# I Green's function and the evolution operator

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1. Jun 17, 2017

### redtree

The Green's function is defined as follows, where $\hat{L}_{\textbf{r}}$ is a differential operator:

\begin{split}

\hat{L}_{\textbf{r}} \hat{G}(\textbf{r},\textbf{r}_0)&=\delta(\textbf{r}-\textbf{r}_0)

\end{split}

However, I have seen the following description of the Green's function (which contradicts the above definition):

\begin{split}

\delta(\textbf{r}-\textbf{r}_0) &= \langle \textbf{r}| \textbf{r}_0 \rangle

\end{split}

Where $\textbf{r}$ is a 4-vector with components of 3-space $\vec{x}$ and 1-time $t$:

\begin{split}

\textbf{r}&= [\vec{x},t]

\end{split}

Such that, where $\hat{U}(t,t_0)$ denotes the time evolution operator, $\hat{U}(t,t_0)\doteq e^{-2 \pi i \omega(t-t_0)}$:

\begin{split}

\langle \textbf{r}| \textbf{r}_0 \rangle&=\langle \vec{x},t| \vec{x}_0,t_0 \rangle

\\

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

\\

&=\langle \vec{x}|e^{-2 \pi i \omega(t-t_0)}| \vec{x}_0 \rangle

\end{split}

Where the Green's function $\hat{G}(\vec{x},t|\vec{x}_0,t_0)$ is defined such that:

\begin{split}

\hat{G}(\vec{x},t|\vec{x}_0,t_0) &\doteq \langle \vec{x}|e^{-2 \pi i \omega(t-t_0)}| \vec{x}_0 \rangle

\end{split}

What am I missing?

2. Jun 18, 2017

### hilbert2

There are Green's functions for many kinds of differential operators, all of them are based on the idea of solving a differential equation with an integral transform where the Green's function is the kernel. In quantum mechanics the Green function is called "propagator" and its physical interpretation is that it gives the probability of a particle moving from point $\mathbf{x}_0$ to point $\mathbf{x}$ during a time interval $t - t_0$. It's also used when forming the path integral representation of quantum dynamics.

So, the definition in (1) is that of a general Green's function while the latter equations describe the Green function in the special case of quantum time evolution.

3. Jun 18, 2017

### redtree

Yes; that is why I assign such importance to understanding Green's functions properly. Whatever the specific application of a Green's function, it should be consistent with the general definition. In this case, the latter definition is not consistent with the first. In definition (1): $\hat{G}(\textbf{r},\textbf{r}_0) = \hat{L}_{\textbf{r}}^{-1} \delta(\textbf{r}-\textbf{r}_0)$ while in the latter definition, $\hat{G}(\textbf{r},\textbf{r}_0) = \delta(\textbf{r}-\textbf{r}_0) = \langle \vec{x} | e^{-2 \pi i \omega(t-t_0)}| \vec{x}_0 \rangle$.

4. Jun 18, 2017

### Orodruin

Staff Emeritus
Because your differential operator is not an operator in space only. You also have the time variable that needs to be included.

5. Jun 18, 2017

### redtree

Yes; that is true, but I don't see how that solves the problem. Could you please explain?

6. Jun 18, 2017

### Orodruin

Staff Emeritus
Have you thought anything about what operator your function is the Green's function for and what differential equation it should satisfy?

7. Jun 18, 2017

### redtree

My motive is to understand the mathematical foundations of the propagator.

In my opinion, solutions of various differential equations via a Green's function approach are really just variations on a theme. For any application, the underlying mathematics should be rigorous and consistent, which brings me back to my original question....

8. Jun 19, 2017

### Orodruin

Staff Emeritus
This really does not answer my question in #6. If you have not thought about it, I suggest that you do. If you have, what are your conclusions?

9. Jun 20, 2017

### redtree

Is this what you mean?

