[NOTES/QFT-04007] Green Function for Schrodinger equation

 $\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}\newcommand{\dd}[2][]{\frac{d#1}{d#2}}$    
$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$    
$\newcommand{\ket}[1]{|#1\rangle}$   
$\newcommand{\bra}[1]{\langle #1|}$   

 I Lesson Overview

Recall and Discuss

II VEV of Time Ordered Product 

• Show that \(\begin{equation} G(x -x{'}, t -t{'}) = \matrixelement{0}{T\psi(x, t)\psi^\dagger (x{'}, t{'}))}{0} \end{equation}\) obeys the equation for Green function of the free particle Schrodinger equation.
• Use expansion of the field operators in terms of free particle wave function \(N \exp(ikx - iE_k t)\), where \(E_k = \frac{\hbar^2k^2}{2m}\). Obtain an explicit expression for this time ordered product as a function of \(x, t, x{'} , t{'}\).
• Have you seen this object before? Where?  

1. The field operator satisfies the equation $$\displaystyle{i\hbar{\partial\over\partial t}\psi(\mathbf x,t) = -{\hbar^2\over2m}\nabla^2\psi(\mathbf x,t)}$$ The function \(G(\mathbf x,t;\mathbf x{'},t{'})\) is \begin{align*} G(\mathbf x,t;\mathbf x{'}, t{'}) =&\langle 0|T\big(\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})\big)|0\rangle\\ =& \langle0|\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'}, t{'})|0\rangle\theta(t-t{'}) +\langle0|\psi^\dagger(\mathbf x{'} t{'})\psi(\mathbf x,t)|0\rangle\theta(t{'}-t) \end{align*} Therefore, \[\begin{eqnarray*} i\hbar{\partial\over\partial t}G(\mathbf x,t;\mathbf x{'},t{'}) &=& \langle0|i\hbar{\partial\over\partial t} \psi(\mathbf x,t)\psi^{\dagger}(\mathbf x{'}, t{'})|0\rangle\theta(t-t{'}) +i\hbar\langle0|\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})|0\rangle \delta(t-t{'})\\ &&+i\hbar \langle0|\psi^{\dagger}(\mathbf x{'}, t{'}){\partial\over\partial t} \psi(\mathbf x,t) |0\rangle\theta(t{'}-t)-i\hbar\langle0|\psi^\dagger(\mathbf x{'}, t{'})\psi(\mathbf x,t)|0\rangle\delta(t{'}-t)\\ && \qquad\qquad \mbox{\text{we have used} \(\quad\dd{t}\theta(t-t{'})=\delta(t-t{'})= -\dd{t} \theta(t{'}-t)\) } \\ &=&-{\hbar^2\over2m}\nabla^2\langle0|\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'}, t{'})|0\rangle\theta(t-t{'})-{\hbar^2\over2m}\nabla^2\langle0| \psi^\dagger(\mathbf x{'}, t{'}) \psi(\mathbf x,t)|0\rangle\theta(t{'}-t)\\ &&+i\hbar\langle0|\big[\psi(\mathbf x,t),\psi^\dagger(\mathbf x{'}, t{'})\big]|0\rangle\delta(t-t{'}) \end{eqnarray*}\] The last term becomes equal time commutator due to $\delta(t-t{'})$ function.\\ Therefore, we get \begin{align*} i\hbar{\partial\over\partial t}G(\mathbf x,t;\mathbf x{'},t{'})+{\hbar^2\over2m}\nabla^2G(\mathbf x,t, \mathbf x{'}, t{'}) = i\hbar\delta(t-t{'})\delta(\mathbf x-\mathbf x^{\,{'}}) \end{align*}
