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Help - Quantum Mechanics

  • Thread starter Fjolvar
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  • #1
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Hello,

I am having some difficulty with my Quantum Mechanics homework so I will post the questions and answer them to the best of my ability, which is still at a beginners level since this subject is new to me. I will update my answers as I figure them out. Any elaboration on the questions or my answers is greatly appreciated. Thank you.

1. Answer the following in brief sentences:

a. The time independent Schrodinger Equation applies to only a particular class of systems. The Hamiltonians of all these systems have a common property. What is this property?

The Hamiltonians do not depend on time.

b. If the Hamiltonian satisfies the equation: H(x) = H(-x), what can we say about the eigenfunctions of the Hamiltonian?
If [tex]\Phi[/tex](x) is a solution, so is [tex]\Phi[/tex](-x).

c. A wave equation is solved and a relationship between the frequency [tex]\omega[/tex] and the wave number k (=2*pi/[tex]\lambda[/tex], with [tex]\lambda[/tex] being the wavelength) is obtained. A wave packet is formed using waves with k values near k0. What is the speed of the wave packet?
I'm not sure how to approach this question.

2. Consider an Infinite one-dimensional potential well:

A particle of mass m is released in the well at time t=0 with its initial state given by the wave function.

a. A measurement of its position is made, what are the possible results for x? For each measured value, what is the probability density?
A measurement for x is possible between its boundaries such as 0<x<a.
I'm not sure how to calculate the probability density.

b. If instead of its position, its energy is measured, What are the possible results for E? For each measured value of E, What is the probability?
I don't know.

c. Does <x (t)> depend on time? Calculate <x (t)>.
No. <x> = a/2

d. Does <E (t)> depend on time? Calculate <E (t)>.
No. E = n^2*pi^2*hbar^2 / 2*m*a^2

e. If the particle is in the ground state and its momentum is measured, what are the possible results of the measurement? What is the associated probability for each measured value?
I don't know.
 
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Answers and Replies

  • #2
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Sorry if there is alot left unanswered, but I'm trying my best..
 
  • #3
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the expectation value of an operator is

[tex]\left\langle\ O \right\rangle = \int\psi^{*}(x)O\psi(x)[/tex]

where [tex]\widehat{x}[/tex] is the position operator
the Hamiltonian is the energy operator and momentum operator for momentum

the possible values are the eigenvalues of the operator e.g. [tex] \widehat{H}\psi = E\psi[/tex]

for an infinite potential well the eigenvalues of the Hamiltonian are [tex]E_{n}=\frac{\hbar^{2}(k_{n})^{2}}{2m}[/tex] and since for an infinite potential well [tex] \widehat{H} = \frac{\widehat{p}^{2}}{2m} \Rightarrow p_{n} = \hbar k_{n}[/tex]

and <E> does not depend on time
 
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  • #4
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To calculate the probability density, do you use the coefficient equation cn = [tex]\int[/tex][tex]\Psi[/tex]n(x)* f(x) dx ?
 
  • #5
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no the probability density is just [tex]\left|\psi(x,t)\right|^{2}[/tex] no integral
 
  • #6
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no the probability density is just [tex]\left|\psi(x,t)\right|^{2}[/tex] no integral
What about the probability for E and p?
 
  • #7
213
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for those you do use the coefficient equation. The probability is the modulus square of the coefficient. Plus [tex] \left\langle E \right\rangle = \sum_{n} \left|c_{n}\right|^{2} E_{n}[/tex] what you have down for that i.e. [tex]E_{n}[/tex] are the possible results of the measurement and these have the same probabilities as the possible values of the momentum
 
  • #8
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Can anyone help me with 1 part c and 2 part e?
 
  • #9
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I'm having a hard time seeing the difference between 2a. and part b. with 2 c. and part d..

It asks the possible results for x and then find <x> along with E and <E>. It seems one is a single measurement while the other is the average of measurements, but I'm not sure what equations are that pair up with these concepts.

I have an exam soon, so any advice is much appreciated. Thanks in advance.
 
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  • #10
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the expectation value of an operator is

[tex]\left\langle\ O \right\rangle = \int\psi^{*}(x)O\psi(x)[/tex]

where [tex]\widehat{x}[/tex] is the position operator
the Hamiltonian is the energy operator and momentum operator for momentum

the possible values are the eigenvalues of the operator e.g. [tex] \widehat{H}\psi = E\psi[/tex]

for an infinite potential well the eigenvalues of the Hamiltonian are [tex]E_{n}=\frac{\hbar^{2}(k_{n})^{2}}{2m}[/tex] and since for an infinite potential well [tex] \widehat{H} = \frac{\widehat{p}^{2}}{2m} \Rightarrow p_{n} = \hbar k_{n}[/tex]

and <E> does not depend on time
you can use what was suggested in the previous post to find <x> and <E>, substituting the operator O with x to find <x> and E to find <E> .. and of course, the wavefunction to be used in this situation should be the wavefunction of an infinite one-dimensional pontential well (as far as I remember the wavefunction for a boundary 0<x<a should be a sine function) .. Good luck!
 
  • #11
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For part 2e. do I just plug in the momentum operator p= hbar /i * d/dx? Or do I use [tex]
\widehat{H} = \frac{\widehat{p}^{2}}{2m} \Rightarrow p_{n} = \hbar k_{n}
[/tex]? and just set n=1 since its in the ground state?
 
  • #12
213
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[tex]p_{n} = \pm\hbar k_{n}[/tex] since the general solution for the infinite potential well is [tex]Ae^{ikx}+Be^{-ikx}[/tex] i.e. a wave with momentum +p and another with -p each with equal probability of a half
 

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