Effect of Spin on Vacuum Energy

[DrInfinity]

TL;DR Summary

What is the effect of spin on a given Quantum Field's Vacuum Energy Density?

In equation (3) of this article https://arxiv.org/pdf/astro-ph/0605418.pdf , on page 3, you can see an expression for the vacuum energy density of a field as a function of its particle spin, particle mass and the wave number k.

It is fairly straight forward to derive the integral portion of this equation which is a function of m and k. However I am unsure about how the spin factor ½ ((-1)^2j) (2j+1) was derived? There is no reference provided in this paper and I could not find anything by doing some search on the web, Arxiv and some books I have. Does anyone know how this factor is derived or know of an authoritative source that explains it?

Thank you.

 

[jedishrfu]

While I can’t answer your question, I did find this article on Wikipedia about zero point energy that talks about the divergence

https://en.wikipedia.org/wiki/Zero-point_energy?wprov=sfti1

you might find why they chose the 1/2 spin factor. They did mention the notion of an infinite number of harmonic oscillators and that may be why.

scroll down to the Quantum Field section.

also I found this one on the Casimer effect

https://en.wikipedia.org/wiki/Casimir_effect?wprov=sfti1

and it mentions the reason for 1/2 is the vacuum has a spin property like any particle

Effects
Casimir's observation was that the https://www.physicsforums.com/wmfapp%3A//host/wiki/Canonical_quantization quantum electromagnetic field, in the presence of bulk bodies such as metals or https://www.physicsforums.com/wmfapp%3A//host/wiki/Dielectric, must obey the same https://www.physicsforums.com/wmfapp%3A//host/wiki/Boundary_value_problem that the classical electromagnetic field must obey. In particular, this affects the calculation of the vacuum energy in the presence of a https://www.physicsforums.com/wmfapp%3A//host/wiki/Electrical_conductor or dielectric.
Consider, for example, the calculation of the vacuum expectation value of the electromagnetic field inside a metal cavity, such as, for example, a https://www.physicsforums.com/wmfapp%3A//host/wiki/Cavity_magnetron or a https://www.physicsforums.com/wmfapp%3A//host/wiki/Microwave https://www.physicsforums.com/wmfapp%3A//host/wiki/Waveguide. In this case, the correct way to find the zero-point energy of the field is to sum the energies of the https://www.physicsforums.com/wmfapp%3A//host/wiki/Standing_wave of the cavity. To each and every possible standing wave corresponds an energy; say the energy of the nth standing wave is

. The vacuum expectation value of the energy of the electromagnetic field in the cavity is then

with the sum running over all possible values of n enumerating the standing waves. The factor of 1/2 is present because the zero-point energy of the n'th mode is

, where

is the energy increment for the n'th mode. (It is the same 1/2 as appears in the equation

.) Written in this way, this sum is clearly divergent; however, it can be used to create finite expressions.

 

[vanhees71]

For an elementary treatment of the Casimir effect, which is related to what's called "vacuum fluctuations" see

C. Itzykson, J.-B. Zuber, Quantum Field Theory, McGraw-Hill Book Company, New York (1980).

 

[timmdeeg]

https://arxiv.org/pdf/hep-th/0503158.pdf

According to Jaffe the Casimir effect doesn't proof "vacuum fluctuations" because this effect can be computed without taking reference to it.

 

[vanhees71]

I know. There are no vacuum fluctuations in the strict sense, because in fact what's called vacuum fluctuations are fluctuations of real charges (as shown in this very important paper by Jaffe).

