# I Green's function and the evolution operator

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1. 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.

2. Jul 18, 2017

### redtree

3. 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.

4. 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?

5. 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.

6. 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.

7. 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.

8. 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.

9. Jul 22, 2017