The Free-Particle Propagator: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 5: Line 5:
<math>S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.</math>
<math>S=\int_{t_0}^{t_N}dt\,\frac{1}{2}m\dot{x}^2.</math>


Note that we renamed <math>t_i\!</math> to <math>t_0\!</math> and <math>t_f\!</math> to <math>t_N;\!</math> the reason for this will become clear shortly.  Let us now discretize the path that the particle takes, so that the intermediate positions are <math>x_1,\,x_2,\,\ldots,\,x_{N-1}.</math>  We discretize the time axis similarly, with a spacing <math>\delta t\!</math> between two subsequent times.  The action may then be written as
Note that we renamed <math>t_i\!</math> to <math>t_0\!</math> and <math>t_f\!</math> to <math>t_N;\!</math> the reason for this will become clear shortly.  Let us now discretize the path that the particle takes, so that the intermediate positions are <math>x_1,\,x_2,\,\ldots,\,x_{N-1}.</math>  We discretize the time axis similarly, with a spacing <math>\delta t\!</math> between two subsequent times, so that <math>x(t_1)=x_1,\,x(t_2)=x_2,\!</math> and so on.  The action may then be written as


<math>S=\tfrac{1}{2}m\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{i})^2}{\delta t}.</math>
<math>S=\tfrac{1}{2}m\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{i})^2}{\delta t}.</math>
Line 11: Line 11:
The kernel now becomes
The kernel now becomes


<math>K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty} A\int_{-\infty }^{\infty } \int_{-\infty }^{\infty } ...\int_{-\infty }^{\infty } exp(\frac{i}{\hbar}\frac{m}{2}\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{{i}})^2}{\varepsilon })dx_{1}...dx_{N-1}</math>
<math>K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty}\left (\frac{m}{2\pi i\hbar\,\delta t}\right )^{N/2}\int_{-\infty }^{\infty }\cdots\int_{-\infty }^{\infty }dx_{1}\ldots dx_{N-1}\,\exp\left [\frac{im}{2\hbar}\sum_{i=0}^{N-1}\frac{(x_{i+1}-x_{{i}})^2}{\delta t}\right ].</math>
We will now evaluate this integral.  Let us first switch to the variables,


It is implicit in the above that <math>x(t_{0})</math> and <math>x(t_{N})</math> have the values we have chosen at the outset. The factor A in the front is to be chosen at the end such that we get the correct scale for U when the limit <math>N\to \infty</math> is taken.
<math>y_{i}=\sqrt{\frac{m}{2\hbar\,\delta t}}x_{i}.</math>  
Let us first switch to the variables


<math>y_{i}=(\frac{m}{2\hbar\varepsilon })^{1/2}x_{i}</math>
We then get


We then want
<math>K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\lim_{N\to\infty}\left (-\frac{i}{\pi}\right )^{(N-1)/2}\int_{-\infty }^{\infty }\cdots\int_{-\infty }^{\infty }dy_{1}\ldots dy_{N-1}\,\exp\left [-\sum_{i=0}^{N-1}\frac{(y_{i+1}-y_{{i}})^2}{i}\right ].</math>


<math>lim {A}'\int_{-\infty }^{\infty } \int_{-\infty }^{\infty } ...\int_{-\infty }^{\infty } exp(-\sum_{i=0}^{N-1}\frac{(y_{i+1}-y_{{i}})^2}{i } dy_{1}...dy_{N-1}</math>  
Although the multiple integral looks formidable, it is not. Let us begin by doing the integral over <math>y_{1}.\!</math>  Considering just the part of the integrand that involves <math>y_{1},\!</math> we get


where
<math>\int_{-\infty }^{\infty}dy_1\,\exp\left [\frac{-(y_{2}-y_{1})^2-(y_{1}-y_{0})^2}{i}\right ]=\sqrt{\frac{i\pi }{2}}\exp\left [-\frac{(y_{2}-y_{0})^2}{2i}\right ].</math>


<math>{A}'=(\frac{2\hbar\varepsilon }{m })^{\frac{N-1}{2}}</math>  
Now let us evaluate the integral over <math>y_{2}.\!</math>  Again considering just the part of the integrand that involves <math>y_{2},\!</math> we get