 

[DrInfinity]

While I can’t answer your question, I did find this article on Wikipedia about zero point energy that talks about the divergence

https://en.wikipedia.org/wiki/Zero-point_energy?wprov=sfti1

you might find why they chose the 1/2 spin factor. They did mention the notion of an infinite number of harmonic oscillators and that may be why.

scroll down to the Quantum Field section.

also I found this one on the Casimer effect

https://en.wikipedia.org/wiki/Casimir_effect?wprov=sfti1

and it mentions the reason for 1/2 is the vacuum has a spin property like any particle

Thank you for your reply. Indeed, the integral in equation (3) is derived using a quantum harmonic oscillator to model vacuum energy. My interest is to understand what effect spin has on the harmonic oscillator, if any, and hence what is the effect on vacuum energy. Also please note that each elementary field (e.g. photon, gluon, electron, quarks, etc.) has a separate contribution to the vacuum energy, each having their own spin contributions as stated in equation (3). I am interested in the derivation of this spin effect stated as ½ ((-1)^2j) (2j+1)

 

[DrInfinity]

For an elementary treatment of the Casimir effect, which is related to what's called "vacuum fluctuations" see

C. Itzykson, J.-B. Zuber, Quantum Field Theory, McGraw-Hill Book Company, New York (1980).

Thank you, I own a copy of this very good book but it does not discuss the spin effect on vacuum energy. At least, I could not see it. If you know where it is in the book, please let me know.

 

[vanhees71]

The only effect of spin is that for any momentum mode you have $(2s+1)$ spin modes, i.e., the only effect is that the zero-point energy is multiplied by this factor.

It's of course a bit a misleading statement in the preprint quoted in #1 that QFT predicts this infinite "vacuum energy". Within special relativity the absolute value of the energy is unphysical (as in Newtonian physics). You can simply subtract any constant from the energy, and the meaning of this quantity for all physical phenomena is unaltered. That's why in standard QFT texts for the free fields you encounter the first "renormalization" argument at precisely that point: You subtract this unobservable zero-point energy and define the vacuum energy to be 0.

Of course in GR the absolute energy density is in principle physically relevant, because it enters the source term of Einstein's field equation of gravitation and thus there's this infamous cosmological-constant problem, which the paper is trying to solve by additional plausible assumptions.

For a general review on the cosmological-constant problem, I think the best is still Weinberg's famous RMP article:

S. Weinberg, The cosmological constant problem, Rev. Mod.
Phys. 61 (1989) 1.
https://link.aps.org/abstract/RMP/V61/P1

 

[DrInfinity]

The only effect of spin is that for any momentum mode you have $(2s+1)$ spin modes, i.e., the only effect is that the zero-point energy is multiplied by this factor.

It's of course a bit a misleading statement in the preprint quoted in #1 that QFT predicts this infinite "vacuum energy". Within special relativity the absolute value of the energy is unphysical (as in Newtonian physics). You can simply subtract any constant from the energy, and the meaning of this quantity for all physical phenomena is unaltered. That's why in standard QFT texts for the free fields you encounter the first "renormalization" argument at precisely that point: You subtract this unobservable zero-point energy and define the vacuum energy to be 0.

Of course in GR the absolute energy density is in principle physically relevant, because it enters the source term of Einstein's field equation of gravitation and thus there's this infamous cosmological-constant problem, which the paper is trying to solve by additional plausible assumptions.

For a general review on the cosmological-constant problem, I think the best is still Weinberg's famous RMP article:

S. Weinberg, The cosmological constant problem, Rev. Mod.
Phys. 61 (1989) 1.
https://link.aps.org/abstract/RMP/V61/P1

Thank you for your answer but my question is not related to the CC problem (I have read Weinberg's article), nor is it about the claimed divergence of the vacuum energy or other aspects of the original paper I referred too. I am only interested in the derivation of equation (3) in the paper I mentioned from first principles! Your first sentence above in not clear about multiplying with the (2s+1), why would that be the case, do you have a clear reference that explains this? Also the multiplication is not by (2s+1), it is by 0.5*(-1)^(2s) * (2s+1), for s representing spin.

 

[samalkhaiat]

Summary:: However I am unsure about how the spin factor ½ ((-1)^2j) (2j+1) was derived?
Thank you.

It is not derived. The equal-time algebra of free, multi-component field $\psi^{\alpha}(x)$ of spin $j$ and iso-spin $t$ is postulated to have the form $$\big[ \psi^{\alpha}(x^{0},x) , \pi_{\beta} (x^{0} , y) \big]_{\pm} = i \ (2j +1)(2t + 1) \ \delta^{\alpha}_{\beta} \ \delta^{3}(x - y).$$ The factor $(2j + 1)(2t +1)$ represents the number of states with spin $j$ and iso-spin $t$. However, if one starts from the usual creation and annihilation operator algebra and derive the above field algebra, the factor $(2j+1)(2t+1)$ arises from the summation over the spin and the iso-spin states included in the plane wave expansion of the solution of the field equation. The factor $(-1)^{2j}$ is their to account for bosons and fermions.

