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As Weinberg exposited in his QFT Vol1, there are two equivalent ways of arriving at the same quantum field theories:

(1). Start with single-particle representations of Poincare group, and then make a multiparticle theory out of it, while preserving principles of causality etc. I would call this approach the second quantization of particles, since second quantization is usually used to emphasize the many-body nature of a theory.

(2). Start with field representations of Poincare group, canonically quantize it, while preserving principles of causality, positive definiteness of energies etc. I would call this approach the quantization of fields, just as everyone else would call it.

Weinberg showed the proof of the equivalence between the above two approaches using some, though not hard, but let's say nontrivial, mathematics. The equivalence seems like a sheer miracle to me, or a complete coincidence. I do not feel that I understand the equivalence with the current state of mind. Is there a way to trivialize the equivalence? Or putting it another way, is there an a priori reasoning to argue, given the two sets of starting points of (1)(2), we have to get the same theory in the end?

Just as a side remark, many have suggested the term "second quantization" should be totally dumped, because it is really just the first quantization of fields. To me however, it still serves some purposes since the equivalence is not transparent.

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  • $\begingroup$ Which pages in Vol 1 are you referring to? $\endgroup$ – Qmechanic Nov 18 '13 at 12:19
  • $\begingroup$ @Qmechanic: chap 2-5 presents perspective (1), chap 7 presents (2) $\endgroup$ – Jia Yiyang Nov 18 '13 at 13:28
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The presumed equivalence between the canonical quantization and the Fock space representation is only a particular case.

The canonical formalism provides only with canonical Poisson brackets. The first step according to Dirac's axioms is to replace the Poisson brackets by commutators and since these commutators satisfy the Jacobi identity, they can be represented by linear operators on a Hilbert space.

Canonical quantization does not specify the Hilbert space.

Finding a Hilbert space where the operators acts linearly and satisfy the commutation relations is a problem in representation theory. This task is referred to as "quantization" in the modern literature.

The problem is that in the case of free fields, this problem does not have a unique solution (up to a unitary transformation in the Hilbert space). This situation is referred to as the existence of inequivalent quantizations or inequivalent representations. The Fock representation is only a special case. Some of the quantizations are called "non-Fock", because the Hilbert space does not have an underlying Fock space structure (i.e., cannot be interpreted as free particles), but there can even be inquivalent Fock representations.

Before, proceeding, let me tell you that inequivalent quantizations may be the areas where "new physics" can emerge because they can correspond to different quantum systems.

Also, let me emphasize, that the situation is completely different in the finite dimensional case. This is because that due to the Stone-von Neumann theorem, any representation of the canonical commutation relations in quantum mechanics is unitarily equivalent to the harmonic oscillator representation. Thus the issue of inquivalent representations of the canonical commutation relations occurs only due to the infinite dimensionality.

For a few examples of inquivalent quantizations of the canonical commutation relations of a scalar field on a Minkowski space-time, please see the following article by: Moschella and Schaeffer. In this article, they construct inequivalent representations by means of Bogoliubov transformation which changes the vacuum and they also present a thermofield representation. In all these representations the canonical operators are represented on a Hilbert space and the canonical commutation relations are satisfied. The Bogoliubov shifted vacuum cases correspond to broken Poincare' symmetries. One can argue that these solutions are unphysical, but the symmetry argument will not be enough in the case of quantization on a general curved nonhomogenous manifold. In this case we will not have a "physical" argument to dismiss some of the inequivalent representations.

The phenomena of inequivalent quantizations can be present even in the case of finite number of degrees of freedom on non-flat phase spaces.

Having said all that, I want nevertheless to provide you a more direct answer to your question (although it will not be unique due to the reasons listed above). As I understand the question, it can be stated that there is an algorithm for passing from the single particle Hilbert space to the Fock space. This algorithm can be summarized by the Fock factorization:

$$ \mathcal{F} = e^{\otimes \mathcal{h}}$$

Where $\mathcal{h}$ is the single particle Hilbert space and $\mathcal{F}$ is the Fock space. As stated before canonical quantization provides us only with the canonical commutaion relations:

$$[a_{\mathbf{k}}, a^{\dagger}_{\mathbf{l}}] = \delta^3(\mathbf{k} - \mathbf{l}) \mathbf{1}$$

At this stage we have only an ($C^{*}$)algebra of operators. The reverse question about the existence of an algorithm starting from the canonical commutation relations and ending with the Fock space (or equivalently, the answer to the question where is the Hilbert space?) is provided by the Gelfand -Naimark-Segal construction (GNS), which provides representations of $C^{*}$ algebras in terms of bounded operators on a Hilbert space.

