A Cat state acting on given Hamiltonian

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Operators act on states, not the other way around, which is a fundamental principle in quantum mechanics. The discussion clarifies that being an eigenstate of one operator does not prevent other operators from acting on that state; instead, they will alter the state. Specifically, the Hamiltonian given, H = (a^†)b + a(b^†), can act on the even coherent state |α⟩ + |-α⟩, leading to a transformation of the state. Understanding how different operators interact with eigenstates is crucial for analyzing quantum systems. This highlights the importance of recognizing the roles of operators and states in quantum mechanics.
deepalakshmi
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Can a Hamiltonian which contain superposition of (a^†)b act on even coherent state
For example if I consider H = (a^†)b+a(b^†), how will it act on even coherent state i.e. |α⟩+|-α⟩?. I know that |α⟩ don't act on (a^†) because |α⟩ is a eigenstate of lowering operator.
 
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deepalakshmi said:
I know that |α⟩ don't act on (a^†) because |α⟩ is a eigenstate of lowering operator.
First, you are saying this backwards: operators act on states, not states on operators.

Second, the fact that a state is an eigenstate of one operator certainly does not mean that no other operators can act on it. All it means is that other operators will change the state.
 
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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