 

[DrInfinity]

Thank you for you reply. So is iso-spin t=0 in the case of the vacuum energy contribution of a specific field? And that is why we get only the (2j+1) term? Also, can you kindly refer me to a reference that has more info on your statement:
"the factor (2j+1)(2t+1) arises from the summation over the spin and the iso-spin states included in the plane wave expansion of the solution of the field equation."

Also, assuming equation (3) is correct,

if we apply it to obtain the vacuum energy density contribution of say the EM field (e.g. with massless (m=0) photon and spin j=1): then the factor in this case would be 1/2 * (-1)^(2*1) * (2*1+1) = 3/2 ? This does not match any reference that I have seen for EM vacuum energy density which has the factor equal 1 !

 

[vanhees71]

It is not derived. The equal-time algebra of free, multi-component field $\psi^{\alpha}(x)$ of spin $j$ and iso-spin $t$ is postulated to have the form $$\big[ \psi^{\alpha}(x^{0},x) , \pi_{\beta} (x^{0} , y) \big]_{\pm} = \frac{i}{2} \ (2j +1)(2t + 1) \ \delta^{\alpha}_{\beta} \ \delta^{3}(x - y).$$ The factor $(2j + 1)(2t +1)$ represents the number of states with spin $j$ and iso-spin $t$. However, if one starts from the usual creation and annihilation operator algebra and derive the above field algebra, the factor $(2j+1)(2t+1)$ arises from the summation over the spin and the iso-spin states included in the plane wave expansion of the solution of the field equation. The factor $(-1)^{2j}$ is their to account for bosons and fermions.

I've never seen such a convention. If you have a field of spin $s$ and isospin $T$ you have fields $\phi_{\sigma_z,t_3}(x)$ with annihilation operators $\hat{a}_{\sigma_z,t_3}(\vec{p})$ for momentum modes (and the corresponding annihilation operators for antiparticles, which obey the same (anti-) commutation rules):
$$[\hat{a}_{\sigma_z,t_3}(\vec{p}),\hat{a}_{\sigma_z',t_3'}(\vec{p}')]_{\mp}=0, \quad [\hat{a}_{\sigma_z,t_3}(\vec{p}),\hat{a}_{\sigma_z',t_3'}^{\dagger}(\vec{p}')]_{\mp}=\delta^{(3)}(\vec{p}) \delta_{\sigma_3 \sigma_3'} \delta_{t_3 t_3'},$$
where I have chosen the non-covariant normalization conditions for convenience. Of course, the upper sign is for bosons (integer $s$) and the lower for fermions (half-integer $s$).

The total unrenormalized and thus divergent Hamiltonian (the easist way to regularize it is to just introduce a momentum cutoff) is then given by
$$\hat{H}=\sum_{\sigma_3,t_3} \int_{\vec{p}^2<\Lambda^2} \mathrm{d}^3 \vec{p} \left [\frac{1}{2} +\hat{N}_{\sigma_3,t_3}(\vec{p}) \right] E_{\vec{p}},$$
where $E(\vec{p})=\sqrt{\vec{p}^2+m^2}$ and $\hat{N}_{\sigma_3,t_3}(\vec{p})=\hat{a}_{\sigma_3,t_3}^{\dagger}(\vec{p}) \hat{a}_{\sigma_3,t^3}(\vec{p})$.

