Calculating the effect of an operator on an arbitrary state

[Fantini]

Hello. I need help with the following:

Suppose a basis set of states $\varphi_i$. Calculate the effect of the operator $\widehat{R} \equiv \Pi_i \left( \widehat{Q} -q_i \right)$ on an arbitrary state $\Psi$, assuming that the equation $\widehat{Q} \varphi_i = q_i \varphi_i$ is satisfied.

I'm convinced that the answer is zero, but I don't know how to manipulate the indices to reflect that. I've calculated explicitly the cases for $n=2$ and $n=3$.

 

[GJA]

Hi Fantini,

It may help to note that all of the operators $Q-q_{i}$ commute; i.e.
$$(Q-q_{i})(Q-q_{j})=(Q-q_{j})(Q-q_{i})$$
If we use this observation, I think we can avoid dealing with messy indices.

Does this help move things in the right direction? Let me know if anything is unclear/not quite right.

 

[Fantini]

Hi GJA. This doesn't really point me somewhere. I'm still lost at how to do the abstract computation. I'm conjecturing the result holds only for a finite index.

 

[Ackbach]

You can write
$$\Psi=\sum_{j=1}^{\infty}a_j \, \varphi_j,$$
since $\{\varphi_j\}$ is a basis. Then compute
$$\hat{R} \, \Psi=\prod_{i=1}^{\infty}(\hat{Q}-q_i) \sum_{j=1}^{\infty}a_j \, \varphi_j.$$
Can you continue?

 

[Fantini]

No, I can't. I tried that, but I don't know how to actually perform the product/distribution to show that it equals zero.

 

[Ackbach]

Ok, so we get
\begin{align*}
\hat{R} \, \Psi&=\prod_{i=1}^{\infty}(\hat{Q}-q_i) \sum_{j=1}^{\infty}a_j \, \varphi_j \\
&= \sum_{j=1}^{\infty}a_j \, \prod_{i=1}^{\infty}(\hat{Q}-q_i)\varphi_j \\
&=\sum_{j=1}^{\infty}a_j \, \prod_{i=1}^{\infty}(\hat{Q} \,\varphi_j-q_i \, \varphi_j) \\
&=\sum_{j=1}^{\infty}a_j \, \prod_{i=1}^{\infty}(q_j \,\varphi_j-q_i \, \varphi_j) \\
&=\sum_{j=1}^{\infty}\prod_{i=1}^{\infty}(q_j -q_i ) \, a_j \, \varphi_j.
\end{align*}
This all assumes the necessary convergence so that we can switch the order of operations. Does the result look like zero?