14
$\begingroup$

Recently, I have heard of extensions of the double slit experiment that propose to use a bacterium as the particle. I've also heard that the double slit experiment, in principle, would still show an interference pattern if it were done with a human or a car, as long as there is no observer.

This made me wonder: if this is true, and I exist in a superposition, do I have a mass which exists in several locations at the same time and with an associated weight of 200lbs in each of my superposition locations?

If that were true and a fact, wouldn't that account for the missing mass which we attribute to dark matter? Also, even if on some other level this is true about particles only, wouldn't this be where we would find a link between quantum mechanics and gravity, linking it with relativity?

Forgive me if this is a silly question as I am almost 63 years old and have been almost 50 years out of school, but I just love to read and think about science and learn as much as I can.

$\endgroup$
  • 1
    $\begingroup$ Hi, I think this is a very interesting and deep question! However, you might want to remove part of the 3rd paragraph, or else some people won't react well to your question. (Especially the people working on quantum gravity...) $\endgroup$ – knzhou Aug 22 '16 at 8:50
  • $\begingroup$ I also rewrote your question a bit to sharpen it. I think I preserved the idea, but feel free to undo any of my changes if you don't agree with them. $\endgroup$ – knzhou Aug 22 '16 at 8:51
  • 2
    $\begingroup$ Some (e.g. Penrose) suspect that wavefunction collapse is a result of space-time not being able to be in a state of superposition. $\endgroup$ – lemon Aug 22 '16 at 13:00
  • $\begingroup$ You wouldn't have a separate 200 lbs in each location (unless you weight 400 lbs, that is) - the total effective mass across all the locations would add up to 200 lbs, because the (norm-squared) components of the wave function add up to 1. Quantum superposition can't give you extra mass for free, so it can't explain the missing mass (or extra mass, depending on your perspective) of dark matter. $\endgroup$ – tparker Aug 22 '16 at 17:39
  • $\begingroup$ The problem I think, is that cars, people and even bacteria are already decohered at least 99.9% $\endgroup$ – RBarryYoung Aug 22 '16 at 17:56
5
$\begingroup$

Since you interact with your environment (you touch and even breath air and you absorb light and emit thermal radiation), any superposition you might briefly find yourself in will quickly decohere, on the timescale between such interactions. Put another way: On a timescale of picoseconds or longer, you are not in a superposition.

To the extent that you may think of a person as being in a superposition, then there is a probability to find that person's mass somewhere and a probability to find it (incredibly slightly---both fall 0.000005 attometer without separating in earth's gravity in one picosecond) elsewhere. That does not double this mass at all but merely, if you will, distributes it in this superposition.

To phrase this as a direct answer to your question's title: No, particle superposition is not reflected in the particle's gravitational footprint. A particle in a superposition still feels the same gravitational force and exerts the same gravitational force on its environment, except for whatever (usually slight) separation or change of mass (internal energy) between the superimposed states has been created.

$\endgroup$
18
$\begingroup$

The simple answer is that we don't know because we have no theory of quantum gravity.

If I interpret your question correctly you're thinking about semiclassical gravity, where matter is quantised and gravity isn't. The Einstein equation becomes:

$$ \mathbf G = \langle \mathbf T \rangle $$

where $\mathbf G$ is the classical curvature, i.e. gravity, and $\langle\mathbf T\rangle$ is the expectation value of the stress-energy tensor. The problem with this is that if the matter, i.e. $\mathbf T$, is in a superposition that causes problems. As far as I know this was first discussed in The necessity of quantizing the gravitational field by Eppley and Hannah. There is a beginner friendly discussion of the issues involved in Sabine Hossenfelder's blog. So by asking about the gravitational field caused by matter that is in a state of superposition you've put your finger right on one of the fundamental problems.

Footnote:

A comment suggests I may have misunderstood the point of your question. Firstly a particle in a superposition of being in two places doesn't have an increased mass.

The particle is described by a wavefunction $\psi$ and the probability of finding the particle in some small volume $dV$ is given by:

$$ P = \psi^*\psi\,dV $$

If we add up all the probabilities (by integrating) we must get the value $1$, because the particle exists somewhere with probability $1$, so we get:

$$ \int\psi^*\psi\,dV = 1 $$

If the particle is in a superposition of being in two different places then its wavfunction is peaked in those two places, but because the probability must add up to one that means the value of $\psi$ in each place must be less (by a factor of about $\sqrt{2}$) than if the particle was in only one place. The density of our particle is:

$$ \rho = m\psi^*\psi $$

where $m$ is the mass of the particle. So if superposition reduces the value of $\psi$ it also reduces the density. No matter how the particle is delocalised, if we calculate the total mass by integrating we get:

$$ M = \int\rho\,dV = \int m\psi^*\psi dV = m \int \psi^*\psi dV = m $$

So the mass of the particle is always just $m$ and is unaffected by being in any form of superposition.

