Why decoherence and entanglement have nothing to do with Schrödinger’s cat

There’s apparently an interpretation of the Schrödinger cat paradox that claims it is explained by quantum decoherence.  I ran across this on Quora where someone had asked about decoherence.  Someone from SLAC gave a mathematically correct answer, but attributed it to decoherence.  But if we use the usual quantum definition of decoherence, then it actually does not explain what happens. To understand all this let’s review coherence and decoherence.

Quantum systems have the ability to exist in superpositional states.  As a simple analogy, suppose we have a bag that contains a bunch of black and white marbles.  Now suppose that we remove a marble from the bag and hold it in our hand, but don’t look at it.  So we don’t (yet) know whether it is a black marble or a white marble.  If the marble were a quantum mechanical object, we would say that, until we open our hand, that marble exists in a ‘coherent’ superposition of black and white states.  Suppose our bag only has two marbles in it.  If we pull out the white marble, we immediately know that the other marble in the bag is black.  We say the marble in our hand is in a ‘pure’ state because when we open our hand to look at it, it is either black or white (e.g. not striped, grey, etc.), i.e. we would say that the black and white outcomes are orthogonal to one another.  So a pure state is one that allows for a coherent superposition of orthogonal states.

But sometimes the world isn’t quite so black and white (pardon the pun).  Instead of marbles, suppose we’re dealing with, say, dogs.  Pure breds would represent the orthogonal states.  So suppose I go to a (strange) shelter where I’m not allowed to see into the pens where they keep the dogs and there is one dog per pen.  Suppose this shelter has only two pure bred dogs – one dachsund and one poodle.  If I choose the first door and it turns out to be a dachsund, then the poodle must be behind the second door.  These are orthogonal states.

But what if the two dogs are mutts?  Maybe the dog behind the first door has some dachsund in him as well as other things like beagle and basset hound.  The same could be true of the dog behind the second door.  He might have some poodle in him, but he could also have some dachsund in him.  Since both dogs could have some dachsund him them, their states are not necessarily orthogonal.  We would say such a state is a ‘mixed’ state.

Another way to look at this is to consider electromagnetic (light) waves.  Imagine that we build a laser in which the light is produced by some atomic transition.  If the light waves that are produced are all perfectly ‘in step’ with one another, then we say that our beam is ‘coherent’ and represents a ‘pure’ state in that the light waves all have the same wavelength, frequency, etc.  If, however, they don’t all have the same frequency, wavelength, etc. then we have a ‘mixed’ state.

Decoherence in quantum systems, then, is the process by which a pure state becomes a mixed state.  What causes decoherence is something I will leave to another blog post.  But suffice it to say that its definition is fairly consistent throughout quantum physics.

So now what about Schrödinger’s cat?  For those unfamiliar with this gedankenexperiment, it goes something like this.  Suppose we have a cat in a box.  Inside the box is a vial of cyanide gas.  Next to this vial is a small hammer attached to a mechanism.  The mechanism is set to trigger the hammer (and thus break the glass, killing the cat) if a certain radioactive isotope decays.  So the cat’s life (or death) is governed by a purely random process.  Before we open the box, the original version of the paradox says that the cat exists in a superposition of alive and dead states.  It shouldn’t be too difficult to see that these are clearly orthogonal states.  You’re either alive or you’re dead.  Ain’t no way to be a mixture of the two.  Let’s assume it’s about a 50-50 shot that the cat is alive/dead.  The state prior to opening the box is

\frac{1}{\sqrt{2}}(|\textrm{alive}\rangle+|\textrm{dead}\rangle).

This is a coherent, pure state since it is composed of orthogonal terms (there is no dispute about this).  When we open the box the cat is either alive or dead.  Let’s be humane and assume it is alive.  Then the state after we open the box is

|\textrm{alive}\rangle.

This is clearly not a mixed state!  Therefore, decoherence does not explain what happens to Schrödinger’s cat!

Update: I forgot to mention that entanglement doesn’t describe this either.  Prior to measurement, the cat’s state is not entangled!  By definition, an entangled state is a non-factorable state.  In other words, you either need to be comparing a single observable for two objects, or two observables for a single object.  Here we have a single observable (life/death) for a single object.

Update II: Matt Leifer pointed out on Google+ that technically decoherence is not merely a transition from a pure state to a mixed state.  It is actually an irreversible transition from a pure state to a mixed state.  I should also point out that decoherence is often cited as one mechanism by which the classical world arises from the quantum world.  Matt and Thaddeus Ladd pointed out a number of other points, but for the purposes of my post (debunking a myth), I disagree with them regarding the importance of those points.

Advertisements

15 Responses to “Why decoherence and entanglement have nothing to do with Schrödinger’s cat”

  1. Heisenberg for one — the Kappellmeister of docoherence — never claimed that decoherence explains QCat. QCat is a Measurement issue.

    If the Geiger counter detects an emission a measurement has been accomplished, following which the hammer strikes the vial of cyanide gas which is then released and the cat dies. You, the experimenter, could be in the bathroom at the time. The information is nonetheless out there. The cat died while you were sitting on the can. You might have a vidicam with a time stamp and when you got back from the bathroom you run the tape and determine that indeed the cat died at that moment.

