The one-dimensional free particle Lagrangian is given by $$ \mathcal{L} = \frac{m}{2}\dot x^2. $$ Since the Lagrangian is translation-invariant, one usually argues that the propagator can only be a function of $|x - y|$. By the same logic, shouldn't one be able to argue that the propagator can only be a function of $|t_f - t_i|$? However, the propagator is given by $$ \sqrt{\frac{m}{2\pi i \hbar (t_f - t_i)} }\exp\left(\frac{im(x_f - x_i)^2}{2\hbar (t_f - t_i)}\right), $$ so my question is why doesn't this logic doesn't hold?
Why isn't the free particle particle a function of the absolute value of the difference of the time?
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4$\begingroup$ $t=t_f-t_i$ is already what you want. However, it is not time-reversal invariant, so it is not $|t|$ $\endgroup$– naturallyInconsistentCommented Jul 8 at 5:43
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$\begingroup$ Yes, but the Lagrangian time-reversal invariant, so shouldn't we also expect the propagator to be time-reversal invariant? $\endgroup$– zeroknowledgeproverCommented Jul 8 at 5:44
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2$\begingroup$ No, that is like saying that the cosine function is reflection invariant, so why isn't its integral, the (negative) sine function, reflection invariant? The propagator has a part that is the integral of the Lagrangian, so this analogy is actually a lot more applicable than meets the eye. $\endgroup$– naturallyInconsistentCommented Jul 8 at 5:49
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$\begingroup$ Oh ok, so the action is 1) time-translation invariant but not invariant under time-reversal 2) space-translation invariant and invariant under space-reflection. And that what makes the difference? $\endgroup$– zeroknowledgeproverCommented Jul 8 at 5:53
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$\begingroup$ Also, rigorously speaking that path integral is not really convergent. One way to force convergence is to sneak in an infinitesimal imaginary part: $\dot{x}^2 \rightarrow x (-\partial_t)^2 x \rightarrow x (-(\partial_t)^2 + i0^{+}) x$ (first step is integration by part). Now that messes up your time reversal property. In a sense, we have to explicitly enforce causality so that that path integral makes any sense at all. $\endgroup$– T.P. HoCommented Jul 8 at 7:10
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1 Answer
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Time evolution in QM is unitary $U(t)^{\dagger}=U(-t)$, so the free propagator is only expected to be invariant under combined time-reversal $t\to -t$ and complex conjugation.
See also this related Phys.SE post.