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A quantum linear harmonic oscillator has a definite non-zero ground state energy $E_0=\frac{1}{2}\hbar\omega\neq 0$. However, in this energy eigenstate, the position and momenta are uncertain and their standard deviations satisfy an uncertainty relation $(\Delta x\Delta p_x)_{|0\rangle}=\frac{\hbar}{2}$. I want to ask whether this uncertainty is related to the fact that the ground state energy is nonzero and if yes, how exactly this value of $E_0$ is obtained?

Can I extrapolate this inference in free quantum field theory (such as free Klein-Gordon theory)? A free KG field has an infinite ground state energy. Can it be attributed to an uncertainty relation acting between the field $\phi(x)$ and the corresponding conjugate momentum operator $\pi(x)$?

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    $\begingroup$ Uncertainty is related—as in all cases except the energy-time relationship—to non-commuting observables (in the case of a massive harmonic oscillator those observables are position and momentum, but there are other was to construct oscillators). $\endgroup$
    – dmckee --- ex-moderator kitten
    Nov 13, 2016 at 19:01
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    $\begingroup$ Define a new Hamiltonian $H' = H - \hbar \omega / 2$. No zero point energy, same uncertainty relation. $\endgroup$
    – Javier
    Nov 13, 2016 at 19:45
  • $\begingroup$ @Javier- I think your definition of new Hamiltonian is artificial because it does not mimic the classical Hamiltonian and is just to hide the zero pint energy under the carpet (though it is often done as in quantum field theory). You cannot deny the existence of a nonzero energy because it has effects. Differences in the energy of vacuum can be measured in field theory. $\endgroup$
    – SRS
    Nov 13, 2016 at 20:28
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    $\begingroup$ @SRS Why would you expect it to mimic the classical Hamiltonian? nature is quantum, not classical. Classical mechanics is constructed as a limit of quantum mechanics, and not the other way around. In principle, $H$ is as good as $H'=H+c$ for any $c$. There is no principle that singles out a preferred origin of energies. The classical expression is no exception: the formula $H=p^2+q^2$ is as good as $H=p^2+q^2+c$, or even $H=(p+iq)(p-iq)$ (where here there is no zero-point energy even after quantisation) [in any case, Javiers point is that the z-p-energy is unrelated to the uncertainty relat] $\endgroup$
    – AccidentalFourierTransform
    Nov 13, 2016 at 20:31
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    $\begingroup$ Nice point. I haven't thought about it that way. $\endgroup$
    – SRS
    Nov 13, 2016 at 20:43

1 Answer 1

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It can easily be seen that there is no direct relationship between the zero point energy and the uncertainty relation. If the Hamiltonian is

$$H = \frac{p^2}{2m} + \frac12 m \omega^2 x^2$$

then defining

$$H' = H - \frac12 \hbar \omega$$

the zero point energy disappears but the uncertainty remains, since adding a constant to the Hamiltonian won't change any observable (as long as we steer clear of gravity).

This is what is usually done in QFT: most books say something along the lines of "the zero point energy is infinite, so let's substract this infinite contribution from the Hamiltonian and be done with it". The commutation relations aren't affected, because they are independent from the Hamiltonian.

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    $\begingroup$ +1 but a slight comment to the last paragraph: the commutation relations are independent of a constant term in the Hamiltonian, but may change if we add field-dependent terms. $\endgroup$
    – AccidentalFourierTransform
    Nov 13, 2016 at 20:52
  • $\begingroup$ @AccidentalFourierTransform: of course, I forgot. Though technically the commutation relations would only change if we add something with $\dot{\phi}$ since that would change the definition of field momentum, right? $\endgroup$
    – Javier
    Nov 13, 2016 at 21:23
  • $\begingroup$ Although there is everything I agree with, I think since in presence of gravity the zero-point energy couples to it, you better not hide it by defining a new Hamiltonian. Look at this MIT lecture on how the bound on the minimum energy results from position momentum uncertainty. youtube.com/watch?v=Oi-JCJePLlc $\endgroup$
    – SRS
    Nov 13, 2016 at 21:31
  • $\begingroup$ @Javier yes, that's what I had in mind (and in constrained systems, the commutation relations are constructed with the Dirac algorithm, which has an explicit input from the Hamiltonian, and so the brackets themselves depend on the actual form of $H$). $\endgroup$
    – AccidentalFourierTransform
    Nov 13, 2016 at 21:43
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    $\begingroup$ @SRS yes, the zpe couples to gravity, but again, there is no fundamental principle that tells us the value of the zpe. Our put it another way, you may couple $H$ to gravity or couple $H+c$ to gravity. Both Hamiltonians give rise to different dynamics: here the $c$ term is not irrelevant any more. But it is still arbitrary: there is no fundamental principle that tells us its value. It is a parameter that we must measure, not calculate. In other words, your $H$ need not be better that Javier's $H'$: it is nature who's gonna decide which works better, and we must resort to experiments to know $\endgroup$
    – AccidentalFourierTransform
    Nov 13, 2016 at 21:50

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