Secondly we should note that no large object is going to survive in a superposition for a detectable time because it interacts with its environment and decoheres. The most massive object for which superposition has been experimentally measured is tiny - around 50 microns in size. So there is no chance of you or a car being in a superposition.

$\endgroup$
13
$\begingroup$

There are several misconceptions here to correct:

  1. Double-slit experiments with humans or cars are impossible in any meaningful sense. If you compute the deBroglie wavelength of humans or cars with feasible momenta, you find that their wavefunction we would naively associate to them won't diffract measurably on any slit you can manufacture since diffraction occurs less the smaller the wavelength is, i.e. you will not get any measurable interference, even if you could guarantee that they behave "quantumly" during the experiment without decohering.

  2. Macroscopic objects like cars or humans do not, for any practical purpose, exist in such superposition states of position. They are constantly interacting with their environment in many ways, forcing them to occupy a state of definite position. Depending on your interpretation of quantum mechanics, you may, as in Many Worlds, assert that some other results of such decohering interactions also "exist", but this "existence" cannot influence "our world" at all, so this is a moot point. Creating superposed position states is impossibly hard, if not outright impossible for objects the size of a human.

  3. On the level of particles, such superpositions are perfectly feasible. However, no additional "dark mass" appears - the total rest mass of all states of a given particle is just the mass of the particle. That you have some probability to find it in one place, and some to find it in another doesn't mean that it is in "both" places before you look. It means the question of where the mass is only has an answer after you look. Of course, it is hard to see how this may be reconciled with gravitation - impossible since we don't know the correct quantum theory of gravity - but that the solution is not to adopt each possible location as exerting gravity can be seen in analogy to a force whose quantum theory we do understand: Electromagnetism.

    Think of an electron, smeared out over space in that "electron cloud" commonly pictured. If you want the classical electrical field it exerts, it is the field of one electron charge. Its charge has not magically multiplied just because it's not certain where it is. Now, if the electron has two disjoint locations where it can be, talking about the classical field becomes harder - but you can either be semiclassical, where you would compute the electric field then from the average position of the electron, or also quantize the electric field, whence you only can talk about expectation values of the electric field to begin with.

  4. Lastly, I want to mention that relativity is not the problem as such - quantum field theory does unify quantum mechanics and special relativity. It is general relativity, hence gravity, whose proper quantum theory is unknown. But, in any case, talking about particles and position states already becomes rather hard and sometimes unhelpful in quantum field theory, so one should not expect that thinking on this non-relativistic quantum mechanical level will result in something that can extend QFT.

$\endgroup$
1
$\begingroup$

I'll try to fill in some of the gap between the OPs question and John Rennie's answer. I will use Schrodinger's Cat to aid explanation without need to go into whether this is what really happens.

You have a quantum particle which is 50% chance spin up, 50% chance spin down:

$$|s\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle$$

The classic entanglement to the cat is naively shown as:

$$\lvert s,cat\rangle \rightarrow \frac{1}{\sqrt{2}}\lvert+,\text{live cat}\rangle + \frac{1}{\sqrt{2}}\lvert-,\text{dead cat}\rangle$$

The immediate concern when you see this is, as the OP describes, that we now have two lumps of cat contributing to gravity.

The simple argument is to show that the energy expectation value (i.e. the average amount of cat) is (approx) unchanged.

$$\langle T\rangle = \frac{1}{2}T_\text{live cat} + \frac{1}{2}T_\text{dead cat} $$

This is the quantity you would plug into John Rennie's equation.

Aside: I personally still find it difficult to think about this question in the Many Worlds Interpretation of QM - it would be good if someone else could suggest a thought experiment to help.

$\endgroup$
  • $\begingroup$ Why would you "follow this logic through to the full Many Worlds Interpretation"? What does that even mean? No interpretation follows purely logically from the formalism of quantum mechanics. $\endgroup$ – ACuriousMind Aug 22 '16 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.