16
$\begingroup$

From Wikipedia, Pauli has said in his Nobel lecture that "It is clear that this zero-point energy has no physical reality". This feels natural - I've always been slightly puzzled by the existence of the zero-point energy $\frac{\omega}{2}$ for the quantum harmonic oscillator. Yet the math seemed 'obvious' to me, until today.

Let $H$, $a$ and $a^\dagger$ be the Hamiltonian, annihilation and creation operators, respectively, defined by:

$$H = \frac{1}{2}(p^2 + \omega^2 x^2)$$ $$a = \frac{x}{\sqrt{2\omega}} + i\sqrt{\frac{\omega}{2}}p.$$

Then one can find the following relations:

$$H = \omega(a^\dagger a + \frac{1}{2})$$ $$H|n\rangle = (n + \frac{1}{2})\omega|n\rangle$$

Where $|n\rangle$ is the $n$'th eigenstate of $H$. This gives a zero point energy (ZPE) of $\omega / 2$. So far so good. Yet today I was reading David Tong's lecture notes on QFT. In the second chapter, he makes the remark that classically:

$$H_{\text{classical}} = \frac{1}{2}(p^2 + \omega^2 x^2) = \frac{1}{2}(\omega x - ip)(\omega x + ip).$$

Let $H'$ be the quantum analogue of the right hand side expression. Then:

$$H' = \frac{1}{2}(\omega x - ip)(\omega x + ip) = \omega a^\dagger a$$ $$H' = H - \frac{1}{2}\omega.$$

Thus yielding a different Hamiltonian operator, due to the fact that $x$ and $p$ do not commute. In particular, the ZPE is zero for $H'$:

$$H'|0\rangle = (H - \frac{1}{2}\omega)|0\rangle = 0.$$

I understand the Hamiltonian operator is ultimately defined as the generator of time evolution. There seems to be no difference between $H$ and $H'$ in that regard. Why is there a preference for $H$ over $H'$ in QM textbooks? And is this notion of strictly non-zero ZPE, which is often linked to the uncertainty principle, incorrect? What are the consequences for our interpretation of QM phenomena if the ZPE is indeed arbitrary?

EDIT: Thanks for all your responses. It's clear to me now that I was confusing zero-point energy and zero-point motion.

Improve this question
$\endgroup$
2

6 Answers 6

Reset to default
16
$\begingroup$

The point of the ZPE is not that it is non-zero, because that could indeed be undone by redefining the potential energy as $$ V\rightarrow \overline{V}=V-E_0 \ . $$ The resulting potential energy has now a somewhat unintuitive dependence on $\hbar$, but the new ZPE is indeed $\overline{E}_0=0$.

However, the important thing is not whether $E_0$ or $\overline{E}_0$ is zero or not, but that the difference $E_0-V_\text{min}=\overline{E}_0-\overline{V}_\text{min}$ is greater than zero, where $V_\text{min}=V(\vec{r}_\text{min})$. This has the important meaning that a particle in an energy eigenstate cannot be strictly localized in the mimimum of the potential (or in its arbitrarily small neighborhood), and this is where the usual arguments with the uncertainty principle come into play. There is obviously no such restriction in classical mechanics, where a particle could be standing still in $\vec{r}_\text{min}$ with zero kinetic energy.

This may not seem such a big deal for the harmonic oscillator, but consider instead a hydrogen atom, where the minimum of the potential energy is either $-\infty$ or a very large negative number, depending on whether a point-like or a small finite nucleus is assumed. The zero line of the energy could be again shifted in any way, but the shift-invariant $E_0-V_\text{min}>0$ means that the electron cannot fall into the nucleus. (This just means that the electron cannot be strictly localized at the nucleus with a delta-function-like density, not that it does not have detection probability at that region; actually, $S$-states always have a non-zero detection probability around the nucleus.)

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ +1 much better physical interpretation than mine $\endgroup$ Commented 20 hours ago
  • $\begingroup$ You're describing zero-point motion (zero-point fluctuations), which is due to uncertainty relations, and thus can't be eliminated in orthodox quantum theory. Zero-point energy refers to energy of the ground state, which can be made zero by using the normally ordered Hamiltonian $H'$. $\endgroup$ Commented 19 hours ago
  • $\begingroup$ Yes the physically meaningful information is that the groundstate has finite $\langle x^2\rangle$ and $\langle p^2\rangle$ $\endgroup$ Commented 18 hours ago
  • 3
    $\begingroup$ @JánLalinský By itself, ZPE is just as meaningless as any other (non-GR) absolute energy, no argument here. What actually has physical relevance is the fact that the ZPE is above the minimum of the potential energy, which is obviously independent of constant shifts. This is all I said. $\endgroup$ Commented 18 hours ago
  • $\begingroup$ You said that, but referred to this idea as ZPE, which is confusing. I think nobody questions validity of uncertainty principle in orthodox QT, and thus zero-point motion/fluctuations in the general sense. But ZPE in EM field is a different idea; it is controversial, and its reality still hasn't been established. $\endgroup$ Commented 18 hours ago
6
$\begingroup$

Why is there a preference for $H$ over $H^\prime$ in QM textbooks?

