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In The Dirac-delta function as an initial state for the quantum free particle, Emilio Pisanty

states that if an object's position wave function, $\Psi(x,0)$, is a delta function (at $t = 0$), then $$ \Psi(x,t)=\begin{cases}\delta(x) & t=0\\ \sqrt{\frac{m}{2\pi\hbar |t|}}e^{-i\mathrm{sgn}(t)\pi/4}\exp\left[i\frac{mx^2}{2\hbar t}\right]&t\neq 0\end{cases} $$ In that case, at $t>0$, the momentum wave function is$$ \Phi\left(k,\, t\right) = \frac{1}{\sqrt{{2\pi}}}\sqrt{\frac{m}{2\pi\hbar t}}e^{-i\mathrm{sgn}(t)\pi/4}\int_{-\infty}^{\infty}\exp\left[i\frac{mx^2}{2\hbar t}\right]e^{ikx}\mathrm{d}x $$

According to Engineering Tables/Fourier Transform Table 2, and if we set $\mathcal{x'}=-x$,$$ \begin{alignat}{7} \Phi\left(k, \, t\right) & ~=~ -\sqrt{\frac{m}{2\pi\hbar t}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \sqrt{\frac{\hbar t}{m}} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\ & ~=~-\frac{1}{\sqrt{{2\pi}}} \, e^{-i\mathrm{sgn}(t)\pi/4} \, \exp\left[{-i\left(\frac{\hbar t k^2}{2m} - \frac{\pi}{4}\right)}\right] \\ & ~=~-\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\frac{\hbar t k^2}{2m} }\right] \\ & ~=~ -\frac{1}{\sqrt{{2\pi}}} \, \exp\left[{-i\omega t }\right]\end{alignat}$$

This seems to indicate that, for $t>0$, both $\Psi^2$ and $\Phi^2$ are constant in space, though $\Psi^2$ diminishes in time. That means that both position and momentum are totally uncertain in space for $t>0$.

Did I get that right?

I know that a position eigenstate is not a solution of Schrodinger's equation and so this is all somewhat outside the scope of orthodox QM, but this does seem to violate the idea that the less certain is one variable, the more certain is it's conjugate variable.

David
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  • Infinity times infinity is still bigger than unity. Your results are perfectly compatible with the uncertainty principle. – Emilio Pisanty May 26 '18 at 18:13
  • @Emilio Is this applicable to the double slit experiment, where we put a which way setting device at each slit (such as a device at one slit that sets electron z spin to up and a device at the other slit that sets electron z spin to down)? I am thinking that such a set up would force the electron into an eigenstate of position at one of the slits, so after that time the position probability would simply be the same everywhere in space and hence there would be no position wave function that is the sum of two radial waves that interfere. – David May 26 '18 at 19:06

1 Answers1

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What you have is a free particle propagator (also called Green's function of the free particle Hamiltonian). This is an auxiliary mathematical concept to express time evolution of a general normalizable $\psi$ function. But propagator itself is not a normalized $\psi$ function, the usual theorems valid for normalized $\psi$ functions do not apply to it.

The Heisenberg uncertainty relations are derived from the commutation relations for normalizable $\psi$ functions only. One cannot apply them to singular objects such delta distribution. Not even expected average value of position

$$ \langle x \rangle = \int\psi^* x\psi dx $$

has a sense for $\psi=\delta(x-x_0)$ (or its time evolved version).

Ján Lalinský
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  • Is this applicable to the double slit experiment, where we put a which way setting device at each slit (such as a device at one slit that sets electron z spin to up and a device at the other slit that sets electron z spin to down)? I am thinking that such a set up would force the electron into an eigenstate of position at one of the slits, so after that time the position probability would simply be the same everywhere in space and hence there would be no position wave function that is the sum of two radial waves that interfere. – David May 26 '18 at 19:04
  • That seems complicated. I think you should post another question about that where you explain in more detail what have you done to analyse this and where is the problem. We should not go into it here in comments. – Ján Lalinský May 26 '18 at 19:21