I'm having trouble understanding why in 3D scattering theory, we choose only waves that "propagate" outwards. In general, what do we mean by outward and inward waves in the quantum mechanical context - I understand what that means for physical waves but what's the interpretation in quantum mechanics?
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2$\begingroup$ How would you arrange an experiment with inward waves? $\endgroup$– John DotyCommented Dec 19, 2022 at 0:35
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$\begingroup$ But what do we mean when we say that a wavefunction e^(ikz) describes an "outward" moving wave in a quantum mechanical sense? In other words, what's the physical meaning of "inwards" and "outwards" quantum wave ? Because if for example we had ψ=e^(ikz) or ψ=e^(-ikz), then their squares are the same, so the probabilities are the same - so what's their difference since we physically care about the probabilities? $\endgroup$– MTYSCommented Dec 21, 2022 at 4:07
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4 Answers
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Incoming/incident waves are usually assumed to be solutions in free space at $t\rightarrow -\infty$ (in the past, before the interaction is important) whereas the outgoing/scattered waves are assumed to be the free space solutions for $t\rightarrow +\infty$ (in the future, after the interaction took place.) This terminology becomes more transparent, if we think about scattering of a classical particle: an electron passing near a nucleus or an asteroid passing near a planet. The reason why we take limits $t\rightarrow \pm \infty$ is that for an infinite range interaction we cannot claim that interaction is limited to a finite interval of time.
It is natural to think of time as flowing from $-\infty$ o $+\infty$, but mathematically we could swap the incoming and outgong solutions and still obtain the correct result - simply because the scattering cross-section is proportional to the square of the matrix element, whereas swapping the solutions is equivalent to simple complex conjugation of this matrix element.
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$\begingroup$ Thank you for your response! I understand what you are saying about swapping incoming and outgoing waves which is the reason I cannot understand what we actually mean by outgoing quantum mechanical wave. What's the "thing" that moves outwards? Is there somekind of velocity? Why wouldn't e^(-ikz) be physically different from e^(ikz)? $\endgroup$– MTYSCommented Dec 21, 2022 at 16:52
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$\begingroup$ @RosTT theres the value of momentum (and hence also velocity) associated with the wave. The sign of the exponent depends in how one defines the momentum operator in QM: $\hat{p}_z= \pm i\hbar \partial_z$ $\endgroup$– Roger V.Commented Dec 21, 2022 at 21:25
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In a scattering problem, the incoming particle comes from some direction, so we model it with a plane wave. The outgoing particle goes (with some probability) in any direction, so we model it with an outgoing spherical wave, usually with an amplitude that depends on direction.
An incoming spherical wave corresponds to a particle whose source is (coherently) in every direction. This is difficult to arrange.
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The essenial problem in QM as far as scattering is concerned is to find the solution to the schrodinger equation. The solution can be interpreted as the probabaility of finding the particle. Now the time factor has already been taken out and we solve for the space part. That is stationary states. The waves may travel inward and then outward what we get is the net result.
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In general, what do we mean by outward and inward waves in the quantum mechanical context <...>?
There are two kinds of velocity in wavelike motion: phase velocity and group velocity. Suppose our waves have linear dispersion relation, so that group velocity equals phase velocity. Then what we see when we watch the evolution of the wavefunction in time is that equal-phase surfaces are concentric spheres that either extend from a point, or converge into it. This wave is then outward or inward, respectively.
Now, if the dispersion relation makes group velocity differ from phase velocity, this may be different: these velocities might even have different directions. In this case it would be more sensible to explicitly consider group velocity (which corresponds to particle velocity in the classical limit), e.g. by forming a (spherically symmetric) wave packet and watch how it moves. If its bulk converges to a point, then the wave is inward, otherwise outward.
Both inward and outward waves can exist, but it's vastly more likely to get outward ones, because to create an inward spherical wave we'd need some kind of a spherical shell emitting this wave inwards, while in the case of scattering the scattering center is usually a point-like secondary source of waves, which, naturally, propagate outwards from it.
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$\begingroup$ Thanks for the response. I get what you are saying but for "physical" waves it is easy to interpret what group and phase velocities are - because ψ represents for example the height which has a physical meaning. In quantum mechanics though, ψ is the amplitude of probability, it is not some physical quantity, so what's the meaning of group/phase velocity? $\endgroup$– MTYSCommented Dec 21, 2022 at 16:47
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$\begingroup$ @RosTT phase velocity is not an observable (for fermions), so it doesn't have physical meaning. Group velocity is the velocity that a wave packet will have as a whole. If your wavefunction defines a bump-like probability density and its spectrum has a well-defined peak at some frequency, then this bump will move with group velocity of the wave at this frequency. This motion influences measurements of particle position. $\endgroup$– RuslanCommented Dec 21, 2022 at 17:56