Mostly just a mix of laziness/convenience and familiarity. Students are much more familiar with $H$, and professors who even pondered about $H^\prime$ tend to decide that it is not worthwhile to waste precious class time on the difference.

In QFT, there is a big difference, because we integrate over all possible $\omega$, and that means that $H$ is not tolerable, for it will give us infinite ZPE. The obvious way out of this conundrum is simply to adopt $H^\prime$ as gospel.

I mean, there is also no good reason to take $H$ in quantum theory in the first place; $H$ is a classical expression that is familiar, and we must always be prepared that quantum theory might force us to consider a replacement that, in the classical limit, reduces to the same thing.

In particular, we might start with wanting to guess a thing that behaves like $H$, because we know that this is the correct classical behaviour. And then we derive that $H$ suffers from this infinite ZPE problem, and the solution is to subtract the offending term by adopting $H^\prime$. We already see that $H^\prime$ leads to the same and correct classical limit, so this replacement should be seen as always good.

However, there might be some evidence that ZPE is real. By that I do not mean the situation in QFT; in QFT the only tolerable choice is $H^\prime$. But, when you combine systems, say, in a crystalline structure you get an approximately QHO interaction between ions, then this is a emergent, non-fundamental thing that arises from more-fundamental interactions. Then the ZPE contribution is not infinite, and there seems to be experimental evidence that ZPE for these cases is really there.

Again, only in the situation where ZPE contribution is not infinite do we really have solidly good evidence that it is really there. See, for example, https://physicsworld.com/a/zero-point-motion-of-atoms-measured-directly-for-the-first-time/ and a lot more will appear in any Google search.

Improve this answer
$\endgroup$
7
  • $\begingroup$ At least the wikipedia page about ZPE mentions that liquid helium does not freeze due to ZPE. $\endgroup$ Commented 23 hours ago
  • $\begingroup$ Yes, I also know that; it is an oft-quoted piece of trivia on the topic. However, it falls into the category that I had explicitly mentioned as the ions thingy. You might want to read dennismoore's take below on how to interpret that much better. $\endgroup$ Commented 20 hours ago
  • $\begingroup$ It makes no sense to split physics into a part with QFT and no crystals or molecules and a part with crystals, molecules and no QFT. $\endgroup$ Commented 6 hours ago
  • $\begingroup$ 'an oft-quoted piece of trivia' Superfluid helium is by no means trivial. Only someone specialised in QFT with no interest in condensed matter would say this. $\endgroup$ Commented 6 hours ago
  • $\begingroup$ @my2cts I worked in condensed matter. $\endgroup$ Commented 6 hours ago
5
$\begingroup$

For single or finite number of harmonic oscillators, it does not matter - both $H$ and $H'$ are valid Hamiltonians with the same consequences. They just define different values for energy (energy has to be defined somehow - in quantum theory, usually we define it based on the eigenvalues of the time-independent Hamiltonian adopted). But only differences of energy can be determined experimentally, and these are the same in both cases. So it's like a gauge freedom in EM theory - the change does not affect physical results that can be checked in experiments.

The first one, $H$, is preferred in textbooks on quantum mechanics for historical and simplicity reasons. It's what we get by straightforward canonical quantization of the usual Hamiltonian function of an harmonic oscillator, when we replace $x,p$ by the corresponding operators. The second one, $H'$, is a bit newer invention that makes the ground state energy equal to zero.

In quantum field theory, e.g., quantum theory of EM radiation, an infinity of harmonic oscillators appears when describing the field, because of no limit to how high wavenumbers are possible in the Fourier expansion of the field. If we base our calculations on $H$, total zero point energy is infinite, and also all other Hamiltonian eigenvalues are infinite, and this breaks the usual mathematics. Even if we try to tweak the theory by imposing a maximum wavenumber possible, or some other softer form of limit that makes the sums finite, the limit has to be so high to make the theory agree with observations that we get zero-point energy contribution that is immense, yet non-detectable. $H'$ makes total zero point energy zero and all its eigenvalues finite, which is an improvement. Thus in quantum theory of radiation, $H'$ is preferable.

When we use $H'$, this does not affect the uncertainty principle. This is because zero-point energy being zero does not mean the zero-point motion is zero.

When we have one harmonic oscillator in its ground state, position and momentum uncertainties obey the Heisenberg uncertainty relation

$$ \Delta x \Delta p~ \geq~ \frac{\hbar}{2}, $$ regardless of which Hamiltonian we use.