The GNS construction starts from a state $\omega$ which is a positive linear functional on the algebra $ \mathcal{A}$ (in our case the algebra is the completion of all possible products of any number creation and annihilation operators).

The second step is choosing the whole algebra as an initial linear space $ \mathcal{A}$. In general, there will be null elements satisfying:

$$\omega (A^{\dagger}{A}) = 0$$

The Hilbert space is obtained by identifying elements differing by a null vector:

$$ \mathcal{H} = \mathcal{A} / \mathcal{N} $$

($\mathcal{N} $ is the space of null vectors).

The inner product on this Hilbert space is given by:

$$(A, B) = \omega (A^{\dagger}{B}) $$

It can be proved that the GNS construction is a cyclic representation where the Hilbert space is given by the action of operators on a cyclic "vacuum vector". The GNS construction gives all inequivalent representations of a given $C^{*}$ algebra (by bounded operators). In the case of a free scalar field the choice of a Gaussian state defined by its characteristic function:

$$ \omega_{\mathcal{F} }(e^{\int\frac{d^3k}{E_k} z_{\mathbf{k}}a^{\dagger}_{\mathbf{k}} + \bar{z}_{\mathbf{k}}a^{\mathbf{k}} }) = e^{\int\frac{d^3k}{E_k} \bar{z}_{\mathbf{k}} z_{\mathbf{k}}}$$

Where $z_{\mathbf{k}}$ are indeterminates which can be differentiated by to obtain the result for any product of operators.

The null vectors of this construction will be just combinations vanishing due to the canonical commutation relations (like $a_1 a_2 - a_2 a_1$). Thus this choice has Bose statistics. Also subspaces spanned by a product of a given number of creation operators will be the number subspaces.

The state of this specific construction is denoted by: $\omega_{\mathcal{F}}$, since it produces the usual Fock space. Different state choices may result inequivalent quantizations.

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  • $\begingroup$ +1, I like the perspective you offered. I still need to think to what extent it answers my question. $\endgroup$ – Jia Yiyang Nov 21 '13 at 1:06
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To me, the equivalence seems obvious. I will try to explain my train of thought.

In both approaches you start with the Poincare group.

Now in the first approach you start by constructing a Fock space of states with the only additional requirement of causality. The result is that your space is created by spacetime-dependent operators (as otherwise, causality could not be enforced), which is to say operator-valued fields. These fields obey canonical (Fock-space creating) covariant (requirement) commutation relations.

In the second approach you start by constructing fields from the poincare representations and then quantize them with the additional requirement of causality. Again, you impose canonical (Fock-space creating) covariant (requirement) commutation relations on your fields.

The difference is in the perspective. While in the first approach, you deem your particle states as more fundamental, seeking an operator that will create your physical space, the second apprach sees the fields as the object of importance, creating the multi-particle states "along the way".

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    $\begingroup$ Thanks for the answer, but your answer is essentially summarizing the math into words. I agree that there are common assumptions, e.g. positive definitess of energies, causality etc, however, I disagree in that I would consider the difference of taking particle or field as the starting point is a huge one, naively I would expect this difference will cause a huge difference in the final form of the theory, but it actually won't. $\endgroup$ – Jia Yiyang Nov 18 '13 at 13:22
  • $\begingroup$ @JiaYiyang The difference is purely mathematical in the end. Both approaches lead to identical physics, to identical descriptions of your system. On the one hand you start with the free particles - plane waves - and conclude that there must be a field in which these waves propagate. On the other hand you start with the field and conclude that the field allows harmonic waves - thus leading to particles. In both approaches you end up with fields and particle states and the same relation between the two. $\endgroup$ – Neuneck Nov 18 '13 at 14:24
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    $\begingroup$ I completely agree, but in your reply the "in the end" is my real issue, I'm just wondering if there is a way to foresee the equivalence in the beginning rather than in the end. $\endgroup$ – Jia Yiyang Nov 18 '13 at 15:57
  • $\begingroup$ Yes, I think it is. This hinges on accepting the equivalence between many-particle states and the operator fields creating them, though - which seems to be exactly what you try to understand. $\endgroup$ – Neuneck Nov 18 '13 at 16:37
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Here I would like to give an attempted answer to this fundamental question, through the following simple example:

