# 8. A setup for Green’s function in quantum mechanics

Some prerequisites: Quantum mechanics, bra-ket notation, basics of Green’s function.

1. Introduction

[NB: In this note, I tried to see how far I understand the basic set-up of Green’s function in quantum mechanics. So it is extremely likely that there are mistakes, misunderstandings, incompletions, etc. Any suggestions for improvements are welcomed.]

Let us consider a classical system, for example, a damped harmonic oscillator. Suppose we exert an external force ${f(t)}$ to the system. The system has no control over ${f(t);}$ it can only respond to the external force. The equation describing the motion would be

$\displaystyle \left(m\frac{d^2}{dt^2} + b\frac{d}{dt} + k\right)x(t) = f(t), \ \ \ \ \ (1)$

where ${m,b,k}$ are constants, and ${x(t)}$ is the position from the equilibrium. One way to solve this equation is to make use of Green’s function. We may view this equation as an inhomogeneous ordinary differential equation. A particular solution can be written as

$\displaystyle x_p(t) = \int_{-\infty}^{\infty}dt' G(t;t')f(t'). \ \ \ \ \ (2)$

Here, ${G(t;t')}$ is called the Green’s function, and it satisfies

$\displaystyle \left(m\frac{d^2}{dt^2} + b\frac{d}{dt} + k\right)G(t;t') = {\delta}(t-t'). \ \ \ \ \ (3)$

So one may view Green’s function as the response to the delta impulse.

Green’s functions also appear in quantum mechanics. For example, a Green’s function ${G(t,\vec{x};t',\vec{x'})}$ satisfies

$\displaystyle \left(i\hbar\frac{{\partial}}{{\partial} t} - \hat{H}\right)G(t,\vec{x};t',\vec{x'}) =i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'}). \ \ \ \ \ (4)$

We know that time-dependent Schrödinger’s equation is given by

$\displaystyle \left(i\hbar\frac{{\partial}}{{\partial} t} - \hat{H}\right)\psi(t,\vec{x}) =0. \ \ \ \ \ (5)$

So based on our experience with Green’s function in classical mechanics, we might try to interpret the quantum mechanical Green’s function ${G(t,\vec{x};t',\vec{x'})}$ as the response to some delta source. This would raise to questions such as “How do we introduce external source into Schrödinger’s equation?”.

It turns out, however, that our viewpoint is in fact incorrect. The quantum mechanical Green’s function ${G(t,\vec{x};t',\vec{x'})}$ is not due to external source. They are defined and widely used for “deterministic” quantum systems.

2. Quantum mechanical Green’s function

Consider a quantum system described by a Hamiltonian ${\hat{H}(\hat{\vec{x}},\hat{\vec{p}})}$ which does not depend explicitly on time. However, the state ${|\psi,t\rangle}$ evolves in time. This is the viewpoint of the Schrödinger’s picture.

The state at the time ${t}$ can be obtained from the state at the time ${t'}$ by using the unitary time evolution operator ${\hat{U}(t,t')}$ such that

$\displaystyle |\psi,t\rangle = \hat{U}(t,t')|\psi,t'\rangle \ \ \ \ \ (6)$

[NB: unitarity of ${\hat{U}(t,t')}$ ensures that the norm of the state does not change under the time evolution]. The evolution can run forward as well as backward in time. If ${t>t',}$ this means that we will be able to know exactly how the system behaves at present time ${t}$ once we know the state of the system at some time ${t'}$ in the past. On the other hand, if ${t this means that we can work out exactly how the system in some time in the past behaves once we know the state of the system at the present time. Similarly, the evolution can also be done between the present and the future.

The ability to predict the state at any time after the state at one time is given is what we mean by “deterministic”. As will be seen shortly, quantum mechanical Green’s function is defined in the deterministic quantum system.

As for the classical system, however, Green’s functions are not defined for the deterministic one. Instead, they should be defined for systems under the influence of external source. These systems are nondeterministic. The introduction of the external source forbids us to know what would happen to the system in the future as there is no information on how the external source would behave. For example, in the next five minutes, I might or might not disturb the damped harmonic oscillator. So nobody (in some case, this includes myself) has any idea how the system would behave five or six minutes later. That is to say, if we allow the external source, we will not be able to predict the future status of the system.

