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How do we get angular dependent wave function from only radially dependent potentials 
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#1
Jan3012, 04:51 PM

P: 146

Hi everyone!
So we're learning about the Hydrogen atom in QM and I'm having trouble reconciling something in my head. We're looking at potentials that are only radius dependent, like the Coulomb potential. Now, I know the math. I see that we assume the wave function can be separated into the product of three single variable function, one for each spherical coordinate, then we apply the Laplacian to it, we do the whole rigmarole for the radial one. I've done it all out. But for the θ,[itex]\phi[/itex] one, we get the Spherical Harmonics. And while I'm not disagreeing with the actual math, from a physical standpoint, I'm confused as to why the probability distribution could have different values at the same radius (but different values of [itex]\theta[/itex] or [itex]\phi[/itex]) if the potential is totally spherically symmetric. The way I'm thinking about it is, nature has no concept of the z axis which the angles [itex]\theta[/itex] and [itex]\phi[/itex] are defined with respect to, so where would spherical harmonics for an atom actually be pointed? The closest classical analogy I can think of is a planet orbiting in the gravitational potential (same as the coulomb, really), but the planet only goes on its specific orbit because of the initial conditions that sent it spinning in the plane it currently orbits in. Can someone help me out? Thanks! 


#2
Jan3012, 05:21 PM

P: 28

Actually, if you sum all the probability distributions, for example, the three distributions from those eigenfunctions for l=1, it gives an actual spherically symmetric function.
Your intuition is good, there is no Z axis. Without any external fields, an electron with l=1 would be 1/3 in each orbital, and that's spherically symmetric. When you fix a magnetic field, for example, and you measure number m, then you get the electron in that orbital. 


#3
Jan3012, 05:35 PM

P: 146

So when we add more electrons in the l = 1 orbital, they keep spreading themselves among the various available spherical harmonics such that the sum is spherically symmetric? Thanks for the response! 


#4
Jan3012, 08:13 PM

P: 430

How do we get angular dependent wave function from only radially dependent potentials
It looks strange, but atoms with partially filled shells are in general not spherically symmetric (e.g., F atom is not). This does not mean that there is a preferred direction; any given F atom will be polarized in some direction (which we could call "z"), but which direction that actually is is random because all the possible polarizations are degenerate in energy. There are not just three such directions, any vector is possible.
So how does this arise from a spherical potential? If you look closely, you see that in order for a wave function to be compatible[1] with the symmetry of the problem (here: the spherical potential) it is not necessary that this wave function has the same symmetry as the problem. It is sufficient if it "transforms" according to one of the irreducible representations of that symmetry group, and not all of those are individually spherically symmetric. [1] That is, that the wave function is an exact eigenfunction of the (symmetric) Hamiltonian 


#5
Jan3012, 08:44 PM

P: 146




#6
Jan3112, 12:16 AM

P: 685

Remember that electrons can have angular momentum in the atom! If angular momentum is 0, of course we get a spherically symmetric probability distribution.
But if the electron has angular momentum it must be, in some sense, rotating, so we have to pick an axis. Our probability distribution should be cylindrically symmetric, but it won't be spherical anymore. 


#7
Jan3112, 12:54 AM

P: 146




#8
Jan3112, 12:58 AM

Sci Advisor
P: 2,820

Essentially initial conditions. When you make a measurement of the atom, you had to choose an axis with which to make that measurement, and thus you "collapse" the atom into a certain state that may or may not be spherically symmetric.



#9
Jan3112, 07:10 AM

P: 28

And, VortexLattice, keep in mind, that when you add electrons, you cannot know "which" electron has m=1 , m=0 or m=1, they spread in such antisymmetric tensor product so that every electron can be in any orbital with the same probability (fermions indistinguishability)
[itex] \psi_1> \otimes \psi_2> \otimes \psi_3>  \psi_2> \otimes \psi_1> \otimes \psi_3> +\psi_2> \otimes \psi_3> \otimes \psi_1>... [/itex] 


#10
Feb612, 01:08 AM

P: 146

[itex]\psi = \frac{Y^{1} _1 + Y^{0} _1 + Y^{1} _1}{3}[/itex] With [itex]Y^{±1} _1 = \mp (3/8\pi)^{1/2} sin(\theta) e^{\pm i\phi}[/itex] [itex]Y^0 _1 = (3/4\pi)^{1/2} cos(\theta)[/itex] So, [itex]\psi[/itex] alone definitely isn't spherically symmetric, if you add it up. But that's not surprising, so I assume you meant the probability distribution [itex]P(\theta, \phi) \propto \left\psi\right^2[/itex] is symmetric, which is what we "see" anyway (wave functions are basically just math, probability distributions are the things we'd actually "measure" in some effect, right?). But if we look at [itex]P(\theta, \phi)[/itex], unless I made an arithmetic error, it isn't spherically symmetric either. You get a bunch of cross terms which would be orthogonal if we were integrating, but we're not. Most of them actually seem to cancel out (and the three of the spherical harmonics themselves each squared and summed actually does cancel out into a constant term), but you're left with two terms that seem to be angularly dependent. I end up getting: [itex]P(\theta, \phi) = \left\psi\right^2 = \left\frac{Y^{1} _1 + Y^{0} _1 + Y^{1} _1}{3}\right^2 = \frac{1}{9}\frac{3}{4\pi}(1  sin^2(\theta)cos(2\phi))[/itex] Which is still not spherically symmetric. Thoughts? 


#11
Feb612, 04:55 PM

P: 28

Hi, VortexLattice
Try with the mixed state [itex]1/3\psi_1 >< \psi_1+1/3\psi_2 >< \psi_2+1/3\psi_3 >< \psi_3 [/itex]and take the trace of the product with [itex]\delta_x >< \delta_x [/itex] Do not try with any pure state ;) 


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