Time derivative of creation/annhilation operators

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Taking the time derivative of creation and annihilation operators in quantum field theory (QFT) is generally considered impossible because these operators exist in momentum space, depending on momentum (p) rather than position (x). The discussion highlights that while the time derivative of a standard annihilation operator, denoted as d hat{a}/dt, equals zero, this does not imply the derivative exists; rather, it suggests that the derivative is undefined. There is some uncertainty about whether exceptions might exist under different circumstances. Overall, the consensus is that the time derivative of these operators cannot be defined within the standard framework of QFT. Understanding this limitation is crucial for accurate interpretations in quantum mechanics.
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Basically is it possible to take a time derivative of a creation/annhilation operator?
 
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I think that's not possible. If you are talking about creation and annhilation operators usually found in QFT, they live in the momentum space, so they don't depend on x, but on p. However, I am not sure if in some other circumstances this could be possible.
 
yeah so for eample \frac{d \hat{a}}{dt} would =0 if a was the standard QFT annihilation op
 
Saying you cannot take the "time derivative" does NOT mean that derivative is 0. It means that the derivative does not exist at all.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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