Does higher Q factor leads to higher resonance frequency which means higher bandwidth

It was real nice and elaborated one.
Since coupling = Qo/Qext
So it means what you are saying as tight coupling that is critically coupled system with Q and Qext having the same value making the coupling equal to 1. In this case energy transfer will be maximum.
For under coupled system Qo<Qext so energy transfer will not be efficient.
But what will be the case if system is over coupled means Qo>Qext ?
What will be the energy transfer scenario?
How can we change Qext as for a given structure at a particular resonance Qo remains fixed so to change the coupling we can only change Qext.

I think that you have to be careful with the difference between Qu (unloaded) and Ql (loaded). Qu means the components are high quality, Ql should be as low as consistent with bandwidth. As Ql decreases, the RF losses decrease because the circulating RF current squared times the loss resistance decreases. It should really be 1 (or a transformer - no Q magnification of the input current).
Critical coupling is just a mathematical nicety, half power loss, double the bandwidth.
For a transmitter the working arrangement would be to have a Qu >100 and a Ql of 10, this would give about a 10% loss of power (as heat in the tuned circuit components). So for a 50 kW carrier power transmitter the losses in the circuit (mainly in the coil) would be 5 kW. If critical coupling was used the loss would be 25 kW, which is a lot of heat.
Frank

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Just remembered, your original question, does Q shift the centre frequency? The answer is yes but only very slightly, its a second order effect because the slope of the curve each side of resonance are assymetrical.
Years ago I was aligning IF strips with a nominal centre frequency of 250 KHZ, but all the circuits were peaked at 249 KHZ. The bandwidth as measure was symetrical about 250 KHZ.
Frank

chuckey said:

Critical coupling is just a mathematical nicety, half power loss, double the bandwidth.
For a transmitter the working arrangement would be to have a Qu >100 and a Ql of 10, this would give about a 10% loss of power (as heat in the tuned circuit components). So for a 50 kW carrier power transmitter the losses in the circuit (mainly in the coil) would be 5 kW. If critical coupling was used the loss would be 25 kW, which is a lot of heat.
Frank

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So yet another Q
Let me define few things so that I may be sure that we are referring to same things
Q
physically measure of power loss in cavity
Mathematically ratio of energy stored in cavity to power dissipated in walls of cavity
Higher Q means good conductor material means less losses so low power required to attain a fixed gradient
Qext
Physically measure of power loss in the external circuit
Mathematically ratio of energy stored in cavity to the power dissipated in the external circuit
Coupling Beeta
Ratio of power loss in external circuit to the power loss in the walls of cavity
Just mathematical manipulation and coupling Beeta becomes ratio of Q with Qext
Now Loaded Q Ql
it is 1/Ql=1/Q+1/Qext
Now kindly explain your statement that Q= 100 and Ql= 10 it is 10% loss of power
As how you can calculate Ql without Qext?

Q
physically measure of power loss in cavity no, its a "goodness" factor of the tuned circuit. On its own, it can only indicate the minimum bandwidth (b/w = Fres/Q).
Higher Q means good conductor material means less losses YES if its Qu so low power required to attain a fixed gradient I don't understand this.
Qext
Physically measure of power loss in the external circuit , No. Power loss is measured in dB
Ql is the measured Q of the loaded circuit. As Qu is normally high or very high, 100 for LC circuits, 2000 for cavities. The effect of Qu can be ignored compared with the Ql for power loss.
Frank