# Average squared Hamiltonian of linear combination of eigenfunctions

As part of a larger problem, I am trying to find the average squared Hamiltonian of a system with eigenfunctions $\psi_{1,1}$, $\psi_{1,2}$, $\psi_{2,1}$, $\psi_{2,2}$ and the following wave function:

$$\Psi\left(\mathbf{r};t=0\right)=c\sum_{j=1}^2 \psi_{ij}\left(\mathbf{r}\right)$$

The problem defines the following operators:

\begin{align*} \hat{H}\psi_{ij} &= iE\psi_{ij} \\ \hat{Q}\psi_{ij} &= jQ\psi_{ij} \end{align*}

where $\{i,j\}\in\mathbb{R}$. I have already calculated that

\begin{align*} p\left(\mathbf{r}\right) &= c\,\left(\langle\psi_{i1}|\psi_{i1}\rangle + \langle\psi_{i1}|\psi_{i2}\rangle + \langle\psi_{i2}|\psi_{i1}\rangle + \langle\psi_{i2}|\psi_{i2}\rangle\right) \\ 1 &= c\,\left(1 + 0 + 0 + 1\right) \\ c &= \frac{1}{2} \end{align*}

and

\begin{align*} \langle\Psi|\hat{H}|\Psi\rangle &= \langle\psi_{i1}|\hat{H}|\psi_{i1}\rangle + \langle\psi_{i1}|\hat{H}|\psi_{i2}\rangle + \langle\psi_{i2}|\hat{H}|\psi_{i1}\rangle + \langle\psi_{i2}|\hat{H}|\psi_{i2}\rangle \\ &= \frac{i}{2}\left(E_{i1}\langle\psi_{i1}|\psi_{i1}\rangle + E_{i2}\langle\psi_{i1}|\psi_{i2}\rangle + E_{i2}\langle\psi_{i2}|\psi_{i1}\rangle + E_{i2}\langle\psi_{i2}|\psi_{i2}\rangle\right) \\ &= \frac{i}{2}\left(E_{i1}\langle\psi_{i1}|\psi_{i1}\rangle + 0 + 0 + E_{i2}\langle\psi_{i2}|\psi_{i2}\rangle\right) \\ &= \frac{i}{2}\left(E_{i1}+E_{i2}\right) \end{align*}

However, I'm not quite sure how to scale it up to $\langle\Psi|\hat{H}^2|\Psi\rangle$. I am putting forward the assumption that $\langle\hat{H}^2\rangle - \langle\hat{H}\rangle^2$ should = 0 since I am working with eigenfunctions, and that therefore

\begin{align*} \langle\Psi|\hat{H}^2|\Psi\rangle &= \,?!? \\ &= \left(\frac{i}{2}\left(E_{i1}+E_{i2}\right)\right)^2 \\ &= \frac{i^2}{4}\left(E_{i1}^2 + E_{i1}E_{i2} + E_{i2}^2\right) \end{align*}

But I am unsure how to prove it, hence the ?!?.

• what is $p(\boldsymbol{r})$? Moreover, is the sum in $\Psi(\boldsymbol{r};t=0)$ only over $j$ or is it also over $i$? If there is no sum over $i$, why to you need this index on your $\psi_{ij}$? – ZeroTheHero Oct 5 '17 at 14:50
• @ZeroTheHero - $p\left(\mathbf{r}\right)$ is the probability at location $\mathbf{r}$, it's just for normalizing. The sum is only over $j$, but according to my peers the $i$ must stay. I think my main source of confusion is the $E$ - my teacher omits the index both in the assignment and his notes, but when I look in the Griffiths it's almost always defined as $E_n$. – AgentRev Oct 5 '17 at 18:00

Actually this is not quite correct. Your $p(r)$ should be $$p(\boldsymbol{r})=cc^*\left( \langle \psi_{i1}\vert\psi_{i1}\rangle + \langle \psi_{i1}\vert \psi_{i2}\rangle +\langle \psi_{i2}\vert\psi_{i1}\rangle + \langle \psi_{i2}\vert \psi_{i2}\right)$$ from which you find that $c=\frac{e^{i\varphi}}{\sqrt{2}}$ for arbitrary phase $\varphi$. You may choose $\varphi=0$ for convenience but you don't have to.
Then, \begin{align} \hat H \left(c\sum_{j=1}^2\psi_{ij}(\boldsymbol{r})\right)&=\left(c\sum_{j=1}^2i E\psi_{ij} (\boldsymbol{r})\right)=i E\left(c\sum_{j=1}^2\psi_{ij} (\boldsymbol{r})\right)\, ,\\ \hat H^2 \left(c\sum_{j=1}^2\psi_{ij}(\boldsymbol{r})\right)= \hat H\left(\hat H \left(c\sum_{j=1}^2\psi_{ij}(\boldsymbol{r})\right)\right)&=i^2 E^2\left(c\sum_{j=1}^2\psi_{ij} (\boldsymbol{r})\right)\, . \end{align} You can use orthogonality to finish the calculation. Note that all your states with same $i$ have the same energy and this should produce a simplified result.
• Thanks for the help. I got confused with $E$ and forgot that $c$ is part of the inner product. The problem also defined $c\in\mathbb{R}^+$, so $cc^* \simeq c^2$. Another peer suggested simply placing the summation directly in bra-kets, so I ended up with this: i.imgur.com/KaYdlun.png – AgentRev Oct 6 '17 at 19:43
• @AgentRev oui c'est exact puisque vos functions d'ondes sont des fonctions propres de l'Hamiltonian avec la meme valeur propre, i.e. la combinaison $\sum_{j=1}^2\psi_{ij}$ est egalement une fonction propre, donc la variance sera $0$. – ZeroTheHero Oct 6 '17 at 19:47