After reading an article on Schrodinger's Cat, it seems that if we take the environment as an observer, that superposition cannot occur because all atomic and subatomic entities would be observed all the time. Thus, something like quantum entanglement cannot occur. So if superposition cannot occur, why is superposition (and by extension quantum entanglement) still part of quantum mechanics?

Updated: In the question How is it possible that quantum phenomenons (e.g. superposition) are possible when all quantum particles are being constantly observed?, the notion of observer is replaced by measurement. In this context, my question would be: if the system (cat) is constantly being measured by the environment (the observer is watching the cat), how can superposition (the cat is in multiple states; i.e., alive and dead) exist in quantum mechanics?

For example, if a photon passes by a heavy particle and splits into an electron and positron, the splitting process is a measurement of the electron and positron to make sure the total spin is 0. I understand the argument that we might not know which has +1/2 and which has -1/2, but the observation/measurement had to be done to make sure we didn't have 3/4 total spin.


marked as duplicate by Peter Shor , ACuriousMind, Martin, Danu, John Rennie quantum-mechanics Feb 11 '16 at 6:58

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    $\begingroup$ Superposition and quantum entanglement can occur. $\endgroup$ – Peter Shor Feb 10 '16 at 0:44
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    $\begingroup$ It is important to consider time scales when thinking about these problems. Superpositions are not maintained indefinitely due to interactions with the environment, but if the dynamics are fast enough, this does not have to be problematic. I'd look a bit into the concept of decoherence to learn more. $\endgroup$ – user129412 Feb 10 '16 at 0:55
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    $\begingroup$ To emphasize @PeterShor's comment, we run experiments that only work if superposition does occur on a regular basis and all kinds of different energy and length scales. Your question is simply predicated on a false premise. $\endgroup$ – dmckee Feb 10 '16 at 2:49
  • $\begingroup$ Entanglement over large distances is only observable because the self-interaction of the electromagnetic field is extremely small. That's why we can see all the way back to the big bang, even though those fields have been propagating for 13.8 billion years. If you try the entanglement trick on charged particles or atoms, you will find that they will show decoherence with the thermal radiation background very quickly, if you don't control the environment very carefully. In essence you are correct that decoherence with the environment is "normal". That's exactly why there is a classical world. $\endgroup$ – CuriousOne Feb 10 '16 at 4:13
  • $\begingroup$ Obviously, every state is a superposition, so presumably what you're suggesting is not that superpositions can't occur, but that only states in some particular basis can occur. This raises the question: What would determine that basis? $\endgroup$ – WillO Feb 10 '16 at 23:11

Superposition isn't some magical fragile state. Let's look at a unit vector $\hat n$ in the plane.

You could write $\hat n$ as the sum of two orthonormal vectors $\hat x$ and $\hat y.$ Maybe $\hat n=\frac{\sqrt 2}{2}\hat x+\frac{\sqrt 2}{2}\hat y.$ And you could say that $\hat n$ is a superposition of $\hat x$ and $\hat y$ with coefficients $\frac{\sqrt 2}{2}$ and $\frac{\sqrt 2}{2}$ since $$\hat n=\frac{\sqrt 2}{2}\left(\hat x\right)+\frac{\sqrt 2}{2}\left(\hat y\right)$$

But you could also say that $\hat n$ is a superposition of $\frac{\sqrt 1}{2}\hat x+\frac{\sqrt 3}{2}\hat y$ and $\frac{-\sqrt 3}{2}\hat x+\frac{\sqrt 1}{2}\hat y$ with coefficients $\alpha=\frac{\sqrt 2+\sqrt 6}{4}$ and $\beta=\frac{\sqrt 2-\sqrt 6}{4}$ since $$\hat n=\alpha\left(\frac{\sqrt 1}{2}\hat x+\frac{\sqrt 3}{2}\hat y\right)+\beta\left(\frac{-\sqrt 3}{2}\hat x+\frac{\sqrt 1}{2}\hat y\right)$$

The word superposition is just a fancy word for saying the state is a linear combination of some other states. But in general any state is equally good. You could have even said $\hat n$ was a linear combination of $\frac{\sqrt 2}{2}\hat x+\frac{\sqrt 2}{2}\hat y$ and $\frac{\sqrt 2}{2}\hat x-\frac{\sqrt 2}{2}\hat y$ with coefficients 1 and 0.

So why do we bring up the word superposition at all? If you interact with an observer or do a measurement, there is a basis that is natural to that observer or measurement, so writing your state as a superposition of those states is helpful.

So if you have an environment, then it could impose its basis as one that is relevant to the object. But it's not like there are some states that are superposition states and the other states aren't. There are just states. Like there are just vectors. There aren't some vectors that are sums of others vectors and then special vectors that aren't sums of other vectors.

Now another important thing to note is that an interaction takes time. And that if you want to avoid your environment making interactions and changes to your stuff (because you want it to do its own things, the stuff you want it to do) then you do need to shield it from the environment to give it a chance to have time to do its thing before the environment makes it do something else.

If you had a system that was constantly being measured it would not have time to evolve away from the basis states. This is the Quantum Zeno effect. And it's not normal, the measurements take time and the natural evolution takes time.

And not everything in life is a measurement. A measurement starts out as an entanglement that then has the different branches and couple to thermodynamic degrees of freedom. So you aren't a measurement just because a heavier particle is nearby.

  • $\begingroup$ My understanding of superposition is that all solutions to your vector equations are equally valid so all solutions are possible. But an observer will note that x and y are some specific value and so the "wave equation collapses" to just that specific solution. Are you saying that one observer's x and y can be different from another observers x and y? $\endgroup$ – ToaTerra Feb 12 '16 at 22:15
  • $\begingroup$ @ToaTerra Correct $\endgroup$ – Timaeus Feb 13 '16 at 0:59

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