Timeline for Two coherent state input to a beam splitter
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2023 at 9:24 | comment | added | Shoaib Mirzaei | You used Displacement operator $D(\alpha) = exp(\alpha a^\dagger)$ to convert $|\alpha>$ into $|0>$, but $D(|\alpha>) = exp(\alpha a^\dagger - \alpha^* a)$. what happened to annihilation operator? | |
| Aug 5, 2023 at 15:27 | comment | added | SolubleFish | If $A$ is a matrix or an operator, its exponential is defined by the Taylor series : $\exp(\hat A) = \sum_{n=0}^{+\infty} \frac{1}{n!} \hat A^n$. Since we have $\left(\hat A^n\right)^\dagger =\left(\hat A^\dagger\right)^n$, we get $\exp\left(\hat A^\dagger\right) = \exp\left(\hat A^\dagger\right)$. (Note that this work for any analytic function whose Taylor coefficients are real.) | |
| Jul 26, 2023 at 15:36 | comment | added | Shoaib Mirzaei |
@SolubleFish, Thanks for this brilliant answer, but can you describe how you put T and T^\dagger inside exponential?
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| May 12, 2022 at 5:07 | vote | accept | asd.123 | ||
| May 10, 2022 at 17:19 | comment | added | SolubleFish | Yes, I took the liberty to abuse notation and write $|0\rangle$ in place of $|0\rangle \otimes |0\rangle$ for the ground state. | |
| May 10, 2022 at 16:48 | comment | added | asd.123 | Thanks a lot, but I have tried to replicate your work and realized that it should be $|\alpha\rangle \otimes|\beta\rangle=\exp \left(-\frac{|\alpha|^{2}+|\beta|^2}{2}\right) \exp \left(\alpha \hat{a}_{i}^{\dagger}+\beta \hat{b}_{i}^{\dagger}\right)|0\rangle\otimes|0\rangle$ for second line, can you check it? | |
| May 10, 2022 at 16:00 | history | answered | SolubleFish | CC BY-SA 4.0 |