# Identical particles and separating the Schrodinger equation

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1. Mar 6, 2017

### mangojuice14

1. The problem statement, all variables and given/known data
Two identical particles, each of mass m, move in one dimension in the potential
$$V = \frac{1}{2}A(x_1^2+x_2^2)+ \frac{1}{2}B(x_1-x_2)^2$$
where A and B are positive constants and $x_1$ and $x_2$ denote the positions of the particles.
a) Show that the Schrodinger equation is separable in the variables $x_1+x_2$ and $x_1-x_2$. Find the eigenvalues and the corresponding eigenfunctions.
b)Discuss the symmetry of the eigenfunctions with respect to particle exchange.

2. Relevant equations
Schrodinger equation:
$$HΨ = EΨ$$
Hamiltonian for identical particles
$$H = \frac{-ħ}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(x_1^2+x_2^2)+ \frac{1}{2}B(x_1-x_2)^2$$

3. The attempt at a solution
I plugged in the Hamiltonian into the Schrodinger equation, and separated variables by setting $Ψ=X(x_1+x_2)Y(x_1-x_2)$, replaced Ψ with the product and divided by $XY$. I end up with something I can't really solve. I'm mostly not sure if my separation method is correct, I'm confused on what it means by 'separable in the variables $x_1+x_2$ and $x_1-x_2$' rather than just $x_1$ and $x_2$ for example. Thanks

2. Mar 6, 2017

### Staff: Mentor

Define two new variables, say $R \equiv x_1+x_2$ and $r \equiv x_1-x_2$ and try and rewrite the Hamiltonian in terms of $R$ and $r$.

3. Mar 6, 2017

### mangojuice14

Rewriting the Hamiltonian in terms of those new variables, I get
$$H = \frac{-ħ^2}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ^2}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(R^2-2x_1x_2)+ \frac{1}{2}Br^2$$
This seems incorrect though because of the factor of $-2x_1x_2$ which is due to $R^2 = x_1^2 + x_2^2+2x_1x_2$. Also, how would I go about changing the terms $\frac{d^2}{dx_1^2}$ and $\frac{d^2}{dx_2^2}$ in terms of R and r?

4. Mar 6, 2017

### Staff: Mentor

There is also a factor $x_1x_2$ that can appear from $r^2$.

You have to use the multivariable chain rule.

5. Mar 6, 2017

### mangojuice14

Oh right! So I got
$$H = \frac{-ħ^2}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ^2}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(\frac{R^2}{2}+\frac{r^2}{2})+ \frac{1}{2}Br^2$$
for the Hamiltonian.
For the $\frac{d^2}{dx_1^2}$ part I get that:
$$\frac{d^2}{dx_1^2}=\frac{d^2}{dR^2}(\frac{dR}{dx_1})^2=\frac{d^2}{dR^2}$$
but I could have also used $r$ so am I just allowed to pick one thereby combining the two derivative terms (since the answer is the same for $\frac{d^2}{dx_2^2}$) or am I doing something wrong?

6. Mar 7, 2017

### Staff: Mentor

The $\frac{d^2}{dx_1^2}$ operator will act on a function of $R$ and $r$, which both depend on $x_1$.

Have a look at https://www.math.hmc.edu/calculus/tutorials/multichainrule/

7. Mar 7, 2017

### mangojuice14

Oh I see, I think I got it. Thanks!