The quantum harmonic oscillator Hamiltonian can be written as:
H = (2mω^2)x^2 + (1/2)mω^2p^2
In terms of creation and annihilation operators, the Hamiltonian can be expressed as:
H = (a†a + 1/2)ω
Given [a,a†] = 1, we can use the above expression for H to show that [a,H] = -ωa.
Suppose that |ψ⟩ is an eigenstate of H with eigenvalue E. We can derive the eigenvalue of H for the state |a⟩ as follows:
H|a⟩ = (a†a + 1/2)ω|a⟩
= (a†aω + 1/2ω)|a⟩
= (ωa†a + 1/2ω)|a⟩
= ωa†a|a⟩ + 1/2ω|a⟩
= ω|a⟩ + 1/2ω|a⟩
= (ω + 1/2ω)|a⟩
= (3/2ω)|a⟩
Therefore, the eigenvalue of H for the state |a⟩ is (3/2ω).
The ground state |0⟩ is characterized by the condition a|0⟩ = 0. This means that the annihilation operator acting on the ground state gives zero, indicating that there are no excitations or quanta present in the ground state.
To deduce the energy eigenvalue of the ground state, we can substitute a|0⟩ = 0 into the expression for H:
H|0⟩ = (a†a + 1/2)ω|0⟩
= (a†aω + 1/2ω)|0⟩
= (ωa†a + 1/2ω)|0⟩
= ωa†a|0⟩ + 1/2ω|0⟩
= ω|0⟩ + 1/2ω|0⟩
= (ω + 1/2ω)|0⟩
= (3/2ω)|0⟩
Therefore, the energy eigenvalue of the ground state is (3/2ω).