# Harmonic oscillator energy difference between $(n+\frac{1}{2})h \omega$ and $(n+\frac{1}{2})\hbar \omega$

When I was studying the Harmonic Oscillator using the Schrödinger equation, I was told in lectures to pay attention to the units.

There were 2 different equations given for the Energy of a Harmonic Oscillator:

$$(n+\frac{1}{2})h \omega$$

$$(n+\frac{1}{2})\hbar \omega$$

The trouble is that I can't understand the difference between using $$h$$ and $$\hbar$$ i.e. $$\frac{h}{2 \pi}$$ which yield $$s^{-1}$$ and $$rad \ s^{-1}$$ respectively.

Is there a reason for this? Why do we use two different equations with different units for the same observable?

• Well, the difference is that only the second equation is right. The first equation indicates somebody making a mistake. Just plug in $h = 2 \pi \hbar$ and it's clear they're different. – knzhou Jan 1 at 13:39

The difference is one is wrong (probably a typo). The two formulas are: \begin{align} E&=\hbar\omega\left(n+\frac{1}{2}\right)\\ &=hf\left(n+\frac{1}{2}\right). \end{align} The relationship between them is that $$\hbar=\frac{h}{2\pi}$$ and $$\omega=2\pi f$$. We use two different forms because angular frequency, $$\omega$$, appears naturally in a lot of equations when you use sine and cosine that take radians. The preference for them comes from the fact that the formula for their derivatives, the thing you do to calculate rates of change, are particularly simple.
So, bottom line, there's a preference for $$\omega$$ because it cuts down on factors of $$2\pi$$ in equations, and using $$\hbar$$ removes even more of them.
• so if the fundamental vibrational frequency is quoted as $\frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}$ that means that this is equal to $f$? – David Smith Jan 1 at 13:54
• @DavidSmith Yep. The angular frequency is $\omega=\sqrt{\frac{k}{\mu}}$. – Sean E. Lake Jan 1 at 13:55