Does e^{-ix} Have a Limit as x Approaches Infinity?

[jimmycricket]

What is the limit of $e^{-ix}$ as x tends to infinity?

 

[Mark44]

What is the limit of $e^{-ix}$ as x tends to infinity?

What does $e^{-ix}$ represent? IOW, for a given x value, what does $e^{-ix}$ evaluate to?

 

[jimmycricket]

$cos(x) - isin(x)$

 

[Ray Vickson]

$cos(x) - isin(x)$

OK, so what do YOU think the limit should be, if anything?

 

[jimmycricket]

Well this is exactly my problem, I don't know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I don't know if this is now the right place to ask this but if anyone can help that would be great.

 

[mathman]

I can't comment on the physics question. However the original math question is answerable - there is no limit. You can envision it geometrically as being points on the unit circle in the complex plane. As x becomes infinite the point simply goes around the circle indefinitely.

 

[Ray Vickson]

Well this is exactly my problem, I don't know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I don't know if this is now the right place to ask this but if anyone can help that would be great.

Do you know what it means when we say that a function, f(x), has a limit as x → ∞? Never mind the "epsilon-delta" stuff; just give an intuitive description.

Alternatively, think of the graph y = cos(x). Do the y-values settle down to a fixed value as x becomes larger and larger?

 

[Matterwave]

Well this is exactly my problem, I don't know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I don't know if this is now the right place to ask this but if anyone can help that would be great.

A (infinite) square well potential should not have waves outside of the well. Outside the well, the wave function should just be 0. A finite square well can have a non 0 wave function outside the well, but they should exponentially decay instead of oscillate (assuming a bound state). Recheck your answers.

If you are dealing with scattering states, then the wave function must be a wave-packet, not a plane wave since plane waves are not normalizable.

 

[Mark44]

What does $e^{-ix}$ represent? IOW, for a given x value, what does $e^{-ix}$ evaluate to?

$cos(x) - isin(x)$

No, I was looking for a more specific answer, which @mathman gave you in post #6. In my question I specified "for a specific x value," so your response should have taken that into account.