Harmonic oscillator outside classically allowed region

Click For Summary
In the discussion about the harmonic oscillator's ground state, the main focus is on calculating the probability of finding the particle outside the classically allowed region. The probability is derived by subtracting the integral of the wave function from zero to the classical energy limit from the total probability of 1/2. However, the integral of e^(-a*x^2) from 0 to the classical energy cannot be solved analytically. Participants suggest using numerical methods or consulting error function tables for accurate results. Ultimately, numerical solutions are necessary for this type of probability calculation.
yukikokami
Messages
6
Reaction score
0
the problem is as follows: in the ground state of the harmonic oscillator what is the probabilty of finding the particle outside the classically allowed region. where the classically allowed nrg is given by E=(1/2)m*omega^2*a^2 (where a is the amplitude).

were given that psi(x)=(m*omega/pi*h-bar)^(1/4)*(2^n*n!)^(-1/2)*H(zeta)*e^(-zeta^2/2)
where H(zeta) depends on n, but for this problem n=0

so i figured that if the probabilty of this from 0 to infinity (which is 1/2) then subtract the probability from 0 to the classical nrg, then the remainer should be the probablity of finding it outside the said classical region. right?

but that leads to the problem of finding the integral of e^(-a*x^2) from 0 to classical nrg... and this doesn't seem possible... is it?

cheers!
 
Physics news on Phys.org
yukikokami said:
the problem is as follows: in the ground state of the harmonic oscillator what is the probabilty of finding the particle outside the classically allowed region. where the classically allowed nrg is given by E=(1/2)m*omega^2*a^2 (where a is the amplitude).

were given that psi(x)=(m*omega/pi*h-bar)^(1/4)*(2^n*n!)^(-1/2)*H(zeta)*e^(-zeta^2/2)
where H(zeta) depends on n, but for this problem n=0

so i figured that if the probabilty of this from 0 to infinity (which is 1/2) then subtract the probability from 0 to the classical nrg, then the remainer should be the probablity of finding it outside the said classical region. right?

but that leads to the problem of finding the integral of e^(-a*x^2) from 0 to classical nrg... and this doesn't seem possible... is it?

cheers!

It's not possible analytically, it must be done numerically. Look up a table of the "error function" (be careful, depending on the convention used in the book you pick up, you might have to do some work to convert th eresult of the table into the form you need) or use Maple/Mathematica.

Patrick
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K