Difference between revisions of "Quantum Qual Problems"

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<math>w(z) = E\sin(cz) </math><br>
 
<math>w(z) = E\sin(cz) </math><br>
  
Also, w(a)=0 which gives <math>A\sin(cx)=0, cx=\phi n</math>
+
Also, w(a)=0 which gives <math>A\sin(cx)=0, cx=\pi n</math>

Revision as of 02:57, 16 August 2007

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

  • 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 [math]a=b=c[/math] and find the first 6 eigenenergies of the problem in terms of the box side length ([math]a[/math]), the particle mass ([math]M[/math]) 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.)

  • [math] [- \frac{h^2}{2m}\Delta^2 + V]W(x,y,z)=E W(x,y,z) [/math]

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)

[math]\frac{d^2 w(x)}{dx^2} - c^2 w(x) = 0[/math](1)
The same will be for y and z.

Solution of equation (1) is following
[math]w(x) = A\sin(cx)+B\cos(cx)[/math]
[math]w(y) = C\sin(cy)+D\cos(cy)[/math]
[math]w(z) = E\sin(cz)+F\cos(cz)[/math]

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

[math]w(x) = A\sin(cx) [/math]
[math]w(y) = C\sin(cy) [/math]
[math]w(z) = E\sin(cz) [/math]

Also, w(a)=0 which gives [math]A\sin(cx)=0, cx=\pi n[/math]