I Green's function and the evolution operator

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1. Jul 6, 2017

redtree

I think I have it. For a differential operator $\hat{L}_t = \partial_t$, the general form of a the Green's function is as follows:

\begin{split}

\hat{L}_{t} \hat{G}(t,t_0)&=\delta(t-t_0)

\end{split}

Such that:

\begin{split}

\hat{L}_{t} u(t)&=f(t)

\end{split}

Where:

\begin{split}

u(t)&=\int \hat{G}(t,t_0) f(t_0) dt_0

\end{split}

In the context of the Green's function, one can constrain time $t$ via the Heaviside step function as follows:

\begin{split}

\hat{L}_{t} \left[\theta(t-t_0) g(t-t_0)\right]&=\delta(t-t_0)g(t-t_0) + \theta(t-t_0) \acute{g}(t-t_0)

\end{split}

I then constrain $g(t-t_0)$ as follows:

g(t-t_0) =

\begin{cases}

1 & \quad \text{if } t=t_0 \\

\in \mathbb{C} & \quad \text{if } t \neq t_0 \\

\end{cases}

Such that:

\begin{split}

\hat{L}_{t} \left[\theta(t-t_0) g(t-t_0)\right]&=\delta(t-t_0)+ \theta(t-t_0) \acute{g}(t-t_0)

\end{split}

Given:

\begin{split}

\left(\hat{L}_{t}+h(t-t_0)\right) \left[\theta(t-t_0) g(t-t_0)\right]&=\delta(t-t_0)

\end{split}

I note:

\begin{split}

h(t-t_0) \left[\theta(t-t_0) g(t-t_0)\right]&=-\theta(t-t_0) \acute{g}(t-t_0)

\end{split}

Such that:

\begin{split}

h(t-t_0) &=-\frac{\acute{g}(t-t_0)}{g(t-t_0)}

\end{split}

Next, assuming $g(t-t_0) = \hat{U}(t,t_0)$ where $\hat{U}(t,t_0)=e^{- i \omega (t-t_0)}$:

\begin{split}

h(t-t_0) &=-\frac{- i \omega e^{- i \omega (t-t_0)}}{e^{- i \omega (t-t_0)}}

\\

&=- i \omega

\end{split}

Such that:

\begin{split}

\left(\hat{L}_{t}- i \omega\right) \left[\theta(t-t_0) \hat{U}(t,t_0)\right]&=\delta(t-t_0)

\end{split}

Furthermore, multiplying both sides by $\delta(\vec{x}-\vec{x}_0)$:

\begin{split}

\left(\hat{L}_{t}- i \omega\right) \left[\theta(t-t_0) \hat{U}(t,t_0)\delta(\vec{x}-\vec{x}_0)\right]&=\delta(t-t_0)\delta(\vec{x}-\vec{x}_0)

\end{split}

Which can be rewritten:

\begin{split}

\left(-i \partial_t- \omega\right) \left[\theta(t-t_0) \hat{U}(t,t_0)\delta(\vec{x}-\vec{x}_0)\right]&=-i \delta(t-t_0)\delta(\vec{x}-\vec{x}_0)

\end{split}

This is the same equation you wrote using Dirac notation above, where:

\begin{split}

\hat{H}&= -\omega

\\

\hat{G}(\vec{x},\vec{x}_0,t,t_0)&=\theta(t-t_0) \hat{U}(t,t_0)\delta(\vec{x}-\vec{x}_0)

\end{split}

If $\hat{H}= -\omega$ is not clear, I can easily prove it. The second equality should be obvious.

Thus:

\begin{split}

\left(-i \partial_t+\hat{H}\right) \hat{G}(\vec{x},\vec{x}_0,t,t_0)&=-i \delta(t-t_0)\delta(\vec{x}-\vec{x}_0)

\end{split}

Furthermore this derivation seems to answer my earlier question, such that:

\begin{split}

\hat{U}(t,t_0)\delta(\vec{x}-\vec{x}_0) &= \langle \vec{x}| \hat{U}(t,t_0)|\vec{x}_0 \rangle

\end{split}

2. Jul 7, 2017

Orodruin

Staff Emeritus
Correct up to here. However, I suggest removing the hat from $G$, it is not an operator, it is a function of two variables.