Back to our story, let us write down the quantum mechanical Green’s function:

$\displaystyle G(t,\vec{x};t',\vec{x'}) =\langle\vec{x}|{\Theta}(t-t')\hat{U}(t,t')|\vec{x'}\rangle, \ \ \ \ \ (7)$

where ${{\Theta}}$ is Heaviside step function. This is the probability amplitude of the transition from the state ${|\vec{x'}\rangle}$ at time ${t'}$ to the state ${\langle\vec{x}|}$ at the time ${t}$ such that transition should be from an earlier time to a later time. The probability amplitude for the transition from a later time to an earlier time is zero. [NB: I do not know why we do not make the full use of the fact that the system is deterministic. Does this mean that, for some practical physical reason, we do not expect the system to always remain deterministic???]

In order to determine the differential equation for ${G(t,\vec{x};t',\vec{x'}),}$ let us consider the time-dependent Schrödinger’s equation

$\displaystyle i\hbar\frac{{\partial}}{{\partial} t}|\psi,t\rangle = \hat{H}|\psi,t\rangle. \ \ \ \ \ (8)$

Substituting eq.(6) into this equation, we obtain the equation for ${\hat{U}(t,t'):}$

$\displaystyle i\hbar\frac{{\partial}}{{\partial} t}\hat{U}(t,t') = \hat{H}\hat{U}(t,t'). \ \ \ \ \ (9)$

Now consider

$\displaystyle \begin{array}{rl} \displaystyle i\hbar\frac{{\partial}}{{\partial} t}({\Theta}(t-t')\hat{U}(t,t')) &\displaystyle = i\hbar{\delta}(t-t')\hat{U}(t,t')+{\Theta}(t-t')\hat{H}\hat{U}(t,t')\\ &\displaystyle = i\hbar{\delta}(t-t')+\hat{H}({\Theta}(t-t')\hat{U}(t,t')). \end{array} \ \ \ \ \ (10)$

Acting by ${\langle\vec{x}|}$ on the left and by ${|\vec{x}\rangle}$ on the right gives

$\displaystyle \begin{array}{rl} \displaystyle i\hbar\frac{{\partial}}{{\partial} t}(G(t,\vec{x};t',\vec{x'})) &\displaystyle = i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'})+\langle\vec{x}|\hat{H}({\Theta}(t-t')\hat{U}(t,t'))|\vec{x'}\rangle. \end{array} \ \ \ \ \ (11)$

Next, by using

$\displaystyle \langle\vec{x}|\hat{H}(\hat{\vec{x}},\hat{\vec{p}}) =\hat{H}(\vec{x},-i\hbar\vec{\nabla}_x)\langle\vec{x}|, \ \ \ \ \ (12)$

we obtain

$\displaystyle \begin{array}{rl} \displaystyle i\hbar\frac{{\partial}}{{\partial} t}(G(t,\vec{x};t',\vec{x'})) &\displaystyle = i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'})+\hat{H}(\vec{x},-i\hbar\vec{\nabla}_x)\langle\vec{x}|({\Theta}(t-t')\hat{U}(t,t'))|\vec{x'}\rangle\\ &\displaystyle = i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'})+\hat{H}(\vec{x},-i\hbar\vec{\nabla}_x)G(t,\vec{x};t',\vec{x'}). \end{array} \ \ \ \ \ (13)$

That is,

$\displaystyle \left(i\hbar\frac{{\partial}}{{\partial} t}-\hat{H}(\vec{x},-i\hbar\vec{\nabla}_x)\right)G(t,\vec{x};t',\vec{x'}) = i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'}). \ \ \ \ \ (14)$

So the transition amplitude ${G(t,\vec{x};t',\vec{x'})}$ is the Green’s function of the time-dependent Schrödinger’s equation

$\displaystyle \left(i\hbar\frac{{\partial}}{{\partial} t}-\hat{H}(\vec{x},-i\hbar\vec{\nabla}_x)\right)\psi(t,\vec{x}) =0. \ \ \ \ \ (15)$

For example, for the Hamiltonian

$\displaystyle \displaystyle \hat{H}(\hat{\vec{x}},\hat{\vec{p}}) =\frac{\hat{\vec{p}}^2}{2m} +V(\hat{\vec{x}}), \ \ \ \ \ (16)$

the Green’s function satisfies

$\displaystyle \displaystyle \left(i\hbar\frac{{\partial}}{{\partial} t} +\frac{\hbar^2\vec{\nabla}^2}{2m} -V(\vec{x})\right)G(t,\vec{x};t',\vec{x'}) = i\hbar{\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x'}). \ \ \ \ \ (17)$