that there was a slight of hand involved, even in the nonrelativistic case. Deriving the path integral nonrelativistically involves sort of dubious steps such as:

$\frac{1}{2\pi} \int_{-\infty}^{+\infty} dk e^{i (k x - \frac{\hbar}{2m} k^2 t)} = \sqrt{\frac{m}{2\pi i \hbar t}} e^{\frac{i t}{\hbar} \frac{m}{2} \frac{x^2}{t^2}}$

which is, strictly speaking, only valid if $t$ has a negative imaginary part.

 

[Demystifier]

strictly speaking, only valid if t has a negative imaginary part.

Yes, but this imaginary part can be arbitrarily small, and in the weak sense it is valid even when the imaginary part is zero.

 

[vanhees71]

Sure, you have to treat the distributions properly, if you can. In relativistic QFT we can't to a full extent (except in perturbation theory). For Gaussian path integrals the heat-kernel method is nice, if you really need the prefactor.

 

[atyy]

I guess what makes me skeptical about not having to use imaginary time is that as far as I understand, all existing rigorous constructions of relativistic quantum field theories have used the Euclidean path integral.

 

[vanhees71]

Hm are there any rigorous constructions (except for the free particle) (except in lower spacetime dimensions)? If the Euclidean theory is well defined then so should also be the Minkowskian one. In practice the analytic continuation from imaginary to real time can be very non-trivial.

 

[Demystifier]

Hm are there any rigorous constructions (except for the free particle) (except in lower spacetime dimensions)?

I would add - except with too much supersymmety.

 

[atyy]

The only existing rigourous constructions (for non-free QFT) are in lower spacetime dimensions. They use the Euclidean path integral to help convergence, and then show the Osterwalder-Schrader conditions are satisfied which guarantees that the Euclidean path integral corresponds to a field theory that obeys the Wightmann axioms.

 

[vanhees71]

I would add - except with too much supersymmety.

What do you mean by that? Are there rigorous models in 1+3 dimensions of interacting particles with supersymmetry?

 

[Demystifier]

What do you mean by that? Are there rigorous models in 1+3 dimensions of interacting particles with supersymmetry?

What I meant is that, for some supersymmetric theories in more than 1+3 dimensions, some exact non-perturbative solutions are known. But I am not an expert in this, so don't ask me much more!