7
$\begingroup$

In the context of canonical quantization, the ground state/zero-point energy of a harmonic oscillator is $$ E_0 = \frac{1}{2}\hbar \omega. $$ The vacuum is alleged to be permeated with this zero-point energy of quantum fields, hence causing the exorbitantly large cosmological constant issue.

However, the zero-point energy can be normal ordered away (Wick ordering). And path integral formulation automatically takes care of normal ordering. In other words, path integral formulation (and normal ordering canonical quantization) doesn't suffer from the zero-point energy issue.

The Casimir effect is usually cited as an evidence of zero-point energy. Nevertheless, a 2005 paper (https://arxiv.org/abs/hep-th/0503158v1) states that "Casimir forces can be computed without reference to zero-point energies...The Casimir force is simply the (relativistic, retarded) van der Waals force between the metal plates".

So is zero-point energy a spurious artifact of canonical quantization?

Note that in the absence of zero-point energy, the cosmological constant problem still stands because of the vacuum energy shift resulted from spontaneous electroweak symmetry breaking.

Improve this question
$\endgroup$
2
  • $\begingroup$ Consider a Hamiltonian $H = \alpha x^2 / 2 + \beta y^2/2$ with $[x,y]=i \gamma$. It turns out that $\hbar \omega = \gamma \sqrt{\alpha \beta}$ and the zero point motion is $x_0^2 \equiv \left \langle 0 | x^2 | 0 \right \rangle = (1/2)\gamma \sqrt{\beta / \alpha}$. Knowing that the motion is harmonic, the maximum value of $x$ is $x_\text{max}^2 = 2 x_0^2$, so the energy when $x$ is at it's maximum is $E = (1/2) \alpha x_\text{max}^2 = (1/2) \gamma \sqrt{\alpha \beta} = (1/2) \hbar \omega$. (continued) $\endgroup$
    – DanielSank
    Jun 29, 2018 at 21:17
  • 2
    $\begingroup$ (continued) This is why people like to say that the zero point energy $(1/2) \hbar \omega$ is responsible for the zero point motion $x_0$. Note, however, that the results quoted in the previous comment do not in any way rely on writing the Hamiltonian as $H = \hbar \omega ( 1/2 + a^\dagger a)$. In other words, we can compute the zero point motion without having to ever see that the Hamiltonian has a form where there's an obvious zero point energy. (See here for derivation of the results used in the previous comment). $\endgroup$
    – DanielSank
    Jun 29, 2018 at 21:20

3 Answers 3

Reset to default
2
$\begingroup$

This is the data that I use to convince my students. It is the result of a measurement of the velocity of argon atoms. The method is inelastic neutron diffraction, no need to go into details, but one can read those in Fradkin et al, 1994.

In this figure, $mv^2/(2k_B)$ is plotted as a function of temperature. At higher temperature this approaches $3/2\ T$ according to classical physics and equipartition. At low temperature the derivative decreases, in agreement with the third law of thermodynamics that $c_p$ must go to zero.

enter image description here.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Excellent post. I have a question. The prediction that $c_p$ goes to zero, so that the eqyipartituon theorem does not hold at low T is a classical prediction. What is the impact of quantum physics? $\endgroup$
    – my2cts
    Aug 25, 2019 at 19:37
  • 1
    $\begingroup$ @my2cts The third law (Nernst's law) may be classic in the sense that it is classical phenomenological thermodynamics, but quantum statistics is required to explain it. Classical Boltzmann statistics gives equipartition. $\endgroup$
    – user137289
    Aug 25, 2019 at 19:47
  • 1
    $\begingroup$ Love this post, although how does this in any way adress the zero poinr energy of the vacuum? $\endgroup$
    – lalala
    Aug 26, 2019 at 5:30
  • $\begingroup$ There must be something misleading about that plot. It is not possible to extract energy from a quantum system in its ground state, so how is the neutron scattering experiment sensitive to the ground state energy? I think there are hijinx afoot here and I do not trust the interpretation suggested by this answer. $\endgroup$
    – DanielSank
    Sep 18, 2019 at 4:04
  • $\begingroup$ DanielSank The measured points start at 20 K. The energy loss spectra give information about the average $v^2$ in the solid, look at the paper for the nasty detail. But @lalala is right that this was not about the vacuum, sorry for not reading the question past the header. $\endgroup$
    – user137289
    Sep 18, 2019 at 8:43
0
$\begingroup$

Zero point energy is real but indeed the radiative form of van der Waals force between objects. As far as I know it still has to be worked out how the two approaches, which lead to the same result, are compatible. The resulting Casimir force is a general effect, not only relevant for - idealised - metal plates but to all material objects.

Improve this answer
$\endgroup$
-1
$\begingroup$

From the vacuum state in classical physics leads to zero point energy in modern physics.

In classical and modern physics, one cannot create a space, where there is nothing. That is, it is impossible to create "nothing".

Classical physics uses pistons and cylinders to create a vacuum, but in that space, there is heat radiation. See here: https://overunity-generator-guide.blogspot.com/2019/09/the-classical-vacuum-zero-point-energy.html

Modern physics, trendy science (quantum physics) cannot deny the zero point energy, nor can today's technology create something "nothing".

The article on zero point energy on Wikipedia mentions the Ether theory a lot. This proves that Ether Physics has an importance in zero point energy. See here: https://en.wikipedia.org/wiki/Zero_point_energy

Clearly, the zero point energy is real.

Report: Point energy is not mined in Ether. See here: https://overunity-generator-guide.blogspot.com/2019/08/zero-point-energy.html

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.