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I am doing some revision on theoretical physics, specifically propagator theory. This is talking about how to work out the probability amplitude at some time $t_{f}$ and position $x){f}$, given that we know what happens at $(t,x)$.

Working in natural units, and we can rearrange the Schrodinger equation to get an operator.

$$ i\partial_{t_f} \psi(t_f,x_f) = \hat{H}(t_{f},x_{f})\psi(t_f,x_f) $$ $$ [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\psi(t_f,x_f) = 0 $$

and then we can define an operator $\hat{O}$ as: $$ \hat{O}\psi(t_f,x_f) = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\psi(t_f,x_f) = 0. $$

$\psi$ is arbitrary, therefore we can write $\hat{O}$ as

$$ \hat{O} = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]$$

My question is what I would get if I multiplied $\hat{O}$ with a theta function $\theta(t_{f}-t)$, where $\theta = 1$ for $t_{f} \geq t$ and $\theta = 0$ for $t_{f} < t$

$$\hat{O}\theta(t_{f}-t) = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\theta(t_{f}-t)$$

I know that since a theta function only changes gradient at $t_{f} = t$, I will get a Dirac-delta function, but what about when it multiplies with $\hat{H}(t_{f},x_{f})$

$$i\partial_{t_{f}}\theta(t_{f} - t) = i\delta(t_f - t) $$

$$ - \hat{H}(t_{f},x_{f})\theta(t_{f} - t) = ?$$

I want to know this result because solving the propagation equation is done by:

$$ \hat{O}(t_f,x_f)\theta(t_{f} - t)\psi(t_f,x_f) = \int dx [\hat{O}(t_f,x_f) i G(t_f,x_f; t,x)]\psi(t,x) $$

Where G is a Green Function

which is apparently just:

$$i\delta(t_f - t)\psi(t_f,x_f) =\int dx [\hat{O}(t_f,x_f) i G(t_f,x_f; t,x)]\psi(t,x)$$

Compactdrive
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  • TDSE is automatically true for any wave function ?? – Qmechanic Apr 05 '22 at 15:09
  • @Qmechanic Yes, because this equation doesn't require that $\psi(t,x)$ is an eigenstate of $\hat{H}$. If I had required $E_n$ values then it wouldn't be true for any wave equation – Compactdrive Apr 05 '22 at 15:17
  • But you can write down plenty of functions $\psi(t,x)$ that don't satisfy the time-dependent Schrödinger equation. Here's one: $\psi(t,x) = (t - x) e^{-x^2/2} \sin t$. You could apply the operator $\hat{O}$ to it, of course, but it wouldn't have $\hat{O} \psi = 0$ and therefore it wouldn't satisfy the TDSE. – Michael Seifert Apr 05 '22 at 15:32
  • @MichaelSeifert Let me clarify then, $\psi$ is arbitrary, we can choose any valid $\psi$ function. – Compactdrive Apr 05 '22 at 15:34
  • So by "valid" and "arbitrary" you mean you can choose any $\psi$ such that $\hat{O} \psi = 0$? – Michael Seifert Apr 05 '22 at 15:37
  • @MichaelSeifert Yes, more specifically we need $\psi$ to be a wave function. – Compactdrive Apr 05 '22 at 15:38
  • Your "?" amounts to $-\hat H(t_f,x_f)$ for $t_f\geq t$ and 0 otherwise. Why don't you follow the mainstream definitions and procedures for propagators, e.g. in WP? – Cosmas Zachos Apr 05 '22 at 21:33
  • @CosmasZachos This is what I thought it would be too, but then why is there no $\hat{H}$ dependence in the final term, why is only the delta function left? Also these were the definitions in lecturer's notes and I couldn't figure out how he got them. – Compactdrive Apr 05 '22 at 23:03
  • lLinked. – Cosmas Zachos Apr 06 '22 at 03:12

1 Answers1

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Indeed, θ commutes with $\hat H$, trivially; isn't that your teacher's point?

The point is that you just commute the θ past $\hat O$ so as to leave the latter annihilate its wavefunction kernel, $$ \hat{O} = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})] \qquad \leadsto \\ \hat{O}\theta(t_{f}-t) = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\theta(t_{f}-t)\\ = i\delta(t_f - t) +\theta(t_{f}-t)\hat{O}~~, $$ so that $$ \hat{O}(t_f,x_f)\theta(t_{f} - t)\psi(t_f,x_f)= i\delta(t_f - t) \psi(t_f,x_f) \\ =i \int\!\! dx ~~\delta(t_f - t) \delta(x_f - x) \psi(t,x) \\ =\int\!\! dx ~~ [\hat{O}(t_f,x_f) i G(t_f,x_f; t,x)]\psi(t,x) .$$

NB Clarification following comments

You seemed uneasy about the standard formal relation $[∂_{},(_−)]=(_−)$ the whole construction hinges on. It indicates the derivative acts on , but then it survives to further keep on acting on everything else that follows it, including , e.g., , by the chain rule of derivation! That is to say, $$[∂_{},(_−)] = ∂_{}((_−) )- (_−)∂_{} \\ =(_−) .$$

Cosmas Zachos
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  • I understand that $\hat{O} \psi (t_f,x_f)$ = 0. So if I have this correct, you're saying $\hat{O}(t_f,x_f)\theta (t_f - t) \psi(t_f , x_f) = i\delta(t_f - t) \psi(t_f , x_f) + \hat{O} \theta (t_f - t) \psi(t_f , x_f)$ , and the term $\hat{O} \theta (t_f - t) \psi(t_f , x_f) = \theta (t_f - t) \hat{O} \psi(t_f , x_f) = 0$ – Compactdrive Apr 06 '22 at 16:01
  • O doesn't, but H does commute with θ... – Cosmas Zachos Apr 06 '22 at 16:25
  • Maybe I'm missing something but what does $-\hat{H}(t_f , x_f)\theta(t_f - t)$ actually equal? The second and third lines of your equation implies it's equal to $\theta(t_f - t)\hat{O}$, and I understand that if this term acts on a wave function $\psi(t_f,x_f)$, it equals zero, I just want to know how you got that O to be on the end, I can understand if it was $-\theta(t_f - t)\hat{H}$, but not O. – Compactdrive Apr 07 '22 at 08:46
  • Do you understand that [H, θ]=0 but [O, θ]=iδ, and why? – Cosmas Zachos Apr 07 '22 at 10:08
  • $i [\partial_{t_{f}},\theta(t_{f} - t) ]= i\delta(t_f - t) $, instead of what you wrote… – Cosmas Zachos Apr 07 '22 at 10:16