Commit 6cc51ef6 by T. van der Sar

### Update 4_ZPFs.md - typo

parent ba114c2e
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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 $$\hat{H}=\hat{p}^2/2m+0.5m\omega_0^2\hat{x}^2$$, 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). ... ...
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