Semi-infinite potential well

  • #26
PeroK
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Sorry, I corrected.

It's
$$k=\frac{-\pi}{4a}=\frac{3\pi }{4a}$$

As an aside, a mathematical subtlety. If you have:
$$\tan ka = -1$$
And you know that ##ka > 0##, then:
$$ka \ne \arctan(-1)$$
The function ##\arctan## has a range of ##(-\frac \pi 2, \frac \pi 2)##. And ##ka## cannot be in that range. You can, however, simply look for the appropriate value of ##ka## based on the work you did on the problem to see that ##ka = \frac{3\pi }{4}##. In other words, it was important that you already knew that ##ka## had approximately this value (to the nearest ##\pi##).

Note that if you were looking for a different energy level, you may well have had to choose ##ka = \frac{7\pi }{4}##, for example.
 
Last edited:
  • #27
etotheipi
As an aside, a mathematical subtlety. If you have:
$$\tan ka = -1$$
And you know that ##ka > 0##, then:
$$ka \ne \arctan(-1)$$
The function ##\arctan## is defined on ##(-\frac \pi 2, \frac \pi 2)##. And ##ka## cannot be in that domain. You can, however, simply look for the appropriate value of ##ka## based on the work you did on the problem to see that ##ka = \frac{3\pi }{4}##. In other words, it was important that you already knew that ##ka## had approximately this value (to the nearest ##2\pi##).

Note that if you were looking for a different energy level, you may well have had to choose ##ka = \frac{11\pi }{4}##.

I suppose ##ka = \arctan(-1) + n\pi## where ##n \in \mathbb{Z}## might be a good way of writing it!
 
  • #28
PeroK
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I suppose ##ka = \arctan(-1) + n\pi## where ##n \in \mathbb{Z}## might be a good way of writing it!
That's how to write it formulaically, yes.
 

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