Wednesday, April 22, 2015

wavefunction - Shape of Hydrogen atom in excited state with nonzeros angular momentum


Solution of Schrödinger's equation for an Hydrogen atom is well known: Ψn,l,m(r,θ,ϕ)=Nernr1Rln(r)Pml(cosθ)eimϕ.


If we interested with state with nonzeros angular momentum (say, l=1), the wave functions are: Ψ2,1,0=142πa3/20ra0er2a0cosθΨ2,1,±1=18πa3/20ra0er2a0sinθe±iϕ


This wave functions descrie the state with certain projection of angular momentum to some Cartesian axes, i.e. it is an eigenfunctions functions of an operator ˆLz. So, if we switch on the magnetic (or electric) field, the atom will be in some state of three possible, and the shape of its cloud will have some form that will not be spherically symmetric (for example, look here here).




But if there is no such direction, then what will be the shape of the hydrogen atom?





In chemistry, not the wave functions themselves are used, but their linear combinations, which are real:


pz=Ψ2,1,0=142πa3/20ra0er2a0cosθpx=12(Ψ2,1,+1+Ψ2,1,1)=182πa3/20ra0er2a0sinθcosϕpy=i2(Ψ2,1,+1Ψ2,1,1)=182πa3/20ra0er2a0sinθsinϕ


These functions are no longer the eigenfunctions of the operator ˆLz and describe states with an indefinite projection of the angular momentum. But what exactly do they describe? If there is no dedicated direction, then all three functions px,py,pz must be equiprobable, that is, the state of the electron must be a linear combination of these three functions, and the shape of the cloud must therefore be spherically symmetric. How, then, can the directional valence be explained to form a chemical bond?



Answer



Just to be clear, even without a magnetic field, the eigenstates of the hydrogen atom are simply not all spherically symmetric. The space of eigenstates of a given energy is spherically symmetric. Spherical symmetry means that if we transform an eigenstate by rotating it, it will give another eigenstate, not that each eigenstate is spherically symmetric.


Now, if we take a hydrogen atom at a particular energy, and then let it sit in a spherically symmetric environment with which it somehow interacts weakly for a long time, it should end up in a spherically symmetric superposition of eigenstates of the energy we started with, i.e. it should end up looking like a sphere. If we then measure the angular momentum we are equally likely to find it pointing in any direction.


The wavefunctions you mentioned as used in chemistry -- let's focus on the two ''new'' ones px and py -- describe energy eigenstates with total angular momentum 1 and moreover angular momentum 0 in the x and y direction respectively (you missed a factor of i in the definition of py btw). The simplest way to see this is to consider the space of states of total angular momentum 1 as a three dimensional vector space (this makes sense as long as you only look at the angular momentum operators). In this vector space, the angular momentum operators act like matrices of the form Lx(00000i0i0),Ly(00i000i00),Lz(0i0i00000)

and the Ψ2,1,x are represented by Ψ2,1,0(001),Ψ2,1,1(1i0),Ψ2,1,1(1i0).
Now if you take the linear combinations of Ψ2,1,±1 that you write, you can easily see that they are in the null space of Lx and Ly respectively.


As soon as you put more than one hydrogen atom in your universe, there can be a preferred direction (as there definitely is in molecules), and this will put your hydrogen atom in a particular not-spherically-symmetric eigenstate, which allows for the typical picture of a valence bond.


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