2. The field $\psi(\mathbf x,t)$ obeys free particle equation therefore we write it as a superposition \begin{align*} \psi(\mathbf x,t)= \iiint^{\infty}_{-\infty} {d^3k\over(2\pi)^{3/2}} \exp(i\mathbf k\cdot\vec{\mathbf x}-iE_kt/\hbar)\, a(\mathbf k) \end{align*} of free particle solutions \(\displaystyle {1\over(2\pi)^{3/2}} \exp(i\mathbf k\cdot\mathbf x-iE_kt/\hbar)\). Now \begin{eqnarray*} \psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})&=&\iiint_{-\infty}^\infty{ d^3k\over(2\pi)^{3/2}} e^{i\mathbf k\cdot\mathbf x-iE_kt/\hbar} a(\mathbf k)\times \iiint_{-\infty}^\infty{d^3q\over(2\pi)^{3/2}} e^{-i\mathbf q\cdot\mathbf x{'} +i E_qt{'}/\hbar} a^\dagger(\mathbf q) \end{eqnarray*} Therefore, the vacuum expectation value of the time ordered product becomes \begin{align*} \langle0|T&\big[\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})\big] |0\rangle\\&= \iiint_{-\infty}^\infty {d^3k\over(2\pi)^{3/2}} \iiint^\infty_{-\infty} {d^3q\over(2\pi)^{3/2}} e^{i(\mathbf k\cdot\vec{\mathbf x}-E_kt/\hbar)} e^{-i(\mathbf q\cdot \mathbf x{'}-E_qt/\hbar)}\\ &\times\big\{ \langle0|a(\mathbf k)a^\dagger(\mathbf q)|0\rangle\theta(t-t{'})+\langle0|a^ \dagger(\mathbf q) a (\mathbf k)|0\rangle\theta(t{'}-t)\big\} \end{align*} The last term vanishes because $a|0\rangle=0$. Also \begin{eqnarray}\nonumber \langle0|a(\mathbf k)a^\dagger(\mathbf q)|0\rangle\theta(t-t{'}) &=&\langle0|\big[a(\mathbf k),a^\dagger(\mathbf q)\big] |0\rangle\theta(t-t{'})+\langle0|a^\dagger(\mathbf q)a(\mathbf k)|0\rangle \theta(t -t{'})\\ &=&\delta(\mathbf k-\mathbf q)\theta(t-t{'}) \end{eqnarray} Substituting in Eq.(1) and using $\delta(\mathbf k-\mathbf q)$ to do the $\mathbf q$ integrals gives \begin{align*} \langle0|T\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})|0\rangle= \theta(t-t{'}) \iiint_{-\infty}^\infty {d^3k\over(2\pi)^3}\exp\Big[\mathbf k\cdot(\mathbf x-\mathbf x{'}) - {i\hbar^2\over2m} {(t-t{'})\mathbf k^2\over\hbar}\Big] \end{align*} The integrals over $\mathbf k$ are now Gaussian and can be done by completing the sequences. This gives the answer for the vacuum expectation value of the time ordered product as \begin{align*} \langle0|T\psi(\mathbf x,t)\psi^\dagger(\mathbf x{'},t{'})|0\rangle = \theta(t-t{'}) \left[{m\over2\pi i\hbar(t-t{'})}\right]^{3/2}\exp\left({ im(\mathbf x-\mathbf x{'})^2\over2\hbar(t-t{'})}\right) \end{align*}
3. This expression is just the Green function for free particle wave function in Schrodinger quantum mechanics. $$ \psi(\mathbf x,t)=\iiint d\mathbf x{'} G(\mathbf x,t;\mathbf x{'}, t{'})\psi(\mathbf x{'}, t{'})d\mathbf x{'} $$

 III EndNotes

Green function The Green function for a partial differential equation \begin{equation}L G( \mathbf x,t;\mathbf x_0,t_0) = \delta(\mathbf x-\mathbf x_0)\end{equation} will depend on \(t-t_0\) if the operator \(L\) does not depend on time explicitly. In this case the solution can be obtained in terms of eigenfunctions of \(L\), as you have seen in quantum mechanics. The Green function can be written as \begin{equation}G(\mathbf x,t; \mathbf x,t_0)= \theta(t-t_0))K(\mathbf x,t; \mathbf x_0,t_0)\end{equation} and \(K(\mathbf x,t;\mathbf x_0,t_0)\) obeys the equation \begin{equation}L K(\mathbf x,t;\mathbf x,t_0)=0, \qquad\text{where }K(\mathbf x,t;\mathbf x_0,t_0)\big|_{t=t_0}=\delta(\mathbf x-\mathbf x_0)\end{equation} As you would realize that the corresponding objects in quantum mechanics and field theory discussed in this lesson are just special cases for \(L=H\).