**Required math: calculus**

**Required physics: 3-d Schrödinger equation**

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.1.

The Schrödinger equation that we’ve looked at so far involves the wave function for a single particle moving in a potential. To extend this to multi-particle systems, we need to make the wave function and the potential functions of the positions of all the particles and the time. Thus the Schrödinger equation for a system of particles becomes

Needless to say, finding solutions of this equation for even as few as 2 particles is extremely difficult. In one case, however, we can make some progress. In a 2-particle system, if the potential is not time-dependent and depends only on the separation of the two particles, we can fiddle with it a bit and produce a simpler form.

First, we define the centre of mass

If we also introduce the reduced mass

then we get

In the coordinates and , we can find the gradient operators. We use a dummy function to give the gradient something to operate on. We’ll consider the component and use the chain rule (remember that and are independent vectors, each with 3 components, so there is a total of 6 independent position variables):

The relations for the other two components are similar, so dropping the test function , we get for the gradients:

To get the Laplacian operators, we differentiate the component expressions above.

The combination of these two expressions that appears in the Schrödinger equation is, after cancelling terms and putting the remaining terms over common denominators:

The calculations for the other two components are similar, so the final result is:

We can now try the usual technique of separation of variables, so we try

We get, after substituting and dividing through by :

As usual, the terms involving each of the variables and must separately be equal to constants, so we get

The first equation is that of a free particle with mass , while the second is that of a particle with mass moving in a potential . Thus the system separates into one equation for a free particle with the total mass and a position at the centre of mass and another for a single particle with the reduced mass . The total energy is .

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mosaic wolfNow that makes it perfectly clear. I got the free particle confused with the other one moving within the potential. Thax for the simple explanation.

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