I Quick question regarding creation/annihilation operators

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The discussion centers on the mathematical treatment of creation and annihilation operators in quantum mechanics. It confirms that an annihilation operator acting on kets can function as a creation operator on bras, highlighting the duality in their roles. The participant seeks clarification on the proper mathematical approach to using these operators, specifically regarding complex conjugation and their directional action. Additionally, it notes the convention of using Latin letters for these operators, rather than Greek letters. Understanding these principles is essential for correctly applying creation and annihilation operators in quantum mechanics.
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Given the matrix element of an annihilation operator, can we consider it to be a creation operator acting to its left?
I'm reading a textbook and it states the following:

Screenshot 2025-04-09 121748.png


For whatever reason latex previewing is not working for me right now.. but the way I was thinking about this was to take the complex conjugate of the entire left, act with the ensuing creation operator on |x_1 ... x_N-1> to get |x_1...x_N>, and then take the complex conjugate of <y_1...y_N | x_1....x_N>* to get back the final result.

Is that the right way to think about it mathematically? And can I just generally think about it as a creation operator acting to its left?
 
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MaestroBach said:
Is that the right way to think about it mathematically? And can I just generally think about it as a creation operator acting to its left?
Yes an annihilation operator for the kets can act as a creation operator on the bras. Also note that you would usually use just a Latin letter like ##\hat{a}, \hat{b},\hat{c},\hat{f}##, rarely we use ##\hat{\psi}## for annihilation/creation operators.
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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