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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.

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    $\begingroup$ if the constant depends on $x$, it's not a constant... $\endgroup$ – AccidentalFourierTransform Jul 22 '17 at 19:11
  • $\begingroup$ @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... $\endgroup$ – Wojciech Morawiec Jul 22 '17 at 19:19
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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 $\mathrm{e}^{\pm ikx}$.

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  • $\begingroup$ Hello David, and welcome to Physics Stack Exchange! Thanks for your answer, hope to see you around the site! $\endgroup$ – heather Jul 22 '17 at 22:37
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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 $sgn(x)e^{\sqrt{\epsilon~+~x^2 }/a}$.

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  • $\begingroup$ Minor nitpick: The function is perfectly continuous at $x=0$, its derivative, however, is not. $\endgroup$ – Wojciech Morawiec Jul 22 '17 at 20:12
  • $\begingroup$ Ouch, that is what I meant. I will revise. $\endgroup$ – Lawrence B. Crowell Jul 22 '17 at 23:47
  • $\begingroup$ sgn(x) depends on x. Therefore this isn't an eigenfunction. -1. $\endgroup$ – Emilio Pisanty Jul 23 '17 at 0:30
  • $\begingroup$ But that only pertains to the one point $x~=~0$ $\endgroup$ – Lawrence B. Crowell Jul 23 '17 at 1:43
  • $\begingroup$ No, it doesn't. Eigenfunctions should return multiplied by a constant. sgn(x) is not a constant. $\endgroup$ – Emilio Pisanty Jul 23 '17 at 11:09

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