Trying to write falling rod energies as Hamiltonian

[masterkenichi]

"Consider an infinitely sharp pin of mass M and height H perfectly balanced on its tip. Assume that the mass of the pin is all at the ball on the top of the pin. Classically, we expect the pin to remain in this state forever. Quantum mechanics, however, predicts that the pin will fall over within a finite amount of time. This can be shown as follows:
Show that the total energy of the pin (aka. the Hamiltonian) can be expressed in the
form:
E = Ap^2 − Bx^2
if we assume that x << H. p is the momentum of the pin and x is the lateral displacement of the head of the pin. Find expressions for A and B."


My attempt:
When the pin head moves laterally a distance x, it will have lost some gravitational energy equal to E_g = mg\sqrt{H^2-x^2}.
My first shot at this is to write E = P^2/(2m) + mgy, where y= \sqrt{H^2-x^2}; however, I don't think this can be reduced to the form called for. Suggestions?

 

[Heirot]

You should do a Taylor approximation of the expression y= \sqrt{H^2-x^2}. For x/H << 1, sqrt{H^2-x^2} = H sqrt{1-(x/H)^2} = H (1 - (1/2)(x/H)^2 + ...). The first term is a constant, and the second is the one you need.

 

[masterkenichi]

Thank you very much for your help, but I'm a little uncertain what you mean by

The first term is a constant, and the second is the one you need.

This approximation results in the term E = (1/2m)p^2 + mgy, where y = H-x^2/{2H}, which I can't write in the form Bx^2. Is the constant term is not relevant?

 

[gabbagabbahey]

Thank you very much for your help, but I'm a little uncertain what you mean by



This approximation results in the term E = (1/2m)p^2 + mgy, where y = H-x^2/{2H}, which I can't write in the form Bx^2. Is the constant term is not relevant?

Shouldn't you have:
[tex]\Delta E = \frac{p^2}{2m} + mgy[/tex]
And assuming conservation of Energy:
[tex]\Delta E=0[/tex]

 

[Heirot]

This approximation results in the term E = (1/2m)p^2 + mgy, where y = H-x^2/{2H}, which I can't write in the form Bx^2. Is the constant term is not relevant?

Hamiltonian is always uncertain up to an additive constant (it is because the equations of motion involve only the derivatives of Hamiltonian). Just remember that you are free to choose any point to be reference point and have energy E=0.