Graduate Green's function for the wave function

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To solve the equation HΨ = iħ∂Ψ/∂t, the Green's function G(t,t₀) must satisfy the equation (H - iħ∂/∂t)G(t,t₀)Ψ(t₀) = -iħδ(t-t₀). The transition from this equation to the final wave function Ψ(t) = G(t,t₀)Ψ(t₀) requires removing Ψ(t₀) from the differential equation and applying the condition that G(t,t₀) = 0 for t < t₀. This condition ensures that G(t,t₀) approaches 1 as t approaches t₀ from the right. The discussion emphasizes that these steps are crucial for satisfying the Schrödinger equation and deriving the wave function.
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We want to solve the equation.
$$H\Psi = i\hbar\frac{\partial \Psi}{\partial t} $$ (1)

If we solve the following equation for G

$$(H-i\hbar\frac{\partial }{\partial t})G(t,t_{0}) \Psi(t_{0}) = -i\hbar\delta(t-t_{0})$$ (2)

The final solution for our wave function is,

$$\Psi(t) = G(t,t_{0})\Psi(t_{0})$$ (3)I don't understand the steps. How do we get from (2) to (3) ?
 
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First: (2) is not the differential equation for the Green's function, you need to remove the ##\Psi(t_0)##.

Second: The differential equation on its own does not specify the Green's function, you need to add the condition that ##G(t,t_0) = 0## for ##t < t_0##. This will imply that ##G(t,t_0) \to 1## as ##t \to t_0^+##. With that, you will find that (3) satisfies the Schrödinger equation with ##\Psi(t) \to \Psi(t_0)## as ##t \to t_0^+##.

If you have access to my book, this is discussed in section 7.2.1 for a one-dimensional ODE, but it generalises directly to your case.
 
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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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