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Not seeing the action of a free particle in the Path Integral Formulation 
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#1
Dec2112, 06:49 PM

P: 12

In the very first example of Feynman and Hibb's Path Integral book, they discuss a free particle with
[tex]\mathcal{L} = \frac{m}{2} \dot{x}(t)^2[/tex] In calculating it's classical action, they perform a simple integral over some interval of time [itex]t_a \rightarrow t_b[/itex]. [tex] S_{cl} = \frac{m}{2} \int_{t_b}^{t_a} \dot{x}(t)^2 \; dt = \frac{m}{2} \frac{(x_b  x_a)^2}{t_b  t_a}[/tex] I don't see how that result follows! Is there some nifty integration by parts that I'm missing? 


#2
Dec2112, 07:55 PM

P: 32




#3
Dec2212, 05:07 AM

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P: 2,440

The "classical action" in the path integral means the actual value of the classical action functional for the trajectory of the particle, i.e., the solutions of the equations of motion with the boundary conditions at hand. To evaluate the propagator in position representation [itex]U(t_a,x_a;t_b,x_b)[/itex] you need to solve the equation of motion, which is for the free particle
[tex]m\ddot{x}=0[/tex] with the boundary conditions [tex]x(t_a)=x_a, \quad x(t_b)=x_b.[/tex] The general solution is of course [tex]x(t)=a t+b[/tex] with the two integration constants chosen such that the boundary conditions are fulfilled, i.e., [tex]x_a=a t_a+b, \quad x_b=a t_b +b,[/tex] which leads to [tex]a=\frac{x_ax_b}{t_at_b}, \quad b=\frac{x_b t_ax_a t_b}{t_at_b}.[/tex] The action along the trajectory thus is [tex]S[x_{\text{xl}}]=\int_{t_b}^{t_a} \mathrm{d} t \frac{m}{2} \dot{x}^2 = \int_{t_b}^{t_a} \mathrm{d} t \frac{m}{2} a^2 = \frac{m}{2} a^2(t_at_b)=\frac{m}{2} \frac{(x_ax_b)^2}{t_at_b}.[/tex] So there is a sign error in your result. For more on the evaluation of the path integral, see my QFT script, where in the first chapter I deal with nonrelativistic quantum theory in the pathintegral formalism: http://fias.unifrankfurt.de/~hees/publ/lect.pdf 


#4
Dec2212, 07:13 AM

P: 1,020

Not seeing the action of a free particle in the Path Integral Formulation
perhaps the integration is from t_{a} to t_{b}.



#5
Dec2212, 08:43 AM

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P: 2,440

No, you have to read the propagator from right to left (as the usual Smatrix element is [itex]S_{fi}[/itex]), i.e., [itex]U(t_a,x_a;t_b,x_b)[/itex] is the transitionprobability amplitude for the particle, starting at [itex]x_b[/itex] at time [itex]t_b[/itex] to a [itex]x_a[/itex] at time [itex]t_a[/itex]. So you have to integrate from [itex]t_b[/itex] to [itex]t_a[/itex].



#6
Dec2312, 03:49 AM

P: 1,020

I checked the book and it is right as op has written.
@vanhees71How much sure are you that particle goes from b to a and not from a to b. 


#7
Dec2312, 05:22 AM

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Let's do the calculation of the freeparticle propagator in the operator formalism, which is much more convenient than the pathintegral calculation.
The propagator is defined as the solution of the timedependent Schrödinger equation (initialvalue problem): [tex]\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t,x;t',x') \psi(t',x').[/tex] Let's use the Schrödinger picture, where the time dependence is fully at the state vectors, i.e., [tex]\psi(t) \rangle=\exp[\mathrm{i} \hat{H}(tt')] \psi(t') \rangle.[/tex] The observable operators and thus also their eigenvectors are timeindependent. Thus we have [tex]U(t,x;t',x')=\langle x\exp[\mathrm{i} \hat{H}(tt')]x' \rangle.[/tex] For the free particle [tex]\hat{H}=\frac{\hat{p}^2}{2m}[/tex] and thus it's convenient to write this in terms of the momentumeigenstates, for which we know [tex]\langle{x}{p}\rangle=u_{p}(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p)[/tex] and the completeness relation [tex]\int_{\mathbb{R}} \mathrm{d} p p\rangle \langle p=1.[/tex] Inserting this completeness relation in the expression for the propagator, we find [tex]U(t,x;t',x')=\int_{\mathbb{R}} \mathrm{d} p \langle x\exp[\mathrm{i} \hat{H}(tt')]p \rangle u_{p}^*(x') = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{2 \pi} \exp \left [\frac{\mathrm{i}}{2m} p^2 (tt') \right ] \exp[\mathrm{i} p(xx').[/tex] To make sense out of this integral, we have to regularize this expression by substituting [tex](tt') \rightarrow tt'\mathrm{i} \epsilon, \quad \epsilon>0.[/tex] Doing the Gaussian integral and letting [itex]\epsilon \rightarrow 0^+[/itex] afterwards yields [tex]U(t,x;t',x')=\sqrt{\frac{m}{2 \pi \mathrm{i}(tt')}} \exp \left (\frac{\mathrm{i} m(xx')^2}{2 (tt')} \right ).[/tex] The sign is thus clearly as specified in my previous posting. In the pathintegral evaluation of the propagator the exponential is given by the classical action. This is the simple part of the path integral approach. The somewhat tedious part is to get the prefactor which is the path integral over all paths with the homogeneous boundary conditions [itex]x(t)=x(t')=0[/itex]. For this calculation you have to go back to the descretized form of the path integral, evaluate the multidimensional Gauß integral and then take the continuum limit. You find the calculation (for the only slightly more difficult case of the harmonic oscillator) in my QFT manuscript, http://fias.unifrankfurt.de/~hees/publ/lect.pdf 


#8
Dec2312, 07:43 AM

P: 1,020

it's all green function derivation is o.k..But I was saying that xx' appears with a square term so it does not matter for whether xx' or x'x.But with time suppose you write ψ(t',x') in place of ψ(t,x) and ψ(t,x) in place of ψ(t',x').
ψ(t',x')=∫dx U(t',x';t,x)ψ(t,x) so now t't would appear.All this means that t' is later than t.There is no subtle problem with this. 


#9
Dec2312, 09:28 AM

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Sure, but the sign is determined uniquely by the choice of the sign in the timeevolution operator. The usual convention since the very beginning of quantum theory is the one given in my previous posting. If you interchange the time arguments, of course you get the opposite sign.
It's also clear that the time dependence cannot be symmetric unter exchange of the time, because intrinsically you introduce a direction of time, given by the causal sequence of events: You prepare the system in a certain state at a certain time and then determine its evolution in the future. The same time, the Hamiltonian as the operator representing energy must be bounded from below, and thus timereversal invariance must necessarily be represented by an antiunitary rather than a unitary operator. This can be seen on the example of this propagator too: If you interchange [itex]t[/itex] and [itex]t'[/itex] (i.e., initial and final time), you must take the conjugate complex to get the same propagator as before. This complex conjugation is due to the antiunitary nature of the timereflection operation. This is different for parity or space reflection. Here, the Heisenberg algebra forces a unitary reprsentation of this symmetry, and that's why the propagator is symmetric under exchange of [itex]x[/itex] and [itex]x'[/itex]. 


#10
Dec2412, 02:39 AM

P: 1,020

I fully agree.I have a question what is time reversal matrix for dirac eqn.



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