We know that the Dirac function $$\delta(a)=\lim_{a \rightarrow 0} \delta_{a}(x)$$ can be written as an infinitesimally narrow Gaussian: $$ \delta_{a}(x) := \frac{1}{\sqrt{2\pi a^2}}e^{-x^2/2a^2}$$

Our professor told us that for any value $a>0$, the physical position eigenfunction is $$\psi_{x_0}(x)\cong N_1\delta_a(x-x_0).$$

How can I show that $\psi_{x_0}$ is a physical position eigenfunction?

  • $\begingroup$ What is $N_1$?? And what does $\psi_{x_0}$ represent? $\endgroup$ – Danu Feb 24 '14 at 14:55
  • $\begingroup$ Perhaps the question is: show that it could be a valid wavefunction? Maybe he wants you to show that it is normalizable, finding $N_1$ at the same time. $\endgroup$ – garyp Feb 24 '14 at 14:59
  • $\begingroup$ More on position measurements: physics.stackexchange.com/q/92869/2451 $\endgroup$ – Qmechanic Feb 24 '14 at 15:06
  • $\begingroup$ What do you mean by "physical wavefunction?" The way you define $\psi_{x_0}$ (as being proportional to the delta function centered at $x_0$), it's not an element of the Hilbert space, so in that sense it's not "physical." Is this what you're referring to? $\endgroup$ – joshphysics Feb 24 '14 at 19:03

The ostensibly sad reality is that any such $\psi_{x_o}$ isn't really a position eigenfunction. Try acting on it with the position operator. You simply don't get $x_o$ times the wavefunction.

That said, there are important and meaningful ways in which it is nearly a position eigenfunction. Notice that such a Gaussian is extremely narrow for very small values of $a$. Thus you can approximate multiplication by $x$ (which is the action of the position operator in the position basis) by multiplication by $x_o$. In the only places it matters (where the wavefunction isn't exceedingly small), the position is very nearly $x_o$, so it doesn't do too much damage to just pretend like it really is $x_o$.

Indeed, when we take the limit $a \to 0$ it is an exact result. The trouble is that the state becomes unnormalizable in this case. Sure, it's a delta function so it's area is still just one, but the area under its absolute square--what we're really interested in--is infinite. This shouldn't bother you too much though, because these true position eigenstates (delta functions in the position basis) form a complete basis for the space of physical wavefucntions. Any such state $\psi(x)$ can be represented as a superposition of position eigenstates:

$$ \psi(x) = \int dx'\, \psi(x') \delta(x-x') = \int dx'\, \psi(x')\psi_{x}(x')$$

Where now $\psi_x(x')$ represents the true (unnormalizable) position eigenstate at position x.

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$\psi_{x_0}$ is the ground state wavefunction of a harmonic potential centered at $x_0$ with a curvature chosen to match $a=\sqrt{\frac{\hbar}{m\omega}}$. Since harmonic potentials (approximately) occur often in real systems, this is "physical".

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