What happened here? Where did $h$ come from? This equation has nothing to do with what you have been doing so far.

What kind of assumption is this and why are you making it? What differential equation are you really trying to solve here?

No you cannot, because it is not true unless $\hat H$ acts on an energy eigenstate, which the $|\vec x\rangle$ states are not. It makes no sense to act with the operator $\hat U$ on a function, in particular a delta function in $\vec x$. What you have written down as $\hat U$ is a function, not an operator, and it is very distinct from the operator $\hat U$ that acts on the state vectors.

It is not, because it is not true. The Green's function you are referring to is the Green's function of a partial differential operator, not of the ordinary differential operator you mentioned in the beginning. Its Green's function is $\theta(t-t_0)$.

You really must learn to differentiate between functions and operators. The LHS here makes no sense whatsoever. The time evolution operator $\hat U$ is an operator that acts on the states and you have just extracted it from the bracket. You could find the appropriate Green's function by Fourier transformation, which leads to an ordinary differential equation for the Fourier modes.

3. Jul 14, 2017

redtree

This is an extension the previous work. Equally, I could have equally subtracted $\theta(t-t_0) \acute{g}(t-t_0)$ from both sides of the previous equation such that:

\begin{split}

\hat{L}_{t} \left[\theta(t-t_0) g(t-t_0)\right]- \theta(t-t_0) \acute{g}(t-t_0)&=\delta(t-t_0)

\end{split}

Then, in order to mirror the form of the Green's function of the Schrodinger equation (though for generalized functions), I rewrite the equation as follows:

\begin{split}

\hat{L}_{t} \left[\theta(t-t_0) g(t-t_0)\right]- \theta(t-t_0) \acute{g}(t-t_0)&=\left( \hat{L}_{t}+\frac{\acute{g}(t-t_0)}{g(t-t_0)} \right)\left(\theta(t-t_0) g(t-t_0)\right)

\\

&=\delta(t-t_0)

\end{split}

To mirror further the form of the Green's function for the Schrodinger equation above, I then define a function $h(t)=\frac{\acute{g}(t-t_0)}{g(t-t_0)}$, such that:

\begin{split}

\left( \hat{L}_{t}+h(t-t_0) \right)\left(\theta(t-t_0) g(t-t_0)\right)&=\delta(t-t_0)

\end{split}

The goal of this exercise is several-fold: 1) to emphasize the dependence of the term $h(t-t_0)$ on $g(t-t_0)$ in a generalized equation of this form; 2) to emphasize that the time constraint $\theta(t-t_0)$ on the function $g(t-t_0)$ (and not the function itself), really acts as the Green's function and that this is true for any retarded function of the form $\theta(t-t_0) g(t-t_0)$; 3) to preserve the standard form of the Green's function in this case, an addition term $h(t-t_0)$ must be added.

The assumption is not necessarily a physical one, but a mathematical one. I want to understand the consequences of the mathematical assumption $g(t-t_0) = e^{-i \omega (t-t_0)}$ in the context of the prior equations.

I note the following:

\begin{split}

\partial_t (u(t-t_0) v(t-t_0) )&= \acute{u}(t-t_0)v(t-t_0) + u(t-t_0)\acute{v}(t-t_0)

\\

\partial_t (u(t-t_0) w(t-t_0) )&= \acute{u}(t-t_0)w(t-t_0) + u(t-t_0)\acute{w}(t-t_0)

\end{split}

Such that:

\begin{split}

\partial_t (u(t-t_0) v(t-t_0) )- u(t-t_0)\acute{v}(t-t_0)&= \acute{u}(t-t_0)v(t-t_0)

\\

\partial_t (u(t-t_0) w(t-t_0) )- u(t-t_0)\acute{w}(t-t_0)&= \acute{u}(t-t_0)w(t-t_0)

\end{split}

Assuming $\acute{u}(t-t_0)v(t-t_0)=\acute{u}(t-t_0)w(t-t_0)$:

\begin{split}

v(t-t_0)&=w(t-t_0)

\end{split}

Thus, given:

\begin{split}

\partial_t\left[ \theta(t-t_0) e^{-i \omega (t-t_0)} \delta(\vec{x}-\vec{x}_0)\right ]-\left(-i \omega \theta(t-t_0)e^{-i \omega (t-t_0)}\delta(\vec{x}-\vec{x}_0)\right)&=\delta(t-t_0) e^{-i \omega (t-t_0)} \delta(\vec{x}-\vec{x}_0)