For the vacuum energy you get
$$E_{\text{vac}}=\langle \Omega|\hat{H}|\Omega \rangle=\frac{1}{2} (2s+1)(2T+1) \int_{\vec{p}^2<\Lambda^2} \mathrm{d}^3 \vec{p} E(\vec{p}).$$

 

[DrInfinity]

I've never seen such a convention. If you have a field of spin $s$ and isospin $T$ you have fields $\phi_{\sigma_z,t_3}(x)$ with annihilation operators $\hat{a}_{\sigma_z,t_3}(\vec{p})$ for momentum modes (and the corresponding annihilation operators for antiparticles, which obey the same (anti-) commutation rules):
$$[\hat{a}_{\sigma_z,t_3}(\vec{p}),\hat{a}_{\sigma_z',t_3'}(\vec{p}')]_{\mp}=0, \quad [\hat{a}_{\sigma_z,t_3}(\vec{p}),\hat{a}_{\sigma_z',t_3'}^{\dagger}(\vec{p}')]_{\mp}=\delta^{(3)}(\vec{p}) \delta_{\sigma_3 \sigma_3'} \delta_{t_3 t_3'},$$
where I have chosen the non-covariant normalization conditions for convenience. Of course, the upper sign is for bosons (integer $s$) and the lower for fermions (half-integer $s$).

The total unrenormalized and thus divergent Hamiltonian (the easist way to regularize it is to just introduce a momentum cutoff) is then given by
$$\hat{H}=\sum_{\sigma_3,t_3} \int_{\vec{p}^2<\Lambda^2} \mathrm{d}^3 \vec{p} \left [\frac{1}{2} +\hat{N}_{\sigma_3,t_3}(\vec{p}) \right] E_{\vec{p}},$$
where $E(\vec{p})=\sqrt{\vec{p}^2+m^2}$ and $\hat{N}_{\sigma_3,t_3}(\vec{p})=\hat{a}_{\sigma_3,t_3}^{\dagger}(\vec{p}) \hat{a}_{\sigma_3,t^3}(\vec{p})$.

For the vacuum energy you get
$$E_{\text{vac}}=\langle \Omega|\hat{H}|\Omega \rangle=\frac{1}{2} (2s+1)(2T+1) \int_{\vec{p}^2<\Lambda^2} \mathrm{d}^3 \vec{p} E(\vec{p}).$$

Thank you for your reply. Your final expression given above is indeed close to equation (3). How does one obtain it, is there a reference that gives more details on this? For example, when you apply the Hamiltonian above on $\Omega$ to obtain the vacuum energy how does one end up with the factor (2s+1)(2T+1), does it come from the summation over all possible spin and iso-spin states? Also, after thinking about it some more, if we only consider elementary fields then iso-spin should have no place in the vacuum energy expression as iso-spin is to categorize different (non-elementary) hadrons which are made up of elementary Quarks/anti-quarks held together with elementary Gluons (e.g. neutron, proton, Pion). This is most likely the reason we do not see the (2T+1) factor in equation (3).

 

[vanhees71]

Yes. I think what I've given is the complete derivation.

Instead of isospin in the SM you have flavor-degrees of freedom. I also forgot the factor $(-1)^{2s} V/(2 \pi)^3$. The sign comes from the commutation or anticommutation of a term of the structure $\hat{a} \hat{a}^{\dagger}$ to $\hat{N}=\hat{a}^{\dagger} \hat{a}$ and the $1/(2 \pi)^3$ comes from $1/(2 \pi \hbar)^3$ and our choice of units where $\hbar=1$; $V$ is the quantization volume. So the correct formula is indeed given by Eq. (3) in the quoted paper though the author didn't bother to write $V$ explicitly.

The correct factor is derived more carefully in Itzykson Zuber.

 

[DrInfinity]

Yes. I think what I've given is the complete derivation.

Instead of isospin in the SM you have flavor-degrees of freedom. I also forgot the factor $(-1)^{2s} V/(2 \pi)^3$. The sign comes from the commutation or anticommutation of a term of the structure $\hat{a} \hat{a}^{\dagger}$ to $\hat{N}=\hat{a}^{\dagger} \hat{a}$ and the $1/(2 \pi)^3$ comes from $1/(2 \pi \hbar)^3$ and our choice of units where $\hbar=1$; $V$ is the quantization volume. So the correct formula is indeed given by Eq. (3) in the quoted paper though the author didn't bother to write $V$ explicitly.

The correct factor is derived more carefully in Itzykson Zuber.

