Is the moon there when nobody looks? Reality and the quantum theory

This document contains comments about the article Is the moon there when nobody looks? Reality and the quantum theory by N. David Mermin in Physics Today volume 38, April 1985, pages 38-47. A reformatted version of this is available at: or at

The reason of this document are comments in the sci.physics.reasearch Quantum weirdness? It's all in your mind.
For more information about that same subject go here: Quantum weirdness? It's all in your mind.

The title of the article is:
"Is the moon there when nobody looks"
The moon exists. This is the observable reality, which everybody accepts.
A much more interesting question is:
Does a photon always exists, if it moves freely?
The next sentence in the article is (modified):
"Einsteins opinion was that the in order to explain physical phenomena no spooky actions are required (For example: communication larger than the speed of light c). The article tries to prove that he was wrong. "
In order to demonstrate the article uses an apparatus C which transmits in each test two photons.
The experiment becomes easy to understand if you assume that both photons are completely identical, except that they move in different directions.
The two photons are measured by two identical detectors A and B. Each detector has a switch with three directions: 1,2 and 3 Each detector can measure the polarization direction of the photon: +x, -x, +y, -y, +z and -z
In order to measure the x direction, the switch is in position 1. To measure y, the switch is 2. To measure z, the switch is 3.
When the detector measures a photon in the + direction a green lamp flashes. In the - direction you get a red light.

In the original article 45 individual tests are performed.
The results are summarized in the following table:

11GG 2 11RR 1
12GR 3 12RR 1 12RG 3
13GR 1 13GG 2 13RR 1 13RG 1
21GR 2 21RR 2
22RR 2 22GG 2
23GR 3 23RR 1
31RR 2 31GR 1 31RG 4
32RR 1 32RG 2 32GR 1
33RR 4 33GG 3

The first row shows the results for detector A with the switch in position 1.
The second row shows the results for detector A with the switch in position 2.
The first column shows the results for detector B with the switch in position 1.
The second column shows the results for detector B with the switch in position 2.
In the above table there are certain results in Yellow. This are the results where the switches of detector A and detector B are in identical positions. What the results show that in those cases always both lamps have the same colour.
In the other 6 cases the switches are different. In those cases you get 4 different colour combinations.

Let us repeat the above experiment but in a slightly different manner.
  • First you do the experiment with 1000 tests and both switches are in position 1.
    What you get is a patern like RR GG GG RR RR GG RR etc. In total 498 times RR and 502 times GG.
    In the article this is called the combination 11RR and 11GG.
    What the results show is that if you compare different tests you cannot predict the outcome of the next test. In fact the outcome of each test is completely random.
    On the other hand if you compare the results of each test than both emitted photons are identical (have the same spin). This is because always when one detector shows the green lamp also the other detector/lamp is green. Or both are red.
  • Next you change the switch to 2. The results are 504 times 22RR and 496 times 22GG. Also here the same explanation.
  • Next you try switch 3. The results 501 33RR and 499 33GG.
    This means that the spin of photon pairs, produced in this experiment, are identical in all 3 directions x, y and z.
  • Next you change the experiment such that both detectors A and B use a different switch. For example 1 and 2. You again use 1000 tests. However the pattern will be different. You get something like:
    In total you get 250 times RR, 245 times GG, 252 times RG and 253 times GR That means roughly each combination 25% times.
    To explain that is rather easy.
    With detector A (switch = 1) you measure the x direction and in 50% you measure +x (Green) and in 50% -x (Red)
    In the case +x (Green) than for detector B (switch = 2) you measure in 50% +y (Green) and 50% -y (Red). In total you get 25% GG and 25% GR.
    In the case -x (Red) than for detector B (switch = 2) you measure in 50% +y(Green) and 50% -y (Red). In total you get 25% RG and 25% RR.
    This means that in all the cases with different switches the chance for each combination is 25%.

The important question is: is this in agreement with actual observations. I don't know, but the article gives the impression that it has be done and that it is not a thought-experiment.
And if that is the case than the explanation lies completely in the details of the reaction, inside apparatus C, which produces the two identical photons. As a consequence, the whole explanation does not require any form of spooky action at a distance nor faster than light communication and the doubt by Einstein is correct.

In the original article by DM the two emitted photons are identical(spin).
A new experiment whould be to have a different reaction inside apparatus C such that the two photons have opposite spins.
In that case when two switches have the same position (both 1) you get:
or approximate 498 RG and 502 GR.
In the case the two switches have different positions the same results as before apply: you get 4 combinations RG GG GR and RR, with each 25% chance.

Reflection part 1 - Thought Experiment

The original article starts with a thought experiment. To do science with thought experiments is very tricky. What you should do is to perform first experiments and observe and describe the results and see if they can be explained using accepted concepts.
As I already mentioned I do not know where you can observe the above experiments. The article mentions at page 10 the experiments of Aspect and his colleagues at Orsay . However I get the impression that those experiments are different.

