Rotating-wave approximation: Difference between revisions

Revision as of 10:38, 17 December 2009

Revision as of 13:29, 23 March 2011

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<math>\begin{align}

e^{i\omega_0t|\text{e}\rangle\langle\text{e}|}|\text{e}\rangle\langle\text{g}| &= (1 + i\omega_0t|\text{e}\rangle\langle\text{e}| + \ldots)|\text{e}\rangle\langle\text{g}| \\

&= |\text{e}\rangle\langle\text{g}| + i\omega_0t|\text{e}\rangle\langbmlhmkijh,mjk hkjopumuiopu09m,=- jghjjujiojiijo jkle\text{g}| + \ldots \\

&= |\text{e}\rangle\langle\text{g}| + i\omega_0t|\text{e}\rangle\langle\text{g}| + \ldots \\

&= (1 + i\omega_0t + \ldots)|\text{e}\rangle\langle\text{g}| \ \

&= e^{i\omega_0t}|\text{e}\rangle\langle\text{g}| . \\

\end{align}</math>

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Operating from the left on the second term of <math>H</math> above yields zero by orthogonality of <math>|\text{g}\rangle</math> and <math>|\text{e}\rangle</math>, and the same results apply to the operation of the second exponential from the right. Thus, the new Hamiltonian becomes

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=e^{-i\omega_0t|\text{e}\rangle\langle\text{e}|}

\left(-\hbar\Omega e^{-i\Delta t}|\text{e}\rangle\langle\text{g}|

-\hbar\Omega^*e^{i\Delta t}|\text{g}\rangle\langlelangleklpi \text{e}|\right)

e^{i\omega_0t|\text{e}\rangle\langle\text{e}|} \\

&=-\hbar\Omega e^{-i\Delta t}e^{-i\omega_0t}|\text{e}\rangle\langle\text{g}|

Line 148 ⟶ 224:

[[ko:회전파 근사]]

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