\\

\bigwedge \partial_t\left[ \theta(t-t_0) \langle \vec{x}|\hat{U}(t,t_0) | \vec{x}_0 \rangle \right ]-\left( \theta(t-t_0)\langle \vec{x}|\acute{\hat{U}}(t,t_0) | \vec{x}_0 \rangle\right)&=\delta(t-t_0)\langle \vec{x}|\hat{U}(t,t_0) | \vec{x}_0 \rangle

\end{split}

Given $u(t-t_0)=\theta(t-t_0)$, $v(t)=\delta(\vec{x}-\vec{x}_0) e^{-i \omega (t-t_0)}$ and $w(t-t_0)=\langle \vec{x}|\hat{U}(t,t_0)|\vec{x}\rangle$:

\begin{split}

\delta(\vec{x}-\vec{x}_0) e^{-i \omega (t-t_0)} &=\langle \vec{x}|\hat{U}(t,t_0)|\vec{x}\rangle

\end{split}

Where: (if necessary I can prove this too)

\begin{split}

-\omega &=\hat{H}

\end{split}

4. Jul 14, 2017

Orodruin

Staff Emeritus
I am sorry, I give up. You are mixing concepts wildly and do not really seem to understand what you are doing. Essentially everything you wrote in your last post is wrong. Your assumptions seem to suggest that you are guessing wildly. Instead of taking one step at a time and asking about whether it makes sense or not, you start by making a wrong turn and then spend a long post trying to push your assumption through.

As I said in the previous post: No. You cannot. It makes no sense. Equation (40) makes no sense. You have not even defined $\omega$ properly. Your solution has nothing to do with the Green's function I provided you with earlier in this thread.

5. Jul 16, 2017

redtree

I am sorry for providing you such frustration. Nevertheless, I do appreciate the efforts you are making in engaging in this discussion. From my perspective, I am merely trying to understand the Green's function from a purely mathematical perspective. Regarding equation (40), I am not the only one with this understanding. Please see the following link: http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pthic02.pdf, particularly equation (2.12). I would also note that the author shares my understanding of the relationship between $\omega$ and the Hamiltonian. Perhaps, this reference will lessen your frustration.

6. Jul 16, 2017

Orodruin

Staff Emeritus
Equation (2.12) has nothing to do with what you have been presenting here. It is a statement about the matrix elements of the potential operator. The author can do what he does here since $V$ is a function of $x$ only and not the momentum. Therefore that relation holds for the position eigenstates. You cannot use the same kind of argumentation for the Hamiltonian, which generally contains both the momentum and position operators.

Also, please refer explicitly to where the author allegedly agrees with your statement that $\hat H = -\omega$. As with your misinterpretation of Equation (2.12), I find it very likely that you are not really understanding what the author wants to convey.

7. Jul 18, 2017

redtree

8. Jul 18, 2017

Orodruin

Staff Emeritus
It does not have much to do with what you provided. They are using energy eigenstates, i.e., eigenstates of the Hamiltonian, not the position eigenstates you are trying to use. In the end, they end up with a Green's function that contains a sum over all energy eigenstates.

9. Jul 18, 2017

redtree

What?!? My derivation is almost exactly the same as the one in last the paper cited. The paper starts with the evolution operator as I do. Then, it considers the first-order Green function in $t$ as I do. Next, it uses that result to expand the equation into position space, the difference being that the author utilizes an intermediate step with $\sum_j u_j(\vec{r}) u_j^*(\vec{r})$ where $\sum_j u_j(\vec{r}) u_j^*(\acute{\vec{r}})=\delta(\vec{r}-\acute{\vec{r}})$, which I do not use.

It's not clear to me why you consider my derivation as using position eigenstates and this one as using energy eigenstates. For instance, as the derivation makes very clear, $\delta(\vec{r}-\acute{\vec{r}})$ may be considered as a function of energy eigenstates.

Does not the most recent paper cited make just such an argument for the full Hamiltonian?

10. Jul 18, 2017

Orodruin

Staff Emeritus
If your derivation was the same you would get the same result and you don't.