    It would get interesting only if you next lift the lid of the box and discover the cat alive.

  2. Is Schrodinger’s cat an example of decoherence in quantum mechanics?…

    I respectfully disagree with one claim in the answer given by Jay Wacker: decoherence does not explain Shrödinger’s cat. Decoherence in quantum systems is the process by which a system transitions from a pure state to a mixed state. Prior to measureme…

  3. I have to point out that your understanding of the mixed state and decoherence is flawed. Mixed states do not really exist. (You can never, for instance, take an electron, or a cat or whatever and say that it is in a mixed state \rho). It is just a formalism which is used to describe the probabilities of various outcomes after a measurement is made.

  4. […] same consistent definition). This has particular relevance to my recent posts on decoherence and the Schrödinger cat paradox.  As Hongwan Liu pointed out on Quora, while the latter paradox was originally formulated […]

  5. What makes you so certain that |alive> and |dead> are orthogonal states? If a state is an eigenstate of the time evolution operator, that means that a system in that state will stay in that state forever. Clearly |dead> is that sort of thing, because once the cat’s dead, it stays dead forever. |alive> also is clearly not an eigenstate of the time evolution operator, because if enough time passes you will eventually be in the |dead> state. Decoherence must occur at some point because once you arrive in the |dead> state that transition is irreversible. So these two states aren’t both eigenstates of the same operator.

    Also I think it is relevant to look at what |alive> and |dead> mean on a smaller scale. |dead> clearly means that all the cells in your body are dead and they’re staying that way. But |alive> doesn’t mean all the cells in your body are alive and staying that way forever. Many kinds of cells in the body are constantly dying (skin cells for example) and being replaced by new cells via mitosis. So if we think of |dead> as a product state of |dead cell xxx> states, then there will be some non-zero overlap when projected onto the |alive> state because even when you are alive, there are cells in your body that are dead. So |alive> and |dead> are not orthogonal.

  6. Solomon Freer Says:

    Why do you assume that since you believe there’s a 50-50 chance that the cat is alive or dead means that the cat is in the state:
    1/sqrt(2)*(|Alive>+|Dead>)?
    This seems arbitrary when the state 1/sqrt(2)*(|Alive>-|Dead>) or 1/sqrt(2)*(|Alive>+i*|Dead>) or 1/sqrt(2)*(|Alive>-i*|Dead>) would all give the same 50-50 probabilities, since all these states lie on the x-y plane of this cat’s supposed Bloch sphere. Your choice of phase is arbitrary, not stemming from any information from the experiment.

    I believe that this is because the Schrodinger’s cat picture is misunderstood as a coherent superposition, when really the cat is in a mixed state (one not representable on the Bloch sphere – it has no phase) of simply 50-50 alive/dead. The cat is never in a superposition of alive and dead, our information is simply too limited to know what state it is in. It’s density matrix would be [0.5,0;0,0.5] (off-diagonals=zero), a maximally mixed state.

    I’d be very interested to hear your thoughts on this.

    • quantummoxie Says:

      Well, yeah, but the point is that the probabilities are all the same for those states. So, at least operationally, it doesn’t matter.

      But, the fact is that I agree with you — it is in a maximally mixed state to begin with. That is precisely why decoherence doesn’t apply: it’s already in the mixed state!

  7. Author wrote: “Prior to measurement, the cat’s state is not entangled! By definition, an entangled state is a non-factorable state. In other words, you either need to be comparing a single observable for two objects, or two observables for a single object. Here we have a single observable (life/death) for a single object.”

    But shouldn’t it be this?
    The formulation |cat = a|life + b|death obscures the nonlocal entanglement. The cat state must be written as |cat = a|livecat |undecayed nucleus + b|deadcat| decayed nucleus.

    excerpt from :http://physics.uark.edu/hobson/pubs/13.07.PRA.revised.pdf

    • quantummoxie Says:

      Yeah, that’s how the state *should* be written. My point was that they didn’t know it at the time Schrödinger came up with the idea.

  8. Parallel universes do offer a possible explanation. Assuming there really is no actual collapse of the wave function, then decoherence is basically hiding all other possibilities (that actual do occur) at the macro-level as a result of “observation” which at the micro-level is caused by ANY interaction of particles which cascades up and across the macro-level. So the cat is dead and alive at the same time just not in the universe you’ve been “cloned” to.

    • quantummoxie Says:

      True. I actually understand the attractiveness of MWI. But it still needs some experimental support for me to buy it completely.

  9. The stability of superposed cat states is dependent on them remaining unmeasured. The purpose of the box is to keep the cat from being measured. But no box is a perfect isolator, the cat still exists in an environment, whatever its state. So the utility of decoherence here is that it happens spontaneously as a result of the cat being in an environment, like every other physical thing that exists. The environment is effectively acting as an observer, a “measurement device.” In a sense, the box cannot stay “closed” for any time greater than a Planck time, so the cat cannot remained unmeasured, it cannot stay in superposed states due to the measurement action of its environment, which causes spontaneous decoherence. Bottom line, to exist is to be measured.

Comment (obtuse, impolite, or otherwise "troll"-like comments may be deleted)

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: