55

In Schrodinger's Cat thought experiment, why doesn't the cat itself qualify as an observer?

Reading through the replies there seem to be two suggestions for what can take the role of observer:

  1. any "large" body

  2. any "living" thing (or should that be "conscious"?)

10 Answers10

77

The point is, it has made you think about the issue. Whereas we all might agree a hydrogen atom is not an observer and a human is an observer, the case of a cat is not so clear. The point of the thought experiment is to expose problems with the Copenhagen interpretation - which it does very successfully.

37

This isn't actually a mystery and the answer is fairly simple: real quantum physics is equivalent to many worlds quantum physics. There really isn't an alternative explanation available.

Yes, I know there are a number of other interpretations available. But I was specific in my language. Copenhagen, for example, doesn't explain anything. It just says "this is what appears to happen and there is no explanation."

As it turns out, the other interpretations do this as well. Bohm, for example, 'explains' how there is actually no such thing as superpositions (by placing all the mathematics of a superposition in to the 'pilot wave' -- mathematically equivalent to superpositions but not called that) but doesn't even attempt to solve any of the big problems of quantum physics, such as your question (know as the observation problem) or how quantum physics can be squared with locality. Or put another way, Bohm 'explains' something that needed no explanation (superpositions--which are in fact real) but doesn't even attempt to explain / solve any of the problems of quantum physics.

This is true for every single 'interpretation' of quantum mechanics except one: many worlds.

Many worlds isn't merely another interpretation. It's an actual explanation of why quantum mechanics behaves so counter intuitively to us. And it does it by simply taking the theory seriously and refusing to compromise on it. It literally add nothing at all to the theory. It just says "well, if we take the theory seriously, what does that mean?" This is the complete opposite of every other 'interpretation' all of which add extra stuff to the theory that serves no purpose but to not take the original theory to it's own logical conclusions. That is why I say Many Worlds Quantum Physics is Quantum Physics. All the others are Quantum Physics plus some extra unnecessary stuff that just complicate the theory for no particular reason.

So how does many worlds quantum physics explain Schrodinger's Cat? When the quantum event takes place that both kills and doesn't kill the cat, reality superpositions and there is then a dead cat in one world and a live cat in another. When you walk over and open the box, you then superposition. One version of you in one world sees the dead cat and one version of you sees the live cat.

"Observation" plays no role in many worlds quantum physics except in the sense that it explains why observation seems to cause the wave function to collapse -- namely that you become part of the overall super position. In fact, it never collapses at all.

If you are curious to explore this answer further than I can explain here, look but the books by David Deutsch (creator of quantum computational theory). Both of his books, The Beginning of Infinity and The Fabric of Reality, explain this in much better detail.

Also, look up the Elitzur–Vaidman bomb tester experiment on Wikipedia and spend some time thinking about how many worlds would explain what is going on and see if any of the other interpretations even attempts to. It is much weirder than Schrodinger's Cat and much harder to explain away. For example, the wave function 'collapses' even when there is no observation at all! (Because the observation of the bomb exploding takes place in a different world. The world where you don't see the bomb explode never has any observation take place in 50% of the cases.)

26

Schrodinger used the hypothetical cat to illustrate what he thought was the absurdity of assuming that wave-function collapse only occurred when there was an 'observer' making a measurement. The point he was making was the common-sense notion that collapse of the wave function would occur when a particle interacted with any large body, whether it was a measuring device or not.

As for what constitutes 'an observer', an observation is nothing more than some interaction at a microscopic scale that has a quantitative consequence. When you observe an emission spectrum, your eye is registering light that has been split through a prism; in other words the effect of photons interacting with the electric field due to the distribution of atoms within the glass. The interaction between the photons and the atoms within the glass takes place regardless of whether anyone or anything is 'observing'- the prism of the spectrometer acts as the 'large body' you mention in your question.

12

To build upon Bruce's answer, what happens according to Schrödinger's equation is the following:

The initial quantum state of the system (after putting the cat in the box) is

$$|\text{cat alive}⟩ |\text{you don't see the cat}⟩$$

As the Geiger counter measures whether the radioactive atom decayed or not, and kills or spares the cat accordingly, this quantum state evolves unitarily into

$$\frac{|\text{cat alive}⟩ + |\text{cat dead}⟩}{\sqrt{2}} |\text{you don't see the cat}⟩$$

which is of course equivalent to

$$\frac{|\text{cat alive}⟩ |\text{you don't see the cat}⟩ + |\text{cat dead}⟩ |\text{you don't see the cat}⟩}{\sqrt{2}}$$

As you open the box, this quantum state again evolves unitarily (i.e. without any kind of "collapse") into

$$\frac{|\text{cat alive}⟩ |\text{you see a live cat}⟩ + |\text{cat dead}⟩ |\text{you see a dead cat}⟩}{\sqrt{2}}$$

This quantum state is called an entangled state because it cannot be decomposed into a product of two quantum states describing you and the cat respectively, i.e.

$$(\alpha_1|\text{cat alive}⟩ + \alpha_2|\text{cat dead}⟩) (\beta_1|\text{you see a live cat}⟩ + \beta_2|\text{you see a dead cat}⟩)$$

where $|\alpha_1|^2 + |\alpha_2|^2 = |\beta_1|^2 + |\beta_2|^2 = 1$. This is what we mean when we say that you become entangled with the cat.

