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*}


\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 ?!?.

  • $\begingroup$ 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 $\psi_{ij}$? $\endgroup$ – ZeroTheHero Oct 5 '17 at 14:50
  • $\begingroup$ @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 $E_n$. $\endgroup$ – 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.

  • $\begingroup$ 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 $\endgroup$ – AgentRev Oct 6 '17 at 19:43
  • 1
    $\begingroup$ @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 $0$. $\endgroup$ – ZeroTheHero Oct 6 '17 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.