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For a inhomogeneous differential equation of the following form $$\hat{L}u(x) = \rho(x) ,$$ the general solution may be written in terms of the Green function, $$u(x) = \int dx' G(x;x')\rho(x'),$$ such that $$\hat{L}G(x;x') = \delta(x-x') .$$

In this case, I don't have difficulty in understanding why the above equality should be met.

However, in a homogeneous case, $$\hat{L}u(x) = 0$$ the solution can be written using the propagator $$u(x) = \int dx' K(x;x')u(x')$$ that satisfies (according to books) $$\hat{L}K(x;x') = \delta(x-x').$$

I am not able to understand this fact because, if you inserted $u(x)$, written in terms of the propagator $K(x;x')$, into this differential equation, you would expect $\hat{L}K(x;x')=0$.

This issue emerged during studies of the Green function in many-body quantum mechanics, e.q. Zagoskin or Bruus, Flensberg.

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  • $\begingroup$ e.q. Zagoskin p.11 where $\hat{L} = i\hbar\partial_t - H(x,\partial_x,t)$ $\endgroup$
    – WoofDoggy
    Sep 29, 2015 at 22:06
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    $\begingroup$ This answer might help to clear up your confusion. $\endgroup$
    – Winther
    Sep 29, 2015 at 22:10

1 Answer 1

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Your question has been answered again and again, and again, albeit indirectly and elliptically--I'll just be more direct and specific. The point is you skipped variables: in this case, t, and so the expression you wrote ("according to books, $\hat{L}K(x;x') = \delta(x-x')~$"), is nonsense, as you already properly found out; unless you included t in the generalized coordinates, but then again the convolution preceding it is not right.

It is all a deplorable misunderstanding, caused by sloppy language in the community. The WP article gets it right. The retarded Green's function G is the inverse of $\hat{L}=i\hbar\partial_t-H$, $$ \hat{L}G(x,t;x',t') = \delta(t-t')\delta(x-x')~, $$ and it is not exactly the propagator. (Together with its advanced twin, they comprise a hyperformal unitary time evolution operator, of no practical concern here.)

Consider $$G\equiv \frac{1}{i\hbar} \theta(t-t') K(x,t;x',t'), $$ and posit $K(x,t;x',t)=\delta(x-x')$. In that case, the propagator K turns out to be the kernel (null eigenfunction) of $\hat{L}$, simply because the time derivative in $\hat{L}$ acting on the step function $\theta(t-t')$, yields a $\delta (t-t')$ and hence a $\delta(x-x')$ when acting on K by the above posit. These then cancel the two deltas on the r.h.side, and leave only $$ \theta(t-t'\!) ~~ \hat{L}K(x,t;x',t') = 0 $$ behind.

So K is the fundamental solution of the homogeneous equation, the real TDSE (recall you never wish to solve the inhomogeneous TDSE!); and all works out.

Without loss of generality, take $t'=0$, so write $K(x,t;x')$. Consequently $$ \psi(x,t)=\int dx' K(x,t;x') u(x') $$ is a null eigenfunction of $\hat{L}$ with initial condition $\psi(x,0)=u(x)$, which, in turn, justifies the posit.

That is, the above integral is the most general superposition of the fundamental solutions corresponding to all possible I.C.s

Illustrating the above with the free particle, $$ \left(i\hbar\partial_t + \frac{\hbar^2 \partial^2_x}{2m}\right) K(x,t;x')= 0 $$ yields $$ K(x,t;x')=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dk\,e^{ik(x-x')} e^{-\frac{i\hbar k^2 t}{2m}}=\left(\frac{m}{2\pi i\hbar t}\right)^{\frac{1}{2}}e^{-\frac{m(x-x')^2}{2i\hbar t}}, $$ which satisfies the I.C. posit.

You appear to have verified the solution of the above homogeneous equation already, $$ \left ( i\hbar\left( -\frac{1}{2t} +\frac{m(x-x')^2}{2i\hbar t^2}\right) +\frac{\hbar^2}{2m}\left ( -\frac{m}{i\hbar t } - \frac{m^2(x-x')^2}{\hbar^2 t^2} \right) \right ) K =0, $$ alright.

It is also straightforward to likewise find the propagator for the oscillator (quadratic potential), so then solve for the magnificent Mehler kernel (1866), $$ K(x,t;x')=\left(\frac{m\omega}{2\pi i\hbar \sin \omega t}\right)^{\frac{1}{2}}\exp\left(-\frac{m\omega((x^2+x'^2)\cos\omega t-2xx')}{2i\hbar \sin\omega t}\right) ~,$$ which, analytically continued, precedes this QM problem by more than half a century...

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