# B How decoherence destroys superpositions

1. May 17, 2017

### jaydnul

If the particle is undisturbed in the two slit experiment, an interference pattern will show up. The weird quantum behavior here is that the particle can have "negative probabilities" associated with its position and momentum, thus they can cancel each other and form the pattern we see.

When the particle interacts with the environment (decoheres), it is said that superposition disappears. Does this mean that the particle is still described by a probability distribution of outcomes, but those outcomes now look much more classical and can't interfere? So really when you say superposition disappears, you mean the negative amplitudes disappear, yes?

2. May 17, 2017

### Staff: Mentor

It cannot. All probabilities are positive.
Amplitudes can be negative, but they are not observable.
Not "still". After decoherence you can start interpreting individual amplitudes (e. g. from the two slits) as probabilities. Before this interpretation is not possible.

3. May 18, 2017

### thephystudent

Perhaps you are referring to the Wigner quasi-probability distribution W(x,p). This one is negative in some regions of phase-space corresponding to superpositions (hence the quasi) but in such a way that the observable probabities
$$|\psi (x)|^2=\int dp W(x,p)$$ and $$|\psi (p)|^2=\int dx W(x,p)$$
are positive anywhere. W itself is a mathematical construction/interpretation, but it not observable itself.
When there is decoherence, W itself becomes positive everywhere, making it a mathematically 'legit' probability distribution, though again only the above marginals are physically observable.

4. May 18, 2017

### bhobba

I have some dim memories of that distribution, but without doubt probabilities in QM can never, ever be negative. Its utterly impossible as the Kolmogorov axioms easily show. In fact, mathematically, its known these days QM is in fact a generalized probability model - the simplest one after ordinary probability theory:
https://arxiv.org/abs/1402.6562

What happened is during the early days of QM when they applied the Klein Gordon (KG) equation they sometimes got negative probabilities which confounded the early pioneers. After a while what was really going on became apparent - it was positive probabilities of antiparticles. Nowadays its all part of a more comprehensive theory, Quantum Field theory, that explains it all from first principles.

The interesting thing about the KG equation is its the most general relativistic equation you can write for a single valued fields. Of course what that single value is isn't spelled out, but its pretty easy to see it must be complex by simply solving it. In fact it's the most general equation for a spin zero particle - but that is a whole new discussion - you can find the full detail here:

There is a lot more that can be said about this interesting equation - but it way off topic - start as new thread if interested.

To answer the OP's original question what happens is a superposition is converted to a mixed state by decoherence. A lot more can be said, but is beyond a B level thread. I have tried to explain more at the B level many times, but failed utterly so won't even try. At the B level get a hold of the following:
https://www.amazon.com/Where-Does-Weirdness-Go-Mechanics/dp/0465067867

Thanks
Bill

Last edited: May 18, 2017
5. May 18, 2017

### stevendaryl

Staff Emeritus
Maybe the original poster was getting probability and probability amplitudes confused. Interference effects involve cancellation between probability amplitudes, which can be positive, negative or even complex.

6. May 18, 2017

### jaydnul

I marked it as "B" but I have a bachelor's in physics and took quantum mechanics... its just been a while. I referred to the amplitudes as "negative probabilities" because of a recent explanation I read that said "quantum mechanics is basically probability theory with negative probabilities allowed".

So decoherence effectively turns the amplitudes into probabilities? My question is about the use of the word "classical". After decoherence, the particle can still only be predicted to land in a certain location based on the probabilities given by the decohered wavefunction, but those probabilities are now spread out in a manner that looks much more classical (as opposed to the particle just stays at the value it was measured as when initially decohered).

Edit: let me try to clear up my question. After decoherence, is the particle still in a superposition of the new classical probabilities, or is the particle definitely in one of those states, and the probabilities just reflect how much we know (theoretically we could know exactly).

Last edited: May 18, 2017
7. May 18, 2017

### Staff: Mentor

I think it is confusing to call it superposition here.

8. May 18, 2017

### jaydnul

Let's assume the Schrodinger wave collapse interpretation. The quantum particle entangles with the measurement device and they become one quantum system together. The particle will take single value at the moment of entanglement (the probability of outcomes being proportional to the square of the amplitudes of its wavefunction). Now that they are entangled, does the particle stay at that collapsed value for the rest of time? Does it become a classical particle in that sense?

9. May 18, 2017

### Staff: Mentor

It does not - because we get decoherence as soon as that would start happening, and with wavefunction collapse that means one option is chosen at random.
Its future depends on the setup. We have to use quantum mechanics again to predict what it does in the future - but not starting from the collapsed state as initial state.

10. May 18, 2017

### jaydnul

That's where I am confused. So the decoherence (measurement) prevents the particle from becoming entangled with the measuring device? I thought entanglement happened as a result of decoherence.

11. May 18, 2017

### Staff: Mentor

One of the reasons I don't like collapse interpretations. You have to define an arbitrary point where collapse happens.

Without collapses, the particle just gets entangled with the measurement device. Nothing magical happens.

12. May 18, 2017

### jaydnul

Edit for clarity: If a particle has become entangled with the measuring system, what happens when the measurement system tries to measure it again at a later time. Will have the same value as before?

13. May 18, 2017

### Staff: Mentor

If nothing happens in between: yes.

14. May 18, 2017

### stevendaryl

Staff Emeritus
No, what turns amplitudes into probabilities is just squaring. What decoherence does is to destroy interference between alternatives.

