Difference between revisions of "Quantum Qual Problems"

From New IAC Wiki
Jump to navigation Jump to search
Line 53: Line 53:
  
  
*  <math>E = \frac{\pi^2 \hbar^2 n^2}{2M^2 a^2}</math>, where <math>n^2=n_x ^2</math>
+
*  <math>E = \frac{\pi^2 \hbar^2 n^2}{2M^2 a^2}</math>, where <math>n^2=n_x ^2 + n_y ^2 + n_z ^2</math>

Revision as of 05:08, 16 August 2007

1.) Given a quantum mechanical particle of mass M confined inside a box of sides a,b,c. The particle is allowed to move freely between 0<x<a,0<y<b and 0<z<c.

  • Use the time-independent Schrodinger equation for this problem to obtain the general form for the eigenfunctions of the particle
  • Now apply boundary conditions to obtain the specific eigenfunctions and eigenenergies for this specific problem.
  • Assume a=b=c and find the first 6 eigenenergies of the problem in terms of the box side length (a), the particle mass (M) and standard constants. What are their quantum number? Make a sketch of the eigenvalue spectrum, a table listing these eigenenergies and the quantum numbers of all the states that correspond to them.


Solution:


2.)


  • [22MΔ2+V]W(x,y,z)=EW(x,y,z)

In our case, using separation of variables, we will get 3 differential equations for x, y and z. W(x,y,z)=w(x)w(y)w(z)

d2w(x)dx2m2w(x)=0(1)
The same will be for y and z.

Solution of equation (1) is following
w(x)=Asin(mx)+Bcos(mx)
w(y)=Csin(ky)+Dcos(ky)
w(z)=Esin(qz)+Fcos(qz)


  • Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have

w(x)=Asin(mx)
w(y)=Csin(ky)
w(z)=Esin(qz)

Also, w(a)=0 which gives Asin(ma)=0,m=πnxa. For y component Csin(kb)=0,k=πnyb and for z Esin(qc)=0,q=πnzc

A, C and E are normalization constants

1A2=sin2(πnx/a)dx=a2, limits are from 0 to a.

The eigenfunction for each component will be

w(x)=2asin(πnxx/a)
w(y)=2bsin(πnyy/b)
w(z)=2csin(πnzz/c)

The eigenenergies

Enx=π22nx22Ma2, Eny=π22ny22Mb2, Enz=π22nz22Mc2
Total energy is sum of these energies.


  • E=π22n22M2a2, where n2=n2x+n2y+n2z