# What does it mean “differentiation with respect to the coordinates of particle 1 or 2”?

I was reading Introduction to Quantum Mechanics by Griffiths. In Chapter 5, Identical Particles, I came across the notation $$\nabla_1$$ and $$\nabla_2$$. Griffiths writes that it means "differentiation with respect to the coordinates of particle 1 or particle 2", which I am unable to understand. Could anyone please help me out with it?

Let $$\mathbf{r}_i = x_i \hat x + y_i \hat y + z_i \hat z$$ label the coordinates of particle $$i$$. $$\nabla_i$$ therefore means $$\hat x_i \frac{\partial}{\partial x} + \hat y\frac{\partial}{\partial y_i} + \hat z \frac{\partial}{\partial z_i}$$.

• In Hamiltonian operator Laplasian operator $\nabla^2$ is used, meaning that it is second order derivatives over position vector – Agnius Vasiliauskas Oct 30 at 14:36

Hamiltonian operator is defined as sum of kinetic and potential energies operators of system: $${\displaystyle {\hat {H}}={\hat {T}}+{\hat {V}}}$$

In many particles system :

$${\hat {T}}=\sum _{n=1}^{N}{\hat {T}}_{n}, \\ {\displaystyle \hat{V}=V(\mathbf {r} _{1},\mathbf {r} _{2},\ldots ,\mathbf {r} _{N},t)}$$

Putting all this together and with exact particle kinetic energy expression, we get :

$$\hat {H} = \sum _{n=1}^{N}\left({-{\frac {\hbar ^{2}}{2m_{n}}}\nabla^{2}_{n}}\right) + V(\mathbf {r} _{1},\mathbf {r} _{2},\ldots ,\mathbf {r} _{N},t)$$

Because you have only 2 particles, this simplifies to :

$$\hat {H_{2}} = {-{\frac {\hbar ^{2}}{2m_{1}}}\nabla^{2}_{1}} {-{\frac {\hbar ^{2}}{2m_{2}}}\nabla^{2}_{2}} + V(\mathbf {r} _{1},\mathbf {r} _{2},t)$$

It might help to think of this problem in just one dimension first: Suppose particle 1 is located at the position $$x_1$$, and that particle 2 is located at the position $$x_2$$.

The key idea here is that even though there is only 1 dimension, you need to describe the position of two particles by using a coordinate system with two coordinates (one for each particle).

If you think of each particle separately, imagining they don't interact at all, you would think that they should each have some wavefunction $$\psi_j(x_j)$$ for $$j\in\{1,2\}$$ which each satisfy the free time-independent Schrodinger equations $$- \frac{\hbar^2}{2m} \frac{d^2\psi_j(x_j)}{dx_j^2} = E \psi_j(x_j) \ .$$

Instead if you think of them as a combined system, there is now an overall wavefunction $$\Psi(x_1,x_2)$$ which describes the two particles, satisfying an equation $$- \frac{\hbar^2}{2m} \frac{d^2\Psi(x_1,x_2)}{dx_1^2} - \frac{\hbar^2}{2m} \frac{d^2\Psi(x_1,x_2)}{dx_2^2} = E \Psi(x_1,x_2) \ .$$ This is easy to solve because you can use the method of seperation of variables to reduce the above equation to the previous ones.

Finally, adding an interaction means to include some potential $$V(x_1, x_2)$$ where $$- \frac{\hbar^2}{2m} \frac{d^2\Psi(x_1,x_2)}{dx_1^2} - \frac{\hbar^2}{2m} \frac{d^2\Psi(x_1,x_2)}{dx_2^2} + V(x_1,x_2) \Psi(x_1,x_2) = E \Psi(x_1,x_2) \ .$$

For example, you could make $$V(x_1,x_2)$$ the Coulomb potential where $$V(x_1,x_2) \propto \frac{1}{|x_1 - x_2|}$$ and now the particles are talking to each other (although obviously the equation is now harder to solve).

When working in 3D you just swap $$x_1 \to \mathbf{r}_1$$ and $$x_2 \to \mathbf{r}_2$$ in the obvious way, and you need to track a total of 6 coordinates now to full describe the positions of each of the two particles.