Quantization of real Klein-Gordon field (sign issues)

[The Tortoise-Man]

I have a pretty naive question about quantization of real Klein-Gordon (so scalar) field $\hat{\phi}(x,t)$.
The most conventional form (see eg in this one ; but there are myriad scripts) is given by

$\hat{\phi}(x,t)= \int d^3p \dfrac{1}{(2\pi)^3} N_p (a_p \cdot e^{i(\omega_pt - p \cdot x)}+b_p^\dagger \cdot e^{-i(\omega_pt - p \cdot x)})$

where $N_p$ is a momentum-dependent normalization factor, and $\omega_p= \sqrt(p^2+m^2)$, $a_p$ annihilation and $b_p^\dagger$ creation operators.

My question is what is/are the reason(s) such that $a_p$ is "weighted" with positive exponential factor $e^{i(\omega_pt - p \cdot x)}$ and $b_p^\dagger$ with negative exponential $e^{-i(\omega_pt - p \cdot x)}$ and not another way around?

I'm seeking for two reasonings, the pure formal ("sober" calculation), but also heuristical/physical (intuitive interpretation of this factor)

 

[vanhees71]

First of all it's utmost crucial to start with the right definitions concerning these signs! You must have an annihilation operator in front of the positive-frequency and a creation operator in front of the negative-frequency modes:
$$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} N_p \left \{ \hat{a}(\vec{p}) \exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] + \hat{b}^{\dagger}(\vec{p}) \exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] \right \}.$$
This is correct in you anonymous manuscript.

From the Lagrangian,
$$\mathcal{L}=(\partial_{\mu} \phi^*)(\partial^{\mu} \phi)-m^2 \phi^* \phi$$
you get the canonical field momenta
$$\Pi(x)=\frac{\partial L}{\partial \dot{\phi}}=\dot{\phi}^*, \quad \Pi^*(x)=\frac{\partial L}{\partial \dot{\phi}^*}.$$
and from this the equal-time canonical commutator relations
$$[\hat{\phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]= [\hat{\phi}^*(t,\vec{x}),\hat{\Pi}^*(t,\vec{y})] =\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y})$$
and all other equal-time commutation relations between field operators and canonical field momentum operators vanishing, imply that for the most convenient choice of the normalization
$$N_p=\frac{1}{\sqrt{(2 \pi)^3 2\omega_p}}$$
commutation relations of the creation and annihilation operators read
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=[\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{p}')] = \delta^{(3)}(\vec{p}-\vec{p}').$$
The vacuum state is defined by
$$\hat{a}(\vec{p}) |\Omega \rangle=\hat{b}(\vec{p}) |\Omega \rangle=0.$$
From the classical Hamiltonian and applying normal ordering to the corresponding expression for the quantized version you get the positive semidefinite Hamilton operator
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_p [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})].$$
At the same time the field fulfills the microcausality condition
$$[\hat{\phi}(x),\hat{\phi}(y)]=0 \quad \text{for} \quad (x-y)^2<0 \quad (\text{i.e., space-like separated arguments}).$$
That positive definiteness of the energy together with the microcausality constraints explains why you must choose these signs and Bose-Einstein quantization (i.e., assuming commutation rather than anti-commutation relations for the canonical field and field-momentum operators) in the mode decomposition.

If you do the same exercise with the Dirac field, you have to use Fermi-Dirac quantization to get a microcausal theory with positive semidefinite Hamiltonian. That are special cases of the general spin-statistics theorem.

 

[The Tortoise-Man]

Could you elaborate how to deduce from requirements on positive definiteness of the energy eigenvalues and microcausality that the coefficients/exponents in

$$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} N_p \left \{ \hat{a}(\vec{p}) \exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] + \hat{b}^{\dagger}(\vec{p}) \exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] \right \}.$$

are choosen in the "right" way? Why another choice would lead to a contradiction?

The positive definiteness of energy tells just that $\langle s | \hat{H} | s \rangle >0$ for every state $| s \rangle >0$. How does it help here?

 

[vanhees71]

Could you elaborate how to deduce from requirements on positive definiteness of the energy eigenvalues and microcausality that the coefficients/exponents in
are choosen in the "right" way? Why another choice would lead to a contradiction?

The positive definiteness of energy tells just that $\langle s | \hat{H} | s \rangle >0$ for every state $| s \rangle >0$. How does it help here?

