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In classical (say Hamiltonian) mechanics, adding a constant energy offset has no effect on the dynamics of the physical system. One way to understand this might be to understand that Hamilton's equations of motion which determine the time evolution of the system variables depend only on derivatives of the Hamiltonian, and the derivative of a constant is zero.

In the Heisenberg picture we get the Heisenberg equations of motion in place of the Hamilton's equations (they are the exact same save for hats on the variables). By the same argument a constant offset to the energy has no effect on the dynamics.

In the Schrodinger picture we know that an energy offset simply adds a global time-evolving phase to every state. This global phase has no observable effects.

Main Question:

How then can any argument be made that zero point energy has an physical consequences?

Note 1:

I would say that the following Hamiltonians result in the exact same physics:

\begin{align} \hat{H} =& \hbar \omega \left(\hat{a}^{\dagger}\hat{a}+ \frac{1}{2}\right)\\ \hat{H} =& \hbar \omega \hat{a}^{\dagger}\hat{a}\\ \hat{H} =& \hbar \omega \left(\hat{a}^{\dagger}\hat{a}+ 1,000,000\right) \end{align}

Note 2:

Note that I distinguish between zero point energy and zero point motion. I do believe that zero point motion is physical and has observable consequences. I also believe that the theory predicts zero point motion even in absence of including zero point energy in the Hamiltonian. See Consistent, complete, and generalized description of the quantum harmonic oscillator

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    $\begingroup$ Consider General Relativity. $\endgroup$ – G. Smith Feb 12 at 18:36
  • $\begingroup$ @G. Smith in General relativity do absolute energy levels matter? If so, then why is it so dramatically different than previous classical theories and many quantum theories? $\endgroup$ – Jagerber48 Feb 12 at 19:49
  • $\begingroup$ I don't understand your question. You argue perfectly well that a constant shift of the Hamiltonian has no physical consequence, and then you ask us to refute you? $\endgroup$ – Ryan Thorngren Feb 12 at 19:52
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    $\begingroup$ On the right side of the Einstein field equations is the energy-momentum tensor. If the energy density in this tensor could be adjusted up and down by an arbitrary constant, the curvature tensor on the left side would change, which would not make sense because curvature is a measurable geometrical quantity. So, yes, in GR absolute energy matters. $\endgroup$ – G. Smith Feb 12 at 21:38
  • $\begingroup$ @G.Smith For others, Adding on to what you said above, any absolute energy change in GR is effectively a cosmological constant. $\endgroup$ – R. Rankin Feb 13 at 3:38
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I think what is typically meant by "zero-point energy" in this context is actually zero-point kinetic energy. Exactly as you say, a constant offset $E_0$ in the Hamiltonian $H = H_0 + E_0 1\!\!1 $ has no consequences on the dynamics, and classically speaking, it corresponds to picking a different reference point for your potential energy function.

The kinetic energy on the other hand can have a non-zero expectation value in the ground state of the system, which is not arbitrary at all. For example, in the case of the Harmonic oscillator, this is $$\frac{1}{2} m \omega_0^2\langle x^2 \rangle =\frac{\hbar\omega_0}{4},$$ which is not at all just a matter of picking potential energy reference. It means that the oscillator is physically moving even at zero temperature due to quantum fluctuations.

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  • $\begingroup$ Perhaps people mean the expectation of the ground state kinetic energy when they talk about zero point energy. If this is the case then I address it in my second note and you address in your answer. However, in my experience people often refer to the factor of $\frac{1}{2}\hbar \omega$ as the zero point energy and this is what I'm asking about. $\endgroup$ – Jagerber48 Feb 12 at 19:46
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Outside of general relativity, shifting the zero point energy doesn't matter. It's changes in the amount of zero point energy that are observable.

For example, suppose you trap bosonic atoms in a harmonic potential and cool them. At a low enough temperature, most of them will condense into the ground state, a result known as Bose-Einstein condensation. But when you turn the trap off, the atoms will start to spread out, as shown at right in this famous picture:

enter image description here

This is because the $\hbar \omega/2$ zero point energy of each atom becomes ordinary kinetic energy.

