Simplifying a two-particle system to a one-body problem

[batclocks]

Homework Statement

Griffiths Introduction to Quantum Mechanics, 3rd ed. Problem 5.1: "A typical interaction potential depends only on the vector displacement r=r1-r2 between the two particles: V(r1, r2) --> V(r). In that case the Schrödinger equation separates, if we change variables from r1, r2 to r and
R=(m1r1+m2r2)/(m1+m2) (the center of mass).

a) Show that r1=R+(μ/m1)r, r2=R-(μ/m2)r, and ∇1=(μ/m2)∇R+∇r, ∇2=(μ/m1)∇R-∇r"

Relevant Equations

∇=i(d/dx)+j(d/dy)+k(d/dz)

The first part is actually fine. You just note that since
r=r₁+r₂
that means
r₁=r+r₂ and r₂=r₁-r
and you substitute that into the center of mass, R, and simplify to get
r₁=R+(μ/m₁)r, and r₂=R-(μ/m₂)r

But the next part is where I'm very confused.
The general idea is that you want to prove
∇₁ = (μ/m₂)∇_R + ∇_r and ∇₂ = (μ/m₁)∇_R - ∇_r

But you really only need to do this for one component of ∇, so you let
r₁ = (x₁, y₁, z₁); r₂ = (x₂, y₂, z₂); R = (X, Y, Z); and
r = (x, y, z)

and this is where I get confused. The solution posted by my professor said that
d/dx₁ = (dX/dx₁)(d/dX) + (dx/dx₁)(d/dx)

This looks vaguely like the chain rule, but it isn't anything I've seen before. He told me that, since x and X are function of x₁, their derivatives need to be taken into account when taking d/dx₁. I don't really know what to make of that. I think I just straight up don't understand the math. He said "that's just how we take derivatives," but I don't think I've ever seen a derivative defined like that before.

 

[BvU]

Hello @batclocks, !

You have a $$\Psi({\bf r}_1, {\bf r}_2) = \Psi(x_1, y_1, z_1,x_2, y_2, z_2)$$ that you now write as
$$\Psi'({\bf R}, {\bf r}) = \Psi'\bigl ({\bf R}(x_1, y_1, z_1,x_2, y_2, z_2), {\bf r}(x_1, y_1, z_1,x_2, y_2, z_2)\bigr )$$

The first component of ##\nabla_1## is ##\displaystyle {\partial \Psi\over\partial x_1}\ ##, ##\ \ ##the first component of e.g. ##\nabla_R## is ##\displaystyle {\partial \Psi'\over\partial R_x}\ ##, ##\ \ ##and -- now comes this chain rule -- you get:
$${\partial \Psi\over\partial x_1} = {\partial \Psi'\over\partial R_x} {\partial R_x\over\partial x_1} + {\partial \Psi'\over\partial r_x} {\partial r_x\over\partial x_1} = (\nabla{\bf R})_x {\mu\over m_2} +(\nabla{\bf r})_x$$

and that's all. What is confusing to ordinary folks like me is the notation: you have to imagine all these things are operators: they work on a function ##\Psi## (c.q. ##\Psi'##) that you have to fill in yourself.

If this still causes trouble, look at ##f(x) = cos( 2x)##, substitute ##y = 2x## and look at ##\displaystyle {d\over dx} g(y) ##

(then you also see why I use a ##\Psi## and a ##\Psi'## )

 

[batclocks]

$${\partial \Psi\over\partial x_1} = {\partial \Psi'\over\partial R_x} {\partial R_x\over\partial x_1} + {\partial \Psi'\over\partial r_x} {\partial r_x\over\partial x_1}$$

Okay, in this case you've written it in a different order and it makes tons of sense. The way I have it written up there is $${\partial \over\partial x_1} = {\partial R_x\over\partial x_1}{\partial \over\partial R_x} + {\partial r_x\over\partial x_1}{\partial \over\partial r_x}$$

But (and I didn't think of it this way until you threw they ##\Psi'## in there) you can just switch the order because it's only multiplication, which is commutative (duh).

And the thing that would lead you to express ##{\partial \over\partial x_1}## operator in this way is the fact that we know we are acting on a ##\Psi'## that is a function of R and r.

Thank you so much for the help!