Here are the rules for computing probabilities in quantum mechanics. If you have an initial state $|I\rangle$ and a final state $|F \rangle$, then you compute the probability amplitude for going from state $|I\rangle$ to state $|F\rangle$ using the Hamiltonian. You get a complex number $T_{IF}$. The probability of going from state $|I\rangle$ to $|F\rangle$ is then $P_{IF} = |T_{IF}|^2$, which is always a positive number (or zero).

Now, what about interference? Well, suppose that there are two paths to get from $I$ to $F$: Via intermediate state $|A\rangle$, or via intermediate state $|B\rangle$. For example, you send an electron through a screen with two slits, and the electron collides with a photographic plate beyond the screen, creating a black dot. You're trying to figure out the probability of getting a dot at a particular location on the plate. The electron could either go via the first slit, or via the second slit.

The amplitude for getting from the initial state $|I\rangle$ to the final state $|F\rangle$ is just the sum of the amplitudes for all the alternative paths. So in the simple case of two alternative intermediate states, we have:

$T_{IF} = T_{IAF} +T_{IBF}$

where $T_{IXF}$ is the probability for getting from I to F via intermediate state X.

Then the probability will be:

$P_{IF} = |T_{IF}|^2 = |T_{IAF}|^2 + |T_{IBF}|^2 + 2 Re ((T_{IAF})^* T_{IBF})$

(This just uses the fact that for two complex numbers A and B, $|A + B|^2 = |A|^2 + |B|^2 + 2 Re(A^* B)$)

If we write $P_{IAF} = |T_{IAF}|^2$ and $P_{IBF} = |T_{IBF}|^2$, then this becomes:

$P_{IF} = P_{IAF} +P_{IBF} + \chi_{AB}$

where $P_{IAF}$ is the probability of going via intermediate state $A$, and $P_{IBF}$ is the probability of going via intermediate state $B$, and $\chi_{AB}$ is the interference term: $2 Re ((T_{IAF})^* T_{IBF})$, which can be either positive or negative. The probabilities behave like classical probabilities if the interference term is zero (or more generally, is unpredictable so that it averages to zero).

Entanglement destroys this interference term.

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15. May 19, 2017

### jaydnul

By nothing, do you mean the particle continues to stay entangled with the device? If so, how does it become unentangled again?

This is the perfect explanation, thank you. Would it be naive to ask how that interference term is destroyed in a conceptual sense, rather than mathematically?

16. May 19, 2017

### stevendaryl

Staff Emeritus
Well, interference only happens between two different intermediate states that lead to the same final state. So for example, in the two-slit experiment you set things up so that there is a source of electrons (or photons, or whatever), then they have to pass through a screen with two slits, and then finally they smash into a photographic plate beyond the screen, making a black dot. So the final state is smashing into a dot on the plate, and the two intermediate states are the two possible slits the electron can pass through. Interference between these two possibilities leads to the distinctive pattern of spots on the plate. So the initial state $I$ is the creation of the electron on one side of the screen, the intermediate states $A$ and $B$ are the electron passing through one or the other slit, and the final state, $F$ is the electron making a dot on the plate.

Now, suppose that immediately after the electron passes through one of the slits, we make a measurement as to which slit it went through. I don't know--maybe take a picture of it (that's not really possible, but say that it is). Then in that case, the two alternative paths produce different final states for the combined system of electron + measurement
1. There is a dot on the screen at some location, and there is a record of the electron going through the first slit, A. Call that state $F_A$.
2. There is a dot on the screen at that location, and there is a record of the electron going through the second slit, B. Call that state $F_B$
For different final states, you don't add the amplitudes and then square, you first square and add the results. So instead of:

$P_{IF} = |T_{IAF} + T_{IBF}|^2$

you get:

$P_{IF} = |T_{IAF_A}|^2 + |T_{IBF_B}|^2$

The interference term is lost, because there is only interference between paths with the exact same final state.

Decoherence is like a measurement, in the sense that the particle interacts with the environment in a way that makes a permanent change to the environment. Maybe the pattern of photons radiated away from the electron. So a different path leads to different final states of the environment (even though, unlike a real measurement, you can't actually compute which path the electron took based on the patterns in the environment) and so there is no interference between the two alternatives.

17. May 19, 2017

### jaydnul

I am still curious how unentanglement (de-entanglement?) happens. When two systems become entangled, what needs to happen to reverse it?

18. May 19, 2017

### Staff: Mentor

By nothing, I mean exactly what I said.

If you measure the momentum of a particle, then let the particle fly through a magnetic field changing its momentum, then measure the momentum again, you will get a different result. This is true even in classical physics and shouldn't be surprising.
If you measure the momentum of a particle, then be careful to avoid anything that would change the momentum, then measure it again, you will get the same result as the first measurement.

Entanglement with things called "measurement devices" is irreversible due to decoherence. Otherwise we don't call the things measurement devices, but see them as part of the quantum system.

19. May 19, 2017

### jaydnul

Would the particle's altered momentum then be predictable by classical mechanics given that you know the strength of the magnetic field and everything else about the system? Or would altering the momentum cause it to slip back into a superposition of different possible momenta?

20. May 19, 2017

### Staff: Mentor

It depends on the system. As an example, you measure the momentum - which means you will have some position uncertainty. The magnetic field could have a different field strength at different locations.