Expected value of Hamiltonian with operators in atomic units Hello I am currently working on a problem sheet for university, and am confused on how to solve it correctly.
We deal with a helium atom and are asked to express the Hamiltonian in atomic units. This is easy as we can use the Born-Oppenheimer approximation:
H=(-1/2)∇₁²-(1/2)∇₂²-(2/r₁)-(2/r₂)+(1/r₁₂)
Now, the wave function of the electrons is seperated into a product of two 1s-functions:

This is understable until this point. Now it is asked to formulate the expected value for the wave function, using these expressions given here:
But I don't really know how to approach this: As the function is expressed as a product of two 1s-function, I don't think we can seperate the integral and formulate it like this. I have tried to write that down, but my problem is how we can deal with the fact that the expressions only deal with single functions. Here is what I tried: I don't know how to incorporate the expressions.

 A: This is, actually, doable. You are on the right path, but your expectation value definition is not correct. Since you have 6 spatial dimensions (3 for each particle), it should be:
$$\begin{align}
\langle \psi(\vec{r}_1, \vec{r}_2) | A | \psi (\vec{r}_1, \vec{r}_2) \rangle 
&= \int \int \psi(\vec{r}_1, \vec{r}_2)^* A \big(\psi(\vec{r}_1, \vec{r}_2)\big)\mathrm{d}^3\vec{r}_1 \mathrm{d}^3\vec{r}_2 \\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \psi(\vec{r}_1, \vec{r}_2)^* A \big(\psi(\vec{r}_1, \vec{r}_2)\big)\mathrm{d}x_1\mathrm{d}y_1\mathrm{d}z_1\mathrm{d}x_2\mathrm{d}y_2\mathrm{d}z_2
\end{align}
$$
Let's look at the first term, $\langle \psi(\vec{r}_1, \vec{r}_2) | -\frac{1}{2} \Delta_1 | \psi (\vec{r}_1, \vec{r}_2) \rangle$. This is:
$$
\left\langle \psi(\vec{r}_1, \vec{r}_2) \middle| -\frac{1}{2} \Delta_1 \middle| \psi (\vec{r}_1, \vec{r}_2) \right\rangle
= \int\int \phi(\vec{r}_1)^* \phi(\vec{r}_2)^* \bigg( -\frac{1}{2} \Delta_1 \bigg)\bigg(\phi(\vec{r}_1) \phi(\vec{r}_2)\bigg) \mathrm{d}^3\vec{r}_1 \mathrm{d}^3\vec{r}_2$$
Notice that the operator does not act on $\phi(\vec{r}_2)$, so we can rewrite it as:
$$
\int\int \phi(\vec{r}_1)^* \phi(\vec{r}_2) \phi(\vec{r}_2)^* \bigg( -\frac{1}{2} \Delta_1 \bigg)\bigg(\phi(\vec{r}_1) \bigg) \mathrm{d}^3\vec{r}_1 \mathrm{d}^3\vec{r}_2$$
We can also reorder the integrals:
$$
\int\int \phi(\vec{r}_1)^* \phi(\vec{r}_2) \phi(\vec{r}_2)^* \bigg( -\frac{1}{2} \Delta_1 \bigg)\bigg(\phi(\vec{r}_1) \bigg) \mathrm{d}^3\vec{r}_2 \mathrm{d}^3\vec{r}_1$$
And since $\vec{r}_1$ is constant with respect to $\vec{r}_2$, we can pull out the appropriate terms:
$$
\int\phi(\vec{r}_1)^*\bigg( -\frac{1}{2} \Delta_1 \bigg)\bigg(\phi(\vec{r}_1) \bigg) \int \phi(\vec{r}_2) \phi(\vec{r}_2)^*  \mathrm{d}^3\vec{r}_2 \mathrm{d}^3\vec{r}_1$$
But now we can do the integral for the variable $\vec{r}_2$! Why? Because it's just the norm of the one-particle wave function, which is $s_2$ in our case. So we are left with:
$$
s_2 \int\phi(\vec{r}_1)^*\bigg( -\frac{1}{2} \Delta_1 \bigg)\bigg(\phi(\vec{r}_1) \bigg) \mathrm{d}^3\vec{r}_1$$
Can you see what is this quantity?