You should do the calculations for yourself. It's not so difficult and relys just on basic properties of the Fourier transform.

What might help though is to introduce the proper eigenmodes of the Klein-Gordon field (which are important building blocks for all other kinds of fields too since after all all free-particle equations must fulfill the Klein-Gordon equation since it fixes simply the eigenvalue Casimir operator, $\hat{p}_{\mu} \hat{p}^{\mu}=m^2$, where $m^2 \geq 0$, which is one piece to define an irreducible unitary representations of the proper orthogonal Poincare group, which defines what we call "elementary particles"):
$$u_{\vec{p}}(x)=\frac{1}{\sqrt{(2 \pi)^3 \omega_p}} \exp(-\mathrm{i} p \cdot x)|_{p^0=+\omega_p}.$$
These mode functions have the important properties
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x \mathrm{i} u_{\vec{p}}(x) \overleftrightarrow{\partial}_t u_{\vec{p}'}(x)=0, \quad \int_{\mathbb{R}^3} \mathrm{d}^3 x \mathrm{i} u_{\vec{p}}^*(x) \overleftrightarrow{\partial}_t u_{\vec{p}'}(x)=\delta^{(3)}(\vec{p}-\vec{p}'),$$
with help of which you can express the $\hat{a}(\vec{p})$ and $\hat{b}(\vec{p})$ in terms of the field operators $\hat{\phi}(x)$ and $\hat{\phi}^{\dagger}(x)$ using the expansion in terms of these mode functions. Here the notation with the double-arrow means
$$A(x) \overleftrightarrow{\partial}_t B(x)=A(x) \partial_t B(x)-(\partial_t A(x)) B(x).$$
The very idea, why you necessarily need a quantum-field theoretical formulation of relativistic QT is the argument that you should have "microcauality", i.e., that local operators, that represent observables, commute with space-like separated arguments. This must hold particularly for the Hamilton density (=energy density), which is a sufficient condition for the S-matrix elements being Poincare covariant and the S-matrix unitary (for $m^2 \geq 0$, for tachyonic irreps you run into trouble with causality as soon as you consider interacting fields). That $\hat{H}$ should be bounded from below is necessary to ensure stability.

Now it turns out that (at least for the free fields) the local observables like energy, momentum, angular momentum, various charge densities are bi- or sesquilinear functions of the field operators. Thus you can realize the microcausality condition with both bosonic (taking the equal-time Poisson brackets of the classical theory to equal-time commutators of the field operators of QFT) or with fermionic (taking the equal-time Poisson brackets of the classical theory to equal-time anti-commutators of the field operators of QFT), and there are topological arguments for the necessity to realize indistinguishability of particles in terms of either bosons and fermion if you have 3 or more spatial dimensions.

Now the recipe to build relativistic QFTs is to first look for unitary irreducible representations of the proper orthochronous Poincare group, generated by local field operators (the latter restiction to be able to realize relativistic causality by using microcausality together with local interactions), which leads to the fundamental properties of these representations described by mass and spin as well as conserved charges.

You can start with classical field theories, but it's quickly clear that you can't find any non-trivial unitary irreps in the sense of "1st quantization" for the proper orthochronous Poincare group, but you can use these fields in the action formalism to "canonically quantize" them. This works straight forward for the cases spin 0 and spin 1/2, i.e., to the Klein-Gordon fields as well as Weyl (and taking also parity and time reversal as probable symmetries into account) and Dirac fermions (also spin 1/2).

For fields with higher spin you need to impose constraints to project out the pieces with a definite spin. E.g., when realizing a massive spin-1 field with a four-vector field $V^{\mu}(x)$ you have to project out the spin-0 part of this field by imposing the condition $\partial_{\mu} V^{\mu}=0$, which leads to trouble in the canonical quantization formalism, but this you can get rid of using some modfications of the canonical-quantization concept, which is only a heuristical method anyway.

For the case $m=0$ it becomes even more complicated since for fields with spin $\geq 1$ you necessarily need to formulate the theory as a gauge theory to avoid "continuous spin-like degrees of freedom", which leads to the conclusion that such massless fields have only 2 helicity degrees of freedom, i.e., having helicity eigenstates (the projection of the total angular momentum of the particle to the direction of its momentum for the momentum eigenmodes) $\pm s$ (where $s$ is the spin).

