Allowed energies for an SHO

1. Aug 9, 2007

Moham1287

Hi all

I was just looking through my notes from my first year of my degree, and I couldn't find a missing bit. I know that Planck's postulate states that the allowed energies of a quantum simple harmonic oscillator are 0, hf, 2hf etc and that by the Schroedinger equation, you get E(n)=(n+1/2) hbar omega, but I can't explain why each of these give different answers. A quick google didn't bring up anything not password protected, and I don't have my textbooks with me at the moment. Anything to clear this up would be much appreciated!
Many thanks.

2. Aug 9, 2007

Staff: Mentor

Planck was dealing with electromagnetic radiation, not the simple harmonic oscillator (mass on a spring and similar things).

3. Aug 10, 2007

olgranpappy

$$\hbar\omega=hf$$

...the remaining difference is just a constant ($$0.5\hbar\omega$$) which can be ignored.

4. Aug 10, 2007

olgranpappy

I.e., the Hamiltonians

$$H=\frac{p^2}{2m}+\frac{kx^2}{2}$$

and

$$H'=\frac{p^2}{2m}+\frac{kx^2}{2}-\hbar\sqrt{\frac{k}{4m}}$$

describe the same physics since they differ only by a constant.... and furthermore, it should be obvious that they are equal in the classical limit.

5. Aug 10, 2007

Gokul43201

Staff Emeritus
There's actually a little known paper by Einstein, where he first figures out that the Harmonic Oscillator needs a non-zero ground state energy, and he sticks in the 0.5 \hbar \omega by hand. Can't recall or pull up the citation. Anyone know what I'm talking about.

6. Aug 11, 2007

premagg

According to schrodinger equation,the particle cannot have zero energy.Because the solutions are (v+1/2)hf,v is vibrational quantum number.If v happens to be -1,then energy is -ve,considering zero potential the KE is -ve and so is momentum.This is not in correspondance with uncertainity principle.

7. Aug 11, 2007

premagg

jtbell is correct also,I am sorry to misunderstand your question.Planck was dealing with the blackbody radiations and schrodnger with quantum mechanical harmonic oscillator.