Special relativity and quantum mech

[bs vasanth]

An atom can exist in two states , ground state of M and excited state of mass M+Δ , it goes to excited state by absorption of an photon, what is the photon frequency in the in the lab frame of reference ( where the atom is initially at rest )?

A simple energy equation gives the answer:
E=( M+Δ -M)$c^{2}$=hγ
γ=Δ$c^{2}$/h
But my intuition says that I am overlooking something and can't be that simple..

 

[PAllen]

An atom can exist in two states , ground state of M and excited state of mass M+Δ , it goes to excited state by absorption of an photon, what is the photon frequency in the in the lab frame of reference ( where the atom is initially at rest )?

A simple energy equation gives the answer:
E=( M+Δ -M)$c^{2}$=hγ
γ=Δ$c^{2}$/h
But my intuition says that I am overlooking something and can't be that simple..

It is that simple.

 

[Meir Achuz]

You should include the recoil of the excited atom in the final state.
Use Energy^2=(M+K)^2=(M+Delta)^2+K^2, and solve for K (the photon energy). (with c=1)

 

[PAllen]

You should include the recoil of the excited atom in the final state.
Use Energy^2=(M+K)^2=(M+Delta)^2+K^2, and solve for K (the photon energy). (with c=1)

Yes, I ignored recoil because its contribution is small for common cases. I, of course, agree with the above formula. Note that even for as extreme a case as a 10 Mev delta for a hydrogen atom, including recoil only changes the required photon energy to about 10.05 Mev. For Sodium, to about 10.002 Mev. For delta of 1 Mev and hydrogen, including recoil would require a 1.0005 Mev photon.

 

[sardar]

I can confirm that your intuition is correct. While the equation you have provided is correct, it does not take into account the principles of special relativity and quantum mechanics. In order to accurately calculate the photon frequency in the lab frame of reference, we must consider the relativistic effects on the energy and momentum of both the atom and the photon.

In special relativity, the energy and momentum of a particle are not independent quantities, but are related through the famous equation E=mc^2. This means that the energy of a particle is not just its rest mass, but also includes its kinetic energy. In the case of the atom, when it absorbs a photon and transitions to the excited state, it gains a certain amount of kinetic energy. This change in energy must be taken into account when calculating the photon frequency.

In addition, quantum mechanics tells us that particles can exist in a state of superposition, meaning they can be in multiple states at the same time. This is true for the atom in this scenario, which can exist in both the ground and excited states simultaneously. This adds another layer of complexity to the calculations and requires us to use the principles of quantum mechanics to accurately determine the photon frequency.

Therefore, while your initial equation may give a rough estimate of the photon frequency, a more accurate calculation must take into account the principles of special relativity and quantum mechanics. This highlights the importance of considering multiple perspectives and theories when approaching scientific problems.