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

[warhammer]

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..

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.

 

[TSny]

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.

 

[warhammer]

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

 

[TSny]

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?

 

[warhammer]

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)..

 

[TSny]

Yes, there are different ways to get the limit. Good.

I think your result that $\psi(0) = \frac{A}{\sqrt {2\pi}}$ is correct.

 

[warhammer]

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..

 

[TSny]

You're very welcome. I'm glad I could help.