Now it turns out that you can fulfill the microcausality condition together with the postive semidefiniteness of the Hamiltonian only if you quantize integer-spin fields as bosons and half-integer-spin fields as fermions, which is the celebrated spin-statistics theorem (Pauli 1940).

It also follows that for each particle there must be also the corresponding anti-particle (which can be both identical if you consider "strictly neutral particles" like photons), and that despite the necessary symmetry under proper orthochronous Lorentz transformations also the "grand reflection" CPT (charge conjugation, parity, time reversal) must be a symmetry.

For a very clear treatment in this pretty heuristic chain of arguments, see

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

Large parts of this are also available for free (legally). E.g.,
https://arxiv.org/abs/1110.5013

For a more systematic approach, considering carefully all unitary irreps. of arbitrary spin etc. see

S. Weinberg, The Quantum Theory of Fields, vol. 1,
Cambridge University Press (1995).

 

[apostolosdt]

[Post resubmitted to come round LaTeX error after an edit.]

vanhees71's post above is complete and there isn't much to add, but I'll try to elaborate a bit more, at your request.

Consider the Hamiltonian (trying to use notation in your post as much as possible),
$$H = \int {d^3p\over (2\pi)^3}\omega_{\mathbf p}\,b_{\mathbf p}^\dagger a_{\mathbf p}.$$
Then, you can easily show that
$$[H,b_{\mathbf p}^\dagger] = \omega_{\mathbf p}b_{\mathbf p}^\dagger,\qquad [H,a_{\mathbf p}] = -\omega_{\mathbf p}a_{\mathbf p}.$$
Next step; you'll need the relations (readily shown from the commutators above),
$$Ha_{\mathbf p} = a_{\mathbf p}(H-\omega_{\mathbf p}), \qquad Hb_{\mathbf p}^\dagger = b_{\mathbf p}^\dagger (H+\omega_{\mathbf p}).$$
Okay, now you might need a bit of algebra to first turn the above expressions into ones valid for any power $n$:
$$H^n a_{\mathbf p} = a_{\mathbf p}(H-\omega_{\mathbf p})^n, \qquad H^n b_{\mathbf p}^\dagger = b_{\mathbf p}^\dagger (H+\omega_{\mathbf p})^n.$$
You'll need this step to eventually get time-involving ladder operators via standard "sandwiching" with exponentials:
$$e^{iHt}a_{\mathbf p}e^{-iHt} = a_{\mathbf p}e^{-i\omega_{\mathbf p}t}\qquad e^{iHt}b_{\mathbf p}^\dagger e^{-iHt} = b_{\mathbf p}^\dagger e^{i\omega_{\mathbf p}t}.$$
Now, after inserting time into $\phi(\mathbf x)$ (Heisenberg picture), you will be able to show that
$$\phi(x) = \int{d^3 p\over(2\pi)^3}N_{\mathbf p}\left(a_{\mathbf p}e^{-ip\cdot x} + b_{\mathbf p}^\dagger e^{+ip\cdot x}\right)$$
where, of course, $p^0 = \omega_{\mathbf p}$. One final thing: $p^0$ is of course always positive, but observe that it appears in exponentials with both $-$ and $+$ signs; in quantum mechanical terms, these are positive- and negative-frequency states, hence the terms.

You can find all this stuff in any standard QFT text; I use Peskin and Schroeder.

 

[vanhees71]

But this Hamiltonian is not even self-adjoint. I have no clue, what you are after here. It's much more "intuitive" to start with canonical quantization as a heuristic principle (as you start with non-relativistic QM using the canonical quantization for non-relativistic point-particles leading to the Heisenberg algebra for position and momentum operators).

The correct (normal-ordered) Hamiltonian for the charged KG field is
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_p (\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})).$$

 

[apostolosdt]

But this Hamiltonian is not even self-adjoint. I have no clue, what you are after here. It's much more "intuitive" to start with canonical quantization as a heuristic principle (as you start with non-relativistic QM using the canonical quantization for non-relativistic point-particles leading to the Heisenberg algebra for position and momentum operators).

