This concerns calculation of Rabi splitting of exciton polaritons when the exciton states are mixed but the mixing is known, i.e., the coefficients of the mixed states are known.(adsbygoogle = window.adsbygoogle || []).push({});

I read from a thesis that the Rabi splitting is proportional to the square root of the oscillator strength. Is this correct? (I haven't found other sources for it.) But it is the Rabi splitting which is an obvious factor in polariton experiments.

Now, heavy hole (HH) excitons couple to light better than light hole (LH) excitons. When they are mixed together, maybe due to reduction in symmetry, we expect the oscillator strength of the mixed state to be modified. According to another thesis, the oscillator strength is proportional to

[tex] f \propto |\langle vac|p\cdot e|i\rangle|^2 [/tex]

the square root of this is the Rabi splitting.

If the mixing between LH and HH is strong, we expect the mixed state to have a smaller oscillator strength due to the weaker LH. Is this right? (I have doubt about it) Or is there a similar level-repulsion for oscillator strength like in energy when there is mixing?

This will all be answered if I do the calculations of f, of course.

Let's say I know the Rabi splitting for both, R1 for HH, R2 for LH. If the mixed state is exactly half LH, and half HH, i.e.,

[tex] |i1\rangle=\frac{|HH\rangle + |hh\rangle}{\sqrt{2}}, |i2\rangle=\frac{|HH\rangle - |hh\rangle}{\sqrt{2}} [/tex]

using |i1>, the oscillator strength should be

[tex] f=|\langle vac|p\cdot e|i1\rangle|^2=\frac{|\langle vac|p\cdot e|HH\rangle + \langle vac|p\cdot e|hh\rangle|^2}{2} [/tex]

But I know,

[tex] \sqrt{f}=|\langle vac|p\cdot e|HH\rangle|=R1 [/tex]

It follows then that the Rabi splitting for the mixed states are

[tex] R_{i1}= \frac{|R1+R2|}{\sqrt{2}},R_{i2}= \frac{|R1-R2|}{\sqrt{2}} [/tex]

This means there is indeed a sort of level-repulsion in oscillator strength.

Are my calculations right?

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# Oscillator strength of mixed LH- and HH-excitons

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