Feynman Path Integrals: Difference between revisions
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<math>\Psi(x_f,t_f)=\int_{-\infty}^{\infty} dx\,K(x_f,t_f;x,t)\Psi(x,t).</math> | <math>\Psi(x_f,t_f)=\int_{-\infty}^{\infty} dx\,K(x_f,t_f;x,t)\Psi(x,t).</math> | ||
As a justification of this method, we will show that it reproduces the [[Schrödinger Equation|Schrödinger equation]]. We will use the method used by Feynman. Let us begin by assuming that the elapsed time <math>t\!</math> is so small, that we may approximate the path integral with a single "time slice" of length <math>\delta t.\!</math> In this case, the kernel is just <math>e^{iS/\hbar},</math> and the action | As a justification of this method, we will show that it reproduces the [[Schrödinger Equation|Schrödinger equation]]. We will use the method used by Feynman. Let us begin by assuming that the elapsed time <math>t\!</math> is so small, that we may approximate the path integral with a single "time slice" of length <math>\delta t.\!</math> In this case, the kernel is just <math>e^{iS/\hbar},</math> and the action is just its average over the time interval times is length: | ||
<math>S=\frac{m(x_f-x)^2}{2\delta t}-V(x_f,t | <math>S=\frac{m(x_f-x)^2}{2\delta t}-V[\tfrac{1}{2}(x_f+x),t]\,\delta t</math> | ||
The kernel now becomes | The kernel now becomes | ||
<math>K(x_f,t_f;x,t)=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\exp\left [\frac{im(x_f-x)^2}{2\hbar\,\delta t}\right ]\exp\left [-\frac{i}{\hbar}V(x_f,t)\,\delta t\right ].</math> | <math>K(x_f,t_f;x,t)=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\exp\left [\frac{im(x_f-x)^2}{2\hbar\,\delta t}\right ]\exp\left [-\frac{i}{\hbar}V[\tfrac{1}{2}(x_f+x),t]\,\delta t\right ],</math> | ||
so that the wave function <math>\Psi(x_f,t_f)=\Psi(x_f,t+\delta t)\!</math> is | |||
<math>\Psi(x_f,t+\delta t)=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\int_{-\infty}^{\infty} dx\,\exp\left [\frac{im(x_f-x)^2}{2\hbar\,\delta t}\right ]\exp\left [-\frac{i}{\hbar}V[\tfrac{1}{2}(x_f+x),t]\,\delta t\right ]\Psi(x,t).</math> | |||
Now we introduce the variable, <math>\xi=x_f-x,\!</math> so that the integral becomes | |||
<math>\Psi(x_f,t+\delta t)=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\int_{-\infty}^{\infty} d\xi\,\exp\left (\frac{im\xi^2}{2\hbar\,\delta t}\right )\exp\left [-\frac{i}{\hbar}V(x_f-\tfrac{1}{2}\xi,t)\,\delta t\right ]\Psi(x_f-\xi,t).</math> | |||
We now expand the wave function in the integral in powers of <math>\xi\!</math> up to second order and the factor involving the potential in powers of <math>\delta t\!</math> up to first order. We also drop all dependence of the potential on <math>\xi.\!</math> The result of this is | |||
<math>\Psi(x_f,t+\delta t)=\sqrt{\frac{m}{2\pi i\hbar\,\delta t}}\int_{-\infty}^{\infty} d\xi\,\exp\left (\frac{im\xi^2}{2\hbar\,\delta t}\right )\left [1-\frac{i}{\hbar}V(x_f,t)\,\delta t\right ]\left [\Psi(x_f,t)-\frac{\partial\Psi(x_f,t)}{\partial x_f}\xi+\tfrac{1}{2}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}\xi^2\right ].</math> | |||
The problem has been reduced to the evaluation of Gaussian integrals. Using the formulas, | |||
<math>\int_{-\infty}^{\infty}dx\,e^{-ax^2}=\sqrt{\frac{\pi}{a}},</math> | |||
<math>\int_{-\infty}^{\infty}dx\,xe^{-ax^2}=0,</math> | |||
and | |||
<math>\int_{-\infty}^{\infty}dx\,x^2e^{-ax^2}=\frac{1}{2a}\sqrt{\frac{\pi}{a}},</math> | |||
we obtain | |||
<math>\Psi(x_f,t+\delta t)=\left [1-\frac{i}{\hbar}V(x_f,t)\,\delta t\right ]\left [\Psi(x_f,t)+\frac{i\hbar}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}\,\delta t\right ].</math> | |||
We now multiply out the right-hand side and retain only terms that are first order in <math>\delta t.\!</math> This gives us | |||
<math>\Psi(x_f,t+\delta t)=\Psi(x_f,t)+\frac{i\hbar}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}\,\delta t-\frac{i}{\hbar}V(x_f,t)\Psi(x_f,t)\,\delta t.</math> | |||
Rearranging, we get | |||
<math>i\hbar\frac{\Psi(x_f,t+\delta t)-\Psi(x_f,t)}{\delta t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}+}V(x_f,t)\Psi(x_f,t).</math> | |||
Finally, taking the limit, <math>\delta t\to 0,\!</math> and renaming <math>x_f\!</math> to <math>x,\!</math> we finally arrive at the familiar [[Schrödinger Equation|Schrödinger equation]], | |||
<math>i\hbar\frac{\partial\Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+}V(x,t)\Psi(x,t).</math> | |||
'''Question:''' The Feynman path integral formulation of quantum mechanics is more complex than | '''Question:''' The Feynman path integral formulation of quantum mechanics is more complex than |
Revision as of 01:32, 17 January 2014
The path integral formulation of quantum mechanics was developed in 1948 by Richard Feynman. The path integral formulation is a description of quantum theory that generalizes the action principle of classical mechanics. It replaces the classical notion of a single, unique trajectory for a system with a sum, or functional integral, over an infinity of possible trajectories to compute a quantum amplitude.