Given a linear differential operator on $\textbf{r}$, i.e., $\hat{L}_{\textbf{r}}$, the Green's function $\hat{G}(\textbf{r},\textbf{r}_0)$ is defined such that:

\hat{L}_{\textbf{r}}\hat{G}(\textbf{r},\textbf{r}_0)=\delta(\textbf{r}-\textbf{r}_0)

Thus:

\hat{L}_{\textbf{r}} u(\textbf{r})=f(\textbf{r})

Where:

u(\textbf{r})= \int G(\textbf{r},\textbf{r}_0) f(\textbf{r}_0) d\textbf{r}_0

Since:

\begin{split}

\hat{L}_{\textbf{r}} u(\textbf{r})&=\hat{L}_{\textbf{r}}\int G(\textbf{r},\textbf{r}_0) f(\textbf{r}_0) d\textbf{r}_0

\\

&=\int \hat{L}_{\textbf{r}}G(\textbf{r},\textbf{r}_0) f(\textbf{r}_0) d\textbf{r}_0

\\

&=\int \delta(\textbf{r}-\textbf{r}_0) f(\textbf{r}_0) d\textbf{r}_0

\\

&=f(\textbf{r})

\end{split}

The Green's function can further considered in the context of the Fourier transform. I begin by noting the following:

\begin{split}

\delta(\textbf{r}-\textbf{r}_0)&=\int_{-\infty}^{\infty} e^{2 \pi i \textbf{k}(\textbf{r}-\textbf{r}_0)} d\textbf{k}

\end{split}

And:

\hat{G}(\textbf{r},\textbf{r}_0)=\int_{-\infty}^{\infty} \hat{G}(\textbf{k},\textbf{r}_0) e^{2 \pi i \textbf{k}\textbf{r}} d\textbf{k}

Such that:

\begin{split}

\hat{L}_{\textbf{r}} \hat{G}(\textbf{r},\textbf{r}_0)&=\delta(\textbf{r}-\textbf{r}_0)

\\

\hat{L}_{\textbf{r}}\int_{-\infty}^{\infty} \hat{G}(\textbf{k},\textbf{r}_0) e^{2 \pi i \textbf{k} \textbf{r}} d\textbf{k}&= \int_{-\infty}^{\infty} e^{2 \pi i \textbf{k} (\textbf{r}-\textbf{r}_0)} d\textbf{k}

\end{split}

Given $\hat{L}_{\textbf{r}}=\frac{\partial}{\partial \textbf{r}}$

\begin{split}

\hat{L}_{\textbf{r}}\int_{-\infty}^{\infty} \hat{G}(\textbf{k},\textbf{r}_0) e^{2 \pi i \textbf{k} \textbf{r}} d\textbf{k}&=\frac{\partial}{\partial \textbf{r}} \left[\int_{-\infty}^{\infty} \hat{G}(\textbf{k},\textbf{r}_0) e^{2 \pi i \textbf{k} \textbf{r}} d\textbf{k}\right ]

\\

&=\int_{-\infty}^{\infty} \hat{G}(\textbf{k},\textbf{r}_0)\frac{\partial}{\partial \textbf{r}}\left [e^{2 \pi i \textbf{k} \textbf{r}}\right ] d\textbf{k}

\\

&=\int_{-\infty}^{\infty} (2 \pi i \textbf{k}) \hat{G}(\textbf{k},\textbf{r}_0)e^{2 \pi i \textbf{k} \textbf{r}} d\textbf{k}

\end{split}

Thus:

\hat{L}_{\textbf{r}} \rightarrow \hat{L}_{\textbf{k}}

Where:

\hat{L}_{\textbf{k}} = 2 \pi i \textbf{k}

Thus:

\int_{-\infty}^{\infty} (2 \pi i \textbf{k}) \hat{G}(\textbf{k},\textbf{r}_0)e^{2 \pi i \textbf{k} \textbf{r}} d\textbf{k}=\int_{-\infty}^{\infty} e^{2 \pi i \textbf{k}(\textbf{r}-\textbf{r}_0)} d\textbf{k}

Such that:

\begin{split}

(2 \pi i \textbf{k}) \hat{G}(\textbf{k},\textbf{r}_0)&= e^{-2 \pi i \textbf{k}\textbf{r}_0}

\end{split}

Where:

\begin{split}

\hat{G}(\textbf{k},\textbf{r}_0)&= \frac{e^{-2 \pi i \textbf{k}\textbf{r}_0}}{(2 \pi i \textbf{k}) }

\end{split}

This can be generalized as follows:

\hat{L}_{\textbf{r}^n}=\frac{\partial^n}{\partial \textbf{r}^n}

Such that:

\hat{L}_{\textbf{k}^n}=(2 \pi i \textbf{k})^n

And:

\begin{split}

\hat{G}(\textbf{k}^n,\textbf{r}_0) &= \frac{e^{-2 \pi i \textbf{k}\textbf{r}_0}}{(2 \pi i \textbf{k})^n}

\end{split}

Thus, the $n$th derivative in position space transforms to multiplication by $\textbf{k}^n$ in wavenumber space, and the $n$th integration in position space transforms to division by $\textbf{k}^n$ in wavenumber space.

