My question is how does quantum decoherence happen. What happens with a quantum system when "observed?" Can you give a mathematical explanation that is simple, precise, and easy to understand?
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2Can you give a mathematical explanation that is simple, precise, and easy to understand? No. – Christoph Dec 03 '14 at 22:58
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@Christoph at least try the best you can! – TanMath Dec 03 '14 at 23:00
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One of the better intuitive explanations of decoherence I've seen is at LessWrong. The site also has a few other articles with various takes on the subject. – Nathan Reed Dec 03 '14 at 23:13
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@NathaNReed but I also want a mathematical explanation at a calculus level ... – TanMath Dec 03 '14 at 23:20
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This question was answered in full, and simply, at the location http://physics.stackexchange.com/questions/150146/is-decoherence-continuous/150175#150175 – Sofia Dec 04 '14 at 00:32
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When a particle passes through an ideal pair of slits, it's wave function can be written in the form $$\psi(x) = \alpha\, \psi_1(x) + \beta \,\psi_2(x)$$ where $\psi_{1,2}(x)$ are the wave-functions you would get by closing either slit 2 or slit 1, respectively. We say that the wavefunction is in a coherent superposition.
What is coherent about this superposition? Well, let's place a detector screen at $x$ and measure the rate at which particles accumulate there. This rate will be proportional to $P(x) = \left|\psi(x)\right|^2$ by the usual rules of quantum mechanics. Coherence comes about because the two components of the wavefunction can interfere: $$P(x) = |\alpha|^2 |\psi_1(x)|^2 + \alpha^*\beta\,\psi^*_1(x) \psi_2(x) + \alpha\beta^*\,\psi_1(x) \psi^*_2(x) + |\beta|^2 |\psi_2(x)|^2$$ Here the middle two terms are interference terms, and give a behaviour that is counter-intuitive from a classical probability point of view. Classically, the rate should just be proportional to the sum of the probabilities of the particle going through either slit 1 or slit 2.
Now suppose our particle met a mote of dust on its way to the screen. The wavefunction of the system, particle + dust, should now be written as $$\psi(x_p,x_d) = \alpha\, \psi_1(x_p) \phi_1(x_d) + \beta \,\psi_2(x_p) \phi_2(x_d)$$ where now $\phi_1(x_d)$ is the wavefunction of the dust mote if the particle went through slit 1, and $\phi_2(x_d)$ is its wavefunction if the particle went through slit 2. Again, we place a detector at $x_p$ to measure the probability of detecting a particle there. The rules of quantum mechanics instruct us to sum over the position of the dust mote when squaring the wavefunction: $$P(x) = \int\!dx_d\, |\alpha|^2 |\psi_1(x_p)|^2 |\phi_1(x_d)|^2+ \alpha^*\beta\,\psi^*_1(x_p) \psi_2(x_p) \phi_1^*(x_d) \phi_2(x_d)\\ + \alpha\beta^*\,\psi_1(x_p) \psi^*_2(x_p) \phi_1(x_d) \phi^*_2(x_d)+ |\beta|^2|\psi_2(x_p)|^2 |\phi_2(x_d)|^2$$ Firstly, the wavefunctions $\phi_{1,2}(x_d)$ are normalized. Secondly, depending on how the dust mote encountered the particle, it may have radically different final wavefunctions, so in general $\phi_1$ and $\phi_2$ do not have much overlap: $\int\!dx_d \phi^*_1(x_d)\phi_2(x_d) \ll 1$. Putting this together, we find $$P(x) = |\alpha|^2 |\psi_1(x_p)|^2 + |\beta|^2|\psi_2(x_p)|^2$$ Compare this probability to the one we found when there was no interaction with the dust mote, and the immediate difference is that the interference terms have vanished. There is no longer a coherent superposition but rather an incoherent sum, and the probabilities add classically. One speaks of decoherence due to an interaction with the environment.
lionelbrits
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do these products $\psi_1(x_p) \phi_1(x_d)$ and $\beta ,\psi_2(x_p) \phi_2(x_d)$ imply a tensor product or a normal one? (why was it allowed there to just multiply the two wavefunctions there when you took dust into account?) Last point I didn't get, why $\phi_1$ and $\phi_2$ do not have much overlap? (should one just regard dust here as noise?) Very last point if I may: what happens to the first expression you gave for $P(x)$ if $\psi_1$ and $\psi_2$ are orthogonal states? (spin up and down e.g.) Then wouldn't the interefence vanish $<\psi_1|\psi_2>=0$? thanks in advance. – user929304 Dec 10 '14 at 12:27
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If you view the positions $x_p$ and $x_d$ as indices on a vector, then you can regard $\psi(x_p)\phi_1(x_d)$ to be the tensor product of the two vectors $\psi(x_p)$ and $\phi_1(x_d)$. Numerically it is just ordinary multiplication. But if you are working with spins or something, then the fact that it is a tensor product is really more manifest.
The reason that the two dust wavefunctions have little overlap is that depending on the slit the particle went through, the two wavefunctions might have very different momenta. So upon collision, the dust will have very different momenta.
– lionelbrits Dec 10 '14 at 12:39 -
1Regarding the orthonormality, I am not integrating over $x_p$, so technically I am not taking that inner product. If you want to phrase it in terms of spin, then you might ask what is the probability of measuring spin up along the x direction, in which case you get those cross terms. If you measure spin in the z direction, you will not. – lionelbrits Dec 10 '14 at 12:42
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@lionelbrits thanks for answering, I honestly think it would have been a lot more clearer had you used the Dirac notation throughout this answer. – user929304 Dec 10 '14 at 14:31
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2@user929304, probably. Beginning students aren't always exposed to it, so I wanted to make it more accessible. Maybe if I have some time, I'll edit it. Worth the effort? – lionelbrits Dec 10 '14 at 18:49
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