The correct (normal-ordered) Hamiltonian for the charged KG field is
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_p (\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})).$$

Perhaps a misunderstanding on my behalf. When I first posted those equations, I used $a_{\mathbf p}^\dagger$, but then trying to comply with the OP's notation, I changed that ladder operator to $b_{\mathbf p}^\dagger$. In the OP, I took $b_{\mathbf p}^\dagger$ to mean a creation operator for the Klein-Gordon field. As for the Hamiltonian I wrote down in the beginning in my post, that is just the one left from
$$H=\int{d^3p\over (2\pi)^3}\omega_{\mathbf p}\left(a_{\mathbf p}^\dagger a_{\mathbf p} + {1\over 2}\left[a_{\mathbf p},a_{\mathbf p}^\dagger\right]\right)$$
after one ignores the infinite constant term. Nothing fancy, therefore, just plain Klein-Gordon QFT stuff. But thanks for your remark, appreciated.

 

[apostolosdt]

The correct (normal-ordered) Hamiltonian for the charged KG field is
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_p (\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})).$$

I now notice that you're referring to the "charged KG field." I assume you mean the complex Klein-Gordon. I limited my explanation to real field only, as that was the question in the OP. So, no contradiction I guess.

 

[Demystifier]

I have a pretty naive question about quantization of real Klein-Gordon (so scalar) field $\hat{\phi}(x,t)$.

Your field is actually not real, but complex. It would be real (or hermitian, to be more precise) if $b_p$ was equal to $a_p$.

My question is what is/are the reason(s) such that $a_p$ is "weighted" with positive exponential factor $e^{i(\omega_pt - p \cdot x)}$ and $b_p^\dagger$ with negative exponential $e^{-i(\omega_pt - p \cdot x)}$ and not another way around?

I'm seeking for two reasonings, the pure formal ("sober" calculation), but also heuristical/physical (intuitive interpretation of this factor)

I'll give you a heuristic reason, formal ones have been given above. In non-relativistic QM, the wave function of the particle is proportional to $e^{-i\omega_pt}$, not to $e^{i\omega_pt}$. So when you create a particle, you want to reproduce a wave function of this form. That's why you multiply $e^{-i\omega_pt}$ with the creation operator, not with the annihilation operator.

 

[vanhees71]

I now notice that you're referring to the "charged KG field." I assume you mean the complex Klein-Gordon. I limited my explanation to real field only, as that was the question in the OP. So, no contradiction I guess.

For the neutral KG field you have $\hat{b}(\vec{p})=\hat{a}(\vec{p})$. Then the Hamiltonian is of course
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_p \hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
There are no wave functions but field operators (in the Heisenberg picture)!

 

[The Tortoise-Man]

For the case $m=0$ it becomes even more complicated since for fields with spin $\geq 1$ you necessarily need to formulate the theory as a gauge theory to avoid "continuous spin-like degrees of freedom", which leads to the conclusion that such massless fields have only 2 helicity degrees of freedom, i.e., having helicity eigenstates (the projection of the total angular momentum of the particle to the direction of its momentum for the momentum eigenmodes) $\pm s$ (where $s$ is the spin).

What do you mean precisely by "continuous spin-like degrees of freedom" and why one "wants" to avoid them?

 

[vanhees71]

This is all revealed by a careful analysis of the irreducible representations of the Poincare group a la Wigner (1939).

The "heuristic upshot" is that an irreducible representation of the proper orthochronous Poincare group can be built with momentum-"spin"-energy eigenstates. The unitary group operators must act transitive on these states, i.e., you can transform to any momentum eigenstate of eigenvalue $\vec{p}$ to any other (on the "mass shell", $p^0=\sqrt{m^2+\vec{p}^2}$ using a "standard transformation" acting on a state with "standard momentum" $\vec{p}_0$. The mass is a Casimir operator, $\hat{p}_{\mu} \hat{p}^{\mu}$, and thus it's one of the parameters that determine the irrep. So is the "spin" $s \in \{0,1/2,1,\ldots, \}$.

Now for massive particles you can choose $\vec{p}_0=0$, i.e., you have the rest frame of the particle as a physically distinguished frame, and you can get all other possible eigenvalues $\vec{p}$ on the mass shell by a corresponding Lorentz boost with velocity $\vec{v}=\vec{p}/E_{\vec{p}}=\vec{p}/\sqrt{m^2+\vec{p}^2}$. The spin of the particle is determined by the usual angular-momentum representations in the eigenspace of $\vec{p}=0$, and then you define the entire momentum-spin eigenbasis by boosting these vectors with the corresponding Lorentz boost.