In other words, value of zero-point energy (zero or non-zero) does not in any way confirm/deny validity of the uncertainty principle.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 'But only differences of energy can be determined experimentally, so it's like a gauge freedom in EM theory, it does not affect physical results that can be checked in experiments.' This is incorrect. Consider an atom. Electromagnetically we can only measure its transition energies, but by gravity we can measure its rest energy. If gravity is included absolute energies have physical meaning. $\endgroup$ Commented yesterday
  • 1
    $\begingroup$ @my2cts quantum theory does not cover gravitational effects of systems it describes, so lack of experimental evidence for strong gravity of EM field near its ground state does not imply that quantum theory of this field has to be low; the two theories and energy concepts are just too different. So the gravity argument is problematic. Rest energy of an atom can be measured also electromagnetically, by measuring its motion in known electric or magnetic field. But zero point energy is an issue only for fields. Finite number particle systems can have zero point energy. $\endgroup$ Commented 23 hours ago
  • $\begingroup$ You have a good point about measuring the ratio of charge to inertial mass by electromagnetic force. However, your argument that we should distinguish between different fields is problematic. Any quantum harmonic oscillator has ZPE. Weighing such a system with sufficient accuracy in principle allows to determine the ZPE of the system. Quantum field theory is based on the quantum harmonic oscillator. All QHEs are equal. $\endgroup$ Commented 21 hours ago
  • $\begingroup$ @my2cts A field harmonic oscillator is physically very different from a particle harmonic oscillator. Also mathematics gets different. There are formal similarities, but we should not overstate them. $\endgroup$ Commented 21 hours ago
  • $\begingroup$ @my2cts I agree that every QT harmonic oscillator has zero point motion in its ground state, this is due to uncertainty principle. But Hamiltonian and energy of field is best defined so that ground state has zero energy. This reduces the number of problems with the theory. $\endgroup$ Commented 21 hours ago
2
$\begingroup$

Is zero-point energy a mathematical artifact?

Generally, unless you have already chosen some conventional zero of your energy, the energy can always be shifted by a constant value, since it is only energy differences that matter. So, in this sense, yes, it is a "mathematical artifact."

Why is there a preference for H over H′ in QM textbooks?

Simplicity of presentation, convention, and other opinion-based reasons.

What are the consequences for our interpretation of QM phenomena if the ZPE is indeed arbitrary?

Nothing, really. You can shift the energy by a constant, as explained above.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 'it is only energy differences that matter' Consider my comment to @JanLalinsky's answer. $\endgroup$ Commented yesterday
1
$\begingroup$

While the zero point energy in an isolated system is not observable, differences in zero point energy certainly are observable. For example, if you are able to slowly change the frequency $\omega$ of your oscillator, the zero point energy will change. If you start with the ground state for a particular value of $\omega$, and then increase $\omega$, the resulting state will now have a higher energy than the ground state of the new oscillator. This is a reflection of the fact that the zero point energy of the original oscillator was nonzero and is true even if you make the change very slowly.

There is an important effect in Quantum Field Theory called the Casimir effect, which is also due to difference in ground state energy. Basically, if you take a box in empty space and try to push on it's walls to make the box smaller, you will feel a very slight resistance because you are increasing the ground state energy of the vacuum within the box.

Improve this answer
$\endgroup$
4
  • $\begingroup$ > "the zero point energy will change" Only if we define energy so that the ground state energy is $\frac{1}{2}\hbar \omega$. If we define energy by the normally ordered Hamiltonian $H'$, then zero point energy is 0, and it does not change when $\omega$ changes. $\endgroup$ Commented 20 hours ago
  • 1
    $\begingroup$ The Casimir force calculation based on zero-point energy works only in the idealized case of boundary conditions appropriate for perfect conductor. Real Casimir forces are between material bodies, and are due to real EM interactions. See arxiv.org/abs/hep-th/0503158 $\endgroup$ Commented 19 hours ago
  • $\begingroup$ @JánLalinský I agree. This implies that the quantum harmonic oscillators also must be based on real physical interactions. This solves the infinite ZPE problem. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ This is my unpublished solution of the cosmological constant problem or vacuum catastrophe. en.wikipedia.org/wiki/Cosmological_constant_problem $\endgroup$ Commented 13 hours ago
0
$\begingroup$

H' is incorrect. The ground state of the harmonic oscillator will still have kinetic and potential energy and the sum of these energies will still be $\hbar \omega/2$. These energies contribute to inertial and gravitational mass, but this would be practically impossible to demonstrate experimentally. Also, the zero point energy gives rise to the Casimir Polder effect. This effect is very real and needs to be taken into account in the design of MEMSs (micro-electromechanical systems).

In QFT an infinite continuum of degrees of freedom is postulated, to be compared with a finite number of DoF in condensed matter. This causes a quasi infinite vacuum energy, only limited if the universe is indeed finite. This infinite ZPE can of course not be real. One way to solve this is to use H', but then you lose very real physics. The alternative is to consider only those harmonic oscillators that correspond to actual system DoF, as in condensed matter physics.

As to the question whether the Casimir-Polder effect is due to 'real' interactions or ZPE, I believe it is the same thing. The boundary conditions of the ZPE calculation are just an abstraction of metal plates. Boundary conditions that apply to all energies are unphysical. Physicists should not entertain $\it two$ superficially entirely different explanations for $\it one$ phenomenon, depending on the relevant field of physics.

This analysis solves the infamous Cosmological constant problem, which is part of the list of unsolved problems in physics.

/whining However, I have no expectation that a manuscript with this analysis will ever be accepted by $\it any$ journal. /whining

Improve this answer
$\endgroup$

Your Answer

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

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.