P(x)dx of a one-dimensional harmonic oscillator

In summary, the conversation is discussing the probability of a harmonic oscillator's displacement and the total energy of the oscillator. The probability of the oscillator's displacement is expressed in terms of the phase angle and the area in phase space. The conversation also mentions using the phase space method to solve the problem, but the person is having trouble setting up the necessary equations. The solution to the problem is dx/(Pi Sqrt(A^2 - x^2)) when summing W(Phi) over all angles.
  • #1
Buce
2
0
The displacement of a harmonic oscillator is given by

x = A Cos(wt + Phi)

The phase angle phi is equally likely to have any value in the range 0 < Phi < 2Pi, so the probability W(Phi) that Phi lies between Phi and Phi + delta-Phi is delta-Phi/(2Pi). For a fixed time t, find the probability P(x)dx that x lies between x and x+dx by summing W(Phi)delta-Phi over all angles for which x lies in this range. Express P(x) in terms of A and x.

I know that for a harmonic oscillator, the total energy is

E = p^2/(2m) + kx^2/2

where the momentum p is

p = mx' = -mAwSin(wt + Phi)

I can solve the problem using px phase space, where the area between E and E+dE forms a elliptical band of area s, and P(x)dx = (dp/dE)(dE/ds)dx, but I'm only supposed to do it that way for the second part of the problem (which I won't reproduce here). If I did that part correctly, the answer is dx/(Pi Sqrt(A^2 - x^2)), but I'm having a hard time setting up the equations needed to evaluate an integral with respect to the phase angle.
 
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  • #2
i've got the exact reverse problem. i can get your result by summing W(phi) over all angles for which x lies, but I'm off by a factor of 1/(mw^2A) when i try the ratio of areas in phase-space. let's help each other out
 

Related to P(x)dx of a one-dimensional harmonic oscillator

1. What is the significance of the "P(x)dx" in the one-dimensional harmonic oscillator equation?

The "P(x)dx" term in the equation represents the probability of finding the harmonic oscillator at a specific position, within the interval between x and x+dx. It is a measure of the likelihood of the oscillator being at a particular location.

2. How is the "P(x)dx" term related to the wave function of the harmonic oscillator?

The wave function of the harmonic oscillator, denoted by Ψ(x), is related to the probability density function, P(x), through the equation P(x) = |Ψ(x)|^2. This means that the square of the magnitude of the wave function gives the probability density at a given position.

3. What is the physical interpretation of the "P(x)dx" term in the context of a harmonic oscillator?

The "P(x)dx" term represents the probability of finding the harmonic oscillator at a specific position, x, within a small interval of width dx. This can be thought of as the chance of observing the oscillator at a particular point as it undergoes oscillations.

4. How does the "P(x)dx" term change with respect to time in a one-dimensional harmonic oscillator?

As time passes, the probability density function, P(x), will change as the wave function of the oscillator evolves. This change is governed by the Schrödinger equation, which describes the time evolution of the wave function. Therefore, the "P(x)dx" term will also change with time.

5. How is the "P(x)dx" term used in calculating the expectation value of the position for a harmonic oscillator?

The expectation value of the position, denoted by ⟨x⟩, is calculated by integrating the product of the position operator, x, and the probability density function, P(x), over all possible positions. This integral can be expressed as ⟨x⟩ = ∫xP(x)dx. Therefore, the "P(x)dx" term is essential in determining the expectation value of the position for a harmonic oscillator.

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