Commit 6cc51ef6 authored by T. van der Sar's avatar T. van der Sar
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Update - typo

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......@@ -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
$$, we can derive that
\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})
\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).
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