Question

I just need a little guidance.
(see attached)

Answer 1

Join the club. This is a standard "problem" in intro quant mechs, i think.
I'd guess at a value for a, the width of the well. I'd make it "simple" -> 1 is pretty simple BUT REMEMBER that you've set a to 1.
the variable x now has to be between zero and 1.
go in, say, 0.1 steps for x and crunch the numbers either/or on a calculator, in your head (hmm headache), or on Xcel spreadsheet.
if on the spreadsheet, there's a button which "sketches" graphs based on data selected.
any screaming and shouting is acceptable at this point.
Work out what "NORMALISATION" means. Possibly something to do with 1, AGAIN.
So in terms of trigonometry, it could be that the wave function normalises to a value of 1.
That may have something to do with the complex conjugate of the wave function and then an integration over some distance (possibly associated with the number 1 again), the integral having a value of .... 1
If you follow this verbiage ...
I wish you the very best of luck waiting for "the penny to drop into a jackpot"
http://perendis.webs.com

Answer 2

well, my top-tip for just sketching the thing -- use the wave number (#^.^#)



\(k = \dfrac{3 \pi }{2 a} = \dfrac{2 \pi}{\lambda}\)

\(\lambda = \dfrac{4a}{3}\)

then \(\Psi(0) = \dfrac{1}{\sqrt{a}} \)

so you've got 3/4th of a wavelength either side of x = 0 and its fixed at the walls....all the trappings of a nice standing wave

for second bit, yep, normalise with \(\Psi^2 \), they've given the trig shortcut. should give 1 as the answer, or so they're telling us

Answer 3

Wow that's great help. Thanks @IrishBoy123 .

Answer 4

does this look right?

Answer 5

i can't draw to save my life so i won't comment on the artwork (!!!)

but as long as it's this i think it makes sense

Answer 6

hope that's not overdoing it, drawing-wise

:-)

Answer 7

it looks like good work in progress ... maybe needs a vertical scale ? It is zero at the potential well boundaries which is (sort of) what you'd expect.
it's also maximum at the centre which is "sort of" where you'd expect a "quantum particle" (which you can't see etc etc etc) to spend "most of it's time".
The next thing is to say "what IS the maximum ?" And that's maybe where the NORMALISATION bit kicks in. Maybe if that were on the graph, as well as every other bell and whistle, you'd get a meagre 4 marks. I can tell this is a maths related question, just by the meanness of the mark scheme.
Shamrock/Rembrandt/Michelangelo/Leonardo is at it again.

Answer 8

Sorry about this net one ...
drawing-wise = wise drawing ...

Answer 9

osprey beak flash ... the NORMALISATION bit is to set the arithmetic such that the PROBABILITY that the particle will be SOMEWHERE in the potential well is 1. After all, it's got infinite sides, so there's "no escape" ?
If it had FINITE sides (Kronnig-Penney development) then the particle-wave could "leak out" under the ideas of QUANTUM TUNNELLING, and EVANESCENT WAVE in optics.
flutter flutter flutter

Answer 10

Yet another one ... normalisation normally involves integrating the product of the wave function with its complex conjugate since it's a complex number.
complex number could be cos angle + i sine angle.
cos is real, sine is "imaginary".
I think that in this case, it's cos angle ONLY, so that the imag bit "drops out".
this would help to "explain" the other part of the q in the post.
cos times cos is cos squared ... etc

Answer 11

Thanks for the help on the normalisation part @osprey, I completely forgot to square the wave function before integrating.