Quantum mechanics, harmonic oscillator

[fluidistic]

Homework Statement

Consider a classical particle in an unidimensional harmonic potential. Let A be the amplitude of the oscillation of the particle at a given energy. Show that the probability to find the particule between $x$ and $x+dx$ is given by $P(x)dx=\frac{dx}{\pi \sqrt {A^2-x^2}}$.
1)Graph P(x) and compare it with the probability to find a quantum particle between x and dx for the first state of the unidimensional harmonic quantum oscillator.
2)What is the probability to find the classical and quantum particles in some position between 0 and A/2?
3)Graph the density of probability for the state n=10 of the harmonic quantum oscillator and compare it with the classical density of probability of the particle for this same energy.

Homework Equations

Details in next section.
$T=\frac{2 \pi}{\omega}$.

The Attempt at a Solution

So my strategy to show the mathematical relation of P(x) is to write find what percentage of its time the particle passes from x to dx compared to half its period (the particle "runs" a distance of A in that time).
So I start with $x(t)=A\cos (\omega t)$. I want to get dt in function of dx. So I find t in function of x first. This gives me $t= \left [ \arccos \left ( \frac{x}{A} \right ) \right ] \frac{1}{\omega}$.
Now I take the differential $d/dx$ in both sides of this equation:
$\frac{dt}{dx}= \left ( \frac{1}{\omega} \right ) \frac{d}{dx}\left [ \arccos \left ( \frac{x}{A} \right ) \right ]=-\frac{1}{\omega} \left [ \frac{1}{\sqrt {A^2-x^2}} \right ]=-\frac{1}{\omega} \left ( \frac{1}{\sqrt {A^2-x^2}} \right )=-\frac{T}{2\pi} \left ( \frac{1}{\sqrt {A^2-x^2}}\right ) \Rightarrow dt =- \frac{T}{2 \pi} \left ( \frac{1}{\sqrt {A^2-x^2}} \right ) dx$.
But I know that if $t_1-t_0=T/2$, $x_0=0$ and $x_1=A$ can be associated to $t_0$ and $t_1$ respectively. So that I can write $\frac{T}{2}=-\frac{T}{2\pi} \int _0 ^A \frac{dx}{\sqrt {A^2-x^2}} \Rightarrow 1=-\frac{1}{\pi} \cdot \frac{\pi}{2}=-\frac{1}{2}$ which is obvisouly wrong.
I don't where I went wrong. I hope someone can help me.

 

[ehild]

.
But I know that if $t_1-t_0=T/2$, $x_0=0$ and $x_1=A$ can be associated to $t_0$ and $t_1$ respectively. So that I can write $\frac{T}{2}=-\frac{T}{2\pi} \int _0 ^A \frac{dx}{\sqrt {A^2-x^2}}$

You set t₀=0 and t₁=T/2. What is cos(wt) at t=0 and and at t=T/2?

ehild

 

[fluidistic]

You set t₀=0 and t₁=T/2. What is cos(wt) at t=0 and and at t=T/2?

ehild

Oh nice... Now I see my error. I get 1 for both. So rather than t=T/2 I must pick t=T/4.
Now I reach 1=-1 though.
$T/4=-\frac{T}{2\pi} \int _0 ^A \frac{dx}{\sqrt {A^2-x^2}} \Rightarrow 1=-1$.

Edit: Probably I should inverse the limits of integration... but still I don't really understand why it's not the same.

 

[fluidistic]

I'm somehow at a loss.
I know that in a time t=T/4, the particle runs a distance A.
In a time t=dt the particle runs a distance in terms of an [arctan (fraction involving a square root ) ] and differential x. I'm not sure about this though.
I've been 2 more hours on this problem, trying to reach the result but I don't get it.
I tried dt/(T/4) to get the probability for the particle to be between x and x+dx but this didn't work well. My answer differs from a factor of -1/2 from the answer if I do this.
I tried other stuff but senless I think. I'm lost.