As another example, consider the Casimir effect, an attractive force between two metal plates. The fact that the total zero point energy of all the modes in the electromagnetic field is infinite is completely irrelevant. The effect comes about because moving the plates changes the frequencies $\omega_i$ of some of the modes, and therefore changes the total zero point energy, which is the sum $\sum_i \hbar \omega_i/2$. Since the energy changes with the positions of the plates, you get a force between them.

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  • $\begingroup$ Your first point about atoms in the ground state of a harmonic trap can be explained without reference to zero point energy. If you prepare atoms with a spatial pattern matching that of the ground state of a harmonic trap (whether you use a harmonic trap to do it or not) and then expose them to the free space kinetic Hamiltonian then we expect them to fly away according to the Schrodinger equation. Of course their dynamics are different under the new Hamiltonian? $\endgroup$ – Jagerber48 Feb 12 at 20:07
  • $\begingroup$ Regarding the Casimir effect I don't understand it as well but there seem to references in which the Casimir effect is explained with no reference to vacuum energy: See en.wikipedia.org/wiki/… $\endgroup$ – Jagerber48 Feb 12 at 20:07
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    $\begingroup$ @jgerber It is a universally true principle in physics that anything can be explained without referring to any specific other thing, though. You can also explain everything in quantum mechanics without ever using the Schrodinger equation, but that doesn't mean it's a useless idea, because for many things the easiest description is through the Schrodinger equation. Same for zero point energy. $\endgroup$ – knzhou Feb 12 at 20:11
  • $\begingroup$ @jgerber Derivations of the Casimir effect "without" zero point energy just are doing the same derivation with different words. It's like saying states don't exist because you can do a QM calculation in Heisenberg picture. It's all just slightly different ways of describing the exact same theory; no feature of the theory can be weirder than any other part if they all get you to the exact same place. $\endgroup$ – knzhou Feb 12 at 20:12
  • $\begingroup$ @knzhou > "just are doing the same derivation with different words". Not the calculations for finite dielectric constant systems based on retarded van der Waals forces by Lifshitz and others, which result in more general formulae than the Casimir zero-point energy method can provide. Casimir's method works only in the idealized case of perfectly conducting plates as a lucky math trick, for real plates of finite conductivity its results are incorrect. See Grundler G., The Casimir-Effect: No Manifestation of Zero-Point Energy, arxiv.org/abs/1303.3790 $\endgroup$ – Ján Lalinský Feb 12 at 21:31
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A constant zero-point term in total energy expression changes nothing about internal dynamics of the system. It does affect its behaviour in external field though. Adding a positive total energy $\Delta E$ to a system means this system has then greater inertial mass by $\Delta m = \Delta E/c^2$.

So in theory one could test the prediction that there is a zero-point energy term proportional to $\hbar \omega$, by making careful measurements of weight of harmonic oscillators of known same composition, differing in the oscillation frequency.

Similar mass change effect is known to exist due to chemical bonds, and zero-point energy is of similar order of magnitude, so maybe it can be measured as a mass effect too.

For example, by changing isotopes one can change mass of nuclei in a lattice without changing chemical bonds too much, or only in a controlled accountable way. Then natural frequencies of oscillation in the lattice change and associated zero-point energy changes as well. Maybe the associated change in inertial mass or weight can be measured.

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The zero point energy should be the source of gravity and of space curvature. This leads to the vacuum catastrophy. Your eq. 3 makes thus problem even worse if that is possible. The zero point energy is also believed to cause the Casimir-Polder effect. https://en.m.wikipedia.org/wiki/Cosmological_constant_problem#:~:text=In%20cosmology%2C%20the%20cosmological%20constant,suggested%20by%20quantum%20field%20theory.

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You are correct that a constant energy shift does not affect the dynamics of the quantum harmonic oscillator. However, relative energies do matter. It is conventional (at least in non-relativistic quantum theory) to define the energy of the free zero-momentum eigenstate as zero, and define other energies relative to that energy.

Suppose we start with a free particle (in other words, $H=P^2/(2m)$). Then the eigenstate of $P$ with eigenvalue zero also has zero energy. (Technically, this state is not normalizable and not a proper member of the Hilbert space, but we can still get proper states with arbitrarily small energy.) Now, suppose we turn on a potential so that $H$ is now the harmonic oscillator Hamiltonian. The zero point energy will have to be greater than the energy of the free particle. Now if we added a constant shift to $H$, then we could make the zero point energy equal to $0$. However, then the energy of an unmoving free particle would be negative.

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