Although the multiple integral looks formidable, it is not. Let us begin by doing the
<math>\int_{-\infty }^{\infty}dy_2\,\exp\left [-\frac{(y_{3}-y_{2})^2}{i}-\frac{(y_{2}-y_{0})^2}{2i}\right ]=\sqrt{\frac{2i\pi }{3}}\exp\left [-\frac{(y_{3}-y_{0})^2}{3i}\right ].</math>
<math>y_{1}</math> integration. Considering just the part of the integrand that involves <math>y_{1}</math>, we get


We now continue to do this until all of the <math>y_i\!</math> have been integrated out.  At the <math>k^{\text{th}}\!</math> step (i.e., integrating out <math>y_k\!</math>), the integral that we evaluate and the solution are


<math>\int_{-\infty }^{\infty }exp\left (\frac{-1}{i}(y_{2}-y_{1})^2+(y_{1}-y_{0})^2\right )dy_{1}=(\frac{i\pi }{2})^{1/2}e^\frac{-(y_{2}-y_{0})^2}{2i}</math>
<math>\int_{-\infty }^{\infty}dy_k\,\exp\left [-\frac{(y_{k+1}-y_{k})^2}{i}-\frac{(y_{k}-y_{0})^2}{ki}\right ]=\sqrt{\frac{ki\pi }{k+1}}\exp\left [-\frac{(y_{k+1}-y_{0})^2}{(k+1)i}\right ].</math>


Consider next the integration over yr. Bringing in the part of the integrand involving
Combining all of these results together, we find that the kernel is
<math>y_{2}</math> and combining it with the result above we compute next


<math>K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\lim_{N\to\infty}\frac{1}{\sqrt{N}}\exp\left [-\frac{(y_{N}-y_{0})^2}{Ni}\right ],</math>


<math>(\frac{i\pi}{2})^{1/2} \int_{-\infty }^{\infty }e^\frac{-(y_{3}-y_{2})^2}<math>{i}e^\frac{-(y_{2}-y_{0})^2}{2i}dy_{2}=</math>  
or, rewriting in terms of <math>x_N=x_f\!</math> and <math>x_0=x_i,\!</math>


<math>(\frac{(i\pi )^{2}}{3})^{\frac{1}{2}}e^\frac{-(y_{3}-y_{0})^2}{3i}</math>
<math>K(x_{N},t_{N};x_{0},t_{0})=\lim_{N\to\infty}\sqrt{\frac{m}{2\pi i\hbar N\,\delta t}}\exp\left [-\frac{m}{2\hbar iN\,\delta t}(x_f-x_i)^2\right ].</math>
 
Since we divided the time interval up into equal amounts, we note that <math>N\,\delta t=T=t_f-t_i.\!</math>  We may now take the limit, finally obtaining the free-particle propagator,
 
<math>K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar(t_f-t_i)}\exp\left [-\frac{m}{2\hbar i(t_f-t_i)}(x_f-x_i)^2\right ].</math>

Revision as of 02:38, 17 January 2014

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian , it describes how a state evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

We will now evaluate the kernel for a free particle. In this case, the action is just

Note that we renamed to and to the reason for this will become clear shortly. Let us now discretize the path that the particle takes, so that the intermediate positions are We discretize the time axis similarly, with a spacing between two subsequent times, so that and so on. The action may then be written as

The kernel now becomes

We will now evaluate this integral. Let us first switch to the variables,

We then get

Although the multiple integral looks formidable, it is not. Let us begin by doing the integral over Considering just the part of the integrand that involves we get

Now let us evaluate the integral over Again considering just the part of the integrand that involves we get

We now continue to do this until all of the have been integrated out. At the step (i.e., integrating out ), the integral that we evaluate and the solution are

Combining all of these results together, we find that the kernel is

or, rewriting in terms of and

Since we divided the time interval up into equal amounts, we note that We may now take the limit, finally obtaining the free-particle propagator,

Failed to parse (syntax error): {\displaystyle K(x_{N},t_{N};x_{0},t_{0})=\sqrt{\frac{m}{2\pi i\hbar(t_f-t_i)}\exp\left [-\frac{m}{2\hbar i(t_f-t_i)}(x_f-x_i)^2\right ].}