Probability, QM, harmonic oscillator, comparison with classical

[fluidistic]

Homework Statement

I must calculate the probability that the position of a harmonic oscillator in the fundamental state has a greater value that the amplitude of a classical harmonic oscillator of the same energy.

Homework Equations

$\psi _0 (x)=\left ( \frac{m \omega}{\pi h } \right ) ^{1/4} \exp \{ -\frac{m\omega x^2}{2\hbar} \}$
$E_0=\frac{\hbar \omega}{2}$.
$|\psi _0 (x)|^2=\psi ^* _0 (x) \psi _0(x) =\left ( \frac{m \omega}{\pi h } \right ) ^{1/2} \exp \{ -\frac{m\omega x^2}{\hbar} \}$

The Attempt at a Solution

Classically the Hamiltonian is $H=\frac{\dot x ^2 m}{2} + \frac{m \omega ^2 x^2}{2} =\frac{\hbar \omega}{2}$. Where $x(t)=A\cos (\omega t)$ and $\dot x (t)=-A\omega \sin (\omega t)$. Replacing these into the Hamiltonian yields $A =\sqrt{\frac{\hbar}{m\omega }}$.
I'm asked to calculate $P=1-\int _0 ^A |\psi _0 (x)|^2dx=1-\int _0 ^{\sqrt{\frac{\hbar}{m\omega }}} \left ( \frac{m \omega}{\pi h } \right ) ^{1/2} \exp \{ -\frac{m\omega x^2}{\hbar} \} dx$
After the change of variables $\alpha = \sqrt{\frac{m\omega }{\hbar}}x$, I get that $P=1-\frac{1}{\sqrt \pi } \int _0 ^1 \exp (-\alpha ^2 ) d\alpha =1-\frac{1}{\sqrt \pi } \approx 0.73$.
First off, I can't believe this result. It doesn't depend on the mass of the particle oscillating, this goes against my intuition... I would think that the smaller the mass is, the greater the probability to find it farther than the classical amplitude is. And for m tends to infinity I would have thought to reach classical mechanics, i.e. a probability of 0 to find it farther than A.
Also 73% seems way too big!

So I'm wondering where I went wrong. Thank you.
P.S.:I haven't found my mistake(s) when writing this thread, after rechecking the algebra.

 

[fzero]

I'm asked to calculate $P=1-\int _0 ^A |\psi _0 (x)|^2dx=1-\int _0 ^{\sqrt{\frac{\hbar}{m\omega }}} \left ( \frac{m \omega}{\pi h } \right ) ^{1/2} \exp \{ -\frac{m\omega x^2}{\hbar} \} dx$
After the change of variables $\alpha = \sqrt{\frac{m\omega }{\hbar}}x$, I get that $P=1-\frac{1}{\sqrt \pi } \int _0 ^1 \exp (-\alpha ^2 ) d\alpha =1-\frac{1}{\sqrt \pi } \approx 0.73$.
First off, I can't believe this result. It doesn't depend on the mass of the particle oscillating, this goes against my intuition... I would think that the smaller the mass is, the greater the probability to find it farther than the classical amplitude is. And for m tends to infinity I would have thought to reach classical mechanics, i.e. a probability of 0 to find it farther than A.
Also 73% seems way too big!

First of all, you're only integrating over $x>0$. If you include the other half of the domain, the probability is a bit smaller.

Second, the classical limit doesn't just involve the mass of the particle. The classical limit corresponds to the range of parameters where the spacing between energy levels approaches zero, since the energy of a classical system is continuous. This is the limit of large quantum numbers, so the ground state is really the least classical state.

 

[fluidistic]

First of all, you're only integrating over $x>0$. If you include the other half of the domain, the probability is a bit smaller.

Second, the classical limit doesn't just involve the mass of the particle. The classical limit corresponds to the range of parameters where the spacing between energy levels approaches zero, since the energy of a classical system is continuous. This is the limit of large quantum numbers, so the ground state is really the least classical state.

