Uncertainty in particle position

MaxJ · Aug 13, 2024

For this problem,

Problem text: a particle is in the state psi(x) = sqrt(2/a) sin(pi x/a) for 0 <= x <= a, and 0 otherwise; asks which option gives the uncertainty Delta x in position.

My solution is

Typeset derivation of position uncertainty for the wavefunction psi(x) = sqrt(2/a) sin(pi(x-a)/a) on 0 to a, zero otherwise, showing the expectation values of x and x squared and the trigonometric identity for sin squared.

Handwritten-style typeset derivation of the expectation value <x> using integrals from 0 to a, including a cosine term, an integration by parts step, and the evaluated result a/2 for the first integral.

Handwritten-style typeset derivation computing the position uncertainty of a particle: <x> = a/2, <x^2> = a^2/3, and Delta x = sqrt(a^2/3 - a^2/4) = a/(2 sqrt 3).

However, the my answer for ##\Delta x## is not correct. The correct answer is ##\Delta x = 0.18a##. Does someone please know what I have done wrong?

Kind wishes

Orodruin · Aug 13, 2024

MaxJ said:

what I have done wrong?

You have hand waved "zero by symmetry" where what you have waved away is not zero by symmetry. Why do you think it should be zero and what symmetry would there be?

MaxJ · Aug 13, 2024

Orodruin said:

You have hand waved "zero by symmetry" where what you hade waved away is not zero by symmetry. Why do you think it should be zero and what symmetry would there be?

Be blessed Sir.

I have doubt what you mean in regards to symmetry. Can you explain some more?

Orodruin · Aug 13, 2024

MaxJ said:

Be blessed Sir.

I have doubt what you mean in regards to symmetry. Can you explain some more?

You were the one trying to invoke symmetry arguments so you should be able to tell me why you think the integral should be zero "due to symmetry".

Uncertainty in particle position

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