To emphasize, all of this is based on the understanding of the Green's function in definition 1, i.e., $\hat{G}(\textbf{r},\textbf{r}_0) = \hat{L}_{\textbf{r}}^{-1} \delta(\textbf{r}-\textbf{r}_0)$. Unfortunately, I still don't see how it answers my question. Can you be clearer in your guidance?

10. Jun 21, 2017

### Orodruin

Staff Emeritus
I am talking about the other Green's function you described. What sort of problem do you use it to solve?

Hint: Not all differential equations are differential equations in space only.

11. Jun 21, 2017

### redtree

The equations I presented are sufficiently general to work in Minkowski spacetime where $\textbf{r} = [\vec{x},i t]$, such that in flat space, $\textbf{r}^2 = \vec{x}^2-t^2$.

Nevertheless, I think I understand your question, which interestingly, leads to the second part of question. I was going to save it for another thread, but no need now.

I consider the case of a 3-space vector $\vec{x}$ that is a function of time, such that:

\begin{split}

\vec{x}&\doteq \vec{x}(t)

\\

\vec{x}_0&\doteq \vec{x}(t_0)

\end{split}

I note the following:

\begin{split}

\delta(\vec{x} - \vec{x}_0)&=\langle \vec{x} | \vec{x}_0 \rangle

\end{split}

Where:

\begin{split}

\langle\vec{x}|\psi\rangle &=|\psi(\vec{x})\rangle

\end{split}

Such that:

\begin{split}

\delta(\vec{x} - \vec{x}_0)&=\langle \vec{x} | \vec{x}_0 \rangle

\\

&=\sum_n \langle \vec{x} |\psi_n \rangle \langle \psi_n | \vec{x}_0 \rangle

\\

&=\sum_n \langle \psi_n(\vec{x}_0) | \psi_n(\vec{x}) \rangle

\end{split}

Furthermore, given $\delta(\vec{x} - \vec{x}_0)= \delta(\vec{x}_0 - \vec{x})$:

\begin{split}

\langle \vec{x}_0 | \vec{x} \rangle&=\langle \vec{x} | \vec{x}_0 \rangle

\\

\sum_n \langle \vec{x}_0 |\psi_n \rangle \langle \psi_n | \vec{x} \rangle&=\sum_n \langle \vec{x} |\psi_n \rangle \langle \psi_n | \vec{x}_0 \rangle

\\

\sum_n \langle \psi_n(\vec{x}) | \psi_n(\vec{x}_0) \rangle&=\sum_n \langle \psi_n(\vec{x}_0) | \psi_n(\vec{x}) \rangle

\end{split}

Assuming $\hat{G}(\vec{x},t|\vec{x}_0,t_0)\doteq\langle \vec{x}| \hat{U}(t,t_0)\vec{x}_0\rangle$:

\begin{split}

\hat{G}(\vec{x},t|\vec{x}_0,t_0)&\doteq\langle \vec{x}| \hat{U}(t,t_0) \vec{x}_0\rangle

\\

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

\\

&=\sum_n \langle \vec{x}|\psi_n \rangle \hat{U}^{\dagger}(t,t_0)\langle \psi_n |\vec{x}_0\rangle

\\

&=\sum_n \langle \psi_n(\vec{x}_0) | \hat{U}^{\dagger}(t,t_0)| \psi_n (\vec{x})\rangle

\\

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

\end{split}

Given $\psi(\vec{x})=\hat{U}(t,t_0)\psi(\vec{x}_0)$:

\begin{split}

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

\\

&=\langle \vec{x}|\vec{x}\rangle

\\

&=\delta(\vec{x}-\vec{x})

\\

&=\delta(0)

\\

&=1

\\

&=\hat{G}(\vec{x},t|\vec{x}_0,t_0)

\end{split}

What am I missing?

