Friday 17 July 2020

quantum mechanics - Probability of an hydrogen electron staying in a certain orbital


Suppose, an election of a hydrogen atom has energy corresponding to $n=2$, where $n$ is the principal quantum number.


Is it possible to find the probability of the electron staying in $2s$, $2p_x$,$2p_y$ or $2p_z$ orbital?


Are these probabilities equal, as they are have equal energy?



Answer



The question asked in your comment is different from the question I answered so I'll put a new answer here to your question:




The atom was in ground state and it is given enough energy for n=2. What are the probabilities that now the electron will be found in 2s, 2px, 2py, or 2pz orbital? Is it possible to measure this analytically?



The answer is yes. This can be deteremined analytically. Often we have a cartoon picture in our heads where an atom "magically" absorbs a photon and moves into an excited state. This leaves out many of the important details.


The reason light interacts with atoms is because the electric field of the light pushes around the electron (and thus changes the shape of the electron wavefunction). This means that if you shine on light at a frequency which the electron will respond to you can push the electron from being in one spatial pattern (the ground state, for example) to being in another spatial pattern (an excited state, for example).


Consider the $p_x$ state. In this state the electron is kind of stretched in the $x$ direction. Well this means that you can get from the ground state to this state by shining on light which is linearly polarized along the $x$ direction. The electric field acts to stretch the atom along the $x$ direction. Similarly for the $y$ and $z$ directions.


So this means that which excited state the atom is driven to depends on the geometry of the electric field driving the atom as well as the geometry of the particular ground and excited states under consideration.


In the dipole approximation it is impossible to drive transitions from $1s$ to $2s$ because the states have the parity.


I haven't given you an analytic formula to calculated the different excited state fractions given a particular incident electric field so I haven't fully answered your question. I don't have time for that now but most atom/quantum optics textbooks will have a description of how to calculate the coupling strength between an optical field and an atomic transition, including clebsch-gordon coefficients and polarization vectors which capture the geometric effects I described earlier.


I like this textbook available online by Steck


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...