# Momentum eigenfunctions

Is $e^{-\frac{|x|}{a}}$ an eigenfunction of momentum?

If we apply the momentum operator $\hat{P}=-i\hbar\frac{\partial }{\partial x}$ we get:

$$-i\hbar\frac{\partial }{\partial x}e^{\frac{|x|}{a}}=\cases{i\hbar e^{-x/a} \ \ \ \ (x>0)\\-i\hbar e^{x/a}\ \ \ (x<0)}$$

Which is a constant times the function, however the constant depends on $x$ and so I would say it is not an eigenfunction.

• if the constant depends on $x$, it's not a constant... Commented Jul 22, 2017 at 19:11
• @AccidentalFourierTransform Well, sgn(x) is constant on the left, it's constant on the right and whatever happens in the middle can be attributed to experimental error... Commented Jul 22, 2017 at 19:19

It is not an eigenstate of momentum in infinite 1D space [$x\in(-\infty;\infty)$]. The reason is that the absolute value is not analytic in that space. You can think of it in general terms: Let $\psi(x)=A\mathrm{e}^{if(x)}$, with dimensionless $A$ and $f(x)$. Then $$-i\hbar\partial_x\psi(x)=\hbar A\frac{\partial f(x)}{\partial x}\mathrm{e}^{if(x)}=C\psi(x)\Leftrightarrow C=\hbar \frac{\partial f(x)}{\partial x}.$$

If $f$ is not differentiable everywhere inside the domain, then $\psi$ isn't an eigenfunction. You could redefine your domain to be only half space [$(-\infty,0]$ or $[0,\infty)$], but then what you have is simply $\mathrm{e}^{\pm ikx}$.

• Hello David, and welcome to Physics Stack Exchange! Thanks for your answer, hope to see you around the site! Commented Jul 22, 2017 at 22:37

It is an eigenfunction of $p~=~-i\hbar\partial_x$ except at $x~=~0$. The derivative of the function $f(x)~=~e^{|x|/a}$ jumps discontinuously at $x~=~0$. We may approximate this with $f(x)~=~e^{\sqrt{\epsilon~+~x^2 }/a}$, for $\epsilon$ small. The derivative with respect to $x$ is $$\frac{d}{dx}~=~\frac{x}{\sqrt{\epsilon~+~x^2}}e^{\sqrt{\epsilon~+~x^2 }/a}.$$ In the limit $\epsilon~\rightarrow~0$ this is $\frac{x}{|x|}e^{\sqrt{\epsilon~+~x^2 }/a}$ or $sgn(x)e^{\sqrt{\epsilon~+~x^2 }/a}$.

• Minor nitpick: The function is perfectly continuous at $x=0$, its derivative, however, is not. Commented Jul 22, 2017 at 20:12
• Ouch, that is what I meant. I will revise. Commented Jul 22, 2017 at 23:47
• sgn(x) depends on x. Therefore this isn't an eigenfunction. -1. Commented Jul 23, 2017 at 0:30
• But that only pertains to the one point $x~=~0$ Commented Jul 23, 2017 at 1:43
• No, it doesn't. Eigenfunctions should return multiplied by a constant. sgn(x) is not a constant. Commented Jul 23, 2017 at 11:09