# Quantizing Newton

1. Jun 13, 2003

### Tom Mattson

Staff Emeritus
Many moons ago, when I used to teach Modern Physics to engineering undergrads, I told them that to get to quantum mechanics from classical mechanics, we had to start from the Hamiltonian as opposed to Newton's laws. I told them that the reason was that Newton's second law for a particle, namely:

&Sigma;F=m(d2x/dt2)

presupposed that we could know the trajectory x(t) of a particle, which violates the uncertainty principle. However, lately I am thinking that that is not the reason at all. The Hamiltonian is expressed in terms of the coordinates and momenta, which we are also not allowed to know simultaneously.

So why not try to quantize Newton?

I want to look at an example first, and then go into generalities later. Let's look at the simple harmonic oscillator in 1D along the x axis.

Writing the RHS in terms of momentum, Newton sez:

-kx=(dp/dt)

Now quantize:

-kx=(d/dt)(-i*hbar)(d/dx)
-(i*hbar)(d/dt)(d/dx)+kx=0

Now we have an operator. Let it act on a wavefunction &Phi;(x,t)

-(i*hbar)(d/dt)(d&Phi;(x,t)/dx)+kx&Phi;(x,t)=0

I just thought of this like 5 minutes ago, so I haven't worked it out yet. But what I want to explore is:

1. Is this mathematically valid?
2. If so, what is the relationship between this wavefunction and the Schrodinger wavefunction?
3. Has this been worked out before?
4. Might there be some advantage to this approach?

edit: fixed an omission, and a typo

Last edited: Jun 13, 2003
2. Jun 13, 2003

### lethe

an interesting idea.

i no of no explanation why you have to start from hamilton s equations instead of newton s equations.

however, i tried solving your equation by seperation of variables, and i got something that looked completely unlike what i know the correct solution shuold look like. i got the time component goes as exp(kt), which is odd, and i did get a gaussian for the space component, which is what the ground state looks like. i only get one solution, and no quantization.

but still, i m not sure why it should make a difference. after all, classically, the two approaches are equivalent, right? they encode the same information....

3. Jun 15, 2003

### jeff

Okay, how does this grab you:

We know that because the harmonic oscillator is a conservative system, it's states are determined by the value of a single observable, and any one of position, momentum or energy will do. By taking P=-ihd/dx you've chosen to work in the x-representation.

So what's the action of d/dt on &Phi; in this representation? Not the hamiltonian H(x)=-(1/2m)d2/dx2+(1/2)kx2, because &Phi; doesn't satisfy the shrodinger equation. Yet the evolution of all quantum mechanical systems are governed by their hamiltonians.

Last edited: Jun 16, 2003