# What is the interpretation of $|\langle AB\rangle|$? [closed]

what is the interpretation of $|\langle AB \rangle|$ ? since $|\langle AB \rangle|^2 = \langle AB \rangle [\langle AB \rangle]^\dagger$ , could $|\langle AB \rangle| = \sqrt(|\langle AB \rangle|^2)$ ?

PS. : feel free to correct me

## closed as unclear what you're asking by Emilio Pisanty, probably_someone, Norbert Schuch, John Rennie, AccidentalFourierTransformAug 14 '18 at 19:59

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• Your question is rather unclear - please use clearer notation, using MathJax wherever necessary. – Emilio Pisanty Aug 14 '18 at 15:01
• Should this be migrated to math? – Aaron Stevens Aug 14 '18 at 17:37
• What is $\langle AB \rangle$? Is it the expectation value of the product of two observables? Is it a misnotated inner product of two states (which should be written $\langle A | B \rangle$)? More generally, are $A$ and $B$ states, operators, observables, or something else entirely? – probably_someone Aug 14 '18 at 17:41
• yes it is the expectation value of the operator AB , calculated on some system state $|\psi \rangle$ . when we generally type $\langle AB \rangle$ , it generally means the expectation value of the product of two operators A and B – gautam bhuyan Aug 15 '18 at 7:45

$| \langle AB \rangle|=|\langle \Psi|AB|\Psi \rangle|$ for some operators $A,B$ acting on state $|\Psi \rangle$. Using for example a position representation of $A,B$ we can use a wave function i.e. $\Psi= \langle q|\Psi \rangle$ where $|q \rangle$ is a position eigenket (in some dimension, nothing is specified so it is hard to imagine the dimensionality of the problem 1D, 3D etc.). Then we get:
$| \langle A_{pos}B_{pos} \rangle|=|\int\Psi^* A_{pos} B_{pos}\Psi d^3x|$ where $A_{pos}$is the position representation of the operator $A$. If $A$ already is in position representation, i.e. $\hat{x}=x,\hat{p}=-i\hbar \partial_x$ for 1D then you could say $A_{pos}=A$ and ignore the first part.
• But $|\langle AB \rangle |$ is not same as the expectation of the operator AB calculated on some system state $| \psi \rangle$ , it should mean something else right ? – gautam bhuyan Aug 15 '18 at 7:35
• thats's why i am asking how do i interpret this , as modulus of some complex number or square root of the $l_2$ norm , i.e , the square root of $||AB |\Psi \rangle||^2$ ? – gautam bhuyan Aug 15 '18 at 11:15
• how can $|\langle AB \rangle |$ be equal to $\langle AB \rangle$ ? – gautam bhuyan Aug 15 '18 at 11:24
• That is only true if $AB$ is Hermitian, so that the expectation value is real! – Dani Aug 15 '18 at 11:31