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Apr 24, 2023 at 22:21 comment added Ghoster Even with the negative sign, the question still doesn’t make sense. $Ne^{-\gamma r}$ is not the wave function of a free particle.
Apr 24, 2023 at 22:18 comment added Ghoster Does this minus sign impact the calculations in any way? Yes. It makes them give sensible (i.e., finite) results.
Apr 24, 2023 at 21:52 history edited Zorbakk CC BY-SA 4.0
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Apr 24, 2023 at 18:06 comment added Zorbakk I just edited my post: the wave function did indeed contain a minus sign in the exponential. I must've missed it while writing this post somehow. Does this minus sign impact the calculations in any way?
Apr 24, 2023 at 18:03 history edited Zorbakk CC BY-SA 4.0
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Apr 24, 2023 at 18:00 vote accept Zorbakk
Apr 24, 2023 at 17:41 history edited Qmechanic CC BY-SA 4.0
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Apr 24, 2023 at 17:32 comment added Ghoster Are you sure the wave function wasn’t specified as $Ne^{-\gamma r}$? As written, the problem doesn’t make sense.
Apr 24, 2023 at 16:49 comment added FlatterMann While answers of the form "plane waves are not normalizable" are technically correct, you still have to learn how to deal with plane waves because they are actually far more important in physics than bound solutions. See e.g. physics.stackexchange.com/q/165373 for a few pointers of how to deal with the problem. The "obsession" of non-relativistic quantum mechanics with finite, nicely normalizable systems is more of an educational artifact: it's relatively easy to teach compared to the real problem.
Apr 24, 2023 at 16:34 answer added J. Murray timeline score: 1
Apr 24, 2023 at 14:56 history edited Qmechanic
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Apr 24, 2023 at 12:57 vote accept Zorbakk
Apr 24, 2023 at 14:32
S Apr 24, 2023 at 12:27 review First questions
Apr 24, 2023 at 13:01
S Apr 24, 2023 at 12:27 history asked Zorbakk CC BY-SA 4.0