Group-theoretically the action of the rotation group on the eigenspace of $\vec{p}=0$ is called "the little group" of the corresponding representation, i.e., the subgroup of the proper orthochronous Lorentz group, which leaves this momentum eigenvalue unchanged. In the massive case, we have chosen $\vec{p}=0$ as the standard momentum, and this momentum is of course invariant under all rotations. So the "little group" in the massive case is SO(3) (or rather its covering group SU(2)), which is why you get the usual $(2s+1)$ eigenvalues $-s,-s+1,\ldots,s-1,s$ for the spin-$z$ component as in non-relativistic physics.

Now comes the massless case. There the standard momentum cannot be $0$, because the energy-momentum eigenvalues are light-like, $q_{\mu} q^{\mu}=0$. You can choose the standard momentum in an arbitrary inertial reference frame to be of the form $(1,0,0,1)$, and you can reach all $\vec{p}$ values on the lightcone by corresponding boosts.

The key issue is now the little group, i.e., the subgroup of the proper orthochronous Lorentz group, that keep this standard momentum invariant. It's very easy to see that this is not the entire rotation group anymore but only the rotations around the $z$-axis. But that's not the entire little group. There are also socalled "null rotations". As it turns out this is a group that is isomorphic to the translation group of the Euclidean plane, i.e., you get "continuous spin degrees of freedom", except if you choose only representations, for which the "null rotations" are realized trivially, and this can be realized with local fields only if you assume gauge invariance, i.e., that you identify fields, which differ only in a gauge transformation. The only "spin-like degree of freedom" then left is due to the rotations around the $z$ direction (i.e., the direction of the momentum for an arbitrary momentum eigenstate), i.e., the helicity. As it turns out that for $s=0$ of course you have no additional spin degree of freedom and for $s \in \{1/2,1,\ldots \}$ you have, no matter which value $s$ takes, there are only two eigenvalues (and not $(2s+1)$ as in the massive case!) for the spin-$z$ component operator (in the eigenspace of the standard momentum $\vec{p}=(0,0,1)$). That's the helicity, i.e., when boosted to an arbitrary momentum $\vec{p}$ that's the projection of the total angular momentum to the direction of $\vec{p}$, and the helicity eigenvalues are only $\lambda=\pm s$. For photons (spin $s=1$) you have only two helicity eigenvalues, $\pm 1$, i.e., only two transverse polarization-degrees of freedom (the helicity eigenstates represent left- and right-circular polarization states).

The simple reason for not considering representations with non-trivially relized null-rotations is that we have no empirical evidence for any particles which have a continuous spin-like or polarization-like degrees of freedom.

 

[The Tortoise-Man]

As it turns out that for $s=0$ of course you have no additional spin degree of freedom and for $s \in \{1/2,1,\ldots \}$ you have, no matter which value $s$ takes, there are only two eigenvalues (and not $(2s+1)$ as in the massive case!) for the spin-$z$ component operator (in the eigenspace of the standard momentum $\vec{p}=(0,0,1)$).

This part I not understood. Why forthe case $s \neq 0$ there are only exactly two eigenvalues, independently of $s$ (as long as it's not zero), and not $(2s+1)$ (as I naively had expected)?

 

[vanhees71]

To be more precise for $s \in \{1/2,1,3/2,\ldots \}$ the representation of the little group, i.e., the rotations around the direction of the momentum (since the null rotations are assumed to be represented trivially) must be for either $\lambda=+s$ or $\lambda=-s$, i.e., each of these two possibilities forms an irreducible representation of the proper orthochonous Poincare group.

However, if you want to describe theories, where space reflection (parity) is a symmetry you need both signs $\lambda=\pm s$.

 

[The Tortoise-Man]

...think, I have to take some time to ponder a bit about it.

But coming back to the original question/problem. I still haven't explicitly written out the complete calculation in detail to check why in the expansion of $\phi(x,t)$ the creation operator $\hat{b}^{\dagger}(\vec{p})$ has to be weighted by $\exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})]$ (resp. annihilation operator $\hat{a}(\vec{p})$ by $\exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})]$) and not the other way around, but pondered a bit how I could attack it.

You suggesting in #No 2 to check it using two things: (1) positive semidefiniteness of $\hat{H}$ and (II)microcausality condition.