 

[ehild]

You integrated the time from t=0 to t=T/2.
At t=0, x = Acos(w*0)=A, as cos(0)=1
At t=T/2, x==Acos(w*T/2) Acos(pi)=-A, as cos(pi)=-1.
You have to integrate from x=A to x=-A.

By the way, you can choose x=Asin(wt) and then you do not have problems with the minuses.

ehild

 

[fluidistic]

You integrated the time from t=0 to t=T/2.
At t=0, x = Acos(w*0)=A, as cos(0)=1
At t=T/2, x==Acos(w*T/2) Acos(pi)=-A, as cos(pi)=-1.
You have to integrate from x=A to x=-A.

By the way, you can choose x=Asin(wt) and then you do not have problems with the minuses.

ehild

Yes I realized this. What I did is integrating dt from 0 to T/4 and the expression with dx from A to 0. Now I get 1=1 as it should.
But I'm totally lost after this.
I thought that the probability would be given by $dt/(T/4)=\frac{4dt}{T}=-\frac{2}{\pi} \frac{dx}{\sqrt {A^2-x^2}}$, which is as I said in my previous post, different from the answer by a factor -1/2.

 

[ehild]

The probability of finding the particle between x and x+dx is the length of the time interval during the particle is in the desired x interval, divided by the time period. Draw a cosine curve from t=0 to T. How long is the particle between x and x+dx? Is it only one interval? Do the relative signs of dx and dt count?

ehild

 

[fluidistic]

The probability of finding the particle between x and x+dx is the length of the time interval during the particle is in the desired x interval, divided by the time period. Draw a cosine curve from t=0 to T. How long is the particle between x and x+dx? Is it only one interval? Do the relative signs of dx and dt count?

ehild

Oh thank you, that's right.
I drew the sketch and the particle passes twice at every position (but one), the sign doesn't matter and I reach the desired result.
Now trying to solve 1, 2 and 3.
For the quantum oscillator, I think that $\Psi (x)=(km)^{1/8} \hbar ^{-1/4}\pi ^{-1/2} e^{-\frac{(km)^{1/2}}{2\hbar}x^2}$. I reached this by solving part 1) of https://www.physicsforums.com/showthread.php?t=442195. (that was 1 year ago but I redid all the algebra now).
So that $\int _{x}^{x+dx}|\Psi (x)|^2dx= (km)^{1/4}\hbar ^{-1/2} \pi ^{-1} e^{-\frac{\sqrt {km}}{\hbar }x^2}dx$.
What I expect to see in the graphs is that P(x) will probably be slightly bigger on the edges of the potential for the quantum oscillator compared to the classical one.
I'll try to see this on a graph but I'm not even 100% sure what I expect is right.


Edit: I attach the graph of both functions for arbritrary constants. I had to play for several minutes with the coefficients to get such a graph. I am quite surprised especially for the quantum oscillator graph (blueish line). The reddish line is the classical oscillador.

 

[ehild]

For the quantum oscillator, I think that $\Psi (x)=(km)^{1/8} \hbar ^{-1/4}\pi ^{-1/2} e^{-\frac{(km)^{1/2}}{2\hbar}x^2}$.

Are you sure? It is so easy to Google it... Does the first state mean n=0 or n=1?

You need to compare the same oscillators with the same energy. Choose the constants accordingly. Write the probabilities in terms of ω, and choose the amplitude of the classical oscillator so its energy is $\frac{ωh}{2\pi }(n+1/2)$.


ehild

 

[fluidistic]

Are you sure? It is so easy to Google it... Does the first state mean n=0 or n=1?

You need to compare the same oscillators with the same energy. Choose the constants accordingly. Write the probabilities in terms of ω, and choose the amplitude of the classical oscillator so its energy is $\frac{ωh}{2\pi }(n+1/2)$.ehild

Ok I've progressed on this subject. First state means n=0.
For the quantum oscillator, I know that $E= \left ( n+\frac{1}{2} \right ) \hbar \omega$.

 

[ehild]

And what is the energy of the classical oscillator?
n=0 means no real oscillation. Te particle is nearly at its equilibrium position. If you choose a classical particle with mass much larger than that of an atom, it is confined in a very small interval around x=0.

ehild