Expected value of variance of Hamiltonian in coherent states

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SUMMARY

The discussion focuses on calculating the expected value of the variance of energy in coherent states using the non-hermitian and non-commutative raising and lowering operators, denoted as ##a^\dagger## and ##a##. The user expresses uncertainty about their calculations involving the Hamiltonian operators, specifically 2 and

. The community emphasizes the importance of the commutation relation between these operators, which is crucial for simplifying the expression ##\langle \alpha | (a^{\dagger} a)^2|\alpha \rangle## to ##|\alpha^*\alpha|^2 \langle \alpha | \alpha \rangle##.

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  • Understanding of coherent states in quantum mechanics
  • Familiarity with raising and lowering operators (##a## and ##a^\dagger##)
  • Knowledge of commutation relations in quantum mechanics
  • Basic proficiency in calculating expected values in quantum states
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  • Study the properties of coherent states in quantum optics
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  • Explore the derivation of the commutation relation [##a, a^\dagger##] = 1
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Quantum physicists, students studying quantum mechanics, and researchers focusing on coherent states and operator algebra in quantum theory.

graviton_10
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Homework Statement
Find the variance of the energy in coherent state |ɑ>.
Relevant Equations
<ΔH> = <ɑ| HH |ɑ>
I am trying to find the expected value of the variance of energy in coherent states. But since the lowering and raising operators are non-hermitian and non-commutative, I am not sure if I am doing it right. I'm pretty sure my <H>2 calculation is right, but I'm not sure about <H2> calculation.

Here is my solution:
 

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Check the step circled in orange. ##a^\dagger## and ##a## don't commute.
 
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Yes, but how to do it the right way?
 
graviton_10 said:
Yes, but how to do it the right way?
Please post the steps for how you reduced ##\langle \alpha | (a^{\dagger} a)^2|\alpha \rangle## to ##|\alpha^*\alpha|^2 \langle \alpha | \alpha \rangle##. That way, we can help you see where you made a mistake.
 
So, I used the fact that the commutator of a and a dagger is 1. Does it look good now?
 

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That looks good.
 

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