[Intro QM] Verification requested on possible solution for a question

  • #1
warhammer
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Homework Statement:
Q1 (i) Given φ(k) = \frac{A/Δk} where (k0 − 0.5Δk)< k < (k0 +0.5Δk) & φ(k) is 0 elsewhere. A, k0 and Δk are constants.
Show that ψ(x) = A * \frac{sinb/b} *exp(ik0x) where b =0.5Δkx. Also find ψ(0).

(ii) A particle freely moves in an one dimensional box of length L. The probability of finding the particle between x and x + dx inside the box is P(x)dx = Cdx, 0 ≤ x ≤ L where C is a constant. It is 0 elsewhere. Find C, < x > and < x^n >
Relevant Equations:
ψ(x)= \frac{1/√(2pi)} \int_-∞^∞ φ(k) exp(ikx) dx\

<expectation value of an observable>= \int ψ(x)*(observable)ψ(x)
Below I have attached an image of my possible solution. I have replaced all the relevant limits. For some reason, I am getting the final value for (i) part as ψ(x)= with an additional √2pi in the denominator. Have I made any errors or is it fine if I take it within the constant A..

IMG_20220428_024753_884.jpg


In addition I have also posted images of the solution for the (ii) part where I am obtaining ψ(0)=0 and the expectation values. I shall be extremely indebted if someone can have a look at my solutions and verify if they are fine, or otherwise, graciously provide guidance so that I may be able to correct them.

IMG_20220428_024819_487.jpg

IMG_20220428_024907_630.jpg
 

Answers and Replies

  • #2
TSny
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Your evaluation of ##\psi(x)## looks correct. There are several different conventions for the Fourier transform that differ in factors of ##2 \pi## or ##\sqrt{2 \pi}##. For example, see section A.1 here.

Calculating ##\psi(0)## by letting ##x = 0## leads to ##\frac{0}{0}## which is ambiguous. Instead, take the limit of ##\psi(x)## as ##x## approaches zero.

Your work for part 2 looks good.
 
  • #3
warhammer
138
28
Your evaluation of ##\psi(x)## looks correct. There are several different conventions for the Fourier transform that differ in factors of ##2 \pi## or ##\sqrt{2 \pi}##. For example, see section A.1 here.

Calculating ##\psi(0)## by letting ##x = 0## leads to ##\frac{0}{0}## which is ambiguous. Instead, take the limit of ##\psi(x)## as ##x## approaches zero.

Your work for part 2 looks good.
Thank you for your guidance sir.

Sir upon using the limit here, since x --> 0, and b is related to x directly, so it should also tend to zero; therefore we simply get ##\frac{A}{√2π}## as the answer for ##\psi(0)## . I hope this is now correct
 
  • #4
TSny
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Sir upon using the limit here, since x --> 0, and b is related to x directly, so it should also tend to zero; therefore we simply get ##\frac{A}{√2π}## as the answer for ##\psi(0)## .
It's not clear how you deduced that ##\frac {\sin b}{b}## → 1 as ##x## → 0. Did you use l'Hôpital's Rule?
 
  • #5
warhammer
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It's not clear how you deduced that ##\frac {\sin b}{b}## → 1 as ##x## → 0. Did you use l'Hôpital's Rule?
No sir. Instead I used the Trig identity for Limits: ## Lt b->0 \frac{sinb}{b} = 1## using the assumption that since ##b =0.5Δkx## ; b-->0 for x-->0 (the proof for the same coming from Sandwich Theorem if I recall correctly)..
 
  • #6
TSny
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Yes, there are different ways to get the limit. Good.

I think your result that ##\psi(0) = \frac{A}{\sqrt {2\pi}}## is correct.
 
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Likes Lnewqban and warhammer
  • #7
warhammer
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Yes, there are different ways to get the limit. Good.

I think your result that ##\psi(0) = \frac{A}{\sqrt {2\pi}}## is correct.
Thank you so much sir for your prompt responses. I'm highly obliged, you took out so much time to carefully evaluate my solution..🙏🏻🧡
 
Last edited:
  • #8
TSny
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You're very welcome. I'm glad I could help.
 

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