The classical path is the path that minimizes the action.
This formulation has proved crucial to the subsequent development of theoretical physics, since it is apparently symmetric between time and space. Unlike previous methods, the path-integral offers us an easily method by which we may change coordinates between very different canonical descriptions of the same quantum system.
For simplicity, the formalism is developed here in one dimension.
In the path integral formalism, we start by writing the amplitude for a particle at position Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i\!} at time Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_i\!} to move to a position Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_f\!} at time Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_f\!} as a path integral. This path integral is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K(x_f,t_f;x_i,t_i)=\langle x_f,t_f|\hat{U}(t_f,t_i)|x_i,t_i\rangle=\int D x(t')\,e^{iS[x(t')]/\hbar}, }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S[x(t)]\!} is the action for the the path Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(t')\!} and the integral is defined as
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int D x(t')=\lim_{N\to\infty}\left (\frac{m}{2\pi i\hbar\,\delta t}\right )^{N/2}\left (\prod_{j=1}^{N-1}\int_{-\infty}^{\infty} dx_j\right ),}
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N\!} is a number of "slices" of length Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta t\!} that we divide the time axis up into. Essentially, we define the path integral as a limit of an integral over all possible values of the particle's intermediate positions on its path from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\!} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x.\!}
The action is given by the time integral of the Lagrangian, just as in classical mechanics: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S[x(t')]=\int_{t_i}^{t_f} dt' \mathcal{L}[x(t'),\dot{x}(t'),t'], }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{L}[x(t'),\dot{x}(t'),t']=\tfrac{1}{2}m\dot{x}^2(t')-V(x(t'),t') }
is the Lagrangian.
Our choice of notation for this path integral, is motivated by the fact that it serves as a "kernel" for an integral giving the wave function in terms of This integral is
As a justification of this method, we will show that it reproduces the Schrödinger equation. We will use the method used by Feynman. Let us begin by assuming that the elapsed time is so small, that we may approximate the path integral with a single "time slice" of length In this case, the kernel is just and the action is just its average over the time interval times is length:
The kernel now becomes
so that the wave function is
Now we introduce the variable, so that the integral becomes
We now expand the wave function in the integral in powers of up to second order and the factor involving the potential in powers of up to first order. We also drop all dependence of the potential on The result of this is
The problem has been reduced to the evaluation of Gaussian integrals. Using the formulas,
and
we obtain
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(x_f,t+\delta t)=\left [1-\frac{i}{\hbar}V(x_f,t)\,\delta t\right ]\left [\Psi(x_f,t)+\frac{i\hbar}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}\,\delta t\right ].}
We now multiply out the right-hand side and retain only terms that are first order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta t.\!} This gives us
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(x_f,t+\delta t)=\Psi(x_f,t)+\frac{i\hbar}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}\,\delta t-\frac{i}{\hbar}V(x_f,t)\Psi(x_f,t)\,\delta t.}
Rearranging, we get
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\Psi(x_f,t+\delta t)-\Psi(x_f,t)}{\delta t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x_f,t)}{\partial x_f^2}+}V(x_f,t)\Psi(x_f,t).}
Finally, taking the limit, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta t\to 0,\!} and renaming Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_f\!} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x,\!} we finally arrive at the familiar Schrödinger equation,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial\Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+}V(x,t)\Psi(x,t).}
Question: The Feynman path integral formulation of quantum mechanics is more complex than solving the Schrödinger equation to get the dynamics of a quantum particle, why this formulation is mentioned in the text books and where it may be useful?
Answer: As far as a single particle is concerned it is recommended to use Schrödinger equation of motion. However, to study a many particle system getting dynamics by means of Schrödinger equation is quite complicated and messy (let say sometimes impossible), while the Feynman path integral is a good tool for dealing with many particle problems by defining the field operators. More importantly, the generalization of quantum mechanics to relativistic problem can be done in terms of field theory via Feynman path integral formulation.
Explicit evaluation of the path integral for the harmonic oscillator can be found here File:FeynmanHibbs H O Amplitude.pdf