I see. However the classical limit should involve the mass of the particle, at least, right?
I notice I've made a mistake in the computation, I divided by pi instead of the square root of pi in my last step (numerically), so I should have gotten a probability of 0.52.
But as you pointed out, I forgot to extend the lower limit of the integral to -A. By doing so and by not making the sqrt (pi) error, I reach $P=1-\frac{1}{\sqrt \pi} [\operatorname{erf} (1)-\operatorname{erf} (-1)]\approx 0.049$. Which this time seems too low.

 

[vela]

But as you pointed out, I forgot to extend the lower limit of the integral to -A. By doing so and by not making the sqrt (pi) error, I reach $P=1-\frac{1}{\sqrt \pi} [\operatorname{erf} (1)-\operatorname{erf} (-1)]\approx 0.049$. Which this time seems too low.

There is just no satisfying you, is there?

 

[fluidistic]

There is just no satisfying you, is there?

Well if I'd get a P that depends on A, m and say on omega, I'd be satisfied I guess. As for a "raw" percentage, if P doesn't depend on anything but the amplitude like in my case, I'd guess something between 0.370 and 0.625.

Edit 1: Nevermind, my result doesn't even depend on the amplitude A, so it doesn't even depend on the product of the mass and the frequency.
Edit 2: Discard the Edit 1, it is false.

 

[fzero]

I see. However the classical limit should involve the mass of the particle, at least, right?
I notice I've made a mistake in the computation, I divided by pi instead of the square root of pi in my last step (numerically), so I should have gotten a probability of 0.52.
But as you pointed out, I forgot to extend the lower limit of the integral to -A. By doing so and by not making the sqrt (pi) error, I reach $P=1-\frac{1}{\sqrt \pi} [\operatorname{erf} (1)-\operatorname{erf} (-1)]\approx 0.049$. Which this time seems too low.

0.52 is what I got. The way $\mathrm{erf}(x)$ is usually defined, there's a factor of 2, so that $\mathrm{erf}(\infty)=1$. But the Gaussian integral here is normalized to give 1/2 when integrated from 0 to $\infty$. You integrate from 0 to $x$, so you have $\mathrm{erf}(x)/2$.

If you want to see how the classical limit depends on the parameters, we could try to model the classical state by a superposition of quantum states $\sum_n c_n |n\rangle$. Then comparing the classical energy with the quantum energy yields

$$ m \omega A^2 = \sum_n \hbar c_n \left( n + \frac{1}{2} \right). $$

For fixed $\omega$, for a classical object ($m$ a few grams, $A$ a few cm), the left-hand side will be huge compared to $\hbar$. This means that the sum on the right-hand side must favor large $n$, since the $|c_n| \leq 1$. So a classical-like state is composed of mostly highly excited states. As expected, the energy spacing between highly excited states is extremely small compared to their energies.

 

[fluidistic]

0.52 is what I got. The way $\mathrm{erf}(x)$ is usually defined, there's a factor of 2, so that $\mathrm{erf}(\infty)=1$. But the Gaussian integral here is normalized to give 1/2 when integrated from 0 to $\infty$. You integrate from 0 to $x$, so you have $\mathrm{erf}(x)/2$.

If you want to see how the classical limit depends on the parameters, we could try to model the classical state by a superposition of quantum states $\sum_n c_n |n\rangle$. Then comparing the classical energy with the quantum energy yields

$$ m \omega A^2 = \sum_n \hbar c_n \left( n + \frac{1}{2} \right). $$

For fixed $\omega$, for a classical object ($m$ a few grams, $A$ a few cm), the left-hand side will be huge compared to $\hbar$. This means that the sum on the right-hand side must favor large $n$, since the $|c_n| \leq 1$. So a classical-like state is composed of mostly highly excited states. As expected, the energy spacing between highly excited states is extremely small compared to their energies.

Ok thanks a lot.
I reach a probability of 0.157... I have $P=1- \int _{-1}^1 \frac{1}{\sqrt \pi } \exp (-\alpha ^2 ) d\alpha=1-\frac{1}{2}[\operatorname {erf}(1)- \operatorname {erf} (-1)]\approx 0.157...$

By the way I'm honored to be the first guy who "praised" you with the praise button.