1D Harmonic oscillator

In summary, the probability of finding the particle in the ground state of the new potential is = integral[wavefunction old*wavefunction new]dx.
  • #1
kl14
3
0

Homework Statement


particle in ground state of 1D harmonic oscillator - spring constant is doubled - what is the probability of finding the particle in the ground state of the new potential



Homework Equations


v=1/2kx^2 oscillator potential
wavefunction ground state n=0 = (alpha/pi)^1/4*e^[(-apha*x^2)/2]
alpha = sqroot[(k*m)/hbar^2)]


The Attempt at a Solution



the probability will be = integral[wavefunction old*wavefunction new]dx

initial equation d^2psi/dx^2 + [2m(E-1/2kx^2)/hbar]psi = 0
if you double k in the potential V(x)- the equation is d^2psi/dx^2 + [2m(E-kx^2)/hbar]psi = 0

does this change alpha to 2k? this makes the integral very complex to solve.
or can this be done by changing the wavefunction for the new potential to (alpha/2pi)^1/4*e^[(-apha*x^2)/2]

the change being 2pi - doubling the range of the spring constant?
Thanks!
 
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  • #2
kl14 said:
does this change alpha to 2k?

No. How is alpha related to k? What happens to alpha if k is doubled? Does it stay the same? Double? Or what?
 
  • #3
alpha = sqrt(k*reduced mass/hbar^2)
so doubling k would just be alpha = sqrt(2k*reduced mass/hbar^2)

so if i take the 2 out it is basically just multiplying alpha by sqrt(2). and substituting this into my wave function would give me the wave function for the new potential. does that make sense?
 
  • #4
Yes. So now you have the two wave functions. Earlier, you wrote

the probability will be = integral[wavefunction old*wavefunction new]dx

but this is not quite right. The integral gives you the amplitude; what do you need to do to get the probability?
 
  • #5
kl14 :

Your original quesion was how to solve the integral. Avodyne pointed out an error you did with the sqrt(2).

But to calculate the integral, you get something like:
[tex] k_1 k_2 e^{-a_1 x^2 }\cdot e^{-a_2 x^2} = K \cdot e^{-(a_1 +a_2 )x^2} [/tex]
as integrand, where a_1 is the old number you have in the old wave function and so on. There is not any harder to calc this integral then the ordinary one, just make the substitution (a_1 + a_2) = A, and integrate from x=-inf to +inf; that integral exists in standard integral-tables.
 
  • #6
yes, i realize my error in typing - the probability is the amplitude squared - the value obtained after the integration.
 

1. What is a 1D harmonic oscillator?

A 1D harmonic oscillator is a physical system where a particle is subject to a restoring force that is directly proportional to its displacement from equilibrium. It follows a sinusoidal motion, with the particle oscillating back and forth around its equilibrium position.

2. What is the equation of motion for a 1D harmonic oscillator?

The equation of motion for a 1D harmonic oscillator is given by F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement from equilibrium. This equation is known as Hooke's Law.

3. How does the mass of the particle affect the motion of a 1D harmonic oscillator?

The mass of the particle affects the motion of a 1D harmonic oscillator by changing its period of oscillation. A lighter particle will have a shorter period, while a heavier particle will have a longer period. However, the frequency of oscillation, which is determined by the spring constant, remains constant regardless of the mass.

4. What is the potential energy of a 1D harmonic oscillator?

The potential energy of a 1D harmonic oscillator is given by U = 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. This potential energy function forms a parabolic curve, with the minimum point representing the equilibrium position of the oscillator.

5. How does the amplitude affect the motion of a 1D harmonic oscillator?

The amplitude of a 1D harmonic oscillator is the maximum displacement from equilibrium. It affects the motion of the oscillator by determining the maximum kinetic energy and maximum potential energy of the system. A larger amplitude results in a higher maximum kinetic energy and potential energy, leading to a more energetic and wider oscillation. However, the frequency and period of the oscillator remain constant regardless of the amplitude.

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