I have a question/troubles about this topic.

I define a single and indivisible sample meter equal to 2L. Suppose I have two mirrors between which a beam of photons oscillates with a certain energy per single photon and for each the same. Let's suppose that the two mirrors are exactly 4L apart. I can say that on average position of the photon is in the centre of the mirrors with an error of exactly L. If in these specific conditions of spatial confinement (i.e. measurement) I go to measure the moment of a photon, will it obey or not the Heisenberg uncertainty principle? Is it also in this case not applicable? i.e. by considering $$\Delta x =L$$ and the uncertainity on the frequency could be exactly $$\Delta p= {h \Delta\nu \over c} $$ could I point out that the uncertainity collapses, in this special case, on a frequency-space uncertainity according to following: $$\Delta \nu = {c \over 4\pi \Delta x} $$

In advance I thank you for the answer. I excuse me in advance for my incompetence.


2 Answers 2


Photons are quantum mechanical entities, mirrors are macroscopic entities composed of order $10^{23}$ atoms/molecules with a collective field at the surface. The mirroring from this collective field can be described very well by classical electrodynamics, but does not apply to individual photons and their wave function in the way you examine.

The Heisneberg uncertainty principle (HUP) applies to individual photons, which will interact with the field of the mirror surfaces obeying quantum mechanical equations and the HUP will hold for it. See my answer here to a related question.

The $Δx$ for the photon cannot be the macroscopic dimension of the setup you propose, it will be the quantum mechanical individual photon+field interaction volume.

  • $\begingroup$ if I may, in other words, say that what bounced off the mirror isnt the same photon that was incident upon it. $\endgroup$ Commented Sep 14, 2018 at 16:19
  • $\begingroup$ @ManudeHanoi yes, if you look at the appropriate feynman diagram the "same" is a matter of definition. In general elementary particles with the same parameters, are not distinguishable. $\endgroup$
    – anna v
    Commented Sep 14, 2018 at 17:11

It may come as a shock that the Heisenberg uncertainty principle is not fundamentally a principle of the physical world. It is actually based on more fundamental properties of the mathematics in terms of which quantum mechanics is formulated. Due to this fact, provided that certain conditions apply, it does not really matter whether you consider the light between the mirrors as a single photon or whether you consider it as a classical field. The same principle applies.

Moreover, when there is no nonlinearity in the system, then whether you think of the setup as a classical system or as a quantum system does not matter, because they are described by the same mathematics.$^*$ So, if we ignore issues related to coherence, then a photon would be described by a wavefunction that is identical to classical field.

Now one can specify the wave function of the photon by some function that would be a solution of the Helmholtz equation with the boundary conditions. Then the positon and its uncertainty can be directly obtained from the definition of the wave function. The momentum and its uncertainty follows by computing the Fourier transform of the wave function.

For the case with the two mirrors, the field is periodic. It Fourier transform then produces a discrete spectrum. However, one can still compute an uncertainty in the usual way and it would still obey the Heisenber uncertainty principle.

In response to the request in the comment, I'll give a little more detail on how to do the calculations. It is really not too complicated. The idea is that the product of the width of the wave function and the width of its spectrum is always large than some constant $$ \Delta x \Delta k \geq C . $$ Here, I consider the spatial domain. With the temporal domain, it is kind of obvious that the relationship would be satisfied if the duration of the signal is infinite. So, it simply comes down to calculating the width of the optical pulse frozen in time. So let's fix the time at a point where the pulse is halfway between the mirrors. The width can now be obtained by computing the centralized second moment $$ \sigma_x^2 = \int |\psi(x)|^2 (x-\mu_x)^2 dx , $$ where $\mu_x$ represents the mean location of the wave function. Then we compute the Fourier transform of the wavefunction $$ \Psi(k) = \int \psi(x) \exp(ikx) dx . $$ Note that this is the spatial domain Fourier transform only. It does not contain any information about the temporal behaviour since we've fixed the time to a specific value. A similar calculation then leads to the width of the spectrum $$ \sigma_k^2 = \int |\Psi(k)|^2 (k-\mu_k)^2 dk . $$ The uncertainties are now given by the square roots of the variances $\Delta x = \sigma_x$ and $\Delta k = \sigma_k$.

The values of $\mu_x$ and $\mu_k$ are obtained from the first moments. For instance, $$ \mu_k = \int |\Psi(k)|^2 k\ dk . $$

Let me know if I need to add more detail somewhere.

$^*$ One possible exception is when you are dealing with a complicated quantum states such as when it is entangled, but that is beyond the scope of the current question.

  • $\begingroup$ Could I ask you how would you practically calculate it step by step? Thank you in advance. $\endgroup$ Commented Sep 16, 2018 at 12:38
  • $\begingroup$ By placing ourselves in the hypothesized conditions of confined space we can then write: $$\sigma_x^2 = \int_{-2L}^{2L} |\psi(x)|^2 x^2 \mathrm{d}x$$ What is the value of $\mu_k$? If as you mentioned the Fourier transform is periodic then we can write $\sigma^2_k$ as a sum. How could we close the expression? Are we able to calculate the integrals supposing the light beam to be monochromatic? $\endgroup$ Commented Sep 18, 2018 at 5:43
  • $\begingroup$ Moreover, if you consider that the space as I define it in the hypotheses is quantized then the integrals also become the sum of finite intervals equal to 2 sample meters. $\endgroup$ Commented Sep 18, 2018 at 6:48
  • $\begingroup$ Actually, since I've fixed the time, one does not see the periodic behaviour anymore. The reason for fixing the time is because the signal theoretically lasts for ever so its uncertainty is infinite. The spatial spectrum of the pulse is not a discrete spectrum. So you need an integral. One can however approximate these calculation by discretizing the space and the spectrum if you want. $\endgroup$ Commented Sep 21, 2018 at 12:23
  • $\begingroup$ I've added some clarifications. However, I'm not sure what you mean by how we can "close the expression." $\endgroup$ Commented Sep 21, 2018 at 12:24

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