Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In a paper, I ran into the following definition of the zero point fluctuation of our favorite toy, the harmonic oscillator: $$x_{ZPF} = \sqrt{\frac{\hbar}{2m\Omega}} $$ where m is its mass and $\Omega$ its natural frequency. However, when I try to derive it with simple arguments, I think of the equality: $$E = \frac12 \hbar\Omega=\frac12 m \Omega^2 x_{ZPF}^²$$ (using the energy eigenvalue of the $n=0$ state) giving me:

$$x_{ZPF} = \sqrt{\frac{\hbar}{m\Omega}} $$ differing from the previous one by a factor $\sqrt2$. I am just puzzled, is it a matter of conventions or is there a fundamental misconception in my (too?) naive derivation?

share|improve this question
If you do simple dimensional estimates you should not expect the numerical factors to come out right! –  Fabian Dec 15 '12 at 17:09
Yeah, but I tried to do more that simple dimensional analysis, I wrote down an equality between the energy of the vacuum and the oscillating energy of an harmonic oscillator with $<x^2> = x_{ZPF}^2$ when I should have taken $<x^2> = \sqrt2 x_{ZPF}^2$ and so my question is again is it coming from a convention or is it physically motivated? –  Learning is a mess Dec 15 '12 at 21:59
writing down an equality between the energy of the vacuum and the oscillating energy of an harmonic oscillator is nothing more than dimensional analysis. I hope that you would agree with me that in principle you are supposed to solve the Schrödinger equation to find the ground state wave function! If you do that you will find the expression for $\langle x^2 \rangle$ with all the numerical prefactors. –  Fabian Dec 15 '12 at 22:29
I agree with Fabian: you can get any expectation value you want from the wavefunction (Hermite Polynomial) for your chosen state. Also, when you used $\frac{1}{2}m\Omega^2x_{zpf}^2$ for the energy, didn't you ignore the kinetic term in the Hamiltonian? –  twistor59 Dec 16 '12 at 21:36
Yes but equipartition says that in average the energy is distributed between the potential and the kinetic one. So I should add a factor 1/2 in front of it, when I should, for an unknown reason, add a factor 2 to get the correct formula. –  Learning is a mess Dec 16 '12 at 23:05
show 1 more comment

2 Answers

up vote 2 down vote accepted

I think this is a combination of both a convention and a physical problem. You are equating the energy eigenvalue (ie, the total energy) to an expression that contains only $x_{ZPF}$, and does not contain $p$ at all. In other words, you are equating the total energy to a potential energy. This would be analogous to equating $E_\mathrm{total} = \frac{1}{2}kA^2$ to find the amplitude $A$ of a classical harmonic oscillator. The result is that you are using $x_{ZPF}$ to mean the "amplitude" of the zero-point fluctuation. The true result, as Ondrej Cernotik's answer derives, uses the rms value $x_{ZPF} = \sqrt{\langle\hat x^2\rangle}$. So that's the sense in which it is a convention.

The sense in which it is a real physical problem is that the "amplitude" of a quantum oscillator isn't really a well-defined, measurable thing. The quantum oscillator has a non-zero probability amplitude going all the way out to infinity. The rms value is well-defined and easy to measure. So that's the preferred definition.

share|improve this answer
You could also argue that when defining zero point fluctuations in terms of variance of the position, the value will be smaller than the amplitude. I bet you would find that the variance (or its square root) will be exactly $\sqrt{2}$ times smaller. –  Ondřej Černotík May 24 '13 at 16:20
@OndřejČernotík, that's my guess too. I know that's the case for the classical oscillator, and I suspect someone could prove it for the quantum oscillator. I wasn't confident enough in that claim to include it in my answer, but I might add it after I think about it for awhile. –  Colin McFaul May 24 '13 at 16:34
add comment

You can find the value of zero point fluctuations just by calculating the variance $\langle(\Delta\hat{x})^2\rangle = \langle\hat{x}^2\rangle$ in the vacuum state. You can do this either using the $x$-representation or expressing the $\hat{x}$ operator using creation and annihilation operators. These are usually introduced by $$ \hat{a} = \sqrt{\frac{m\Omega}{2\hbar}}\left(\hat{x}+i\frac{\hat{p}}{m\Omega}\right), $$ so that you get $$ \hat{x} = \sqrt{\frac{\hbar}{2m\Omega}}(\hat{a}+\hat{a}^\dagger). $$ Using this to calculate $\langle\hat{x}^2\rangle = \langle 0|\hat{x}^2|0\rangle$ indeed gives you $$x_{ZPF} = \sqrt{\langle\hat{x}^2\rangle} = \sqrt{\frac{\hbar}{2m\Omega}}.$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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