Here how I tried to approach it up to now: Say we start with general expression

$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} N_p \left \{ \hat{a}(\vec{p}) \exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] + \hat{b}^{\dagger}(\vec{p}) \exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] \right \}$

but at this stage we a priori NOT know if $\hat{a}(\vec{p})$ is creation or annihilation. Then we pick any eigenstate $| \alpha \rangle$ of $\hat{H}$ with eigenvalue $\alpha$, then calculate the bracket$[\hat{H}, \hat{a}(\vec{p})]= -\omega_p \hat{a}(\vec{p})$

and exploit this to show finally that

$\hat{H}(\hat{a}(\vec{p}) | \alpha \rangle)=(\alpha -\omega)((\vec{p}) | \alpha \rangle))$

Conclusion: $\hat{a}(\vec{p})$ lowers the state, so it must be annihilation, and analogously for $\hat{b}^{\dagger}$, so now it's "legal" to call one the annihilation, the other creation and so the problem is solved.But with one aspect in my approach I'm not completely happy (...maybe I missing some inportant issue, where it is unavoidable to make use of (I) and (Ii)): You suggested in the hints that the proof should exploit explicitly the (1) positive semidefiniteness of $\hat{H}$ and (II) microcausality condition, but in my "proof" (...if it is a "correct" one) I nowhere use neither (I), nor (II).

So either I did somewhere in my considerations some mistakes, or you suggested at the beginning another proof strategy how to show with which coefficient the annihilation op should be weighted, and with wich the creation. Do you see what is going wrong in my reasings, ie why I seemingly nowhere have to make usage (I) and (II)? It looks strage to me that seemingly the calculation above goes throught without them?

 

[vanhees71]

But for calculating $[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$ you use the commutation relations for the annihilation and creation operators as well as $\hat{H}$ in terms of these. The commutation relations of course are derived from the equal-time canonical commutation relations for the field operators and $\hat{H}$ via Noether's theorem as the generator for time translations.

 

[The Tortoise-Man]

But for calculating $[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$ you use the commutation relations for the annihilation and creation operators as well as $\hat{H}$ in terms of these. The commutation relations of course are derived from the equal-time canonical commutation relations for the field operators and $\hat{H}$ via Noether's theorem as the generator for time translations.

I see, the commutation relations for $\hat{a}(\vec{p})$, $\hat{b}(\vec{p})$ their daggers (... note that at this stage we a priori not know who is annihilation and who creation) are determined by commutation relations for the field operators, and vice versa. So the microcausality is indeed directly involved and dictate the used equality

$[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$
But where in the reasonings in #No 15 the other ingredient - the positive semidefiniteness of $\hat{H}$ - is going to be exploited?

 

[vanhees71]

Well, just do the entire quantization procedure for spin-0 (Klein-Gordon) fields assuming fermionic anticommutation relations. What follows for the (renormalized) Hamiltonian concerning definiteness?

 

[The Tortoise-Man]

Well, just do the entire quantization procedure for spin-0 (Klein-Gordon) fields assuming fermionic anticommutation relations. What follows for the (renormalized) Hamiltonian concerning definiteness?

A lower bound for energy eigenvalues? :) Or are you referring here to another aspect?

 

[vanhees71]

If you quantize spin-0 fields as fermions, you don't get a positive definite Hamiltonian.

 

[The Tortoise-Man]

If you quantize spin-0 fields as fermions, you don't get a positive definite Hamiltonian.

ah right, that's what you mean, I see. And since it's a "natural physical requirement" that the Hamiltonian should be positive definite, the only "physically reasonable" commutation rule in this case can only be the bosonic one, that's the point, right? So this observation gives here an heuristical answer to my other question: Anti-commutation relation for quantized fieldBut I'm not sure how this reasoning gives an answer to my concern in this thread, namely about which exponential factor

$e^{\pm i(\omega_pt - p \cdot x)}$

in the expansion of the field belongs to/ "weights" the annihilation operator and which the the creation.

The most general ansatz we start naively with is the usual Fourier expansion of our field operator

$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \hat{c}_{\vec{p}}(t) \exp[-\mathrm{i}(\vec{p} \cdot \vec{x})]$

and my problem was how to find out that it can be expressed as

$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} N_p \left \{ \hat{a}(\vec{p}) \exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] + \hat{b}^{\dagger}(\vec{p}) \exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] \right \}$

already knowing that $\hat{a}(\vec{p})$ should be annihilation and $\hat{b}^{\dagger}(\vec{p})$ creation operators, ie why the exp coefficients are exactly given in that way and not the other way.

