I want to know what's the physical meaning of the reduced mass?
if we try to connect it with quantum mechanics is the wave function of the reduced mass of two bodies has any physical meaning?
If you have a system of two "bodies", you can describe it in 2 ways (>2, but let's limit ourselves to these 2 for now): either with separate coordinates for each of the bodies, or with a set of {coordinates of their center of mass} and {relative coordinates of one of these bodies wrt the other one}. In the above it's better to talk about pointlike bodies, to avoid the need to consider other generalized coordinates to describe the "bodies" orientations.
The second of these descriptions makes life simpler in some circumstances, e.g. when the potential energy only depends on this relative distance between the bodies, rather than on the coordinates of those bodies individually -- than you get a free motion in the first of these generalized coordinates (center of mass): corresponding momentum is conserved, and you've got a simplification of the problem (namely, part of the equations of motion solved, you are left with eq. of motion for the relative coordiantes only).
From the point of view of classical mechanics you are good to go with any set of generalized coordinates, so either of these 2 descriptions is perfectly legitimate, and second one gas its advantages as outlined above. Now, going to quantum description is also straightforward from both of these classical descriptions. You can describe your system either with a wavefunction \psi(r1,r2,t), or \psi(rc, \delta r, t), where rc is the vector describing the position of the cenetr of mass, and \delta r = r1-r2. Shroedinger equation is again trivially solved for the dependence of \psi on rc in the case when the potential only depends on \delta r -- so your wavefunction in the second set of coordinates (rc, \delta r) factors into a trivial function of rc (namely, free motion, i.e. plain wave for a state with defined momentum corresponding to rc; so you can loosely forget about this part of the wavefunction), and a (less trivial) function depending on \delta r -- after this simplification the Shroedinger equation reduces to that for \delta r only (so: the problem is again simplified, half degrees of freedom are gone into the trivial part). That reduced wavefunction of \delta r is precisely what you called "wave function of the reduced mass of two bodies" -- and it has a perfectly physical meaning.
For example, a wave function of a hydrogen atom is a superposition of basis wavefunctions, each being a product of \psi_c(r_c) describing arbitrary motion of the "atom as a whole" (= the center of mass of the proton and the electron) and the part desribing relative motion of p and e (say, 1s, 2s, 2p, 3d etc state). In textbooks it is the latter part given most attention -- and it is this "reduced mass wavefunction" you asked about.
Dear Andrey
Do you think there is a deeper relation of reduced mass and the yielded mass from combination of two particles?
Not exactly sure what you call the yielded mass -- if it is the rest mass of the two particle system (taking into account their interaction) than I don't think there is a relation. Such a yielded mass crucially depends on the specifics of their interaction, whereas the reduced mass is a purely kinematic notion.
Dirac's solution of reduced mass-energy lower than free stationary electron has an important physical meaning. It is not considered as a acquired sum of mass energy and negative potential energy but the quantum state of the electron has that energy. This means that electron has assimilated/acquired the negative energy arising from virtual photons of the electromagnetic interaction leading to a combined wave function, which needs equivalent positive energy photons to time evolve it to a higher energy state or free state. Furthermore as that is a lower energy possible state, the electrons have spontaneous tendency to fall into it.
Let us assume we have two bodies, 1 and 2, with masses m1 and m2. if a force F12 acts from 1 to 2, then by Newton's third law an equal force F21 acts, in the opposite direction, from 2 to 1. Denote the magnitude of both forces by F. Then body 1 has an acceleration a1=F/m1 and body 2 has acceleration F/m2. Assume the force is radial (this is important) and attractive (this is just to be specific. If you change all sign, you will get the same result for the repulsive case). So body 1 is accelerated by F/m1 towards body 2, whereas body 2 is accelerated by F/m2 towards body 1. So r12, the distance between body 1 and body 2, decreases with an acceleration given by a1+a2, that is, F(1/m1+1/m2). Now if you define the reduced mass as 1/mu=1/m1+1/m2, then r12 is accelerated by F/mu. So mu is the mass that describes the effect of a force on the interparticle distance. It is less than the mass of a single particle, because the interparticle distance accelerates more than if the force acted on one particle alone.
This is for the physical meaning in classical mechanics. Quantum mechanics is formally similar to classical, so iti s not surprising to find that the same algebra yields the same result. But there is no such easy intuition.
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