# Harmonic Oscillator Problem (Classical, yet strangely quantum-like)

1. Jun 29, 2008

### Domnu

Problem
A harmonic oscillator consists of a mass of $$1 g$$ on a spring. Its frequency is $$1 Hz$$ and the mass passes through the equilibrium position with a velocity of $$10 cm/s$$. What is the order of magnitude of the quantum number associated with the energy of the system?

Solution?
Okay, so the eigenenergies associated with a harmonic oscillator are of the form

$$E_n = \hbar \omega_0 \left (n + \frac{1}{2}\right)$$​

So, here, we have that $$\omega_0 = 2\pi Hz$$. Now, we know that the energy of the system is (nonrelativistically, because the speeds are very small...)

$$1/2 m v^2 = 5 \cdot 10^{-6} J$$​

Now, we equate this with the eigenenergy formula, which gives us that the quantum number, $$n = 7.54594 \cdot 10^{27}$$, which is a ridiculously large number. So the magnitude would be $$\box{27}$$. Now I am guessing that such a large order of magnitude is fine since a one gram object (which is decently sized... say the density of water) in quantum mechanics is massive (like the size of the sun, compared to us). Is my answer, and more importantly, my method, correct?

2. Jun 29, 2008

### G01

Your method is correct and your numbers look fine. (I got 7.8 X10^27, but I rounded a little.)

Good job.

3. Jun 29, 2008

### Domnu

Very good. Thanks for verifying

4. Jun 30, 2008

### Domnu

Associated Problem

Here, let's say that the eigenstate corresponding to this function is $$\varphi_n$$. What is the average spacing between zeroes of an eigenstate with the quantum number of this problem?

Solution
We know that the quantum eigenstate of $$n$$ is represented by

$$\varphi_n (x) = A_n \mathbb{H}_n e^{-\zeta^2 / 2}$$​

where $$\zeta = \beta x$$ and $$\beta^2 = m\cdot \omega_0 / \hbar$$ and $$\mathbb{H}_n$$ denotes the nth Hermite polynomial. Now, we are just trying to find the spacing (in cm) of the zeros of nth Hermite polynomial as n goes to infinity (and of course, we divide this by sqrt(2^n * n! * sqrt(pi)) )... but wouldn't this grow arbitrarily close to zero as n approaches infinity? We can see this, because the roots of the Hermite polynomial get farther apart polynomially, but we are dividing the result by some higher-than-exponential order function (sqrt 2^n ...) So, the spacing in centimeters would be a number very close to zero (it may as well be zero in this case... we are dividing the spacing of the 10^28th Hermite polynomial by ((10^28)! * 2^28)... just as a comment, the number we are dividing by is unfathomably high... just as a statistic, the lower bound for its order of magnitude is (10^27)... not just 27. (oh my gods)

Is my above calculation correct?