Let us here assume that we already know that the field operators are bosonic, so we assume that the field operators satisfy bosonic commutation rules.Now my problem is how to deduce from these informations which exp coefficient weights the creation and which the annihilation.

You gave in #No2 the hint that it can be proved using (I) positive semidefiniteness of $\hat{H}$ and (II)microcausality condition.

In #No15 I tried to solve it, and apparently the crux of the problem lies in the equation

$[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$

(since other steps seemingly go through without problems). So far I understand you correctly, this equation in a consequence of microcausality.

But my question is where the other ingredient you suggested in #No2 to use - the positive semidefiniteness of $\hat{H}$ - goes in the verification that $\exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})]$ belongs to creation op., respectively $\exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})]$ to annihilation operator.

Firstly, is the approach you have in mind in #No2 & #No4 to show it using these two conditions (I) and (II) correctly carried out in #No15 or did you intended there a different strategy?#EDIT/UPDATE: Maybe I understand now what you intended to tell: as remarked before the crux lies in the equation $[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$.

And this follows from bosonic(!) commutation rules. So do I understand the chain of arguments you proposing correctly:
The (I) and (II) assure that the commutation rules MUST the bosonic (otherwise theory non physical), and having established this, this commutation rule implies $[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$.
In other words the contribution by (I) and (II) is "indirectly", namely to set up "correct" commutation rule and from these we deduce
$[\hat{H},\hat{a}(\vec{p})]=-\omega_{\vec{p}} \hat{a}(\vec{p})$
and therefore the claim (compare with #No15).

Is this the argument you intended to point out with neccessarity for (I) and (II) or do I misunderstand your concern?

 

[The Tortoise-Man]

[...] have added an #UPDATE

 

[vanhees71]

It depends upon, how you approach the "field quantization formalism". What I tried to explain in this and the other thread is the pretty hand-waving "canonical-quantization procedure". This is completely heuristic and follows Dirac's idea that you "translate" the Poisson brackets of the Hamilton formalism of classical physics to commutator relations for the corresponding operators in quantum theory.

You can also approach the entire thing in a more abstract way, as Weinberg does in his Quantum Theory of Fields, vol. 1. There he constructs the general unitary representations of the proper orthochronous Poincare group and then chooses the special ones that are realized by local representations on quantum-field operators, imposing the microcausality condition. I think from this he gets the spin-statistics relation from microcausality alone, but I've to check this again carefully.

Here's also a nice paper about this:

https://www.sciencedirect.com/science/article/pii/S1355219803000662

 

[The Tortoise-Man]

It depends upon, how you approach the "field quantization formalism". What I tried to explain in this and the other thread is the pretty hand-waving "canonical-quantization procedure". This is completely heuristic and follows Dirac's idea that you "translate" the Poisson brackets of the Hamilton formalism of classical physics to commutator relations for the corresponding operators in quantum theory.

But just from point of curiously (towards more mathematical viewpoint on this question; say in the flavour of a "thought experiment", 'assuming we know A, what can we deduce from it' and so on), is it true that if we pose as assumpion (say as "axiom" in mathematical sense) that our field $\hat{\phi}(t,\vec{x})$ satisfies the bosonic commutation rules; is then this information completely sufficient to deduce that
in the expansion

$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} N_p \left \{ \hat{a}(\vec{p}) \exp[-\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] + \hat{b}^{\dagger}(\vec{p}) \exp[+\mathrm{i}(\omega_p t - \vec{p} \cdot \vec{x})] \right \}$

where a priori is not known who of $\hat{a}(\vec{p})$ and $\hat{b}^{\dagger}(\vec{p})$ is annihilation and who creation, only from the single assumption on bosonicity of commutation rules for these field operators, which one is annihilation and which one is creation?

 

[vanhees71]

The commutation relations for the creation and annihilation operators follow uniquely from the equal-time commutation relations between field and canonical-field-momentum operators. You can uniquely determine $\hat{a}$ and $\hat{b}$ from the above mode-decomposition ansatz. I've given the corresponding formulae in the other thread.