Difference between revisions of "Quantum Qual Problems"
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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)<br> | 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)<br> | ||
− | <math>\frac{d^2 w(x)}{dx^2} - | + | <math>\frac{d^2 w(x)}{dx^2} - m^2 w(x) = 0</math>(1)<br> |
The same will be for y and z.<br> | The same will be for y and z.<br> | ||
Solution of equation (1) is following <br> | Solution of equation (1) is following <br> | ||
− | <math>w(x) = A\sin( | + | <math>w(x) = A\sin(mx)+B\cos(mx)</math><br> |
− | <math>w(y) = C\sin( | + | <math>w(y) = C\sin(ky)+D\cos(ky)</math><br> |
− | <math>w(z) = E\sin( | + | <math>w(z) = E\sin(qz)+F\cos(qz)</math><br> |
* Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have<br> | * Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have<br> | ||
− | <math>w(x) = A\sin( | + | <math>w(x) = A\sin(mx) </math><br> |
− | <math>w(y) = C\sin( | + | <math>w(y) = C\sin(ky) </math><br> |
− | <math>w(z) = E\sin( | + | <math>w(z) = E\sin(qz) </math><br> |
− | Also, w(a)=0 which gives <math>A\sin( | + | Also, w(a)=0 which gives <math>A\sin(ma)=0, m=\frac{\pi n}{a}</math>, <math>C\sin(kb)=0, k=\frac{\pi n}{b}</math>, <math>E\sin(qc)=0, q=\frac{\pi n}{c}</math> |
Revision as of 03:01, 16 August 2007
1.) Given a quantum mechanical particle of mass
confined inside a box of sides . The particle is allowed to move freely between and .- 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 and find the first 6 eigenenergies of the problem in terms of the box side length ( ), the particle mass ( ) 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.)
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)
The same will be for y and z.
Solution of equation (1) is following
- Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have
Also, w(a)=0 which gives
, ,