Harmonic Oscillator: Evaluating Ground State Probability

In summary, the problem asks for the probability of finding an oscillator in the ground state beyond the classical turning points ±x0. To solve this, we use the normalization constant A and the value of a to find the probability at both tails of the distribution. However, there is an inconsistency in the question and there are two possible interpretations. One interpretation is to find the probabilities at both tails, while the other interpretation is to find one number, P(|x|>x0). The upper limit on the first integral in the first interpretation is incorrect and should be -0.1 instead of 0.
  • #1
glebovg
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Homework Statement



Using the normalization constant A and the value of a, evaluate the probability to find an oscillator in the ground state beyond the classical turning points ±x0. Assume an electron bound to an atomic-sized region (x0 = 0.1 nm) with an effective force constant of 1.0 eV/nm2.

Homework Equations



[itex]\psi(x)=Ae^{-ax^{2}}[/itex], where [itex]A=(\frac{m\kappa}{\pi^{2}\hbar^{2}})^{1/8}[/itex] and [itex]a=\sqrt{m\kappa}/2\hbar[/itex]

The Attempt at a Solution



How do I find m?
 
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  • #2
The mass of the electron? You look it up.
 
  • #3
or you could weigh one …

if you have one on you o:)
 
  • #4
I will try to read the question more carefully next time.
 
  • #5
Am I supposed to find probabilities at both tails of the distribution?
 
  • #6
Yes.
 
  • #7
Apparently there is an inconsistency in the question and there are two interpretations of the question. What would be the other one?
 
  • #8
Beats me.
 
  • #9
My interpretation: find probabilities at both tails of the distribution i.e. [itex]\int^{0}_{-\infty}\psi^{2}(x)\;dx[/itex] and [itex]1-\int^{0.1}_{-\infty}\psi^{2}(x)\;dx[/itex].
 
  • #10
I'd say the problem is asking for one number, P(|x|>x0). The upper limit on your first integral is wrong.
 
  • #11
Do you mean it is wrong according to your interpretation or it is generally wrong? Should it be 0.1?
 
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  • #12
It is generally wrong. Why would the upper limit be 0?
 
  • #13
It should be -0.1, right?
 
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FAQ: Harmonic Oscillator: Evaluating Ground State Probability

1. What is a harmonic oscillator?

A harmonic oscillator is a system in which the force acting on an object is directly proportional to its displacement from its equilibrium position. This results in a periodic motion of the object, with the object oscillating back and forth around its equilibrium position.

2. How is the ground state probability of a harmonic oscillator evaluated?

The ground state probability of a harmonic oscillator can be evaluated using the wave function, which describes the probability amplitude of finding the oscillator in a particular state. The square of the wave function gives the probability of finding the oscillator in that state.

3. What factors affect the ground state probability of a harmonic oscillator?

The ground state probability of a harmonic oscillator is affected by the spring constant, the mass of the object, and the energy of the oscillator. A higher spring constant and lower mass will result in a higher ground state probability, while a higher energy will decrease the probability.

4. How does the ground state probability change as the energy of the oscillator increases?

As the energy of the oscillator increases, the ground state probability decreases. This is because higher energy states have a higher probability of being occupied, leaving less probability for the ground state.

5. What is the significance of evaluating the ground state probability of a harmonic oscillator?

Evaluating the ground state probability of a harmonic oscillator allows us to understand the distribution of energy in the system and make predictions about the behavior of the oscillator. It also has applications in various fields such as quantum mechanics, statistical mechanics, and thermodynamics.

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