12. Jun 21, 2017

### Orodruin

Staff Emeritus
Your states $|\vec x\rangle$ are time independent (they are just a time independent basis of the Hilbert space) and build the state space (let us start working with regular QM), they do not depend on time. You can apply the time evolution operator $U(t,t_0)$ to the state $|\vec x_0\rangle$ to find out what the state is at a later time. This is what the Green's function tells you, the amplitude of evolving $|\vec x_0\rangle$ at time $t_0$ into $|\vec x\rangle$ at time $t$. Note that $\vec x$ and $\vec x_0$ are not connected such that $U(t,t_0) |\vec x_0\rangle = |\vec x\rangle$ (furthermore $\delta(0) \neq 1$). The Green's function in this case is the Green's function of the SchrÃ¶dinger equation and satisfies
$$(-i\partial_t + \hat H) G(\vec x,\vec x_0,t,t_0) = -i \delta(\vec x - \vec x_0) \delta(t-t_0),$$
where $\hat H$ is the position space representation of the Hamiltonian (a bit depending on how you define the RHS of the equation). You can quite easily check that it satisfies this relation if you let $G(\vec x,\vec x_0, t, t_0) = \langle \vec x|U(t-t_0)|\vec x_0\rangle \theta(t-t_0)$ (this is the retarded Green's function and so it is zero if $t_0 > t$).

Note that
$$\partial_t G(\vec x,\vec x_0, t,t_0) = \theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle + \delta(t-t_0) \langle \vec x|\vec x_0\rangle = \theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle + \delta(t-t_0) \delta(\vec x - \vec x_0).$$
The second term here takes care of the inhomogeneity for the Green's function and the first enters the differential equation satisfied by the operator $U(t-t_0)$.

13. Jun 22, 2017

### redtree

Sorry, dumb mistake stating $\delta(0)=1$.

A derivation of the connection between $\vec{x}$ and $\vec{x}_0$:

\begin{split}

\psi_{\vec{x}}(t)

&=\hat{U}(t,t_0)\psi_{\vec{x}}(t_0)

\\

\langle \vec{x} | \psi \rangle &=\langle \vec{x}_{0} |\hat{U}(t,t_0) | \psi \rangle

\end{split}

Such that:

\begin{split}

\langle \psi |\vec{x}\rangle &=\langle \psi|\hat{U}^{\dagger}(t,t_0) | \vec{x}_0 \rangle

\end{split}

Thus:

\begin{split}

|\vec{x}\rangle &=|\hat{U}^{\dagger}(t,t_0) | \vec{x}_0 \rangle

\end{split}

Where is the mistake in this derivation?

Additionally, I'm not clear on how you derived the following:

\begin{split}

\partial_t G(\vec x,\vec x_0, t,t_0) &= \theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle + \delta(t-t_0) \langle \vec x|\vec x_0\rangle

\end{split}

Could you please be more explicit? Thanks.

14. Jun 23, 2017

### Orodruin

Staff Emeritus
Unfortunately, there is not much that is right about it. You are being sloppy with the use of $|\psi\rangle$. The $|\psi\rangle$ on the LHS represents the state at time $t$ and the one on the RHS the state at time $t_0$. Furthermore, it is unclear how you start. Where does the $x_0$ come from in the first place?

I suggest you consider the connection between the position states and a basis of a finite dimensional Hilbert space (ie, a system with a finite number of eigenstates).

Which part is not clear? It is just the product rule for derivatives applied to the previous expression.

15. Jun 27, 2017

### redtree

I see your point on $|\psi \rangle$.

