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I Is the ground state energy of a quantum field actually zero?

  1. Aug 21, 2018 #1
    I start by outlining the little I know about the basics of quantum field theory.

    The simplest relativistic field theory is described by the Klein-Gordon equation of motion for a scalar field ##\large \phi(\vec{x},t)##:
    $$\large \frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0.$$
    We can decouple the degrees of freedom from each other by taking the Fourier transform:
    $$\large \phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t).$$
    Substituting back into the Klein-Gordon equation we find that ##\large \phi(\vec{p},t)## satisfies the simple harmonic equation of motion
    $$\large \frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi.$$
    Therefore, for each value of ##\large \vec{p}##, ##\large \phi(\vec{p},t)## solves the equation of a harmonic oscillator vibrating at frequency
    $$\large \omega_\vec{p}=+\sqrt{\vec{p}^2+m^2}.$$
    Thus the general solution to the Klein-Gordon equation is a linear superposition of simple harmonic oscillators with frequency ##\large \omega_\vec{p}##. When these harmonic oscillators are quantized we find that each has a set of discrete positive energy levels given by
    $$\large E^p_n=\hbar\omega_\vec{p}(n+\frac{1}{2})$$
    for ##\large n=0,1,2\ldots## where ##\large n## is interpreted as the number of particles with momentum ##\large \vec{p}##.

    My question is what about the harmonic oscillator solutions that vibrate at negative frequency
    $$\large \bar{\omega}_\vec{p}=-\sqrt{\vec{p}^2+m^2}?$$

    When these harmonic oscillators are quantized we get a set of discrete negative energy levels given by
    $$\large \bar{E}^p_n=\hbar\bar{\omega}_\vec{p}(n+\frac{1}{2})$$
    for ##\large n=0,1,2\ldots## where ##\large n## can now be interpreted as the number of antiparticles with momentum ##\large \vec{p}##.

    If this is correct then the total energy of the ground state, per momentum ##\large \vec{p}##, is given by
    \begin{eqnarray*}
    \large T^p_0 &=& \large E^p_0+\bar{E}^p_0\\
    &=& \large \frac{\hbar\sqrt{\vec{p}^2+m^2}}{2} + \frac{-\hbar\sqrt{\vec{p}^2+m^2}}{2}\\
    &=& \large 0.
    \end{eqnarray*}

    Thus the total ground state energy, ##\large T_0##, is zero; there is no zero-point energy.

    Does this interpretation of the negative frequency solutions make sense?
     
    Last edited: Aug 21, 2018
  2. jcsd
  3. Aug 21, 2018 #2

    king vitamin

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    The solutions to the Klein-Gordon equation with negative frequency do not have negative energy! You need to go back to the Hamiltonian for the Klein-Gordon theory,
    [tex]
    H = \int d^3 x \left[ \frac{1}{2} \Pi^2 + \frac{1}{2} \left( \nabla \phi\right)^2 + \frac{m^2}{2} \phi^2 \right],
    [/tex]
    and consider the energy for a field
    [tex]
    \phi(\vec{x},t) = \int \frac{d^3 k}{(2 \pi)^3} \left( a(k) e^{i \vec{k} \cdot \vec{x} - i \omega t} + b(k) e^{i \vec{k} \cdot \vec{x} + i \omega t} \right).
    [/tex]
    Here you can see that we're including both the positive and negative frequency solutions to the Klein-Gordon equation. The coefficients in the Fourier expansion are operators. Now do the usual trick of writing these coefficients as ladder operators, and do the same trick for the canonical momentum [itex]\Pi(\vec{x},t)[/itex] such that [itex][\phi(\vec{x},t),\Pi(0,t)] = i \delta^3(\vec{x})[/itex] is satisfied, and then calculate [itex]H[/itex]. You should only find a unique ground state, and all other states have higher energy than this.

    In fact, this is true even without quantum mechanics. Negative frequency solutions to the classical Klein-Gordon equation still result in a positive energy in the classical Klein-Gordon Hamiltonian (show this!). In contrast, a "classical" Dirac Hamiltonian has negative energy solutions which signal that it is sick.
     
    Last edited: Aug 22, 2018
  4. Aug 22, 2018 #3

    bhobba

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    To answer the title of the question look into something called normal ordering. The energy of the vacuum is a big fat zero.

    That one fooled me to until l read a proper book on QFT a few years ago now.

    Thanks
    Bill
     
  5. Aug 22, 2018 #4

    A. Neumaier

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    Is the ground state energy of a quantum field actually zero?

    In the relativistic case, yes, by Lorentz covariance.

    In the nonrelativistic case, it is a matter of convention since there energies are determined only up to an arbitrary constant shift.
     
  6. Aug 22, 2018 #5

    Demystifier

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    It's not that simple:
    https://arxiv.org/abs/hep-th/0204048
     
  7. Aug 22, 2018 #6

    Demystifier

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    Are you saying that normal ordering is the only correct ordering? And which book was that?

    If you read e.g. Bjorken and Drell, you will see that vacuum energy can be any number, the effect of which is to modify the phase of the scattering amplitude without changing any measurable quantity.
     
  8. Aug 22, 2018 #7

    A. Neumaier

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    It is that simple, in spite of your reference.