Because what you do when you assume that $\langle \vec x|\hat U|\vec x_0\rangle = \langle \vec x| e^{-i\omega (t-t_0)}|\vec x_0\rangle$ is essentially that $|\vec x_0\rangle$ is an energy eigenstate.

If you do not understand this, please show or give reference to the derivation of equation 41. In general, the Green's function will not be proportional to a delta function except for at $t = t_0$.

Things might get clearer for you if you work with a concrete example. Try finding the Green's function of the free particle in one dimension.

11. Jul 22, 2017

redtree

Quick correction from my previous derivation: Sign error by me, such that $\omega = \frac{\hat{H}}{\hbar}$, which for natural unts, $\omega = \hat{H}$.

I will go through the derivation.

Equation 17.21:

\begin{split}

| \psi_t \rangle &= e^{-\frac{i}{\hbar} \hat{H} (t-\acute{t})} | \psi_{\acute{t}} \rangle

\end{split}

Given:

\begin{split}

\omega &= \frac{\hat{H}}{\hbar}

\end{split}

\begin{split}

| \psi_t \rangle &= e^{-i \omega (t-\acute{t})} | \psi_{\acute{t}} \rangle

\end{split}

Which is my version of the time evolution operator, $\hat{U}(t,\acute{t})$.

Equation 17.26:

\begin{split}

\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\delta(t-\acute{t})

\end{split}

Where:

\begin{split}

G_j^{+}(t,\acute{t})&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t})

\end{split}

Such that:

\begin{split}

\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) \left(-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)

\\

&=\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)

\end{split}

And:

\begin{split}

\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=\delta(t-\acute{t})

\\

-i\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})

\\

\left(-i \frac{\partial}{\partial t}- \frac{E_j}{\hbar} \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})

\end{split}

Given $\omega_j=\frac{E_j}{\hbar}$ and $\partial_t = \frac{\partial}{\partial t}$:

\begin{split}

\left(-i \partial_t- \omega_j \right ) \left( e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})

\end{split}

Where:

\begin{split}

e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t})&=G_j^{+}(t,\acute{t})

\end{split}

Such that:

\begin{split}

\left(-i \partial_t- \omega_j \right ) G_j^{+}(t,\acute{t})&=-i \delta(t-\acute{t})

\end{split}

Again, this is the same as my derivation.

Equation 17.28:

\begin{split}

G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j e^{-\frac{i}{\hbar}E_j (t-\acute{t})}\theta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})

\end{split}

Equation 17.31:

\begin{split}

\left(i\hbar \frac{\partial}{\partial t} -\hat{H} \right) G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j \left(i\hbar \frac{\partial}{\partial t} -E_j \right) G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})

\\

&=\sum_j \delta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})

\\

&=\delta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})

\\

&=\delta^4(\vec{x}-\vec{\acute{x}})

\end{split}

Where:

\begin{split}

\sum_j u_j(\vec{r}) u_j^*(\vec{\acute{r}})&=\delta^3(\vec{r}-\vec{\acute{r}})

\end{split}

And:

\begin{split}

G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=\sum_j G_j^+(\vec{r},t;\vec{\acute{r}},\acute{t})

\\

&=\sum_j G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})

\\

&=\sum_j G_j^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})

\\

&=G^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})

\\

&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})

\\

&=-\frac{i}{\hbar} e^{-i \omega (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})

\end{split}

Which is the SAME expression for the Green function that I derived (assuming some algebra with $i$ and $\hbar$, which I demonstrate above). In other words, the result is the same. Trying to distinguish my derivation from the one in the paper by asserting one involves position and the other energy in order to accept one and not the other is simply not tenable.

12. Jul 22, 2017

Orodruin

Staff Emeritus
Regardless of the sign, it is simpy not true. It is only true on a degenerate subspace of the Hamiltonian, which you generally do not have.

13. Jul 22, 2017

Orodruin

Staff Emeritus
It is very tenable. You get different results and your's is wrong. Not only are you assuming that the position eigenstates are energy eigenstates (they are not), you are also asserting that they all have the same energy.. The Green's function for $t > t'$ is not proportional to a delta function in space.

14. Jul 22, 2017

Staff: Mentor

Thread closed for moderation.