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In a double slit experiment done with particles having mass (say electrons), the results are inferred as being caused by probability waves. The wavelength of these waves is dependent upon the velocity (momentum) of the particles( de broglie wavelength = h/p). Now if the observer is in another frame in a way that the velocity of the observer's frame of reference is equal and in the same direction to the velocity of particles; the velocity of the particles w.r.t. the observer will be zero. Zero momentum means there should be no debroglie waves. But the slits and the screen do have momenta. Does the observer still see the same pattern on the screen? If yes, then how? If not, then doesn't it violate the first postulate of relativity?

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    $\begingroup$ Comment to the question (v2): Note that the Schrödinger equation is not relativistically invariant, so apparent violations of relativity should be taken with a grain of salt. I'm looking forward to a fuller answer. Nice question and welcome to Physics.SE! $\endgroup$
    – rob
    Oct 14, 2015 at 17:23
  • $\begingroup$ Secondly here the observer and the projectile electrons are moving with same velocity in same direction. How is it really possible to get the pattern of electron for the observer? $\endgroup$ Oct 14, 2015 at 18:04
  • $\begingroup$ @RajeshSardar Exactly. Now suppose there is a positive ion at the centre of the screen. For an observer at rest w.r.t. the screen will observe the ion to neutralize because there is a peak at the centre of the screen during the double slit experiment. But for the observer moving with the electrons, if the wave pattern is not observed, there will be no electron reaching the centre. So the ion will not neutralize. Now charge is an invariant. How can these two observers have different observations of charge at the centre despite being in inertial frames... $\endgroup$ Oct 14, 2015 at 18:06
  • $\begingroup$ @RajeshSardar Respected sir, The frame of references are assigned only to the observers and not experimental setups or events. $\endgroup$ Oct 14, 2015 at 18:10

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It is valid in all frames, but I don't know why you used the special relativity tag. The Schrödinger equation is a Newtonian equation, so it uses Galilean Relativity.

Now let's get to how it works. To make the mathematics simple people assume a single monochromatic plane wave matter wave with a constant phase at each of the slits at any moment. In reality you can't make a perfect matter wave, you could experimentally make a wave packet strongly peaked about some momentum, and the math is very very similar.

If you truly had a single particle wave and you switched Galilean frames then the waves would now be peaked around a zero momentum, but the screen and the barrier with the slits would be moving in the opposite direction as the matter wave originally did in the original frame.

Before the potential was effectively quite large in the fixed location of the barrier except where the slits were and you were computing the wave at the fixed locations on the screen. Now you have a large potential in the moving barrier and are computing the wave's values in the moving locations of the moving screen.

So you could try to rewrite everything like that, but it still missing the essential fact of the Schrödinger equation for a universe with more than a single particle in the universe which is:

The wave function is not a physical field in 3d Euclidian space, it is a function from the 3n dimensional configuration space of the entire system of n particles into the joint spin space of the entire system of n particles.

The most studied example of how the Schrödinger equation actually works is the hydrogen atom studied as a system of two particles, a proton and an electron and ignoring all the spin.

You can have a 3d configuration space $\psi(x_e,y_e,z_e,x_p,y_p,z_p,t)$ and have a potential $V(x_e,y_e,z_e,x_p,y_p,z_p,t)=\frac{-ke^2}{\sqrt{(x_e-x_p)^2+(y_e-y_p)^2+(z_e-z_p)^2}}$ then you can switch coordinates to a relative position $(x_e-x_p,y_e-y_p,z_e-z_p)$ and a center of mass and then look for separable solutions. And what you get is the usual Schrödinger equation single particle hydrogen (with a reduced mass) for the relative position solution, and a free particle for the center of mass solution. That's the most common studied problem and if you truly what to understand the Schrödinger equation you should study it and make sure you get it.

And then the same ideas come up for every problem. For the screen plus electron plus barrier with splits there is a center of mass, and there is a relative coordinate of the electron relative to the center of mass. And when you change frames, it just changes the free particle motion for the center of mass, the relative coordinate is unaffected by your choice of frame.

So the wave function is not a function in space and when people talk about it as if it is they are just talk about one particle's position relative to some other things and since it is a relative coordinate, it isn't affected by changes in frame. So really everything that happens when you change frames happened to a part you were ignoring.

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In a double slit experiment done with particles having mass (say electrons), the results are inferred as being caused by probability waves. The wavelength of these waves is dependent upon the velocity (momentum) of the particles( de broglie wavelength = h/p).

The de Broglie wave of the electron is an effective description of the real wavefunction that describes the system "electron scattering off two slits and falling on a screen". The real wavefunction will be the solution of a differential equation with these boundary conditions.

Now if the observer is in another frame in a way that the velocity of the observer's frame of reference is equal and in the same direction to the velocity of particles; the velocity of the particles w.r.t. the observer will be zero.

So? The observer is not part of the solution of the boundary conditions that give the probabilistic solution. Even for special relativity, there exists the relativistic quantum mechanical equations that have to be solved and the boundary conditions of the setup applied. The observer is not part of the setup.

Zero momentum means there should be no debroglie waves. But the slits and the screen do have momenta. Does the observer still see the same pattern on the screen?

The system "observer moving with the electron velocity" will require a lot of complex transformations for the experimental setup.

If yes, then how?

Fortunately the Lorenz invariance laws assure us that the probabilistic solution will be the same as far as the system "electron scattering off two slits and impinging on screen". The interference pattern will be the same.

Your mistake is in giving too much emphasis on the de Broglie wavelength which is a phenomenological description, not a relativistic one, of something described stringently by the appropriate equations and their solutions.

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The double slit experiment result is valid in the inertial frame that is common to the center of the atom whose paired electron cloud produced any photon(s) passing through the double slit.

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  • $\begingroup$ This answer would benefit from more relevant detail and links to references. $\endgroup$
    – user140434
    Jan 20, 2017 at 2:42
  • $\begingroup$ Except for the quantum effect the double slit experiment is intended to observe, the inertial reference frame referenced would be the same inertial reference frame as the double slit itself. $\endgroup$
    – danshawen
    Jan 21, 2017 at 12:55

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