Thank you very much for your prompt reply. So I think we both agree that the equation (3) is correct, however one has to be careful in applying it blindly, and this is the reason I initially questioned this equation, I explain:
As I stated before in an earlier reply, for the EM field (i.e. the photon bosonic field) with spin j=1, equation (3) results in
\begin{equation}

\hat{\rho}_{vac}= \frac{3}{2} \int_{-\infty}^\infty \frac{d^3k}{(2\pi)^3} \sqrt{\vec{k}^2+m^2}
\end{equation}
which as far as I can see is incorrect and the correct answer should be with a factor of 2/2= 1 and not 3/2.

This is because the photon is massless and can only take on 2 separate spin states or polarization (+1 and -1) and spin state 0 is not physically possible and discarded and hence the (2j+1) is not the correct factor here and rather, it is just (2j) for this special case of photon (and the Gluon field as well as it is also massless with spin=1), however for the Dirac Electron field with spin j=1/2, the (2j+1)=2 produces the correct factor as there are only 2 spin states possible for the electron and positron which are +1/2 and -1/2. Do you agree with my assessment here?

Also, the quantization volume V is not necessary in equation (3) as it gives the Vacuum Energy Density (i.e. E/V), but it is required in your equation where you have given the Vacuum Energy.

Do you know where exactly in Itzykson/Zuber, they discusses this (2j+1) factor? I am very grateful for your help and our exchanges here.

 

[samalkhaiat]

I've never seen such a convention.

Well, you have seen it now! And I can give you a reference if you want. Apart from the wrong half factor (which I corrected), you can always convince yourself by doing the calculations for the simple case of free meson field in the Cartesian representation (iso-spin vector) : $$\vec{\varphi} (0 , x) = \left( \frac{1}{2\pi} \right)^{3/2} \sum_{\tau = -T}^{+T} \int \frac{d^{3}p}{\sqrt{2E}} \left( a_{\tau}(p) \ \eta_{\tau} \ e^{i p \cdot x} + a^{\dagger}_{\tau}(p) \ \eta^{\dagger}_{\tau} \ e^{-i p \cdot x} \right) .$$
The canonical momentum field is
$$\vec{\pi} (0 , \bar{x}) = - i \left( \frac{1}{2\pi} \right)^{3/2} \sum_{ \bar{\tau} = -T}^{+T} \int \frac{d^{3}\bar{p}}{\sqrt{2E}} (E) \left( a_{\bar{\tau}}(\bar{p}) \ \eta_{\bar{\tau}} \ e^{i \bar{p}\ \cdot \ \bar{x} } - a^{\dagger}_{\bar{\tau}}(\bar{p}) \ \eta^{\dagger}_{\bar{\tau}} \ e^{- i \bar{p} \ \cdot \ \bar{x}}\right) .$$
To further simplify the calculation, take the iso-spinors $\eta_{\tau}$ to be the orthonormal unit vectors in iso-space (this is the case for Cartesian pions) and hence $\eta_{\tau} = \eta^{\dagger}_{\tau}$ and $\eta_{\tau} \cdot \eta_{\bar{\tau}} = \delta_{\tau \bar{\tau}}$. Now, evaluate the equal-time commutator, $\big[ \vec{\varphi}(0,x) , \vec{\pi}(0,\bar{x})\big]$, using the usual commutation relations $$\big[ a_{\tau}(p) , a^{\dagger}_{\bar{\tau}}(\bar{p})\big] = \delta_{\tau \bar{\tau}} \ \delta^{3}(p - \bar{p}) ,$$$$\big[ a_{\tau}(p) , a_{\bar{\tau}}(\bar{p})\big] = \big[ a^{\dagger}_{\tau}(p) , a^{\dagger}_{\bar{\tau}}(\bar{p})\big] = 0.$$ You should obtain $$\big[ \vec{\varphi}(0,x) , \vec{\pi}(0,\bar{x})\big] = i (2T + 1) \delta^{3}(x - \bar{x}) .$$

 

[samalkhaiat]

Thank you for you reply. So is iso-spin t=0 in the case of the vacuum energy contribution of a specific field? And that is why we get only the (2j+1) term?

