When talking about the wave-particle duality, teachers and books say that when you send a single photon through a slit, it makes a wave pattern. But if you send that particle through the slit and "you observe it directly", then it appears as a single point (a particle).

What is meant by "observe"? Is that like with your eyeballs? Or is that with some measuring device? It's unclear what that word means, it could mean anything haha.

One reason I am wondering is because, it seems like the act of trying to measure it directly (if "measure with a device" is what is meant by observe) would mean sending out some sort of radiation or particle itself, so yes, that would mess with the experiment. But maybe I just don't quite understand yet, so looking forward to a bit of clarification.


4 Answers 4


I would generally support Ayesha's answer, in that it explains that decoherence is instigated by the microscopic interacting with the macroscopic.

As with many things in Quantum Physics, this is evidently true at the extremes (e.g. putting a detector in the path of a photon in the dual slit experiment), but it is not clear when something is considered macroscopic. For example, somewhere between coupling an atomic spin to a single other atom and coupling to a million other atoms, we (and presumably the universe) say that decoherence has occurred, and the wavefunction has collapsed.

One answer is to look at the Density matrix. Consider that a pair of photons have a 2x2 density matrix. If we entangle the photon polarisations like-for-like, then we have either HH or VV, which would be a density matrix thus:

     H    V
  H  0.5  0
  V  0    0.5

Von Neumann tells us that measurement always increases entropy inside the system, thus decreasing the quantum information contained in it. So even if we perform a measurement that is unrelated, like checking for HV states, we start to corrupt the matrix. So perhaps it becomes:

               H     V
Second     H  0.44   0
           V  0.1    0.46

Suppose we further entangle this pair with another pair (still in its clean state), such that if the second photon of the first pair is V, then so is the first photon of the second pair (and the same for H):

                First pair
                  HH       HV      VH       VV
            HH   0.44                   
Second      HV 
 Pair       VH
            VV            0.1              0.46

If that second pair was a bit sullied, and looked more like:

HH 0.48, HV 0.03, VH 0.07, VV 0.42

Then we have:

       HH          HV        VH       VV
  HH  0.48x0.44            0.48x0  
  HV  0.03x0.44            0.03x0 
  VH             0.07x0.1          0.07x0.46
  VV             0.42x0.1          0.42x0.46

Renormalised (/0.4988):

       HH       HV     VH    VV
  HH  .4234              
  HV  .0265             
  VH          .0140         .0646
  VV          .0842         .3873

We can see how the original purity of the superposition is being polluted by the entropy, and as the space increases with the addition of more particles, this will further erode the original state. Note in particular how the HHHH state has P=42.34%, while the VVVV state has P=38.73%. Clearly this erosion will erode one of the original states more than the other, so after a few interactions we might expect one to disappear below the noise floor, leaving a single preferred state. In the experiment, this would come out as a 'decoherence' in which the increasing interaction with the environment dissipates the original superposition through an ever larger space until random noise takes over.

The signal vs noise metaphor is apt, in that it appears to be the quantum information stored in the states that it eroded. This is the origin of the Quantum Error Correction methods designed to use encodings to sustain a coherent state for longer.

Thus, 'observation' really means the coupling of a quantum state to a much larger system, and in doing so polluting the quantum information stored within with entropy and thus undergoing decoherence.


Suppose we represent the wavefunction of the photon as $\psi_p$ and the wavefunction of the observer as $\psi_o$. As long as the photon and observer do not interact in any way we can write the total wavefunction as a product:

$$ \Psi = \psi_p\psi_o \tag{1} $$

The trouble is that you can't make any measurement of the photon without interacting with it, and as soon as you do this you become entangled with the photon. Entanglement means the total wavefunction of you and the photon is no longer separable into a product as in equation (1).

Exactly what happens next depends on what interpretation of wavefunction collapse you favour. My own preferred interpretation is decoherence/many worlds, and viewed in this way after the interaction the total wavefunction evolves to become roughly separable again so you once again get:

$$ \Psi' = \psi'_p\psi'_o $$

But the wavefunction of the photon has been changed by the interaction, so $\psi_p \ne \psi'_p$, and this will affect the subsequent evolution of $\psi'_p$ as the photon passes through the slits and it will therefore affect the diffraction pattern produced.

Exactly what happens will depend on the strength of the interaction, because this will determine how much the wavefunction is changed. It is possible to make the measurement a very weak interaction, in which case you get a weak measurement and the diffraction pattern wouldn't be affected. However in these sorts of arguments the measurement is usually taken to be one that tells you the particle position, and in that case the interaction would have to be strong enough to significantly reduce the uncertainty in the particle position so it's no surprise that this significantly changes the diffraction pattern.

Note that I haven't assumed anything about the observer other than that it can be described by a wavefunction and that it interacts with the photon. So many systems can be observers - the effect they have just depends on the strength of the interaction.


I'll illustrate by referring to the Schrodinger's cat thought experiment. The experiment consists of a cat in a box, to which is attached a small amount of a radioactive substance. In the course of an hour, the substance may or may not emit a particle. If it does, then a Geiger counter triggers a can of cyanide, killing the cat. After an hour, the cat's wave function has the form $$\psi_{cat} = \frac{1}{\sqrt{2}}\left( \psi_{alive} + \psi_{dead} \right)$$

Now, herein we see the flaw of the term "measurement" - it suggests involvement by a human observer. Prior to your observation of the cat, the cat is neither alive nor dead, but rather a linear combination of the two states, and if the cat is indeed dead, you killed her by looking in the window.

This of course sounds inherently absurd. We resolve the paradox by considering the triggering of the Geiger counter the "measurement," not the intervention of the human observer. A measurement, here, then, consists of an interaction between the microscopic system (the decaying particle) and the macroscopic system (the triggering of the Geiger counter). All in all, "measurement" is a rather poor term, as it inherently suggests the presence of a human observer. Heisenberg suggested the term "event" instead.

(In some formulations of quantum mechanics, it is even possible to dispense with the concept of "measurement" entirely, and instead have the "measuring" apparatus and the system itself be described by a wave function).


To also give a more direct reply to your photon in a slit example, in simple words:

Observing the photon passing through the slit is not just a conceptual act, it means placing a photon detector at the slit and as the photon passes through, its state is measured and since a measurement in quantum mechanics comes at the cost of perturbing the original state of the system, (here meaning whatever mechanism the detector uses, it will have an influence on the state of the photon, e.g. light momentum is transferred to electrons).

Consequently the photon will collapse into one of its eigenstates (on which one will depend on its initial state of superposition) and will not be in an unknown state anymore (i.e. a superposition of states), meaning its state is perfectly known and you will not observe an interference for its intensity spectrum, all the fringes disappear.

Finally let's say if you keep measuring (i.e. in your words observing) the collapsed photon, using the same detector, you will observe the same state for the photon, but if you change detectors (like changing angle of polarizer, simply a different basis) then you may further change the state of the photon, by the same principles.

In order to get used to these ideas, it's best if you start by reading examples of measurement of photon polarization which will also introduce you to Dirac's bra-ket notation in a rather straightforward manner.

  • $\begingroup$ Make sure to read through the other answers as well, as they are well written, mine just gives you an explanation in simplest (friendly) words. $\endgroup$
    – Ellie
    Commented Aug 5, 2014 at 8:36

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