Commit ba114c2e by T. van der Sar

### Update 4_ZPFs.md - added some properties of coherent states

parent c5dcc984
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 ... ... @@ -108,7 +108,12 @@ \$\$ The first term is energy associated with classical motion, and the second comes from quantum fluctuations. So if all of the oscillator eigenstates are stationary, how can we get them to "move"? If we prepare a superposition of two such states, like \$\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\$, then the observable \$\langle x(t) \rangle\$ oscillates in time, due to the relative phase between the two states. So if all of the oscillator eigenstates are stationary, how can we get them to "move"? Suppose we prepare a superposition of two such states, like \$|\Psi(t=0)\rangle=\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\$. This will evolve in time as \$\$ \$|\Psi(t) \frac{1}{\sqrt{2}}(e^{-iE_0t/\hbar}|0\rangle + e^{-iE_1t/\hbar}|1\rangle)\$. \$\$ The relative phase between the two states will cause the observable \$\langle x(t) \rangle\$ to oscillate in time. ## Coherent states ... ... @@ -118,11 +123,30 @@ \$\$ |\alpha\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n\rangle \$\$ where \$\alpha\$ is complex. The coherent state is not an eigenstate of the Schr\"{o}dinger equations, so, generally, it will "move" (\$\langle x \rangle\$ and \$\langle p \rangle\$ oscillate in time). 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}\$. As a function of time, coherent states evolve into new coherent states with an \$\alpha\$ that has the same amplitude but a different phase: \$\alpha(t) = e^{-i \omega t} \alpha\$. The ground state \$|0\rangle\$ is also a coherent state with \$\alpha = 0\$ and \$\langle x \rangle = \langle p \rangle = 0\$. where \$\alpha\$ is complex. The coherent state has the following properties: - The coherent state is *not* an eigenstate of the Hamiltonian. Therefore, generally, it will "move" (i.e., \$\langle x \rangle\$ and \$\langle p \rangle\$ oscillate in time). - 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}^\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 \$\$ \hat{H}=\hat{p}^2/2m+0.5m\omega_0^2\hat{x}^2 \$\$, we can derive that \$\$ \hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}^\dagger+\hat{a}) \$\$ and \$\$ \hat{p} = i\sqrt{\frac{m\omega\hbar}{2}}(\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). - As a function of time, coherent states evolve into new coherent states with an \$\alpha\$ that has the same amplitude but a different phase: \$\alpha(t) = e^{-i \omega t} \alpha\$. - The ground state \$|0\rangle\$ is also a coherent state with \$\alpha = 0\$ and \$\langle x \rangle = \langle p \rangle = 0\$. A coherent state \$\alpha\$ has the following \$\langle x \rangle\$ and \$\langle p \rangle\$: ... ...
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