Yes, they seem to consider an iso-singlet.

Also, can you kindly refer me to a reference that has more info on your statement:
"the factor (2j+1)(2t+1) arises from the summation over the spin and the iso-spin states included in the plane wave expansion of the solution of the field equation."

Do the calculation in #16.
"Methods in Relativistic Nuclear Physics"
M. Danos, V. Gillet & M. Cauvin, North-Holland, 1984.

if we apply it to obtain the vacuum energy density contribution of say the EM field (e.g. with massless (m=0) photon and spin j=1): then the factor in this case would be 1/2 * (-1)^(2*1) * (2*1+1) = 3/2 ? This does not match any reference that I have seen for EM vacuum energy density which has the factor equal 1 !

For massless particles, the number of "spin" states is not $(2j +1)$. Massless particles have only 2 "spin" states.

 

[vanhees71]

Thank you very much for your prompt reply. So I think we both agree that the equation (3) is correct, however one has to be careful in applying it blindly, and this is the reason I initially questioned this equation, I explain:
As I stated before in an earlier reply, for the EM field (i.e. the photon bosonic field) with spin j=1, equation (3) results in
\begin{equation}

\hat{\rho}_{vac}= \frac{3}{2} \int_{-\infty}^\infty \frac{d^3k}{(2\pi)^3} \sqrt{\vec{k}^2+m^2}
\end{equation}
which as far as I can see is incorrect and the correct answer should be with a factor of 2/2= 1 and not 3/2.

This is because the photon is massless and can only take on 2 separate spin states or polarization (+1 and -1) and spin state 0 is not physically possible and discarded and hence the (2j+1) is not the correct factor here and rather, it is just (2j) for this special case of photon (and the Gluon field as well as it is also massless with spin=1), however for the Dirac Electron field with spin j=1/2, the (2j+1)=2 produces the correct factor as there are only 2 spin states possible for the electron and positron which are +1/2 and -1/2. Do you agree with my assessment here?

Also, the quantization volume V is not necessary in equation (3) as it gives the Vacuum Energy Density (i.e. E/V), but it is required in your equation where you have given the Vacuum Energy.

Do you know where exactly in Itzykson/Zuber, they discusses this (2j+1) factor? I am very grateful for your help and our exchanges here.

Of course for a photon, as for any massless particle with non-zero spin you have only 2 helicity degrees of freedom and not $(2s+1)$ (it's not $(2j+1)$, because we use the momentum-spin basis here and $j$ is total angular momentum). Itzykson/Zuber discuss one field-degree of freedom. Has should be clear the "degeneracy factors" are trivial.

 

[DrInfinity]

Of course for a photon, as for any massless particle with non-zero spin you have only 2 helicity degrees of freedom and not $(2s+1)$ (it's not $(2j+1)$, because we use the momentum-spin basis here and $j$ is total angular momentum). Itzykson/Zuber discuss one field-degree of freedom. Has should be clear the "degeneracy factors" are trivial.

Yes I agree. However, in equation (3), $j$ is in fact the spin or momentum-spin basis and not the total angular momentum. I understand that some authors use $s$ for momentum-spin basis but in the paper we are discussing, the author uses $j$ instead.

 

[DrInfinity]

