Heisenberg's uncertianty principle

[Physicsiscool]

Can anyone help me with this question?

A simple harmonic oscillator can be considered to be a small mass "m" on a spring of a force constant "k" when the force is f = -kx and the potential energy is V(x) = 1/2mw(squared)x(squared) where w = square root of (k/m)

Knowing this I need to show that the minimum value for the energy of the ocillator is E (min) = h(bar) * w

 

[dextercioby]

Can anyone help me with this question?

A simple harmonic oscillator can be considered to be a small mass "m" on a spring of a force constant "k" when the force is f = -kx and the potential energy is V(x) = 1/2mw(squared)x(squared) where w = square root of (k/m)

That's correct.

Knowing this I need to show that the minimum value for the energy of the ocillator is E (min) = h(bar) * w

That's wrong,unless you screw up calculations and end up with the wrong formula written above.The right one is,of course,E_{0}=\frac{\hbar\omega}{2}.It's called the vacuum energy of the oscillator and it's a pure quantum effect.
Quantum mechanics books find this by two methons.Using Schroedinger's equation or using creation+annihilation operators.The latter is more elegant.The former requires lots of calculating.
Good luck!

 

[R0man]

/2

Sure, I can help explain Heisenberg's uncertainty principle in relation to the simple harmonic oscillator.

First, it's important to understand that the uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to know both the precise position and momentum of a particle at the same time. In other words, the more precisely we know the position of a particle, the less we know about its momentum, and vice versa. This is represented by the famous equation:

Δx * Δp ≥ h/2π

where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.

Now, in the case of the simple harmonic oscillator, the position of the mass is given by x = A sin(wt), where A is the amplitude of the oscillation and w is the angular frequency. The momentum is given by p = m*w*A*cos(wt).

Using the uncertainty principle, we can say that the product of the uncertainties in position and momentum must be greater than or equal to h/2π.

Δx * Δp ≥ h/2π

Substituting in the equations for position and momentum, we get:

Δx * m*w*A*cos(wt) ≥ h/2π

Now, the maximum value for cos(wt) is 1, which occurs when wt = 0. This means that the minimum uncertainty in momentum occurs at the point when the mass is at its maximum displacement, A.

So, we can rewrite the equation as:

Δx * m*w*A ≥ h/2π

Recall that w = √(k/m), so we can substitute that in:

Δx * m*w*A ≥ h/2π

Δx * m*√(k/m)*A ≥ h/2π

Simplifying, we get:

Δx * √(k*m) ≥ h/2π

Now, we know that the potential energy of the oscillator is given by V(x) = 1/2*m*w^2*x^2. We can rearrange this equation to solve for w and substitute it into our inequality:

w = √(k/m)

w^2 = k/m

√(k*m) = w

Substituting into our inequality, we get:

Δ