If two operators $\hat{a},\hat{b}$ have their position representations as $a, b$ ( both $a, b$ are operators but here they act on wavefunctions rather than kets) then is it true in general that the operator $\hat{a}\hat{b}$ , which is a product of the two operators has its position representation as the product of the position representation of the individual operators i.e is the position representation of $\hat{a}\hat{b}$ given as $a b$ ?
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1$\begingroup$ This reads like a homework question and we do not solve those. What are your thoughts on this? Have you tried solving it? Where do you get stuck? $\endgroup$– infinitezeroCommented Mar 28, 2022 at 7:47
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$\begingroup$ Yes, a representation is supposed to obey the same rules of algebra. $\endgroup$– Qmechanic ♦Commented Mar 28, 2022 at 7:47
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$\begingroup$ @Infintezero,I'll update my attempts. $\endgroup$– KashmiriCommented Mar 28, 2022 at 8:06
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1 Answer
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Let the wavefunction of $|\psi\rangle$ be $\psi (x)$
Let $a, b$ be the position representation of $A, B$.
Then we've $b \psi (x)= \langle x | B|\psi\rangle$ .
Let $f(x)=\langle x | B|\psi\rangle$ be wavefunction of $|f\rangle$
Then we've $a f (x)= \langle x | A|f\rangle$
Then $\begin{aligned} a b \psi(x) &=a\langle x|\hat{B}| \psi\rangle=a f(x) \\ &=\langle x|\hat{A}| f\rangle \\=\langle x|\hat{A}| \hat{B} \mid \psi\rangle \end{aligned}$