 |
|
| Science Forum Index » Mathematics Forum » Angular Momentum Operator |
|
Page 1 of 1 |
|
| Author |
Message |
| Farooq W |
 Posted: Sun Jul 16, 2006 1:44 pm |
|
|
|
Guest
|
I would share to share that J.Chem.Ed has an excellent Living Textbook
"Quantum States of Atoms and Molecules" which can be accessed
online.
http://jchemed.chem.wisc.edu/JCEDLib/LivTexts/index.html
I am using this for self studying to grasp the basics. However there is
a question for which need some push. Chapter 3, problem 3.15
http://jchemed.chem.wisc.edu/JCEDLib/LivTexts/pChem/JCE2005p1880_2LTXT/QuantumStates/Chapterwindows/Chapter3window.htm
Q (sic): Write the wavefunction of an electron moving with an electron
moving in the z direction with energy of 100 eV. The form of the
wavefunction is exp (ikx). You need to find the value for k. Obtain the
electron's momentum by operating on the wavefunction with the
momentum operator.
I have tried to do this way, but since there are no answers to check
with, I am not satisfied (thinking of some misconception here).
Symbols: k - wavevector and "i" is the sqrt(-1), psi = Greek
letter psi.
Let us assume psi(z) = exp (ikz)
Momentum operator on psi(z) gives = kh/2pi * exp(ikz)
Which I assume means that eigenvalue pz = hk/2pi
The value of 100 eV then converted to 1.16 x 10^-17 J and plugged in
kinetic energy T, expression
T = pz^2/ 2 m = (hk)^2 / 2m to obtain k = 5 x 10 ^10
This is "the" point which seem uncomfortable... I am mixing up a
classical expression T = p^2/2 m with a QM one; pz = hk/2pi to
determine the value of k. Is it okay to do that? Is there an
alternative way to do that?
Secondly, there is another question of evaluating an integral where
psi(x)* psi(x) are involved. Whereas psi(x) = sin kx. What would the
conjugate of a real trigonometric function i.e. what should be psi*(x)
?
Regards,
M. Farooq |
|
|
| Back to top |
|
|
|
| Mark Kness |
| Posted: Sun Jul 16, 2006 2:06 pm |
|
|
|
Guest
|
Farooq W wrote:
[quote:23e8816296]Secondly, there is another question of evaluating an integral where
psi(x)* psi(x) are involved. Whereas psi(x) = sin kx. What would the
conjugate of a real trigonometric function i.e. what should be psi*(x)
[/quote:23e8816296]
If there is no imaginary part, then taking the complex conjugate (what
I assume you mean by *) means doing nothing, so psi(x)* and psi(x) are
exactly the same.
The rest of your stuff seemed ok to me, although I didn't check the
numbers. The 'quantum mechanical' way to think of the kinetic energy
is as an operator equal to p^2/2m, and since p = -i * hbar * d/dx, then
p^2 = - hbar^2 * d^2/dx^2, so kinetic energy = (-hbar^2/2m) * d^2/dx^2. |
|
|
| Back to top |
|
|
|
| Salmon Egg |
| Posted: Sun Jul 16, 2006 3:02 pm |
|
|
|
Guest
|
On 7/16/06 10:06 AM, in article
1153069573.254027.44220@m79g2000cwm.googlegroups.com, "Mark Kness"
<mkness21@yahoo.com> wrote:
[quote:2580422332]
Farooq W wrote:
Secondly, there is another question of evaluating an integral where
psi(x)* psi(x) are involved. Whereas psi(x) = sin kx. What would the
conjugate of a real trigonometric function i.e. what should be psi*(x)
If there is no imaginary part, then taking the complex conjugate (what
I assume you mean by *) means doing nothing, so psi(x)* and psi(x) are
exactly the same.
The rest of your stuff seemed ok to me, although I didn't check the
numbers. The 'quantum mechanical' way to think of the kinetic energy
is as an operator equal to p^2/2m, and since p = -i * hbar * d/dx, then
p^2 = - hbar^2 * d^2/dx^2, so kinetic energy = (-hbar^2/2m) * d^2/dx^2.
For real x, sin(x) is its own conjugate. If in doubt, go back to sin(z) =[/quote:2580422332]
(exp(j*z)-exp(-j*z)/(2*j*z).
Bill
-- Ferme le Bush |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Thu Nov 26, 2009 2:26 am
|
|