Well, you have seen it now! And I can give you a reference if you want. Apart from the wrong half factor (which I corrected), you can always convince yourself by doing the calculations for the simple case of free meson field in the Cartesian representation (iso-spin vector) : $$\vec{\varphi} (0 , x) = \left( \frac{1}{2\pi} \right)^{3/2} \sum_{\tau = -T}^{+T} \int \frac{d^{3}p}{\sqrt{2E}} \left( a_{\tau}(p) \ \eta_{\tau} \ e^{i p \cdot x} + a^{\dagger}_{\tau}(p) \ \eta^{\dagger}_{\tau} \ e^{-i p \cdot x} \right) .$$
The canonical momentum field is
$$\vec{\pi} (0 , \bar{x}) = - i \left( \frac{1}{2\pi} \right)^{3/2} \sum_{ \bar{\tau} = -T}^{+T} \int \frac{d^{3}\bar{p}}{\sqrt{2E}} (E) \left( a_{\bar{\tau}}(\bar{p}) \ \eta_{\bar{\tau}} \ e^{i \bar{p}\ \cdot \ \bar{x} } - a^{\dagger}_{\bar{\tau}}(\bar{p}) \ \eta^{\dagger}_{\bar{\tau}} \ e^{- i \bar{p} \ \cdot \ \bar{x}}\right) .$$
To further simplify the calculation, take the iso-spinors $\eta_{\tau}$ to be the orthonormal unit vectors in iso-space (this is the case for Cartesian pions) and hence $\eta_{\tau} = \eta^{\dagger}_{\tau}$ and $\eta_{\tau} \cdot \eta_{\bar{\tau}} = \delta_{\tau \bar{\tau}}$. Now, evaluate the equal-time commutator, $\big[ \vec{\varphi}(0,x) , \vec{\pi}(0,\bar{x})\big]$, using the usual commutation relations $$\big[ a_{\tau}(p) , a^{\dagger}_{\bar{\tau}}(\bar{p})\big] = \delta_{\tau \bar{\tau}} \ \delta^{3}(p - \bar{p}) ,$$$$\big[ a_{\tau}(p) , a_{\bar{\tau}}(\bar{p})\big] = \big[ a^{\dagger}_{\tau}(p) , a^{\dagger}_{\bar{\tau}}(\bar{p})\big] = 0.$$ You should obtain $$\big[ \vec{\varphi}(0,x) , \vec{\pi}(0,\bar{x})\big] = i (2T + 1) \delta^{3}(x - \bar{x}) .$$

ok thank for this reply but I am unsure how this is related to my original question? I was looking for a derivation of equation (3) in the mentioned paper.

 

[samalkhaiat]

I am unsure how this is related to my original question?

As I told you before, the factor $(2j +1)(2T+1)$ (in your “original question” and in the derivation I asked you to do) comes from $\sum_{\sigma = -j}^{+j}\sum_{\tau = -T}^{+T}$.

I was looking for a derivation of equation (3) in the mentioned paper.

Vanhees71 has already done that for you in #12.

 

[DrInfinity]

Yes, they seem to consider an iso-singlet.

Do the calculation in #16.
"Methods in Relativistic Nuclear Physics"
M. Danos, V. Gillet & M. Cauvin, North-Holland, 1984.For massless particles, the number of "spin" states is not $(2j +1)$. Massless particles have only 2 "spin" states.

Thank you, I will check the reference your provided.

 

[DrInfinity]

As I told you before, the factor $(2j +1)(2T+1)$ (in your “original question” and in the derivation I asked you to do) comes from $\sum_{\sigma = -j}^{+j}\sum_{\tau = -T}^{+T}$.

Vanhees71 has already done that for you in #12.

Yes Vanhees71 and yourself have both put me on the right path so thank you both. I will derive equation (3) and post the derivation here for everyone. However, please note that the (2T+1) iso-spin factor you have both mentioned has nothing to do with vacuum energy which arises from elementary fields only!

 

[vanhees71]

Well, you can have more intrinsic quantum numbers than spin. A very important one is isospin ;-).

 

[DrInfinity]

As I told you before, the factor $(2j +1)(2T+1)$ (in your “original question” and in the derivation I asked you to do) comes from $\sum_{\sigma = -j}^{+j}\sum_{\tau = -T}^{+T}$.

Please note that there is no (2T+1) in my "original question" so while you and Vanhees71 have provided a hint as how to proceed with the derivation of vacuum energy, the derivation provided is not entirely correct and must be done for elementary quantum fields only that do not include any concept of the iso-spin T. As I mentioned before, the iso-spin concept and categorization is useful in nuclear physics and when analyzing hadrons such as mesons. I am assuming that both yourself and Vanhees71 are nuclear physicist ;-)

Vanhees71 has already done that for you in #12.

 

[DrInfinity]

Well, you can have more intrinsic quantum numbers than spin. A very important one is isospin ;-).

yes but mostly useful for Hadrons ;-)