Green's function and the evolution operator

[redtree]

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

\begin{equation}

\begin{split}

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

\end{split}

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

\begin{equation}

\begin{split}

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

\end{split}

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

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}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{equation}

\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}

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

\begin{equation}

\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}

\end{equation}What am I missing?

 

[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.

 

[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, $\hat{G}(\textbf{r},\textbf{r}_0) = \delta(\textbf{r}-\textbf{r}_0) = \langle \vec{x} | e^{-2 \pi i \omega(t-t_0)}| \vec{x}_0 \rangle$.

 

[Orodruin]

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

 

[redtree]

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

 

[Orodruin]

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

 

[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...

 

[Orodruin]

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...

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?

 

[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:

\begin{equation}

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

\end{equation}Thus:

\begin{equation}

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

\end{equation}Where:

\begin{equation}

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

\end{equation}Since:

\begin{equation}

\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}

\end{equation}

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

\begin{equation}

\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}

\end{equation}
And:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}Thus:

\begin{equation}

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

\end{equation}Where:

\begin{equation}

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

\end{equation}Thus:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

\end{equation}Where:

\begin{equation}

\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}

\end{equation}This can be generalized as follows:

\begin{equation}

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

\end{equation}Such that:

\begin{equation}

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

\end{equation}And:

\begin{equation}

\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}

\end{equation}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?

 

[Orodruin]

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.

 

[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{equation}

\begin{split}

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

\\

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

\end{split}

\end{equation}I note the following:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Such that:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}
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{equation}

\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}

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

\begin{equation}

\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}

\end{equation}
What am I missing?

 

[Orodruin]

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 Schrödinger 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 $U(t-t_0)$.

 

[redtree]

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 $\delta(0) \neq 1$).

Sorry, dumb mistake stating $\delta(0)=1$.A derivation of the connection between $\vec{x}$ and $\vec{x}_0$:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Thus:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}

Where is the mistake in this derivation?Additionally, I'm not clear on how you derived the following:

\begin{equation}

\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}

\end{equation}Could you please be more explicit? Thanks.

 

[Orodruin]

Where is the mistake in this derivation?

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).

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

\begin{equation}

\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}

\end{equation}Could you please be more explicit? Thanks.

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

 

[redtree]

I see your point on $|\psi \rangle$.The derivative is still not clear to me. See the following:

\begin{equation}

\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}

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

 

[Orodruin]

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

 

[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{equation}

\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}

\end{equation}Or is it the following?:

\begin{equation}

\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}

\end{equation}

 

[Orodruin]

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.)

 

[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{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

\end{equation}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{equation}

\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}

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

 

[Orodruin]

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 Schrödinger equation for $t > t_0$, which you already know as that was our starting assumption that gave us the solution $G = \theta(t-t_0)\langle\vec x | U(t-t_0)|\vec x_0\rangle$.

 

[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{equation}

\begin{split}

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

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

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

\end{split}

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

\begin{equation}

\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}

\end{equation}I then constrain $g(t-t_0)$ as follows:

\begin{equation}
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}

\end{equation}Such that:

\begin{equation}

\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}

\end{equation}Given:

\begin{equation}

\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}

\end{equation}I note:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

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

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}Which can be rewritten:

\begin{equation}

\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}

\end{equation}This is the same equation you wrote using Dirac notation above, where:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}Furthermore this derivation seems to answer my earlier question, such that:

\begin{equation}

\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}

\end{equation}

 

[Orodruin]

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{equation}

\begin{split}

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

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

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

\end{split}

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

\begin{equation}

\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}

\end{equation}

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

\begin{equation}
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}

\end{equation}Such that:

\begin{equation}

\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}

\end{equation}

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

Given:

\begin{equation}

\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}

\end{equation}

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

I note:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

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

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

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}Which can be rewritten:

\begin{equation}

\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}

\end{equation}This is the same equation you wrote using Dirac notation above, where:

\begin{equation}

\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}

\end{equation}If $\hat{H}= -\omega$ is not clear, I can easily prove it.

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.

The second equality should be obvious.

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 $\theta(t-t_0)$.

Thus:

\begin{equation}

\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}

\end{equation}Furthermore this derivation seems to answer my earlier question, such that:

\begin{equation}

\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}

\end{equation}

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.

 

[redtree]

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

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{equation}\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}\end{equation}
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{equation}\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}\end{equation}
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{equation}\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}\end{equation}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.

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

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.

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.

I note the following:\begin{equation}\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}\end{equation}
Such that:\begin{equation}\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}\end{equation}
Assuming $\acute{u}(t-t_0)v(t-t_0)=\acute{u}(t-t_0)w(t-t_0)$:\begin{equation}\begin{split} v(t-t_0)&=w(t-t_0)\end{split}\end{equation}
Thus, given:\begin{equation}\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}\end{equation}
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{equation}\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}\end{equation}
Where: (if necessary I can prove this too)\begin{equation}\begin{split} -\omega &=\hat{H}\end{split}\end{equation}

 

[Orodruin]

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.

Where: (if necessary I can prove this too)

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.

 

[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.

 

[Orodruin]

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.

 

[redtree]

See http://www.phy.ohiou.edu/~elster/lectures/relqm_17.pdf, equations 17.21 to 17.31. The derivation is essentially the same as the one I provide above. The full Hamiltonian is used, not just the potential energy term.

 

[Orodruin]

See http://www.phy.ohiou.edu/~elster/lectures/relqm_17.pdf, equations 17.21 to 17.31. The derivation is essentially the same as the one I provide above. The full Hamiltonian is used, not just the potential energy term.

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.

 

[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.

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.

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

 

[Orodruin]

My derivation is almost exactly the same as the one in last the paper cited.

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

It's not clear to me why you consider my derivation as using position eigenstates and this one as using energy eigenstates

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.

 

[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{equation}

\begin{split}

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

\end{split}

\end{equation}Given:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}\begin{equation}

\begin{split}

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

\end{split}

\end{equation}
Which is my version of the time evolution operator, $\hat{U}(t,\acute{t})$.Equation 17.26:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Where:

\begin{equation}

\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}

\end{equation}Such that:

\begin{equation}

\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}

\end{equation}And:

\begin{equation}

\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}

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

\begin{equation}

\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}

\end{equation}Where:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}Again, this is the same as my derivation.Equation 17.28:

\begin{equation}

\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}

\end{equation}Equation 17.31:

\begin{equation}

\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}

\end{equation}Where:

\begin{equation}

\begin{split}

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

\end{split}

\end{equation}And:

\begin{equation}

\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}

\end{equation}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.

 

[Orodruin]

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

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.

 

[Orodruin]

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.

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.

 

[PeterDonis]

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