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stackexchange, Upon reading in Modern Quantum Mechanics by Sakurai & Napolitano, I fell upon a calculation of the expectation value of the annihilation/creation operators squared, with respect to some basis. This was in relation to the Simple Harmonic Oscillator. Here, using the simple relation $a^{\dagger}|n> = \sqrt{n+1}|n+1>$, one can evaluate first $$<m|a^{\dagger}|n> = \sqrt{n+1}\delta_{m,\ n+1},$$ by orthogonality of states, and then further $$ <m|a^{\dagger}a^{\dagger}|n> = \sqrt{(n+1)(n+2)}\delta_{m,\ n+2} $$

This is very simple, and just as easily calculated for the $a$ operator. Now, however, I was wondering, whether one could just as well evaluate the first $a^{\dagger}$ as acting on the bra, to equally obtain $$\left(<m|a^{\dagger}\right)\left(a^{\dagger}|n>\right)=\sqrt{m+1}\sqrt{n+1}\delta_{m+1,\ n+1}$$

Or whether one then have to use the conjugate operator, as $a$ and $a^{\dagger}$ are not self-adjoint, thus obtaining $$\left(<m|a\right)\left(a^{\dagger}|n>\right)=\sqrt{m}\sqrt{n+1}\delta_{m-1,\ n+1}$$.

Just by evaluating for $m=n+2$, we gain by evaluating both creation operators on the ket, $$<n+2|a^{\dagger}a^{\dagger}|n> = \sqrt{(n+1)(n+2)},$$ whereas using one on the bra and one on the ket, would give us $0$ by orthogonality. So this hints at using

$$=\left(<n+2|a\right)\left(a^{\dagger}|n>\right)=\sqrt{(n+2)(n+1)}.$$

The book does not mention anything of this, and does not use any parenthesis. It has earlier noted, that using dirac notation, then it is associative, that is it does not matter where to put the parenthesis. Here the operator is non-hermitian, and Sakurai still uses the notation of no parenthesis. I might confuse myself with the notation of dirac, since I am much more used to the usual linear algebraic notation, but it seems to be the physicists prefered choice.

What am I missing here? Thank you in advance!

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  • $\begingroup$ Two useful points would be (1) Bra-ket and operator multiplication is associative and (2) The raising operator acts differently on bras vs kets (as noted in the answer already given). $\endgroup$ – Charlie Sep 27 '20 at 13:45
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Note that from $a^\dagger |n \rangle = \sqrt{n+1} |n+1 \rangle$ it follows $\langle n | a = \sqrt{n+1} \langle n+1 |$ so $a$ acting on the left is essentially a creation operator.

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