Consider a crystal containing $m$ atoms localizing around the lattice sites, and each atom at site $i$ has a classical fields $(x_i,p_i)$(position and momentum), after canonical quantization(first quantization), the classical fields $(x_i,p_i)$ is promoted to operators $(\hat{x}_i,\hat{p}_i)$ with commutation relations $[\hat{x}_i,\hat{p}_j]=i\hbar \delta_{ij}$. Further, introduce the ladder operators $a_i=\frac{1}{\sqrt{2\hbar}}(\hat{x}_i+i\hat{p}_i)$ and they satisfy $[a_i,a_j^\dagger]=\delta_{ij}$. Now it's instructive to use the simultaneous eigenstates $ \mid n_1,\cdots,n_m >_c$ of the operaors $\hat{n}_1,\cdots ,\hat{n}_1$(where $\hat{n}_i=a_i^\dagger a_i$) as the basis of the crystal Hilbert space $V_c$, where $\hat{n}_i \mid n_1,\cdots,n_m >_c=n_i \mid n_1,\cdots,n_m >_c,i=1,2,\cdots,m$.

On the other hand, consider the second quantization of boson particles and let $b_i,b_i^\dagger$ be the annihilation and creation operators, the index $i$ represents the $i$ th single boson state and $i$ runs from $1$ to $m$. The particle number operators are dfined as $\hat{n}_i=b_i^\dagger b_i$, and the occupation basis of the boson Hilbert space $V_b$ is $ \mid n_1,\cdots,n_m >_b$ where $\hat{n}_i \mid n_1,\cdots,n_m >_b=n_i \mid n_1,\cdots,n_m >_b,i=1,2,\cdots,m$.

Finally, let's define a map between the crystal Hilbert space $V_c$ and the boson Hilbert space $V_b$ by making $ \mid n_1,\cdots,n_m >_c= \mid n_1,\cdots,n_m >_b$, which makes an equivalence between atoms and single boson states. And I think this is just the equivalence between canonical quantization of classical fields$(x_i,p_i)$ and second quantization of boson particles $b_i,b_i^\dagger$ as you mentioned.

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From what I understand, you simply can't deal with any non-perturbative physics if you try to second-quantize the single-particle states, because these are a priori perturbative excitations around the vacuum state. You miss, for example, instantons, topological effects... Try to do QCD this way and see how far you get.

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    $\begingroup$ emm, I don't know if QCD is a good counter example, as the current status of QCD only permits a path integral formulation, neither second quantization nor canonical quantization is found to work(yet). $\endgroup$ – Jia Yiyang Nov 18 '13 at 13:15
  • $\begingroup$ but it is a good point, so +1. $\endgroup$ – Jia Yiyang Nov 18 '13 at 13:22
  • $\begingroup$ Is that possible to compute quantities variationally based on the second-quantized formulation? If the Fock space constructed by direct product of single-particle states is complete, it should capture any state, whether within or without the scope of perturbation. $\endgroup$ – user26143 Nov 18 '13 at 14:28
  • $\begingroup$ @JiaYiyang, I think it's really just once you try to expand field operators in terms of creation operators acting on the vacuum that things begin to go awry, which, imo, is after canonical quantization. There are some people doing Yang-Mills in Hamiltonian formalism, e.g., H. Reinhardt, and past work by Karabali, Nair. My Masters thesis was on this, though it's all gone into a black hole now :) $\endgroup$ – lionelbrits Nov 18 '13 at 16:21
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You must think to the fundamental importance of the harmonic oscillator, in the realm of relativist fields.

Restrict ourself here, for simplicity, to a real scalar massless bosonic free relativist field $\Phi(x,t)$, the equation is :

$(\frac{\partial^2}{\partial t^2} - \Delta) \Phi(x,t)=0$

By Fourier transform on spatial coordinates, you got :

$(\frac{\partial^2}{\partial t^2} + k^2) \tilde \Phi(k,t)=0$

where $\Phi(k,t)$ is the spatial Fourier transform of \Phi(x,t). We could have used the notation $\tilde \Phi_k(t)$, with $(\frac{\partial^2}{\partial t^2} + k^2) \tilde \Phi_k(t)=0$.

This shows, that, when we think to a relativist field, it is, in fact (at least for bosonic fields), a collection of independent harmonic oscillators $\tilde \Phi_k(t)$. The $2$ definitions are totally equivalent, and no one is better than the other.

Now , quantizing the field $\Phi(x,t)$, is the same thing, as quantizing the collection of harmonic oscilators $\tilde \Phi_k(t)$. No quantization is better than the other. We know how to do that, by writing:

$\tilde \Phi_k(t) \sim a_k e^{-i|k| t} + a_k^+ e^{+i|k| t}$

with $[a_k, a_k^+]=1$ (here, we made a naive correspondance ignoring that $k$ is a continuous indice)

The harmonic oscilators being independent, and $k$ being a continuous indice,this naturally extends to the known relations $[a_k, a_k'^+]=\delta(k-k')$

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