The derivative is still not clear to me. See the following:

\begin{split}

\frac{\partial}{\partial t} \hat{G}(\vec{x},\vec{x}_0,t,t_0)&=\frac{\partial}{\partial t} \left[\theta(t-t_0) \langle \vec x|U(t-t_0)|\vec x_0\rangle\right]

\\

&=\frac{\partial}{\partial t} \left[\theta(t-t_0) \right ]\langle \vec x|U(t-t_0)|\vec x_0\rangle +\theta(t-t_0) \frac{\partial}{\partial t}\left[ \langle \vec x|U(t-t_0)|\vec x_0\rangle\right]

\\

&=\delta(t-t_0) \langle \vec x|U(t-t_0)|\vec x_0\rangle +\theta(t-t_0) \langle \vec x|\acute{U}(t-t_0)|\vec x_0\rangle

\end{split}

How does $\delta(t-t_0) \langle \vec x|U(t-t_0)|\vec x_0\rangle = \delta(t-t_0) \delta(\vec{x}-\vec{x}_0)$?

16. Jun 27, 2017

### Orodruin

Staff Emeritus
The delta function is non-zero only when $t = t_0$ and $U(0) = 1$.

17. Jun 28, 2017

### redtree

Nice. Got it.

A related question: Given the (non-retarded) Green's function $\hat{G}(\vec{x},\vec{x}_0,t,t_0)=\langle \vec x|U(t-t_0)|\vec x_0\rangle$, is the following true?:

\begin{split}

\langle \vec x|U(t-t_0)|\vec x_0\rangle&= U(t-t_0)\delta(\vec x-\vec x_0)

\end{split}

Or is it the following?:

\begin{split}

\langle \vec x|U(t-t_0)|\vec x_0\rangle&= \delta(\vec x-U(t-t_0)\vec x_0)

\\

&=\delta(U^{\dagger}(t-t_0)\vec x-\vec x_0)

\end{split}

18. Jun 28, 2017

### Orodruin

Staff Emeritus
First, that is the retarded GF. Just without the heaviside function, which is fine as long as you assume $t \geq t_0$.

No, the relations you wrote down make no sense. $U(t-t_0)$ is an operator that acts on states in the Hilbert space. It seems you have trouble distinguishing between the states themselves, their representations, and the bases of those representations. I would suggest reading up on this because as it appears in this thread, you are only guessing.

(Please take this the right way, it is not intended as any sort of personal insult, just an observation of something you seem to have problems with and would benefit from thinking more about instead of using guesswork.)

19. Jul 2, 2017

### redtree

I'm not insulted; I am admittedly much more comfortable in coordinate representation as opposed to Dirac notation; In any case, I appreciate your responses, and if you have any suggested references, I would gladly check them out.

A quick question regarding $\hat{H}$ before I get to my larger question. I recall the following:

\begin{split}

(-i \partial_t + \hat{H})G(\vec x,\vec x_0, t,t_0)&=-i \partial_t G(\vec x,\vec x_0, t,t_0)+ \hat{H} G(\vec x,\vec x_0, t,t_0)

\\

&=-i \delta(\vec x- \vec x_0)\delta(t-t_0)

\end{split}

Such that:

\begin{split}

\hat{H} G(\vec x,\vec x_0, t,t_0)&=-i \delta(\vec x- \vec x_0)\delta(t-t_0)+i \partial_t G(\vec x,\vec x_0, t,t_0)

\\

i\hat{H} G(\vec x,\vec x_0, t,t_0)&= \delta(\vec x- \vec x_0)\delta(t-t_0)- \partial_t G(\vec x,\vec x_0, t,t_0)

\end{split}

Given $\partial_t G(\vec x,\vec x_0, t,t_0) = \theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle + \delta(t-t_0) \delta(\vec x - \vec x_0)$

\begin{split}

i\hat{H} G(\vec x,\vec x_0, t,t_0)&= -\theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle

\\

\hat{H} G(\vec x,\vec x_0, t,t_0)&= i\theta(t-t_0) \langle \vec x|U'(t-t_0)|\vec x_0\rangle

\end{split}

Is that correct regarding $\hat{H}$?

20. Jul 3, 2017

### Orodruin

Staff Emeritus
Yes, but it tells you nothing new. It is just a rewriting of the original equation and tells you that $\langle\vec x | U(t-t_0)|\vec x_0\rangle$ satisfies the SchrÃ¶dinger equation for $t > t_0$, which you already know as that was our starting assumption that gave us the solution $G = \theta(t-t_0)\langle\vec x | U(t-t_0)|\vec x_0\rangle$.

21. 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}

22. 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.

23. 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}

24. 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.

25. 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.