- #1

Irishdoug

- 102

- 16

- Homework Statement:
- Hi, I am self-studying QM and was wondering if you could check to see if my answers are right to an online exam I took. There are no solutions availabel

- Relevant Equations:
- See below

**Homework Statement:**Hi, I am self-studying QM and was wondering if you could check to see if my answers are right to an online exam I took. There are no solutions availabel

**Homework Equations:**See below

**A particle with mass m is in an infinite potential well of length x=0 to x=a**

Q1a. Show explicitly that the wave-function ##\phi_{n}## = ##\frac{\sqrt{2}}{a}## sin(n##\pi##x/a) are eigenfunctions of the Hamiltonian of this system H= ##\frac{\hbar^2}{2m}## ##\frac{d^2}{dx^2}## and determine the corresponding eigen-values.

Q1a. Show explicitly that the wave-function ##\phi_{n}## = ##\frac{\sqrt{2}}{a}## sin(n##\pi##x/a) are eigenfunctions of the Hamiltonian of this system H= ##\frac{\hbar^2}{2m}## ##\frac{d^2}{dx^2}## and determine the corresponding eigen-values.

##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi##

##\frac{\hbar^2}{2m}## ##\frac{d^2 sin(n\pi x/a)}{dx^2}## = ##\frac{\sqrt{2}}{a}## ##\frac{-\hbar^2}{2m}## ##\frac{n \pi x}{a}## = ##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}##

so, ##\frac{\hbar^2}{2m}## ##\frac{d^2\phi}{dx^2}## = E##\phi## where E=##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}##

The eigen-values are thus E= ##\frac{\hbar^2 \pi^2 n^2 \sqrt{2}}{2ma^3}## were n = 0,2,4... as sine is an even function.

**b. What is the physical interpretation of ##\phi^2##?**

This is the probability of finding that particle in a certain location x, where x is between 0 and a.

**c. What is the physical reason to demand that the wavefunction be normalised?**

The physical reason is that QM is probabilistic in nature. A probability cannot be greater than 1. Therefore the wavefunction is normalized, i.e.

##\int_{-\infty}^{\infty} \phi^* \phi dx## = 1

to ensure this probability doesn't exceed one, which would be unphysical.

**2. At time t=0 a particle is in an infinite potential well of width a prepared in a state ##\phi## = Nx(a-x)**

a. Determine N so the wavefunction is properly normalized

a. Determine N so the wavefunction is properly normalized

##\int_{-\infty}^{\infty} \phi^* \phi dx## = 1 --> ##\int_{-\infty}^{\infty} N^2(ax-x^2)^2 dx##

Take N outside, the well is symmetrical so the limits of integration become 0 to a, and the integral is multiplied by 2.

End up with:

2##N^2## (##\frac{a^2a^3}{3} -\frac{aa^4}{2}+\frac{a^5}{5}##) = 2##N^2##*##\frac{a}{30}## so N = ##\sqrt{\frac{15}{a}}##

**b. Although it's not an eigenfunction of the infinite potential well Hamiltonian, this wavefunction can be expanded in the eigenfunctions of this Hamiltonian**

##\psi_{x} = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)##

where ##c_{n}## are the expansion coefficients and ##\phi_{n}## the infinite potential well wavefunctions. Calculate ##c_{1}##.

##\psi_{x} = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x)##

where ##c_{n}## are the expansion coefficients and ##\phi_{n}## the infinite potential well wavefunctions. Calculate ##c_{1}##.

##\phi(x)## = ##\sqrt{\frac{15}{a}} x(a-x) ## ; ##c_{n} = \int_{-\infty}^{\infty} x\phi_{n}##

so, ##c_{n}## = ##\int_{-\infty}^{\infty} \sqrt{\frac{15}{a}} x^2(ax-x^2)##

Assume symmetrical so limits of integration go from 0 to a and integral is muliplied by 2

End up with 2##\sqrt{\frac{15}{a}}(\frac{a^5}{4}-\frac{a^5}{5})## = 2##\sqrt{\frac{15}{a}}(\frac{a^5}{20})## = ##\sqrt{\frac{15}{a}}(\frac{a^5}{10})## = ##c_{1}##

**c. Find the probability that the particle is found in the n=1 state when it's energy is measured.**

For this one I carried out the expectation value for E i.e.

<E> = ##\int_{-\infty}^{\infty} \frac{-\hbar^2}{2m} \frac{d^2}{dx^2}(\sqrt{\frac{15}{a}}x(a-x)dx##

I ended up with ##\frac{a\hbar^2}{m} \sqrt{\frac{15}{a}}##

For the last two questions I was more guessing at the methodology however. Thanks for any input.