Whether the cat "collapses the wavefunction because they're a conscious observer" is a red herring for two reasons:

First, the whole thought experiment can be repeated, as garyp suggested, with a Geiger counter spilling some ink on a piece of paper. Whether anything in the box is "conscious" is irrelevant.

Second, if we take Schrödinger's equation seriously, there is no such thing as collapse. There is only unitary evolution all the way through.

user76284
  • 1,408
2

The original version of the experiment didn't use a cat, but instead a keg of gunpowder:

Einstein raised the prospect of a keg of gunpowder being in a superposition of exploded and unexploded states, and Schrödinger upped the ante with his cat, whose life or death is yoked to a quantum event such as the radioactive decay of an atom. If, as Bohr said, the state of the atom is undetermined (in a superposition) until we look, then so must be the state of the cat. (see https://www.theatlantic.com/science/archive/2018/10/beyond-weird-decoherence-quantum-weirdness-schrodingers-cat/573448/)

So in the original formulation the cat is an observer as a keg of gunpowder is an observer. So any large body can be used as "observer".

But assume for a moment that you are the observer and the poison just stands for some incident in the future leading to your death. This basically means you are in "a superposition of live and death your whole life" (from the point of an outside observer). From your point of view this fancy wording just says that there is a finite probability that you will die eventually.

The question is then the other way round: Can there exist an observer in this universe for whom your life and fate is just a quantum superposition, because he/she/it hasn't measured it (hasn't opened the box), yet? In a classical setting I argue the answer is no, since interactions travel with infinite speed. In a relativistic setting observers outside of your light cone cannot have interacted with your existence, and so might consider your fate still being in a quantum superposition.

asmaier
  • 10,250
1

Malcolm Storey,

This thought experiment is not based on the assumption that a cat is not an observer. You can replace the cat with a human and its conclusions remain the same. The point is that each observer uses a different description of the experiment based on what information is available to him. The observer inside the box, if there is one, knows the result of the measurement (say a spin measurement - let's replace the poison bottle here) so, from its point of view there is no superposition. The outside observer does not know how the measurement came out so he uses a superposition of (spin up + inside observer detecting spin up) and (spin down + inside observer detecting spin down). As far as I know this is the standard interpretation of this experiment.

In my personal opinion, even the outside observer knows, in principle, the result of the measurement inside the box. There is no box that can perfectly isolate its interior. Gravitational, electric or magnetic fields cannot be blocked by any material. So, I think that the "collapse" occurs for both observers at the same time. I think that microscopic superposition (live/dead cats) are not possible for this reason.

Andrei
  • 839
1

Yes, it is, but there is no additional knwoledge anyone will gain from it's observation. Infact if it observes "decay within one hour", then so will you, once you open the box and find the dead cat. If, on the other hand, it observes "no decay", so do you again. Anyway: there is no need for the cat to share it's observation to deepen your knowledge of the system.

denklo
  • 268
0

Taking the superposition of one cat in two states seriously, then one is observing itself alive and the other itself dead. Well, of course not, but take a less drastic example of the poison just tiring the cat and not killing it.

Please note that with this answer I’m not giving any statement about the validity of the cat-in-two-states proposition.

0

Ignoring all the physics (yeah, I know), the cat is purely a device to engage the audience of the thought experiment.

It's not really a cat but some method of displaying state, and a cat, being a common pet, and the state used, being whether it is dead or alive provoke an emotive response, while also being something that is recognisable, both as a subject, but also that the state is easy to determine.

As such, we don't actually treat the cat as a living thinking being, but as a secondary indicator to the state of the unstable atom at the heart of the question.

At the fundamental level, every interaction is an observation - when a photon strikes a single atom, it has been observed.

Baldrickk
  • 425
-9

The cat is an observer only if he is aware of the principles of radioactivity and has intimate knowledge of the experimental set up. Note that it should also be alive. A dead cat cannot be an observer. Without physics knowledge it is just an ignorant victim and qualifies no more as an observer as the box the box itself or even the vacuum. Once you accept absurdity, there is no limit to what you can conclude. Schrodinger made a very good point here.

Note that in the ensemble interpretation no observer is needed. Less fun, I agree.

my2cts
  • 27,443