The free nonrelativistic particle

We begin by considering the “free” nonrelativistic particle in \(d\) dimensions – that is, a particle with no force acting on it. The Hamiltonian is

(288)\[H = \frac{{\vec p}^2}{2m}\]

where \(m\) is the particle mass. The time-dependent Schroedinger equation is

(289)\[- \frac{\hbar^2}{2m} \psi({\vec x}, t) = i \hbar \frac{\del}{\del t} \psi({\vec x}, t)\]

The solutions to the time-independent Schrodinger equation are clearly

(290)\[\psi_E({\vec x}) = \frac{1}{(2\pi\hbar)^{d/2}} e^{i {\vec p}\cdot{\vec x}/\hbar}\]

which has energy \(E = \frac{{\vec p}^2}{2m}\). In this case the energy eigenstates are not quite in the Hilbert space. But we will ignore this complication.

We can treat the time-dependent equation as an initial value problem. At time \(t = t_0\), we rpescribe a wavefunction \(\psi({\vec x}, t_0)\). This can be expressed in a Fourier integral:

\[\psi(x,t_0) = \int \frac{d^d p}{(2\pi \hbar)^{d/2}} e^{ i {\vec p}\cdot{\vec x}/\hbar} {\tilde \psi}({\vec p})\]

Since an energy eigenstate evolves simply as

\[\psi_E({\vec x},t) = e^{-i E(t = t_0)/\hbar} \psi_E({\vec x}, t_0)\]

then the full state \(\psi\) evolves as

\[\psi(x,t) = \int \frac{d^d p}{(2\pi \hbar)^{d/2}} e^{- {\vec p}\cdot{\vec x}/\hbar} e^{- i \frac{{\vec p}^2}{2m} t} {\tilde \psi}({\vec p})\]

An equivalent strategy is to start with the formula

(292)\[\begin{split}\begin{align} \psi({\vec x},t) & = \bra{x}U(t,t_0)\ket{\psi(t_0)} \\ & = \int d^d y \bra{x}U(t,t_0)\ket{y}\bra{y}\ket{\psi(t_0)}\\ & \equiv \int d^d y G_0(x,t; y,t_0) \psi(y,t_0) \end{align}\end{split}\]

where \(G_0(x,t;ymt_0) = \bra{x} U(t,t_0) \ket{y}\) and \(U(t,t_0) = e^{-i H (t-t_0)/\hbar}\). \(G_0\) is called the propagator or Green function. With some work you can show that

\[- \frac{\hbar^2}{2m} {\vec\nabla}^2 G_0 = i \hbar \frac{\del}{\del t} G_0\ ,\]

that is, it solves the Schroedinger equation, and that

\[G_0(x,t;y,t_0) \xrightarrow[t \to t_0]{} \delta^d({\vec x} - {\vec y})\]

Given \(G_0\) you can find a solution for any initial value, by doing the integral in (292).

od find \(G_0\) we insert resolutions of the identity \({\bf 1} = \int d^d p \ket{p}\bra{p}\) and use \(\brket{x}{p} = \frac{1}{(2\pi \hbar)^{d/2}} e^{i{\vec p}\cdot{\vec x}/\hbar}\):

(292)\[\begin{split}\begin{align} G_0(x,t;y,t_0) & = \int d^d p \int d^d p' \brket{x}{p} \bra{p} U(t,t_0) \ket{p'}\brket{p'}{y}\\ & = \int\frac{d^3 p d^3 p'}{(2\pi \hbar)^d} e^{i{\vec p}\cdot{\vec x}/\hbar - i {\vec p}'\cdot{\vec y}/\hbar} e^{-i\frac{{\vec p}^2}{2m\hbar} (t - t_0)} \brket{p}{p'}\\ & = \int \frac{d^d p}{(2\pi \hbar)^d} e^{i {\vec p}\cdot({\vec x} - {\vec y})/\hbar - i \frac{{\vec p}^2}{2m\hbar} t} \end{align}\end{split}\]

Now this is a Gaussian integral. The argument is pure imaginart, but if we let \(t \to t - i\eps\), then the integrand will die off exponentially as \(\sim e^{-\eps p^2 t/(2m\hbar)}\) for large \(p^2\). We can then use the standard formulae for Gaussian integrals

\[\int_{-\infty}^{\infty} d q e^{-a q^2} = \sqrt{\frac{\pi}{a}}\]

To do the integral above we need to complete the square. Writing the argument of the exponential as (setting \(t_0 = 0\); this is just a choice of coordinates)

\[\begin{split}\begin{align} & \frac{- i t}{2m\hbar}\left(p^2 - \frac{2m}{t}{\vec p}\cdot({\vec x - \vec y})\right)\\ & \ \ = - \frac{i t}{2m\hbar}\left(p^2 - \frac{2m}{t}{\vec p}\cdot({\vec x - \vec y}) + \frac{m^2}{t^2}(x - y)^2\right) + \frac{i m}{\hbar t}(x - y)^2\\ & \ \ = - \frac{i t}{2m\hbar}(p - \frac{m}{t}(x - y))^2 + \frac{i m}{\hbar t}(x - y)^2 \end{align}\end{split}\]

We can shift the origin of integration by setting \({\vec p} \to {\vec p} + \frac{m}{t}({\vec x} - {\vec y})\) and performing the Gaussian integrals. The result, when the dust settles, is

(293)\[G_0(x,t; y,t_0) = \left(\frac{m}{2\pi i \hbar (t - t_0)}\right)^{d/2} e^{\frac{m({\vec x} - {\vec y})^2}{2i \hbar t}}\]

This is an exponential which oscillates more slowly at \({\vec x} = {\vec y}\) and more quickly as we go off to infinity. The width of the slowly oscillating region increases as a square root of \(t = t_0\); this is the origin of the spreading of the wavefunction. To get some intuition, note that if we set \(t = - i \tau\), the Schroedinger equation becomes

\[\hbar \frac{\del}{\del \tau} \psi = \frac{\hbar^2}{2m} \nabla^2 \psi \Rightarrow \del_{\tau}\psi = D \nabla^2 \psi\]

where \(D = \frac{\hbar}{2m}\). This latter equation is the diffusion equation (a special case of the Fokker-Planck equation). The interpretation is different: \(\psi\) is a density or a probability distribution (rather than a complex amplitude you square to get thr probability). But it turns out you can get some mileage from the realization that there is a formal relation (in many cases) between diffusion problems (and their underlying stochastic dynamics) and quantum problems.