If an electron is in $s$ state, for example in 1s state for Hydrogen or 5s state for Silver atom, $\ell=0$. So,its total angular momentum $L$ is also equal to 0. So, what is electron actually doing in s state. Is it actually not moving at all. I know that it has got spin, but if it is not moving, doesn't it violates uncertainty principle.
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Qmechanic
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Roshan Shrestha
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$| \psi(t) \rangle = e^{-iHt} | \psi (0) \rangle= e^{-i E_s t} | \psi (0) \rangle $ – user26143 Apr 21 '14 at 10:36
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You are still thinking of the bound electron in a classical sense. The electron is a quantum mechanical "entity", it obeys quantum mechanical laws and the definition os S state identifies the energy level the electron is allowed to occupy. Within the energy level it does not have a position or a trajectory, it has an orbital. The orbital is the probability distribution of the position you would find the electron in, if you did an experiment, the probability of finding it at that (x,y,z).
The shapes of the first five atomic orbitals: 1s, 2s, 2px, 2py, and 2pz. The colors show the wave function phase. These are graphs of ψ(x, y, z) functions which depend on the coordinates of one electron. To see the elongated shape of ψ(x, y, z)2 functions that show probability density more directly, see the graphs of d-orbitals in link above.
Note that in an S state there exists a probability for the electron to pass through the nucleus, but as the nucleus is orders of magnitude in space smaller, this is a very small probability but it is seen in electron capture nuclear reactions.
What angular momentum means in the quantum mechanical regime is that the operator of angular momentum acting on the state function of the atom will give the values of L . The L=0 identifies the energy level where operating with the angular momentum operator
on the state function , the value returned is zero.
anna v
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So, what is electron actually doing in s state. Is it actually not moving at all. I know that it has got spin, but if it is not moving, doesn't it violates uncertainty principle.
The "s state" is just a name for the first eigenfunction $\phi_{100}(\mathbf r)$ of the Hamiltonian of hydrogen type. This function can be used to calculate probability that the electron is at given region or that its momentum $p_x$ has value in a given interval. The theory of $\psi$ functions does not deal with "what electron is actually doing", only with "what the $\psi$ function is doing" and how to use it to find what are the probabilities that the system has such position and such momentum.
To answer "what electron is actually doing", one would need another theory, for example Bohm-de Broglie theory (standing still) or stochastic electrodynamics (moving erratically).
Ján Lalinský
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