# Measuring quantum mechanics changes the answer and energy?

A thought occured to me thinking about the quantum slit experiment. Upon firing an electron or particle it would impart an opposing force on the machine shooting it. The direction if this force should give you the direction your electron should be going and allow you to predict if it went through a slit or not. When you do, does it count as a measurement and stop the interference pattern? Or does the particle need to be measured?

Imagine for a moment it does still create an interference pattern because you didnt measure the particle itself. If you then measure the electron if and when it goes through a slit AFTER you measured the force in the opposite direction, would the energy be at odds compared to a particle going the exact same direction but it wasn't measured (because it would go through both slits and none at the same time and thus have multiple potential directions)?

• In quantum mechanical dimensions there is the Heisenberg uncertainty principle which encapsulates the probabilistic nature of quantum mechanical predictions. Itis the probabilities that have wave behavior, not the particle and the slits themselves. It means that the trajectory of the electron in this single electron at a time hitting dounle slits en.wikipedia.org/wiki/… cannot be defined as a classical trajectory, but as a probable classical trajectory, dependent on boundary conditions in interacting with the two slits. – anna v Jul 19 '18 at 13:35
• I'm guessing you wouldn't be able to measure this within the uncertainties. Ion propulsion and laser propulsion are illustrations that there is scope for some measured changes. – CriglCragl Jul 19 '18 at 13:48
• @Anna v I understand that (kind of). The trajectory of the particle cannot really be defined classically, but the recoil/opposing force the particle had upon being fired at the slits would define it's classical trajectory without changing the particles path through measurement (I assume). If you then measure the particle it changes it's unclassical undefined trajectory to a classical defined trajectory. So would that change the measurement of recoil direction at the machine firing the particle? – Demigan Jul 19 '18 at 15:34
• The firing machine which you hypothesize, is also ruled by quantum mechanics as long as it is firing quantum mechanical entities. For a given single event, there will be the probabilistic uncertainty of "firing" , no classical trajectory can be defined in this sense. everything is probabilistic, although energy and momentum are conserved for each event, but there is the heisenberg uncertainty on position and momentu – anna v Jul 19 '18 at 17:47
• – Stéphane Rollandin Jul 19 '18 at 20:51

## 5 Answers

Even if you know the initial direction of the particle, this does not tell you which slit the particle will go through. The wavefunction (probability wave) of the particle will still follow the Schrodinger equation. For example, the initial condition in solving the Shrodinger equation could be a Gaussian wave packet in space centered at the measured starting position with some standard deviation determined by the device measuring the initial position. The Gaussian will also be determined by the measured momentum (with some magnitude and direction, each with their own associated uncertainty). This wavefunction will then evolve according to the Schrodinger equation.

Getting to the heart of your question: just because we know the initial position and direction of the particle does not mean it follows a well defined trajectory afterwards. This is just not the case according to basic interpretations of quantum mechanics.

• Allright, so the answer is "no, measuring the particle will have no effect on the measurement of the opposing force on the machine"? – Demigan Jul 19 '18 at 15:37
• @Demigan The particle goes through both slits, if it can. See the delayed choice experiments for more on that in action – CriglCragl Jul 19 '18 at 16:11
• @Demigan What do you mean by "measuring the particle"? Do you mean measure the position? If you are measuring the force then don't you have some information about where the particle was located upon force measurement? – Aaron Stevens Jul 19 '18 at 21:21
• @Aaron Stevens I mean the measurement nornally associated with changing it's outcome by determining which slit it goes through. Also wouldnt you be able to extrapolate it's position upon force measurement? The velocity of the particle can be found based on the amount of opposing force on the machine, and the direction of the opposing force gives you the direction of the particle (for both give and take the preciseness of the measurement). Use the time it took for the measurement and you at least narrow the amount of potential positions it could be in. – Demigan Jul 19 '18 at 21:29
• @Demigan You are on the right track by saying "at least narrow the amount of potential positions it could be in." You can calculate the spreading out of the wave function based on the initial conditions. But this is based on the Schrodinger equation, not classical trajectories. – Aaron Stevens Jul 19 '18 at 22:46

There is only interference if the setup is such that it is not possible to determine through which slit the electron went.

Whenever you're analyzing such conceptual problems, it's very useful to try and make a concrete model of the system you're considering. Maybe even multiple models, to better see the common properties.

Suppose you have an apparatus, which fires an electron in one direction, and recoil (or whatever other mechanism) results in a second electron being ejected with the opposite momentum. To be able to use the momentum being opposite to predict where the first electron will be, you need the initial positions to also be accurately defined: otherwise even classically you'll get uncorrelated results.

Now, the two-electron system, being quantum mechanical, obeys Heisenberg uncertainty principle. Since both particles are in a precisely defined location, their total momentum is quite uncertain. This then means that, by measuring the momentum of the second electron, you'll obtain the momentum of the first electron... ± that additional uncertainty. The final result will be that you'll still get garbage results if you try to answer the which-way question for the first electron using this measurement of the second electron's momentum.

If you are able to determine which slit the particle went through, by any means, there is no interference pattern.

In practice, I think it would be difficult to set up the experiment as described. But in principle, yes, if you measured the recoil accurately enough to determine which slit the particle went through, the interference pattern would disappear.

(Although as Aaron's answer points out, knowing the momentum does not necessarily determine which slit the particle goes through. I'm assuming that the momentum is high enough, and the slits far enough apart, that the probability of a particle going through the "wrong" slit is minimal. The probability is never going to be zero, so there will always be some interference, but it could in principle be minimized - and that interference pattern would look different to the original one anyway.)

This (your first paragraph) is the same question/famous objection Einstein posed in the Bohr-Einstein debates: https://en.wikipedia.org/wiki/Bohr%E2%80%93Einstein_debates#Post-revolution:_First_stage

So basically Bohr reply was: You also need to apply QM to the double slit. So to measure the momentum transfer, you have to have a sharp momentum (of the measurement apparatus) first. But then your position is not determined leading to washing out of the interference pattern.

## protected by Qmechanic♦Jul 19 '18 at 19:30

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