    The ground state of a relativistic quantum field theory is the vacuum state, defined as a Poincare invariant state. Hence the 4-momentum ##p## must be Lorentz invariant, which is only possible if ##p=0##. The energy is the 0-component, hence vanishes.
     
    Last edited: Aug 22, 2018
  9. Aug 22, 2018 #8

    bhobba

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    No I am not. What I am saying is the issue can be rectified. I read some overview books on QFT that said its arbitrary the energy you call zero because you only measure differences anyway. So you just set the infinite energy as zero. I thought it totally silly and turned me right off. The only solution was to get a proper book on QFT and I chose An Introduction to Quantum Field Theory by George Sterman as my first book. He explained it on page 44. But overall its not my favorite book - I preferred others like QFT For The Gifted Amateur - which does the same thing. I am reading Strednicki right now and it resolves it by an arbitrary ultraviolet cutoff - see page 24 - I suppose in preparation for the modern effective field theory view. I hope so because that is something I want to understand better.

    Thanks
    Bill
     
    Last edited: Aug 22, 2018
  10. Aug 22, 2018 #9

    Demystifier

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    So do you claim that there is no cosmological constant problem? Another Lorentz invariant value for energy is infinity. Indeed, the vacuum energy-momentum tensor of the vacuum with cosmological constant ##\lambda## is
    $$T_{\mu\nu}=\lambda g_{\mu\nu}$$
    which is Lorentz-invariant as long as ##g_{\mu\nu}=\eta_{\mu\nu}## is Lorentz invariant. The 4-momentum is then
    $$P^{\mu}=\int d^3x T^{\mu}_{0}=\lambda \delta^{\mu}_{0} \int d^3x$$
    where ## \int d^3x=\infty##.
     
  11. Aug 22, 2018 #10

    A. Neumaier

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    The universe is not in a vacuum/ground state; so your question has nothing to do with the topic of the thread.
    In this case, all states must have infinite energy and all states would be ground states! Thus the notion of a vacuum/ground state makes no longer sense.
     
  12. Aug 22, 2018 #11

    Demystifier

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    Are you saying that the cosmological constant problem has nothing to do with the vacuum energy?
     
  13. Aug 22, 2018 #12

    A. Neumaier

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    There is no well-defined notion of vacuum energy in cosmology. Its reality status is similar to that of virtual particles popping in and out of existence for a very short time.
     
  14. Aug 22, 2018 #13

    Demystifier

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  15. Aug 22, 2018 #14

    A. Neumaier

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    Only the discussion in terms of vacuum energy, which is similar to discussions of QFT in terms of virtual particles. Its value is nil, apart from making it seemingly less abstract.

    The cosmological constant is a property of the state of our observable universe, which surely isn't in a vacuum state. Thus it can have nothing to do with the properties of the vacuum state.
     
  16. Aug 22, 2018 #15

    Demystifier

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    If it's true, then it's new and very important. If you are convinced that you are right, then you should publish it.
     
  17. Aug 22, 2018 #16

    bhobba

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    Indeed there is IMHO. Normal ordering is just a way of handling the issue in a way that makes sense - but skirts the main issue - why do we have to resort to it in the first place. Like I said I am hopeful a better understanding of Effective Field Theory on my part will help - at least me anyway. We will see.

    Thanks
    Bill
     
  18. Aug 22, 2018 #17

    A. Neumaier

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    Well, he also talks (on p.3 of his paper) about the ''demonstration in the Casimir effect of the reality of zero-point energies'' and ''the gravitational force between the particles in the vacuum fluctuations'', which is virtual particle nonsense.

    Don't take verbal talk involving bare, virtual stuff too serious - the meat is always only in the (renormalized) formulas. For lack of a good renormalization prescription for quantum gravity we can say very little definite. Weinberg's paper (like much in quantum gravity) is just speculation because we don't have anything better.
     
  19. Aug 22, 2018 #18

    Demystifier

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    Vacuum fluctuations are not nonsense, even if virtual particles are.

    Are you going to teach Weinberg renormalization? :wideeyed:
     
  20. Aug 22, 2018 #19

    A. Neumaier

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    But he talked about ''particles in the vacuum fluctuations'', which is nonsense. The vacuum contains zero particles at all times. This shows that his discussion must be taken with large amounts of grains of salt. When he talks about summing up the zero point energies of modes, if his argument were stringent, it would also apply to QED, where we know that energies are not horrendously large.

    No, but I know that he doesn't know how to renormalize gravity. Currently nobody knows!
     
  21. Aug 23, 2018 #20

    vanhees71

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    I've never understood this argument of yours. You can simply add a constant ##E_0 \hat{1}## to the usually used normal-ordered ##\hat{H}##, and the ground-state energy is ##E_0##, where ##E_0## can take any real value. The ground state is given by
    $$\hat{\rho}=|\Omega \rangle \langle \Omega|,$$
    and that's Poincare invariant, particularly it's invariant under temporal translations, no matter which value of ##E_0## you choose
    $$\exp(-\mathrm{i} \hat{H} t) \hat{\rho} \exp(+\mathrm{i} \hat{H} t)=\exp(-\mathrm{i} E_0 t) \hat{\rho} \exp(+\mathrm{i} E_0 t)=\hat{\rho}.$$
    I don't see, where there's anything forbidden in this argument when used in relativistic QFT. Of course, it holds either in non-relativistic QT.
     
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