Reflection part 2 - Error

The original article is based on 45 random tests. When you do random tests you first have to prove that the tests are random. To bypass that problem I propose only to do the experiment with two conditions: One with both switches identical and one with both switches different.
And to do each 1000 times. This is very important because you have to study accuracy.
In the case of both switches are in the same position, for example, the folling outcome is also possible:
493 * RR, 497 * GG, 2 * RG, 3 * GR, 1 * Rb, 1 * Gb, 1 * bR, 1 * bG and 1 * bb
That means that the sum of RR and GG is 990 and 10 results are "errors". b stands for blank (no light).

In the case that both switches are different the same error 1% error percentage applies. That means in some cases when you get RR it should have been RG or GR.

In case that apparatus C emits two photons with opposite spin the same error percentage applies.
What this means: if you observe that detector A shows a Green light, with both switches in the same position, that you know 100% sure that detector B shows a Red light.

Reflection part 3 - Angle.

The original article at page 5 mentions:
moving the detectors farther away from the source increases the delay between when the button is pushed and when the lights flash.
That seems logical and is in agreement with accepted concepts.
However it is important to do the experiment with the switches in the same position at different distances and observe the error rate.
You should perform the experiment at distances of 100 meter, 200 meter and 300 meter and observe the error rate.

What is also important that the detectors are correctly aligned.
To be specific the +x direction of both detectors should point to the same position in the sky. If they are exactly opposite, than with both switches in selection 1, the 50% RR GG pattern first changes into a 50% GR RG pattern.

Reflection part 4 - Action at a distance.

The most disturbing part of the article is that there is no clear answer on the original question:
Is the moon there when nobody looks?
As I said above: Ofcourse the moon is there somewhere. If I look is of no importance.
An other issue is: What is exactly the Quantum Theory?
At page 3 we read:
The conclusion is quite independent of whether or not one believes that the quantum theory offers a complete description of physical reality.
What does it mean: A complete description? And:
They showed that the quantum-theoretic predictions were indeed obeyed.
And at page 9:
Alas, this explanation –the only one, I maintain, that someone not steeped in quantum mechanics will ever be able to come up with - is untenable.
Unfortunate the article does not explain in detail what Quantum Theory (or quantum mechanics) is, which makes the article difficult to understand.

If you want to give a comment you can use the following form Comment form
Created: 15 June 2013
Updated: 26 May 2017
Updated: 1 September 2023

Back to calling page Schrodinger's Cat 1 Questions
Back to my home page Contents of This Document 31RR 12GR 23GR 13RR 33RR 12RR 22RR 32RG 13GG 22GG 23GR 33RR 13GG 31RG 31RR 33RR 32RG 32RR 31RG 33GG 11RR 12GR 33GG 21GR 21RR 22RR 31RG 33GG 11GG 23RR 32GR 12GR 12RG 11GG 31RG 21GR 12RG 13GR 22GG 12RG 33RR 31GR 21RR 13GR 23GR xyRR xyGG xyGR xyRG 11RR 11GG 11GG 1 2 0 0 3 12GR 12RR 12GR 12GR 12RG 12RG 12RG 1 3 3 7 13RR 13GG 13GG 13GR 13GR 1 2 2 5 21GR 21RR 21GR 21RR 2 2 4 22RR 22GG 22RR 22GG 2 2 0 0 4 23GR 23GR 23RR 23GR 1 3 4 31RR 31RG 31RR 31RG 31GR 31RG 31GR 2 2 3 7 32RG 32RG 32RR 32GR 1 1 2 4 33RR 33RR 33RR 33GG 33GG 33GG 33RR 4 3 0 0 7 15 9 13 8 = 45 There are nine posibilities 11,12,13,21,22,23,31,32 and 33. That means the chance for each is 45/9 = 5. This is okay There are also 4 posibilities RR GG GR and RG. That means the chance of each is 45/4 = 11 RR = 7 + 8 = 15 GG = 7 + 2 = 9 GR = 13 RG = 8 11RR + 22RR + 33RR = 1+2+4 = 7 11GG + 22GG + 33GG = 2+2+3 = 7 11RR + 22RR + 33RR + 11GG + 22GG + 33GG = 7 + 7 = 14 12GR + 13GR + 21GR + 23GR + 31GR + 32GR = 3 + 2 + 2 + 3 + 2 + 1 = 13 12RG + 13RG + 21RG + 23RG + 31RG + 32RG = 3 + 0 + 0 + 0 + 3 + 2 = 8 12RR + 13RR + 21RR + 23RR + 31RR + 32RR = 1 + 1 + 2 + 1 + 2 + 1 = 8 12GG + 13GG + 21GG + 23GG + 31GG + 32GG = 0 + 2 + 0 + 0 + 0 + 0 = 2