17. Two State Systems (continued)

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  • Опубліковано 3 гру 2024

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  • @woodandgears2865
    @woodandgears2865 Рік тому +4

    CC fixes
    5:56 here "phase" should be "face"
    11:58 change what I call the
    33:38 so getting to diagonalize this
    1:04:05 state of light
    1:04:43 in which this molecule turns from one state
    1:07:32 imagine this piece of ammonia traveling through the cavity.
    1:13:34 the time evolution is easy
    1:16:20 time dependent
    1:20:18 the original Hamiltonian to this one
    1:20:21 remove the "6"

    • @mitocw
      @mitocw  Рік тому +1

      Thanks for your time and effort! The caption has been updated.

  • @ricardocesargomes7274
    @ricardocesargomes7274 7 років тому +2

    Beauty Lecture!

  • @zphuo
    @zphuo 6 років тому +1

    @40:00, Why in a strong gradient the |G> state will tend to go to the region of larger |E|?

    • @prasadpawar7027
      @prasadpawar7027 4 роки тому +3

      Since, Force = -Grad(Ground Energy) = -Grad(-με) = μ Grad(ε). This means Force will be in the direction where electric field increases.

  • @not_amanullah
    @not_amanullah 4 місяці тому

    Thanks 🤍❤️

  • @benbald
    @benbald Рік тому

    Why not define the Hamiltonian matrix for the Ammonia molecule to be diagonal, like this:
    | E₀ - Δ 0 |
    | 0 E₀ + Δ |.
    It seems like this would make |↑〉and |↓〉to be the stationary energy eigenstates with the correct eigenvalues and the time evolution would just be phase rotation.
    But, this contradicts what Dr. Zwiebach said when he worked out the time evolution of |↑〉and stated that it rotates between |↑〉and |↓〉at 23 GHz, which clearly isn't stationary. So my question is just why isn't my Hamiltonian valid?

    • @DaveShree
      @DaveShree Рік тому

      Your hamiltonian is valid but it say nothing about probabilty of system in that state. Actually atom flip its state and your amplitude is fixed in time, stationary.

    • @benbald
      @benbald Рік тому

      @@DaveShree I don't understand what you mean. If |↑〉is an eigenstate of the Hamiltonian, and the system is initially in that state, then it will remain in that state forever with probability of 1.

    • @DaveShree
      @DaveShree Рік тому

      @@benbald That is correct, but if you express a state which is sum of eigenstates then system is not identified with one definite energy, that is uncertainity in energy and thus system can't remain in any state for long because there is no certain energy.

    • @benbald
      @benbald Рік тому

      I think the answer to my question is implied by what he says at 14:31. He explains that without Δ, the Hamiltonian doesn't "match the physics" - meaning that |↑〉and |↓〉will be degenerate which isn't allowed in a 1D case like this. So he just puts Δ on the off diagonal terms as a simple first attempt to see what happens. He could have put Δ on the diagonal terms, like I suggested, but that leads to another case that doesn't match the physics because |↑〉and |↓〉would be stationary in that case, and clearly they are not. I think what I missed was the trial and error aspect of this approach.

    • @DaveShree
      @DaveShree Рік тому

      @@benbald Wirh eigenvalues of a system, it remain im that state but how could one explain radiation having mean wavelength at average energy with which you added delta. So system must be in off of average energy like mass spring oscillate to mean position.

  • @davidhand9721
    @davidhand9721 4 роки тому

    It also has a lone pair on the other side; that doesn't affect the dipole? I feel like it ought to.

    • @mississippijohnfahey7175
      @mississippijohnfahey7175 2 роки тому

      Ammonia has one lone pair, which occupies a p orbital in the z direction. When the NH3 molecule is in a full pyramidal configuration, the electron density of that lone pair is almost entirely concentrated on the side of the pyramid far from the three hydrogen atoms. NH3 undergoes rapid inversion, so sometimes the lone pair electron density is equally on both sides on the nitrogen atom, but that is true for an infinitesimally small amount of time. Thus, each individual molecules has an electric dipole moment for the vast majority of time.
      However, a mole of NH3 gas under a neutral external electric field will have no overall dipole, because each molecule's inversions and rotations will cause various dipole states to negate each other in the overall system.
      I still don't quite understand how the MASER functions though. Seems like it's emitting photons of energy equal to the transition energy from the NH3 state that opposes the electric field down to the state that is aligned with the electric field. But the details of the significance of the resonance cavity are still hazy to me. Seems like you'd want a constant electric field, not an oscillating one. But I also am not sure why NMR uses a rotating magnet rather than a static one, and the two situations seem related...

  • @p.s.design4338
    @p.s.design4338 7 років тому +2

    tbh, this lecture is a little messy, without the introduction of Dirac's interaction picture.

    • @davidhand9721
      @davidhand9721 4 роки тому

      Link to introduction of Dirac interaction picture?

    • @sadmansr1054
      @sadmansr1054 3 роки тому

      In my opinion, the interaction picture would make things worse as it would require the knowledge of density matrix and time-dependent perturbation theory. It is better left for 8.06. Also, just for convenience the density matrix is actually introduced in the recitations during the offering of this course, the video of which isn't on UA-cam.

    • @sadmansr1054
      @sadmansr1054 3 роки тому

      @@davidhand9721 en.wikipedia.org/wiki/Interaction_picture

    • @mississippijohnfahey7175
      @mississippijohnfahey7175 2 роки тому +1

      I think it just needs more discussion of the physical picture of how the MASER really functions. Too much math not enough application is the downfall of this whole lecture series