Green's function (part II)


Like I have said in :link: my post about the classical Green’s function:

Green’s function method is a way to decompose a complex inhomogenous source using Dirac delta functions and then combine individually solutions to obtain the true answer.

This time, let’s take a look of the quantum version in the first quantization formalism.

In its time-dependent form, the Schrodinger’s equation reads:

\[i\hbar \frac{\partial}{\partial t} \Psi (\vec r, t) = \hat H \Psi(\vec r, t) \tag{1},\]

where the Hamiltonian operator $\hat H$ is $-\frac{\hbar^2}{2m} \nabla^2 + V(\vec r, t)$.

Write the Schordinger’s equation in full:

\[i\hbar \frac{\partial}{\partial t} \Psi (\vec r, t) = \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec r, t)\right] \Psi(\vec r, t) \tag{2}.\]

Now, this doesn’t look like a linear inhomogeneous equation that we’re used to in the classical Green’s function formulation. However, we can re-write it to the following form:

\[\begin{aligned} \left[i\hbar \frac{\partial}{\partial t} +\frac{\hbar^2}{2m} \nabla^2\right] \Psi (\vec r, t) &= V(\vec r, t) \Psi(\vec r, t)\\ \hat{\mathcal{H}} \Psi (\vec r, t) &= g(\vec r, t), \end{aligned} \tag{3}\]

which looks much like the linear inhomogeneous equation we are after. With only one exception being that the inhomogeneous source $g(\vec r, t)$ is $\Psi(\vec r, t)$ dependent. We can solve this, but we need a recursive solution to find $\Psi(\vec r, t)$.

Or, with Greens function, instead, we can solve the following inhomogeneous equations:

\[\begin{aligned} \left[i\hbar \frac{\partial}{\partial t} +\frac{\hbar^2}{2m} \nabla^2\right] G(\vec r, t;\vec r',t') &= \delta(\vec r -\vec r')\delta(t - t') \end{aligned}. \tag{4}\]

Then the original solution can be found by:

\[\Psi(\vec r,t) = \int G(\vec r, t;\vec r',t') \left[V(\vec r, t) \Psi(\vec r',t') \right] d^3 r'. \tag{5}\]

Notice that in Eq. 5, the solution is also present in the integral which suggests we still need a recursive or iterative method to obtain the answer, but it should be much simpler than the original problem.

Taking a good look at Eq. 5, we see that $G(\vec r, t;\vec r’,t’)$ links $\Psi (\vec r’, t’)$ to $\Psi (\vec r, t)$, effectively, propagates the wavefunction from one state to another.

To further clarify this propagation, recall that due to energy conservation, the time evolution of a wavefunction can be described using the so-called “time evolution operator” $U(t,t’) = e^{-\frac{i}{\hbar}H(t-t’)}$. Expanding the wavefunction in the position representation as: $\Psi(\vec r, t) = \braket{\vec r \vert \Psi(t)}$, we can re-express the wavefunction at position $\vec r$ and time $t$ as:

\[\Psi(\vec r, t) = \bra{\vec r} e^{-\frac{i}{\hbar}H(t-t')} \ket{\Psi(t')}\]

Inserting the closure relation of the position operator $\int \ket{\vec r’}\bra{\vec r’} d^3 r’ = \mathcal{1}$:

\[\Psi(\vec r, t) = \int \bra{\vec r}e^{-\frac{i}{\hbar}H(t-t')}\ket{\vec r'} \bra{\vec r'} \Psi(t') d^3 r'. \tag{6}\]

Comparing Eq. 6 to Eq. 5, we see that:

\[\begin{aligned} G(\vec r, t;\vec r',t') &= \bra{\vec r}e^{-\frac{i}{\hbar}H(t-t')}\ket{\vec r'}\\ &= \bra{\vec r}e^{-\frac{i}{\hbar}H(t)} \cdot e^{\frac{i}{\hbar}H(t')} \ket{\vec r'}\\ &= \braket{\vec r, t\vert\vec r', t'}. \end{aligned} \tag{7}\]

From Eq. 7, we see that we have associated the Green’s function to the probability amplitude of preparing the particle in a state $\ket{\vec r’, t’}$ and later find that that particle in state $\bra{\vec r, t}$.

To give these “states” (which are actually operators) meaning, we need to insert another closure relation: $\sum_n \ket{n}\bra{n} = \mathcal{1}$ where $\ket{n}$ and $\bra{n}$ are eigenvectors of the Hamiltonian. With the closure relation inserted, the Green’s function reads:

\[G(\vec r, t;\vec r',t') = \sum_n \bra{\vec r} e^{-\frac{i}{\hbar}H(t-t')} \ket{n}\braket{n\vert\vec r'}\]

since the Hamiltonian $\hat H$ acts on the eigenvectors and gives us the eigenvalues $E_n$, also noting that acting the position operator $\bra{r}$ on the eigenvectors $\ket{n}$ gives us the eigenfunctions $\psi_n(r)$, we have:

\[\begin{aligned} G(\vec r, t;\vec r',t') &= \sum_n \braket{\vec r \vert n}\braket{n\vert\vec r'} e^{-\frac{i}{\hbar}E_n(t-t')}\\ &= \sum_n \psi_n(\vec r) \psi_n(\vec r') e^{-\frac{i}{\hbar}E_n(t-t')}. \tag{8} \end{aligned}\]

This Greens function satisfies Eq. 4 and can be used to construct our general solution to Eq. 1.

Fourier transforming this Green’s function from time-domain to frequency domain, we get:

\[\begin{aligned} \mathcal{F}[G(\vec r, t;\vec r',t')](\vec r, \vec r', E) &= \int \sum_n \psi_n(\vec r) \psi_n(\vec r') e^{-\frac{i}{\hbar}E_n(t-t')} e^{i \frac{E}{\hbar}}(t-t') d(t-t')\\ &= \int \sum_n \psi_n(\vec r) \psi_n(\vec r') e^{-i\frac{E_n}{\hbar}\textbf{t}} e^{i \frac{E}{\hbar}\textbf{t}} d\textbf{t}\\ &= \int \sum_n \psi_n(\vec r) \psi_n(\vec r') e^{i\frac{E-E_n}{\hbar}\textbf{t}} d\textbf{t}\\ &= \sum_n \psi_n(\vec r) \psi_n(\vec r') \delta(\frac{E-E_n}{\hbar}). \tag{9} \end{aligned}\]

From Eq. 9, We see that when there are poles (divergences) when the energy matches the eigenvalue $E_n$.

Second quantization version, coming soon!

Examples, coming soon!

Author | Chengcheng Xiao

Currently a PhD student at Imperial College London. Predicting electron behaviour since 2016.