Commit 6cc51ef6 authored by T. van der Sar's avatar T. van der Sar
Browse files

Update 4_ZPFs.md - typo

parent ba114c2e
Pipeline #47325 passed with stages
in 6 minutes and 36 seconds
......@@ -129,19 +129,20 @@ where $\alpha$ is complex. The coherent state has the following properties:
- Instead, the coherent state is an eigenstate of the annihilation operator $\hat{a}$.
??? Hint
Recall: the creation operator $\hat{a}^\dagger$ and the annihilation operator $\hat{a}$ have the following properties:
- $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$, and a|0\rangle=0.
- $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$, and $\hat{a}|0\rangle=0$.
- $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$.
- $\hat{a}^\dagger\hat{a}|n\rangle = n|n\rangle$. Therefore, $\hat{a}^{\dagger}\hat{a}$ is called the number operator.
We see that the harmonic oscillator Hamiltonian can be written as $\hat{H}=\hbar\omega(\hat{a}^\dagger\hat{a}+0.5)$. Comparing this to
The number operator enables us to write the harmonic oscillator Hamiltonian as $\hat{H}=\hbar\omega(\hat{a}^\dagger\hat{a}+0.5)$. Comparing this to
$$
\hat{H}=\hat{p}^2/2m+0.5m\omega_0^2\hat{x}^2
$$, we can derive that
\hat{H}=\hat{p}^2/2m+0.5m\omega_0^2\hat{x}^2,
$$
\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}^\dagger+\hat{a})
we can derive that
$$
\hat{x} = x_\text{ZPF}(\hat{a}^\dagger+\hat{a})
$$
and
$$
\hat{p} = i\sqrt{\frac{m\omega\hbar}{2}}(\hat{a}^\dagger-\hat{a})
\hat{p} = i\frac{\hbar}{2 x_\text{ZPF}}(\hat{a}^\dagger-\hat{a})
$$
- There are an infinite number of possible coherent states since $\alpha$ can vary continuously: $\alpha = |\alpha|e^{i \theta}$.
- All coherent states are Heisenberg-limited minimum uncertainty wavepackets that satisfy $\sigma_x \sigma_p = \frac{\hbar}{2}$ (see exercise).
......
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment