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I'm studying about many body quantum physics following 'Introduction to quantum field theory in condensed matter physics' (Bruus). In section 7, it says that Green's function can be interpreted to propagator of wavefunction via:

$$\Psi(\vec{r},\vec{t}) = \int d\vec{r'} \int dt' G(\vec{r}t,\vec{r'}t') \Psi(\vec{r'},t').$$

I tried to insert this into Schrodinger equation to verify this: $[i\partial_t-H_0-V]\Psi(\vec{r},\vec{t})=0$

The definition of the full Green's function is given in the book : $[i\partial_t-H_0-V]G(\vec{r}t,\vec{r'}t')= \delta(\vec{r}-\vec{r'}) \delta(t-t')$

Hence these equations lead to :

$ [i\partial_t-H_0-V] \int d\vec{r'} \int dt' G(\vec{r}t,\vec{r'}t') \Psi(\vec{r'},\vec{t'}) = \int d\vec{r'} \int dt' [i\partial_t-H_0-V] G(\vec{r}t,\vec{r'}t') \Psi(\vec{r'},\vec{t'}) = \int d\vec{r'} \int dt' \delta(\vec{r}-\vec{r'}) \delta(t-t')\Psi(\vec{r'},\vec{t'}) = \Psi(\vec{r},\vec{t})$

which means it's not solution of Schrodinger equation. I'm very confused about this result and I really can't get where I made a mistake. Please let me know why it is wrong and what is the appropriate way to verify that Green's function is propagator.

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2 Answers 2

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The expression in the text is $$\Psi(\mathbf r,t) = \Psi^0(\mathbf r,t) + \int\mathrm d\mathbf r' \int \mathrm dt \ G(\mathbf r,\mathbf r';t,t) V(\mathbf r') \Psi^0(\mathbf r',t')\tag{7.17}$$ where $\Psi^0$ is a solution to $[i\partial_t - H_0]\Psi^0 = 0$. It's not difficult to see that $[i\partial_t - H_0 - V]\Psi^0 = -V\Psi^0$, and that the application of $[i\partial_t-H_0-V]$ to the second term yields $V\Psi^0$. As a result, $\Psi$ obeys the Schrodinger equation as promised.

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  • $\begingroup$ I think you used Green's function "not including perturbation-like potential". I think it is slightly deviated from my question, but it helped my understanding. Thank you. $\endgroup$
    – momo
    Oct 17, 2022 at 12:31
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    $\begingroup$ @momo The perturbation-like potential is $V$. $\endgroup$
    – J. Murray
    Oct 17, 2022 at 13:20
  • $\begingroup$ Oh, my question is about eq.7.21 in the textbook. Does it have different meaning comparing to the definition of Green's function defined above eq.7.14(b)? $\endgroup$
    – momo
    Oct 18, 2022 at 14:08
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Starting from dimensional analysis point of view, something is off in your definitions. Your last equations implies $(i\partial_t - H)\psi = \psi$ which can't be right. To see where the issue is coming from we can go back to your starting point: in the first equation it seemed to be implied that $[G] = [r^{-3} t^{-1}]$, as integrating over space and time returns $\psi$ to itself. However from the definition of the GF you give it has dimensions of $[G] = [r^{-3}]$, as the left-hand-side has $[t^{-1}] [G]$ and the right hand side $[r^{-3} t^{-1}]$.

There are two issues here:

i) this is not how the propagator in non-relativistic QM works. We propagate from point $x$ to $x'$ and from time $t'$ to $t$ by multiplying with $K(x, t ; x', t')$ so when we want to look at the evolution into $\psi(r, t)$ we integrate over all space but not over time $$\psi(r, t) = \int \! dr K(r, t; r', t') \psi(r', t')$$ see here under "Non-relativistic propagators".

ii) the GF is not purely the propagator but rather, the GF is related to the propagator $K$ by $G = - i\theta(t-t')K$ (this is the retarded GF. The advanced one is given by $G^a = i\theta(t'-t)K$). Again, I refer you to the wikipedia link above.

So using both these adjustments, when we want to check Schrödinger equation for GF we need to write $$ -i\theta(t-t')(i\partial_t - H)\psi = -i\theta(t-t')(i\partial_t - H)\int \! dr K(r, t; r', t') \psi(r', t') = (i\partial_t - H) \left[-i \theta(t-t')\int \! dr K(r, t; r' t') \psi(r', t')\right] - \delta(t-t')\int \! dr K(r, t; r' t') \psi(r', t') \\ = (i\partial_t - H)\int\! dr G(r, t; r', t')\psi(r', t') -\delta(t-t')\psi(r, t) = \int \! dr \delta(r-r')\delta(t-t')\psi(r', t') \\ = \delta(t-t')\psi(r, t) - \delta(t-t')\psi(r, t) = 0$$ as required

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    $\begingroup$ Thank you for your answer. I'll carefully read your comment and documents attached. $\endgroup$
    – momo
    Oct 17, 2022 at 12:37

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